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Machines and Mechanisms: Applied
Kinematic Analysis, 4/e, Myszka
Chapter 1
Machines and Mechanisms: Applied Kinematic Analysis, 4/e
David Myszka
© 2012, 2005, 2002, 1999 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved.2
Chap 1 Introduction
Machines and Mechanisms: Applied Kinematic Analysis, 4/e
David Myszka
© 2012, 2005, 2002, 1999 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved.3
1.1 INTRODUCTION
◼ Determine appropriate movement
of the wipers
◼ View range
◼ Tandem or opposite
◼ Wipe angle
◼ Location of pivots
◼ Timing of wipers
◼ Wiping velocity
◼ The force acting on the machine
Machines and Mechanisms: Applied Kinematic Analysis, 4/e
David Myszka
© 2012, 2005, 2002, 1999 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved.4
1.2 MACHINES AND MECHANISMS
◼ Machine
◼ Devices used to alter,
transmit, and direct forces
to accomplish a specific
objective.
◼ Mechanism
◼ Mechanical portion of a
machine that has the
function of transferring
motion and forces from a
power source to an output.
Machines and Mechanisms: Applied Kinematic Analysis, 4/e
David Myszka
© 2012, 2005, 2002, 1999 Pearson Higher Education,
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1.3 KINEMATICS
To illustrate the importance of such analysis, refer to
the lift platform. Kinematic analysis provides insight into
significant design questions, such as:
◼ What is the significance of the length of the legs that
support the platform?
◼ Is it necessary for the support legs to cross and be
connected at their midspan, or is it better to arrange
the so that they cross closer to the platform?
◼ How far must the cylinder extend to raise the
platform to a certain weight.
Machines and Mechanisms: Applied Kinematic Analysis, 4/e
David Myszka
© 2012, 2005, 2002, 1999 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved.6
Dynamics
Machines and Mechanisms: Applied Kinematic Analysis, 4/e
David Myszka
© 2012, 2005, 2002, 1999 Pearson Higher Education,
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Kinematics/Dynamics
◼ Kinematic analysis
◼ Determine
◼ Position, displacement, rotation, speed, velocity, acceleration
◼ Provide
◼ Geometry dimensions of the mechanism
◼ Operation range
◼ Dynamic analysis
◼ Power capacity, stability, member load
Machines and Mechanisms: Applied Kinematic Analysis, 4/e
David Myszka
© 2012, 2005, 2002, 1999 Pearson Higher Education,
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1.4 MECHANISM TERMINOLOGYMechanism
◼ Design synthesis is the process of developing mechanism to satisfy a set of
performance requirements for the machine.
◼ Analysis ensures that the mechanism will exhibit motion to accomplish the
requirements.◼ Linkage or mechanism
◼ Link–rigid body
◼ Primary joint (full joint)
◼ Revolute joint (pin or hinge joint)–
pure rotation
◼ Translation joint (piston or prism
joint)– linear translation
Machines and Mechanisms: Applied Kinematic Analysis, 4/e
David Myszka
© 2012, 2005, 2002, 1999 Pearson Higher Education,
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◼ Higher-order joint (half joint)
◼ Allow rotation and sliding
◼ Cam joint
◼ Gear connection
◼ Simple link
◼ A rigid body contains only two joints
◼ Complex link
◼ A rigid body contains more than two joints
◼ Point of interest
◼ Actuator
Machines and Mechanisms: Applied Kinematic Analysis, 4/e
David Myszka
© 2012, 2005, 2002, 1999 Pearson Higher Education,
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1.5 Kinematic Diagram
Simple Link
Simple Link
(with point of
interest)
Complex Link
Revolute Joint
Machines and Mechanisms: Applied Kinematic Analysis, 4/e
David Myszka
© 2012, 2005, 2002, 1999 Pearson Higher Education,
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Kinematic Diagram
Cam Joint
Gear Joint
Translation Joint
1.6 Kinematic Model
Machines and Mechanisms: Applied Kinematic Analysis, 4/e
David Myszka
© 2012, 2005, 2002, 1999 Pearson Higher Education,
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Machines and Mechanisms: Applied Kinematic Analysis, 4/e
David Myszka
© 2012, 2005, 2002, 1999 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved.14
Machines and Mechanisms: Applied Kinematic Analysis, 4/e
David Myszka
© 2012, 2005, 2002, 1999 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved.15
1.7 MOBILITY
15
◼ Constrained mechanism: one degree of freedom
◼ Locked mechanism: zero degree of freedom
◼ Unconstrained mechanism: more than one degree of freedom
Machines and Mechanisms: Applied Kinematic Analysis, 4/e
David Myszka
© 2012, 2005, 2002, 1999 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved.16
Machines and Mechanisms: Applied Kinematic Analysis, 4/e
David Myszka
© 2012, 2005, 2002, 1999 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved.17
Machines and Mechanisms: Applied Kinematic Analysis, 4/e
David Myszka
© 2012, 2005, 2002, 1999 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved.18
Machines and Mechanisms: Applied Kinematic Analysis, 4/e
David Myszka
© 2012, 2005, 2002, 1999 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved.19
1.8 COMMONLY USED LINKS AND JOINTS
1.8.1 Eccentric Crank
1.8.2 Pin-in-a-Slot Joint
1.8.3 Screw Joint
Machines and Mechanisms: Applied Kinematic Analysis, 4/e
David Myszka
© 2012, 2005, 2002, 1999 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved.20
Machines and Mechanisms: Applied Kinematic Analysis, 4/e
David Myszka
© 2012, 2005, 2002, 1999 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved.21
1.9 SPECLAL CASES OF THE MOBILITY EQUATION
Coincident Joints
• One degree of freedom actually if pivoted links are the same size
Machines and Mechanisms: Applied Kinematic Analysis, 4/e
David Myszka
© 2012, 2005, 2002, 1999 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved.22
Machines and Mechanisms: Applied Kinematic Analysis, 4/e
David Myszka
© 2012, 2005, 2002, 1999 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved.23
1.10 FOUR-BAR MECHANISM, 4-BAR LINKAGE
Machines and Mechanisms: Applied Kinematic Analysis, 4/e
David Myszka
© 2012, 2005, 2002, 1999 Pearson Higher Education,
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Machines and Mechanisms: Applied Kinematic Analysis, 4/e
David Myszka
© 2012, 2005, 2002, 1999 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved.25
Machines and Mechanisms: Applied Kinematic Analysis, 4/e
David Myszka
© 2012, 2005, 2002, 1999 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved.26
1.11 SLIDER-CRANK MECHANISM
Machines and Mechanisms: Applied Kinematic Analysis, 4/e
David Myszka
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1.12.3 Quick-Return Mechanisms
27
1.12.4 Scotch Yoke Mechanism
Machines and Mechanisms: Applied Kinematic Analysis,
4/eChapter 4 Displacement Analysis
Machines and Mechanisms: Applied Kinematic Analysis, 4/e
David Myszka
© 2012, 2005, 2002, 1999 Pearson Higher Education,
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4.4 DISPLACEMENT ANALYSIS
◼ Locate the positions of all
links as driver link is
displaced
◼ Configuration
◼ Positions of all the links
◼ One degree of freedom
◼ Moving one link will
precisely position all
other links
Machines and Mechanisms: Applied Kinematic Analysis, 4/e
David Myszka
© 2012, 2005, 2002, 1999 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved.30
Machines and Mechanisms: Applied Kinematic Analysis, 4/e
David Myszka
© 2012, 2005, 2002, 1999 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved.31
Machines and Mechanisms: Applied Kinematic Analysis, 4/e
David Myszka
© 2012, 2005, 2002, 1999 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved.32
r1
r2
r4
r3r5
r4
Y
X
X1
Y1
1 2 3 4
3 31 1 2 2
1 1 2 2 3 3
1 1 2 3
3 4 5
3 3 4 4 1
3 3 4 4
4
0
cos 5.30
sin 3.2
3, 90, 4.9, 3.3
0
00.8
10.1
+ + + =
+ − + + + =
+ −
= = = =
− + + =
− + + =
− −
=
r r r r
r cr r c
r r s r s
r r r
r r r
r c r c x
r s r s
r
2 equations for 2 unknowns
2 equations for 2 unknowns
(1)
(2)
2 3,
4 1and x
4.1 Vector Analysis of Displacement
Machines and Mechanisms: Applied Kinematic Analysis, 4/e
David Myszka
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Machines and Mechanisms: Applied Kinematic Analysis, 4/e
David Myszka
© 2012, 2005, 2002, 1999 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved.34
1 2 3 4
3 32
2 3 3
2 2 1/2
3
2 3
0
2.71.6 2.30
1.5 2.7 0
(3.9 1.2 )
+ + + =
− − + + + =
= +
r r r r
r cc
s r s
r
solve for and
X
Y
r1
r4
r3
r2
Machines and Mechanisms: Applied Kinematic Analysis, 4/e
David Myszka
© 2012, 2005, 2002, 1999 Pearson Higher Education,
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Analysis of Mechanism Position
Machines and Mechanisms: Applied Kinematic Analysis, 4/e
David Myszka
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1 2 3
1 2 1
1 2
1 2 1
1
1 2 2
1 2
2 2
1 2
0
50 400
50 40 0
240 ,
15 , 255
50 400
50 40 0
r r r
c c d
s s
solve for and d
when rotate
c c d
s s
solve for and d
d d d
+ + =
+ + =
=
=
+ + =
= −
r3
r1
Y
r2
X
Machines and Mechanisms: Applied Kinematic Analysis, 4/e
David Myszka
© 2012, 2005, 2002, 1999 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved.37
Machines and Mechanisms: Applied Kinematic Analysis, 4/e
David Myszka
© 2012, 2005, 2002, 1999 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved.38
1 2 3 4
31 2
1 2 3
1 2 3
1 2 3
0
1512 20 250
12 20 15 0
90 , .
60 , .
r r r r
cc c
s s s
eqs solve for and
eqs solve for and
Calculate the differenc
+ + + =
− + + + =
() =
() =
2 e of in and () ()
r3
r2
Y
Xr4
r1
Machines and Mechanisms: Applied Kinematic Analysis, 4/e
David Myszka
© 2012, 2005, 2002, 1999 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved.39
1 2 3
1 2
1 2
1 2 1 max
1 2 1 min
0
0.5 1.75 10
0.5 1.75
,
,
r r r
c c
s s y
for solve for and y
for solve for and y
+ + =
+ + =
=
= +
r1
r3r2
X
Y
Machines and Mechanisms: Applied Kinematic Analysis, 4/e
David Myszka
© 2012, 2005, 2002, 1999 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved.40
4.7 Limiting Positions and Stroke
Machines and Mechanisms: Applied Kinematic Analysis, 4/e
David Myszka
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1 2 3 4
1 2 1 3 max
1 2 1 3 min
0
, )
, )
r r r r
for solve for and
for solve for and
+ + + =
=
= +
Y
X
r2
r4
r3
r1
Machines and Mechanisms: Applied Kinematic Analysis, 4/e
David Myszka
© 2012, 2005, 2002, 1999 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved.42
1 2 3 4
2 3 1 max 2
2 3 1 min
0
,
,
r r r r
for solve for and
for solve for
+ + + =
= )
= + )
r1
r2r4
Y
r3
X
Machines and Mechanisms: Applied Kinematic Analysis 4/e
Chapter 6 Velocity Analysis
Machines and Mechanisms: Applied Kinematic Analysis, 4/e
David Myszka
© 2012, 2005, 2002, 1999 Pearson Higher Education,
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6.2 LINEAR AND ANGULAR VELOCITY
6.2.2 Linear Velocity of a General Point
Machines and Mechanisms: Applied Kinematic Analysis, 4/e
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1 2 3 4
2 2 3 3 4 4
3 4
0
0
r r r r
r r r
eqs for unknowns and
+ + + =
+ + =
r2r1
r3
r4
X
Y
Machines and Mechanisms: Applied Kinematic Analysis, 4/e
David Myszka
© 2012, 2005, 2002, 1999 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved.46
1 2 3
1 1 2 2
1 2
1 4
1 1 2 4
0
00
5
E
E
r r r
r r
eqs for and
r r r
r r r
+ + =
+ + =
= +
= +
r1
r2
r3
r4
X
Y
Machines and Mechanisms: Applied Kinematic Analysis, 4/e
David Myszka
© 2012, 2005, 2002, 1999 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved.47
1 2 3
1 1 2 2 2
2
1 2
2
2
0
0
5 min ,
r r r
solve for and
r r r
vcrad r
vs
eqs for unknowns and v
+ + =
+ + =
= =
Y
X
r3
r2
r1
Machines and Mechanisms: Applied Kinematic Analysis, 4/e
David Myszka
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1 2 3
1 1 2 2 2
2
1 2
0
0
0
50
r r r
r r r
r
eqs for unknowns and
+ + =
+ + =
=
r2
r1
r3
X
Y
Machines and Mechanisms: Applied Kinematic Analysis, 4/e
David Myszka
© 2012, 2005, 2002, 1999 Pearson Higher Education,
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1 2 3
2 2 3
1 1
1 1 1 2 2
1
1
1
1 2
0
16, 340
3
0
8
8
r r r
r r
eqs for unknowns r and
r r r
cr
s
eqs for unknowns and
+ + =
− = = =
−
+ + =
=
r1
r3
r2
Y
X
Machines and Mechanisms: Applied Kinematic Analysis, 4/e
David Myszka
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6.10 INSTANTANEOUS CENTER OF ROTATION
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6.12 GRAPHICAL VELOCITY ANALYSIS:
INSTANT CENTER METHOD
Machines and Mechanisms: Applied Kinematic Analysis, 4/e
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1 2 3 4
1 1 2 2 3 3
3 1 2
0
0
r r r r
r r r
given solve for and
+ + + =
+ + =
r1
r2r4
Y
r3
X
Machines and Mechanisms: Applied Kinematic Analysis, 4/e
David Myszka
© 2012, 2005, 2002, 1999 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved.53
1 2 3 4
2 2 3 3 4 4
2 3
4 5 6
4 4 5 5
5
0
0
60 .
0
00
r r r r
r r r
rpm solve for and
r r r
r rv
eqs for unknowns and v
+ + + =
+ + =
=
+ + =
+ + =
r1
r2 r4
r3
r5
r4X
r6
Y
Machines and Mechanisms: Applied Kinematic Analysis, 4/e
David Myszka
© 2012, 2005, 2002, 1999 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved.54
1 2 3 4
1 1 2 2 3 3
1 2 3
0
0
r r r r
r r r
given find and
+ + + =
+ + =
r1r2
r3
r4
X
Y
Machines and Mechanisms: Applied Kinematic Analysis, 4/e
Chapter 7 Acceleration Analysis
Machines and Mechanisms: Applied Kinematic Analysis, 4/e
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7.2 LINEAR ACCELERATION
7.2.1 Linear Acceleration of Rectilinear Points
Machines and Mechanisms: Applied Kinematic Analysis, 4/e
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( ) ( ) ( )
1 2 3 4
1 1 2 2 3 3
1 2 3
1 1 1 2 2 2 2 2 3 3 3 3 3
2 3
0
0
0
r r r r
r r r
given find and
r r r r r
solve for and
+ + + =
+ + =
+ + + + =
r2r1
r3
Y
r4
X
Machines and Mechanisms: Applied Kinematic Analysis, 4/e
David Myszka
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pr
r5
( )
( ) ( ) ( ) ( )( )
1 2 3 5
1 1 2 2 3 3 5
1 1 1 2 2 2 2 2 3 3 5 3 3 3 5
p
p
p
r r r r r
r r r r r
r r r r r r r r
= + + +
= + + +
= + + + + + +
The displacement, velocity, and acceleration of point P at the rocker
Machines and Mechanisms: Applied Kinematic Analysis, 4/e
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( ) ( )
1 2 3
1 1 2 2
2
1 1 1 1 1 2 2 2 2 2
2
0
00
00
r r r
r rv
solve for and v
r r r ra
solve for and a
+ + =
+ + =
+ + + + =
r2 r3
r1 X
Y
Machines and Mechanisms: Applied Kinematic Analysis, 4/e
David Myszka
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( ) ( )
( )
1 2 3
1 2
1 1 2 2 2
2
1 2 2
2
1 1 1 2 2 2 2 2 2 2 2 2 2
2
1 1 1
2
0
0
400, ,
0
r r r
solve for and
r r r
vcr solve for and v
vs
r r r r r r
acr
as
+ + =
+ + =
= =
+ + + + + =
+ +
( )2 2 2 2 2 2 2
2
2 0r r r
solve for and a
+ + =
r3r2
r1
Y
X
Machines and Mechanisms: Applied Kinematic Analysis, 4/e
Chapter 9 Cams
Machines and Mechanisms: Applied Kinematic Analysis, 4/e
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9.1 INTRODUCTION
Machines and Mechanisms: Applied Kinematic Analysis, 4/e
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Plate cam
Cylindrical cam
Linear cam
Machines and Mechanisms: Applied Kinematic Analysis, 4/e
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9.2 TYPES OF CAMS
Follower motion
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Follower position
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9.3 TYPES OF FOLLOWERS
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9.11 THE 4-STATION GENEVA MECHANISM
Constant rotation producing index motion
Machines and Mechanisms: Applied Kinematic Analysis, 4/e
Chapter 13 STATIC FORCE ANALYSIS
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13.3 MOMENTS AND TORQUES
X
Y
r
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13.5 FREE-BODY DIAGRAMS
13.5.1 Drawing a Free-Body Diagram
13.5.2 Characterizing Contact Forces
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13.6 STATIC EQUILIBRIUM
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13.7 ANALYSIS OF A TWO-FORCE MEMBER
Machines and Mechanisms: Applied Kinematic Analysis, 4/e
Chapter 14 DYNAMIC FORCE ANALYSIS
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Machines and Mechanisms: Applied Kinematic Analysis, 4/e
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Mechanism Dynamics of a Slider Crank System
1 1 xO xBm X F F = +
1 1 1yO yBm Y F F m g = + −
1 1 1 11 1 1 1 1 1sin cos cos sin
2 2 2 2xO yO xB yB
L L L LI F F F F
= + − − +
FxO
m1g
FxB
FyB
FyO
OB
Consider link OB, BA, and piston each has mass m1, m2, and m3, and
mass moment of inertia I1, I2, I3.
In gravitational field,
(1) Define inertial coordinates and let the mass center of OB be at its
midpoint. Consider a torque Γ is applied at link OB. Draw the
free body diagram of all the rigid bodies by representing all
forces in X-Y components.
(2) Write the equation of motion of all the rigid bodies.
(3) Count the number of equations and the number of unknowns in
(2)
2 2 xA xBm X F F = −
2 2 2yA yBm Y F F m g = − −
2 2 2 22 2 2 2 2 2
2 2 5 5cos sin cos sin
7 7 7 7xB yB xA yA
L L L LI F F F F
= + + −
3 3 xA xAm X N F = −
3 3 3yA yAm Y N F m g = − −
3 3 0 0 0 0xA yA xA yAI N N F F = + + +
1 1 1 2 2 2 3 3 3 xA : X , Y , , X , Y , , X , Y , , F , F , F , F , F , F , N , N (17)xO yO xB yB yA xA yAUnknown
: 9Number of equations
m2g
FyA
FxA FyB
FxB
G
m3g
NyA
NxA
FxA
FyA
AB
Piston A
Review of Applied Mechanics - Dynamics
(Meriam and Kraige, Ed. , 2013)
Chapter 1. Introduction
Engineering Mechanics
Statics
Dynamics
Strength of Materials
Vibration
Statics: ,distribution of reaction force from the applied
force
Dynamics: , x(t)= f(F(t)) displacement as a function of
time and applied force
Strength of Materials: δ = f(P) deflection and applied force on deformable
bodies
Vibration: x(t) = f(F(t)) on particles and rigid bodies78
a r(F +F )=mx
a r(F +F )=0 rF
aF
Newtonian Dynamics‧Kinematics: the relation among
‧Kinetics: the relation between
( ) and ( )x t F t
2
2
( ) ( )( ), ( ), and ( ), and
dx t d x tx t x t x t x x
dt dt
without reference to applied force
Terms to Know‧Reference frame: Coordinate system
‧Inertial System: Newton’s 2nd Law of motion
‧Particle and Rigid body
‧Scalar and Vector
79
Chap. 2 Kinematics of Particles
Rectangular Coordinates
Cylindrical Coordinates
Spherical Coordinates
( , , )x y zr
( , , )r zr
( , , )R r
80
81
fig_05_001
Chap.5 Planar Kinematics
82
fig_05_00
4
( )
=
= +
r
r r
r r r
ω
ω ω ω
displacement
velocity
acceleration
Fixed Axis RotationZ
Y
X
83
sp_05_03_01
2
2 (0.4 0.3 ) 0.6 0.8 m / s
2 (0.6 0.8 ) 4 (0.4 0.3 )
2.8 0.4 m / s
= − + = +
= − + + +
= − +
k i j i j
k i j k i j
i j
r
r
2 rad/s= − k
24 rad/s= + k
2 20.6 0.8 1 m/s = + =
2 22.8 0.4 2.83 m/sa = + =
Sample 5/3
The right-angle bar rotates
clockwise with an angular
acceleration
Write the vector expressions for
the velocity and acceleration of
point A when
[ ]
[ ( ) ]
[ ]
[ ]
=
= +
=
= +
t
n t
a
a a a
r r
r r r
r
84
fig_05_005
General Motion: Rotation + Translation
fig_05_006
85
fig_05_007
Instantaneous Center
86
Sample 5/7
3 0 0 10
0.1732 0.1 0
4 1.732 m/s
= + −
−
= +
A
i j k
v i
i j
2 24 (1.732) 19 4.36 m/sA = + = =v
A Ar r r= = +
10 rad/ s
0.2( cos30 sin30 ) 0.1732 0.1 m
3 m/ sO
−
= − + = − +
=
k
i j i j
v i
ω =
Calculate the velocity of point A
on the wheel without slipping for
the instant.
sp_05_07_0
1
x
ω
A Or r= + = + v ω
87
Sample 5/11
for 0,
p
p
p
r r
r r
r r
r
= +
= +
= = −
= − i
Locate the instantaneous center of
zero velocity and use it to find the
velocity of point A for the position
indicated.
Y
X
point C is the instantaneeous center of zero velocity
from the mass center
88
sp_05_08_0
1
Sample 5/8
1 2 3
1 1 2 2 3 3
1 2
1 2
2
1 2
0 100 (175 50 ) 2 75
100 50
175 150
3 6,
7 7
ω ω
ω ω
ω
ω ω
= + +
= + +
= + − +
− =
− −
= − = −
r r r r
r r r r
k j k i j k i
ω ω ω
r
r1r2
r3
For the velocity at point P
P
1 2
1
2p = +r r r
1 1 2 2
1
2p = + r r rω ω
Crank CB oscillates about C through a limited arc, causing crank OA to oscillate about O. When the linkage passes the position shown with CBhorizontal and OA vertical, the angular velocity of CB is 2 rad/s counterclockwise. For this instant, determine the angular velocities of OAand AB.
89
Sample 5/9
2r
1r
3r
y
x
The common configuration of areciprocating engine is that of theslider-crank mechanism shown. Ifthe crank OB has a clockwiserotational speed of 1500 rev/min,determine for the position where θ
= 60° the velocity of the piston A,
the velocity of point G on theconnecting rod, and the angularvelocity of the connection rod.
90
sp_05_09_01
Sample 5/9
r
r1r2
y
x
1 2
1 1 2 2
1 1 1 1 1
2 2 2 2 2
( )
( )
= +
= +
= +
+ +
r r r
r r r
r r r
r r
ω ω
ω ω ω
ω ω ω
91
sp_05_08_0
1
Sample 5/14
1 2 3
1 1 2 2 3 3
1 1 1 1 1
2 2 2 2 2
3 3 3 3 3
( )
( )
( )
= + +
= + +
= +
+ +
+ +
r r r r
r r r r
r r r
r r
r r
ω ω ω
ω ω ω
ω ω ω
ω ω ωr
r1
r2
r3
1 2
3 3100 ( ) ( 100 ) (175 50 )
7 7
6 6( [( ) (175 50 )] 0 2 (2 75 )
7 7
= + − − + −
+ − − − + +
r k j k k j k i j
)k k i j k k i
21 0.1050 rad/s = − 2
2 4.34 rad/s = −
P
For the acceleration at point P1 1 1 1 1 2 2 2 2 2
1 1( ) ( )
2 2p = + + + r r r r rω ω ω ω ω ω
1 2
1 1 2 2 2
2
1
2
2
1 2
2 2
1 2 1
0
r cos(225 0 )
r sin
( 2 ) (225 225 )
0 r cos 4500
255 r sin 450
1r 450 0, r 450 2
2
1225 r 450 0, 4
2
= +
= = + +
= + +
+ − +
= + + = −
+ = = −
+ − = =
r r r
r r r r
k i j
k i j
1 1 (4)(225) 900A
= = =r r j j
92
Sample 5/17 and 5/18
Y
X
r
1r
2r
counter clockwise
The pin A of the hinged link AC is confined to move in the rotating slot of link OD. The angular velocity of OD is ω=2 rad/s clockwise and is constant for the interval of motion concerned. For the position where θ=45° with
AC horizontal, determine the velocity of pin Aand the velocity of A relative to the rotating slot in OD.
93
1 1 1 1 1 2 2 2 2 2 2
222 21
21 2 2
2 2
2
2
1 2
2
0 ( ) ( )
0 r cos 2252250 0
225 r sin 2250
10 ( 225)4 r ( 225)(2) 0
2
1225 r ( 225)(2) 0
2
4500 2
= = + + + +
− − = + + + + =
−
+ − + + − =
+ + − =
=
r r r r r r
r ,
1
1 1 1 1 1
2
1
16
( )
16 (225 ) (225 )
3600 3600
A
= −
= +
= +
= − +
r r r
k i i
i j
Sample 5/17 and 5/18
94
( )
=
= +
r
r r
r r r
ω
ω ω ω
displacement
velocity
acceleration
With and without Body-Fixed Coordinate
Let
( )
5 , 2 , 6
10
-20 30
= = =
= =
= +
r i k k
r r j
r r r
= i + j
95
fig_05_011
Body-Fixed Coordinates in Rotation
( )
( )
A B
A B
A B
B
= +
= +
= + + +
= + + +
r r
r r
r r
r
ρ
ω ρ+ ρ
ω ω ρ ω ρ ω ρ+ω ρ+ ρ
ω ω ρ ω ρ+ 2ω ρ ρ
Ar
Coriolis Acceleration
( ) 2A B= + + +
r r ω ω ρ ω ρ ρ+ ω ρ
Coriolis acceleration
ρ
96
p_05_161
( sin cos )
2
2 ( sin cos )
2 sin
cor
= − +
=
= − +
=
j k
k j k
i
a
(west ward)
The track provides the necessary westward
acceleration so that the velocity vector is properly
rotated and reduced in magnitude.
For 500 km/h =
(a) Equator,
(b) North pole,
0 0cora = =
5 250090 2(7.292 10 ) 0.0203 m/s
3.6cora −= = =
problem 05/163
B y
z
A
97
z
y
5
2
sin(sin cos )
cos
2
(sin cos )
= 2 sin (Eastward)
500 = 2 (7.292 10 ) sin
36
0.0203sin
cor
νν
ν
k ν
ν
−
= = −
=
= −
−
−
= −
j k
j k
i
i
i (m / s )
r
Cruise Missile on North-to-South Track
98
Chap. 6 Dynamics of Planar Rigid BodyEquation of Motion
‧The resultant of the external forces equals to the inertia at the mass center.
‧The resultant moment about C.G. of the external forces equals to the rate
change of the angular moment about C.G.
G
G G
m =
=
F r
M H
99
Equation of Motion in 2D
( )
( )2
G i i i
i i i
m
m
dm
I
=
=
=
=
H ρ ρ
ρ ω ρ
ω
ω
Angular momentum
c
G
m
I
=
=
F r
M ω
100
Moment about a Fixed Point
0, 0,P = =r ρif P is fixed or if
P G C
G C
m
I m
= +
= +
M H ρ r
ω ρ r
P P PI m = + M ω ρ r
P PI =M ωthen
101
when tipping impends
= =F mg mx
+ (0.4) (0.6) 0 = − + = = GM mg mg I
2
3 =
•
AN
AF
G
0.8m
1.2m
Determine the force P which would
cause the cabinet to begin to tip. What
coefficient μs of static friction is
necessary to ensure that tipping occurs
without slipping?
=AN mg
2
3= =x g g
1( ) (50 10) 60 40= + = + = =P m m x x x g
mg
102
Center of Percussion
or
C
G G
O O
m
M I
M I
=
=
=
F r
ω
The resultant force must pass
thru Q. The sum of the
moment of all forces about the
center of percussion is zero.
ok
ok Radius of gyration about point O
Fixed-Axis Rotation
0QM =
103
Center of Percussion
XR
YR
h
mg
F
y
xo
All forces pass through
0
1
2
X
Y
O
F R mx
mg R my
F h I
y
x
− + =
− + =
=
=
= −
2
3
0x
y
h
R
R mg
=
=
=
21I
3O ml=
104
2
2
120(9.81)(0.2) 20(0.4)
3
36.8rad/s
=
=
=
O OM I
2( 0.2)(36.8) 7.36m/s
x r
x
− =
= − = −
G0.2m 0.2m
t
20(9.81) N
2F
O20(9.81) 2 20( 7.36)
12 49.0
4
24.5N
=
− + = −
= =
= = =
x
A B
F mx
F
F mg
F F F
problem 06/34
The 20-kg uniform steel plate is freely hinged about the z-
axis as shown. Calculate the force supported by each of the
bearings at A and B an instant after the plate is released
from rest in the horizontal y-z plane.
y
x
105
21cos
2 3
cos2
sin2
x
y
R mx
R mg my
Lmg mL
Lx
Ly
=
− =
=
=
= −
2
2
( cos sin )2
( sin cos )2
Lx
Ly
= − −
= −
2
2 2
3(3) cos
2
1
2
3 1 3 3cos sin ; sin
2 2 2
g
L
d d
g g gd
L L L
=
= =
= = =
2( cos sin )2
3 3( cos sin cos )
2 2
9cos sin
4
x
mLR mx
mL g gsin
L L
mg
= = − −
= − −
= −
90, 0 and 45 ,
8x xwhen R R mg = = = = −
(1)
(2)
(3)
(4)
(5)
(1)
mg
X
Y
Ry
Rx
106
2
2 2
( sin cos )2
3 3( sin cos cos )
2 2
3(2sin cos )
4
y
mLR my mg mg
mL g gsin mg
L L
mg mg
= + = − +
= − +
= − +
10,
4
1145 ,
8
590 ,
2
y
y
y
R mg
R mg
R mg
= =
= =
= =
when(2)
2 23 3sin ,
g gc
L L = = Let
The time required to reach vertical position.
( )
( )
12
122
0
sin
1sin
c
dt dc
−
=
=
107
mg
y
O
r GO
mg
O
r GO
22
/ 2
( )2
/ 2
= =
=
= − =
=
O O
y
M I mgr mr
g r
gF my mg O mr
r
O mg
(a)
(b)2 21
( )2
2 / 3
2( )
3
/ 3
= = +
=
= − =
=
O O
y
M I mgr mr mr
g r
gF my mg O mr
r
O mg
problem 06/38
Determine the angular acceleration and the force on the
bearing at O for (a) the narrow ring of mass m and (b) the
flat circular disk of mass m immediately after each is
released from rest in the vertical plane with OC horizontal.
Sample 6/7
108
The slender bar AB weighs 60 lb and moves in the vertical plane, with its
ends constrained to follow the smooth horizontal and vertical guides. If the
30-lb force is applied at A with the bar initially at rest in the position for
which θ = 30°, calculate the resulting angular acceleration of the bar and
the forces in the small end rollers at A and B.
Kinematicscos
sin
x a
y a
a R
=
=
=
When taking point C as instaneous center
a
109
b
mg
C
45A
r I
Ama/( )G A tm a
r=b
G
T
2
2
1( ) ( )( )
2 6 2 2
3
4
1 3( )( )6 42
2 2(12)(9.81) 20.8 N
8 8
G
G
A
G
M I mad
mgb b bmb m
g
b
M I
b gT mb
b
T mg
= +
= +
=
=
=
= = =
problem 06/84
The uniform 12-kg square panel is suspended from
point C by the two wires at A and B. If the wire at B
suddenly breaks, calculate the tension T in the wire
at A an instant after the break occurs.
It was before
cut, i.e., the tension becomes
1/4.
1/ 2 T mg=
110
Sample 6/5
A metal hoop with a radius
r = 6 in. is released from
rest on the 20° incline. If the
coefficients of static and
kinetic friction are μs = 0.15
and μk = 0.12, determine
the angular acceleration α
of the hoop and the time for
the hoop to move a
distance of 10 ft down the
incline.
111
Sample 6/5[ ]x xF ma= sin 20mg F ma− =
[ 0]y yF ma= = cos20 0N mg− =
[ ]GM I =2=Fr mr
0.1710 ,F mg= cos20 0.940 ,N mg mg= =
max[ ]sF N= max 0.15(0.940 ) 0.1410F mg mg= =
max[ ]kF N= 0.12(0.940 ) 0.1128F mg mg= =
x[ ]xF ma= sin 20 0.1128mg mg ma− =20.229(32.2) 7.38 ft/seca = =
[ ]GM I = 20.1128 ( )mg r mr = 20.1128(32.2)7.26 rad/sec
6 /12 = =
21[ ]
2x at=
2 2(10)1.646 sec
7.38
xt
= = =
Assume pure rolling 4 equations for 4 unknowns
,a r=
Check if the assumption valid. The friction force be bounded by N
So it is slipping Solve again the 4 unknowns .= kf N
105
Sample 6/7
2cos30 2 cos30 1.732 ft/secxa a = = = 2sin30 2 sin30 1.0 ft/secya a = = =
[ ]GM I =21 60
30(2cos30 ) (2sin30 ) (2cos30 ) (4 )12 32.2
A B − + =
[ ]x xF ma=
[ ]y yF ma=
6030 (1.732 )
32.2B − =
6060 (1.0 )
32.2A − =
68.2 lbA = 15.74 lbB =24.42 rad/sec =
[ ]CM I m d = + 21 6030(4cos30 ) 60(2sin30 ) (4 )
12 32.2
60 60(1.732 )(2cos30 ) (1.0 )(2sin30 )
32.2 32.2
− =
+ +
4.39 9.94=
24.42 rad/sec =
[ ]y yF ma=[ ]x xF ma=
6060 (1.0)(4.42)
32.2A − =
6030 (1.732)(4.42)
32.2B− =
68.2 lbA =
15.74 lbB =
P B mx− =
A mg my− =
cos sin cos2 2 2
G
L L LP A B I − + =
113
mg
r
x
•
A
N
0 0OM I I M= = = 0s =Hence no friction force and
sinx x AF ma mg mr = = sinA
g
r =
problem 06/82
Determine the angular acceleration of each of the two wheels
as they roll without slipping down the inclines. For wheel A
investigate the case where the mass of the rim and spokes is
negligible and the mass of the bar is concentrated along its
centerline. Also specify the minimum coefficient of static
friction μs required to prevent each wheel from slipping.
sinxa g =
G
114
N
CF
mg
r
x
•G
B
2sin 2
sin2
C C B
B
M I mgr mr
g
r
= =
=
1sin / cos
2
1tan
2
s
s
Fmg mg
N
= =
=
2 sin2
gM I Fr mr
r = =
1sin
2F mg =
problem 06/82
Determine the angular acceleration of each of the two wheels as they roll
without slipping down the inclines. For wheel B assume that the
thickness of the rim is negligible compared with its radius so that all of
the mass is concentrated in the rim. Also specify the minimum
coefficient of static friction μs required to prevent each wheel from
slipping.
cos max
1cos sin max
2
1(cos sin )
2x
mg F
mg mg
a g
= =
− =
= −
115
Inclined Plane : Pure Rolling Motion
yx
mg
FN
cos
sin
0
G
N mg my
F mg mx
FR I
x R
y
− =
− =
= −
= −
=
2 2
2
2
2
2
sin( ) sin
cos
sin ( )
sin
0
G G
G G
G
G G G
G
I ImgF mg
IR mR Im
R
N mg
Img mx F m x
R
I I IxF x
R R R R
mgx
Im
R
y
= =+
+
=
− = − = +
− −= = = −
= −
+
=
check if 2sin cosG
G
IF mg mg
mR I =
+
116
Inclined Plane : Rolling with Slipping Motion
cos
sin
0
G
N mg my
F mg mx
FR I
F N
y
− =
− =
=
=
=
cos
cos sin
( cos sin )
cos
cos
0
G
N mg
x mg mg
mg
F mg
mgR
I
y
=
= −
= −
=
=
=
Let2
GI cmR=
( ) sin , cos1
sin cos , tan1 1
cF mg mg
c
c c
c c
= +
+ +
11 , tan
2
1 1, tan
2 3
0 , tan
c ring
c disk
c punt
=
=
=
for pure rolling
Computation of Multibody Kinematics And Dynamics1.1 Multibody Mechanical Systems
2D Planar Mechanism
117
3D Spatial Mechanism
Computer-Aided Design (CAD)
• Mechanics: statics and dynamics.
• Dynamics : kinematics and kinetics.
• Kinematics is the study of motion, i.e., the study of displacement, velocity, and acceleration, regardless of the forces that produce the motion.
• Kinetics or Dynamics is the study of motion and its relationship with the forces that produce that motion.
118
1.2 Coordinate Systems
2 2 2 2
1 1( ) 2 cos 2 cos 2 cos( ) 0r l s d rl ls rs + + − − + − − =
2 2 2 2
2( ) 2 cos 2 cos 0r l s d rl ds + + − − + =
1 2 3 2 0 + + + − =
1 2 3[ ]T =q
A four-bar mechanism with generalized coordinates q.
degrees of freedom 4 – 3 = 1
4 Coordinates
3 Constraints
119
d
r s
l
1
2
3
=
q
1 2 3
1 2 3
cos cos cos 0
sin sin sin 0
r d s l
r d s
+ + − =
+ + =
3 coordinates
dof 3 2 1= − =
2 constraints
120
2 algebraic constraint equations,
1 1 2 2 3 3 1
1 1 2 2 3 3 2
cos cos cos 0
sin sin sin 0
l l l d
l l l d
+ − − =
+ − − =
four-bar mechanism.
Generalized Coordinates
3 generalized coordinates,
121
1 1 1 2 2 2 3 3 3
Tx y x y x y =q
1 1
1 1
1 1 2 2
1 1 2 2
2 2 3 3
2 2 3 3
3 3
3 3
cos 02
sin 02
cos cos 02 2
sin sin 02 2
cos cos 02 2
sin sin 02 2
cos 02
sin 02
rx
ry
r dx x
r dy y
d sx x
d sy y
sx l
sy
− =
− =
+ − + =
+ − − =
+ − − =
+ − − =
− − =
− =
Cartesian Coordinates
dof 9 8 1= − =
9 coordinates and 8 kinematic constraints
101
Generalized Relative Cartesian
coordinates coordinates coordinates
Number of coordinates Minimum Moderate Large
Number of second-order Minimum Moderate Large
differential equations
Number of algebraic None Moderate Large
constraint equations
Order of nonlinearity High Moderate Low
Derivation of the Hard Moderate Simple
equations of motion hard
Computational Efficient Efficient Not as efficient
efficiency
Development of a Difficult Relatively Easy
general-purpose difficult
computer program
†
Coordinates Systems
123
1.3 Computation Kinematics• A mechanism is a collection of links or bodies kinematically connected to one another.
• An open-loop mechanism may contain links with single joint.
• A closed-loop mechanism is a closed chain, wherein each link is connected to at least two other
links of the mechanism.
Figure (a) Open-loop mechanism-double pendulum
and (b) closed-loop mechanism-four-bar linkage.
124
Figure (a) Single-loop mechanism and (b) multi-loop mechanism.
Single and Multi-Loop Mechanism
• Kinematically equivalent.
A double parallel-crank mechanism and its kinematically equivalent.
Redundant Constraint
125
Quick-return mechanism: (a) schematic presentation and (b) its
equivalent representation without showing the actual outlines.
Kinematics of MechanismMainly composed of revolute joint and translation joint
Example of kinematic pairs: (a) revolute joint, (b) translational joint,
(c) gear joint, (d) cam follower joint, (e) screw joint, and (f) spherical ball joint.
High and Low Pair of Kinematic Joint
126
1 2 3, , , , ,T
i c c c ix y z q
127
2.1 Planar Kinematics in Cartesian Coordinates
P P
i i i
P P
i i i i
s= +
= +
r r
r r A s
The column vector is the vector of body-fixed coordinates
for body in a plane.
[ , , ]T
i c c c ix y qi
is the vector of body-fixed coordinates for body
in three-dimensional space
i
Position vector at global coordinates
X Y−
−Inertial coordinates
Body-fixed coordinates
iA is coordinate transformation matrix
P
ir
ir
P
is P
ir
ir
P
is
2.1 Planar Kinematics in Cartesian
Coordinates
P P
i i i
P P
i i i i
s= +
= +
r r
r r A s
cos sin
sin cosi
i
− =
A
Coordinate transformation matrix
128
cos sin
sin cos
c
c i
xx
yy
− = +
p
ir
ir
p
is
0P P
i i j j+ − − =r s r s
( ,2) 0r P P
i i j j + − − =i jr A s r A s
( ,2)cos sin cos sin 0
sin cos sin cos 0
P P P P
i i i i i j j j j jr
P P P P
i i i i i j j j j j
x x
y y
+ − − − + = + − − − +
Revolute joint at point connecting body and .P i j
2.2 Revolute Joint Constraint Model
129
• A constraint equation describing a condition on the vector of coordinates of
a system can be expressed as follows:
• In some constraints and driving functions, the time variable may appear
explicitly:
• Constraint Jacobian Matrix by differentiating the constraint equations
( ) 0 =q
( ), 0t =q
Constraint Equation
130
1
0
0 often denoted as 0
0
also denoted as 0 or
=
= =
=
+ =
−
( )
, ,
( ) ,
( ) ,
m n
q q
q q q
m
q q q
q
q qq
q + qq q
q q q
q = q q
1 1m n ,q
1 cos sin cos sin 0P P P P
i i i i i j j j j jx x + − − − + =
2 sin cos sin cos 0 + + − − − =P P P P
i i i i i j j j j jy y
Time Derivative of Revolute Joint Constraint
yi+ (x
i
P cosfi-h
i
P sinfi)fi- y
j- (x
j
P cosfj-h
j
P sinfj)f
j= 0
xi- (x
i
P sinfi+h
i
P cosfi)fi- x
j+ (x
j
P sinfj+h
j
P cosfj)f
j= 0
131
1 cos sin cos sin 0 + − − − + =P P P P
i i i i i j j j j jx x
Time Derivative of Revolute Joint Constraint
2 sin cos sin cos 0 + + − − − =P P P P
i i i i i j j j j jy y
2
2
( sin cos ) ( cos sin )
( sin cos ) ( cos sin ) 0
P P P P
i i i i i i i i i i i j
P P P P
j j j j j j j j j j
x x
− + − − −
+ + + − =
2
2
( cos sin ) ( sin cos )
( cos sin ) ( sin cos ) 0
P P P P
i i i i i i i i i i i j
P P P P
j j j j j j j j j j
y y
+ − − + −
− − − + =
xi- (x
i
P sinfi+h
i
P cosfi)fi- x
j+ (x
j
P sinfj+h
j
P cosfj)f
j= 0
yi+ (x
i
P cosfi-h
i
P sinfi)fi- y
j- (x
j
P cosfj-h
j
P sinfj)f
j= 0
or
132
( ) ( )
( ) ( )
( ) ( )
( ) ( )
2 2
2 2
2 2
2 2
2 2
cos sin cos sin
sin cos sin cos
= = s s
P P P P
i i i i i j j j j j
P P P P
i i i i i j j j j j
P P
i i i j j j P P
i i j jP P
i i i j j j
x x x x
y y y y
− − + = − − + +
− − − − − − −
2.3 Translational Joint Constraint Model
Figure Different representations of a translational joint.
133
Figure A translational joint between bodies and .i j
0T
i =n d
0
P P
i jP R P R
i i i i P P
i j
x xx x y y
y y
− − − = −
0 0
( )( ) ( )( ) 0
( ) 0
P P P Q P P P Q
i j i i i j i j
i j i j
x x y y y y x x
+ − − + − − = = − − −
Translational Joint
-
P P
i j
P P
i j
x x
y y
−=
− d
P R
i i
P R
i i
x x
y y
−=
− n
-Q P
i i
Q P
i i
x x
y y
−=
− s
( )( ) ( )( )
( ) 0
= 0
P P
i jT T Q P Q P
i i i i P P
i j
P P P Q P P P Q
i j i i i j i i
x xy y x x
y y
x x y y y y x x
− = = − − = −
− − + − − =
n d s d
134
Constraint Jacobian Matrix
( ) 0
0, 0
( ) 0
( )
q
q q q
q q q
Denote as
=
= =
+ =
−
q
q qq
q q q
q = q q
135
0 0
( )( ) ( )( ) 0
( ) 0
P Q P P P Q P P
i i j i i i j i
i j i j
x x y y y y x x
− − − − − = = − − −
Time Derivative of Translational Joint Constraint
22[( )( ) ( )( )] [( )( ) ( )( )]P Q P Q P Q P Q
i i i j i i i j j i i i j i i i j ix x x x y y y y x x y y y y x x = − − − + − − − − − − − −
136
Driving Link
Figure (a) The motion of the slider is controlled in the direction
and (b) the motion of point is controlled in the direction.
( ) 0ix d t − =
( ) 0P
iy d t − =
P
x −y −
137
A Matlab Program for
Kinematics Analysis
138
3.1 Kinematic Analysis
z1-h
1
groundTreat ground as a rigid
body
139
Example of a four-bar linkage
mm180
mm180
mm260
mm80
=
=
=
=
OC
BC
AB
OA
A
B
CO
y
x
1
1
1
2
2
3
3
3
2
140
1 1
1 1
1 1 2 2
1 1 2 2
2 2 3 3
2 2 3 3
3 3
3 3
8 constraint equations:
40 0
40 0
40 130 0
40 130 0
130 90 0
130 90 0
90 180 0
90 0
1 constraint from driving l
x
y
x x
y y
x x
y y
x
y
− + =
− + =
+ − + =
+ − + =
+ − + =
+ − + =
+ − =
+ =
cos
sin
cos cos
sin sin
cos cos
sin sin
cos
sin
1
ink
2 2 0t − − =
T
1 1 1 2 2 2 3 3 3To solve the 9 equations for 9 unknowns x , y , , x , y , , x , y , =q
Jacobian matrix and velocity
equations
141
1
2
2
2 3
2 3
3
3
1 0 40sin 0 0 0 0 0 0
0 1 40cos 0 0 0 0 0 0
1 0 40sin 1 0 130sin 0 0 0
0 1 40cos 0 1 130cos 0 0 0
0 0 0 1 0 130sin 1 0 90sin
0 0 0 0 1 130cos 0 1 90cos
0 0 0 0 0 0 1 0 90sin
0 0 0 0 0 0 0 1 90cos
0 0 1 0 0 0 0 0 0
x
− −
− − − −
− − − −
− −
1
1
2
2
2
3
3
3
0
0
0
0
= 0
0
0
0
2
y
x
y
x
y
Let to solve for the velocity =q J J βq q
Acceleration equations
142
2
2
2 3
2 3
3
3
1 0 40sin 0 0 0 0 0 0
0 1 40cos 0 0 0 0 0 0
1 0 40sin 1 0 130sin 0 0 0
0 1 40cos 0 1 130cos 0 0 0
0 0 0 1 0 130sin 1 0 90sin
0 0 0 0 1 130cos 0 1 90cos
0 0 0 0 0 0 1 0 90sin
0 0 0 0 0 0 0 1 90cos
0 0 1 0 0 0 0 0 0
x
− −
− − − −
− − − −
− −
21 1 1
21 1 1
2 21 1 1 2 2
2 22 1 1 2 2
2 22 2 2 3 3
2 22 2 2 3 3
3 3
3
3
40cos
40sin
40cos 130cos
40sin 130sin
130cos 90cos
130sin 90sin
90cos
y
x
y
x
y
+
+ = +
+
2
3
2
3 390sin
0
To solve for the acceleration Jq = γ q
Solution procedure
143
Determine with ( ) and ( ) to solve ( )t t tγ q q q
Initialize and start at = 0 t
Call nonlinear solver with initial position to solve ( )tq
Determine Jacobian matrix and with ( ) to solve ( ) t tq q
Assume ( + ) and set it as the new initial position of next iterationt tq
Plot the time response of ( ), ( ) and ( )t t tq q q
Determine the positions of the bars and animate them
m.file (1)1. % Set up the time interval and the initial positions of the nine coordinates
2. T_Int=0:0.01:2;
3. X0=[0 50 pi/2 125.86 132.55 0.2531 215.86 82.55 4.3026];
4. global T
5. Xinit=X0;
6.
7. % Do the loop for each time interval
8. for Iter=1:length(T_Int);
9. T=T_Int(Iter);
10. % Determine the displacement at the current time
11. [Xtemp,fval] = fsolve(@constrEq4bar,Xinit);
12.
13. % Determine the velocity at the current time
14. phi1=Xtemp(3); phi2=Xtemp(6); phi3=Xtemp(9);
15. JacoMatrix=Jaco4bar(phi1,phi2,phi3);
16. Beta=[0 0 0 0 0 0 0 0 2*pi]';
17. Vtemp=JacoMatrix\Beta;
18.
19. % Determine the acceleration at the current time
20. dphi1=Vtemp(3); dphi2=Vtemp(6); dphi3=Vtemp(9);
21. Gamma=Gamma4bar(phi1,phi2,phi3,dphi1,dphi2,dphi3);
22. Atemp=JacoMatrix\Gamma;
23.
24. % Record the results of each iteration
25. X(:,Iter)=Xtemp; V(:,Iter)=Vtemp; A(:,Iter)=Atemp;
26.
27. % Determine the new initial position to solve the equation of the next
28. % iteration and assume that the kinematic motion is with inertia
29. if Iter==1
30. Xinit=X(:,Iter);
31. else
32. Xinit=X(:,Iter)+(X(:,Iter)-X(:,Iter-1));
33. end
34.
35.end 144
m.file (2)
36.% T vs displacement plot for the nine coordinates
37.figure
38.for i=1:9;
39. subplot(9,1,i)
40. plot (T_Int,X(i,:))
41. set(gca,'xtick',[], 'FontSize', 5)
42.end
43.% Reset the bottom subplot to have xticks
44.set(gca,'xtickMode', 'auto')
45.
46.% T vs velocity plot for the nine coordinates
47.figure
48.for i=1:9;
49. subplot(9,1,i)
50. plot (T_Int,V(i,:))
51. set(gca,'xtick',[], 'FontSize', 5)
52.end
53.set(gca,'xtickMode', 'auto')
54.
55.% T vs acceleration plot for the nine coordinates
56.figure
57.for i=1:9;
58. subplot(9,1,i)
59. plot (T_Int,A(i,:))
60. AxeSup=max(A(i,:));
61. AxeInf=min(A(i,:));
62. if AxeSup-AxeInf<0.01
63. axis([-inf,inf,(AxeSup+AxeSup)/2-0.1 (AxeSup+AxeSup)/2+0.1]);
64. end
65. set(gca,'xtick',[], 'FontSize', 5)
66.end
67.set(gca,'xtickMode', 'auto')
145
m.file (3)
68.% Determine the positions of the four revolute joints at each iteration
69.Ox=zeros(1,length(T_Int));
70.Oy=zeros(1,length(T_Int));
71.Ax=80*cos(X(3,:));
72.Ay=80*sin(X(3,:));
73.Bx=Ax+260*cos(X(6,:));
74.By=Ay+260*sin(X(6,:));
75.Cx=180*ones(1,length(T_Int));
76.Cy=zeros(1,length(T_Int));
77.
78.% Animation
79.figure
80.for t=1:length(T_Int);
81. bar1x=[Ox(t) Ax(t)];
82. bar1y=[Oy(t) Ay(t)];
83. bar2x=[Ax(t) Bx(t)];
84. bar2y=[Ay(t) By(t)];
85. bar3x=[Bx(t) Cx(t)];
86. bar3y=[By(t) Cy(t)];
87.
88. plot (bar1x,bar1y,bar2x,bar2y,bar3x,bar3y);
89. axis([-120,400,-120,200]);
90. axis normal
91.
92. M(:,t)=getframe;
93.end
146
Initialization1. % Set up the time interval and the initial positions of the nine coordinates
2. T_Int=0:0.01:2;
3. X0=[0 50 pi/2 125.86 132.55 0.2531 215.86 82.55 4.3026];
4. global T
5. Xinit=X0;
1. The sentence is notation that is behind symbol “%”.
2. Simulation time is set from 0 to 2 with Δt = 0.01.
3. Set the appropriate initial positions of the 9 coordinates which are used to solve nonlinear solver.
4. Declare a global variable T which is used to represent the current time t and determine the
driving constraint for angular velocity.
147
Determine the displacement
10. [Xtemp,fval] = fsolve(@constrEq4bar,Xinit);
a. function F=constrEq4bar(X)
b.
c. global T
d.
e. x1=X(1); y1=X(2); phi1=X(3);
f. x2=X(4); y2=X(5); phi2=X(6);
g. x3=X(7); y3=X(8); phi3=X(9);
h.
i. F=[ -x1+40*cos(phi1);
j. -y1+40*sin(phi1);
k. x1+40*cos(phi1)-x2+130*cos(phi2);
l. y1+40*sin(phi1)-y2+130*sin(phi2);
m. x2+130*cos(phi2)-x3+90*cos(phi3);
n. y2+130*sin(phi2)-y3+90*sin(phi3);
o. x3+90*cos(phi3)-180;
p. y3+90*sin(phi3);
q. phi1-2*pi*T-pi/2];
10. Call the nonlinear solver fsolve in which the constraint equations and initial values are necessary. The
initial values is mentioned in above script. The constraint equations is written as a function (which can
be treated a kind of subroutine in Matlab) as following and named as constrEq4bar. The fsolve finds a
root of a system of nonlinear equations and adopts the trust-region dogleg algorithm by default.
The equation of driving constraint
is depended on current time T
148
Determine the velocity14. phi1=Xtemp(3); phi2=Xtemp(6); phi3=Xtemp(9);
15. JacoMatrix=Jaco4bar(phi1,phi2,phi3);
16. Beta=[0 0 0 0 0 0 0 0 2*pi]';
17. Vtemp=JacoMatrix\Beta;
15. Call the function Jaco4bar to obtain the Jacobian Matrix depended on current values of
displacement.
16. Declare the right-side of the velocity equations.
17. Solve linear equation by left matrix division “\” roughly the same as J-1β. The algorithm adopts
several methods such as LAPACK, CHOLMOD, and LU. Please find the detail in Matlab Help.
a. function JacoMatrix=Jaco4bar(phi1,phi2,phi3)
b.
c. JacoMatrix=[ -1 0 -40*sin(phi1) 0 0 0 0 0 0;
d. 0 -1 40*cos(phi1) 0 0 0 0 0 0;
e. 1 0 -40*sin(phi1) -1 0 -130*sin(phi2) 0 0 0;
f. 0 1 40*cos(phi1) 0 -1 130*cos(phi2) 0 0 0;
g. 0 0 0 1 0 -130*sin(phi2) -1 0 -90*sin(phi3);
h. 0 0 0 0 1 130*cos(phi2) 0 -1 90*cos(phi3);
i. 0 0 0 0 0 0 1 0 -90*sin(phi3);
j. 0 0 0 0 0 0 0 1 90*cos(phi3);
k. 0 0 1 0 0 0 0 0 0];
149
Determine the acceleration20. dphi1=Vtemp(3); dphi2=Vtemp(6); dphi3=Vtemp(9);
21. Gamma=Gamma4bar(phi1,phi2,phi3,dphi1,dphi2,dphi3);
22. Atemp=JacoMatrix\Gamma;
21. Call the function Gamma4bar to obtain the right-side of the velocity equations depended on
current values of velocity.
22. Solve linear equation to obtain the current acceleration.
a. function Gamma=Gamma4bar(phi1,phi2,phi3,dphi1,dphi2,dphi3)
b.
c. Gamma=[ 40*cos(phi1)*dphi1^2;
d. 40*sin(phi1)*dphi1^2;
e. 40*cos(phi1)*dphi1^2+130*cos(phi2)*dphi2^2;
f. 40*sin(phi1)*dphi1^2+130*sin(phi2)*dphi2^2;
g. 130*cos(phi2)*dphi2^2+90*cos(phi3)*dphi3^2;
h. 130*sin(phi2)*dphi2^2+90*sin(phi3)*dphi3^2;
i. 90*cos(phi3)*dphi3^2;
j. 90*sin(phi3)*dphi3^2;
k. 0];
150
Determine the next initial positions 29. if Iter==1
30. Xinit=X(:,Iter);
31. else
32. Xinit=X(:,Iter)+(X(:,Iter)-X(:,Iter-1));
33. end
29.~33. Predict the next initial positions with assumption of inertia except the first time of the loop.
151
Plot time response37.figure
38.for i=1:9;
39. subplot(9,1,i)
40. plot (T_Int,X(i,:))
41. set(gca,'xtick',[], 'FontSize', 5)
42.end
43.% Reset the bottom subplot to have xticks
44.set(gca,'xtickMode', 'auto')
45.
46.% T vs velocity plot for the nine coordinates
47.figure
48.for i=1:9;
37.…
37. Create a blank figure .
39. Locate the position of subplot in the figure.
40. Plot the nine subplots for the time responses of nine coordinates.
41. Eliminate x-label for time-axis and set the font size of y-label.
44. Resume x-label at bottom because the nine subplots share the same time-axis.
47.~ It is similar to above.
152
Animation69.Ox=zeros(1,length(T_Int));
70.Oy=zeros(1,length(T_Int));
71.Ax=80*cos(X(3,:));
72.Ay=80*sin(X(3,:));
73.Bx=Ax+260*cos(X(6,:));
74.…
69. Determine the displacement of revolute joint.
80. Repeat to plot the locations by continue time elapsing.
81. Determine the horizontal location of .
88. Plot .
89. Set an appropriate range of axis.
80.for t=1:length(T_Int);
81. bar1x=[Ox(t) Ax(t)];
82. bar1y=[Oy(t) Ay(t)];
83. bar2x=[Ax(t) Bx(t)];
84. bar2y=[Ay(t) By(t)];
85. bar3x=[Bx(t) Cx(t)];
86. bar3y=[By(t) Cy(t)];
87.
88. plot (bar1x,bar1y,bar2x,bar2y,bar3x,bar3y);
89. axis([-120,400,-120,200]);
90. axis normal
91.
92. M(:,t)=getframe;
93.end
OA
OCBCABOA and , , ,
153
Time response of displacement
1x
1
1y
2x
2
2y
3x
3
3y
t
154
Time response of velocity
1x
1
1y
2x
2
2y
3x
3
3y
t
155
Time response of acceleration
1x
1
1y
2x
2
2y
3x
3
3y
t
156
A Slider Crank SystemKinematic Modeling
Figure Kinematic modeling of a slider-crank mechanism.
1 1 10.0, 0.0, 0.0x y = = =
4 4 3 3
3 3 2 2
0 0 0 0
2 2 1 1
0.0, 0.0, 200.0, 0.0
300.0, 0.0, 100.0, 0.0
100.0, 0.0, 0.0, 0.0
A A A A
B B B B
= = = − =
= = = − =
= = = =
4 4 1 1
0 0
4 4 1 1
0.0, 0.0, 100.0, 0.0,
100.0, 0.0, 0.0, 0.0
A A B B
C C
= = = − =
= = = =
2 5.76 1.2 0.0t − + =
driving constraint
rigid body C starts at 330° with 1.2 rad/sec speed.
Ground
Revolute joint
translational joint
157
1 1 1 1 1
1 1 1 1 1
1 1 4
1 1 4
4 4 4
4 4 4
1 1 4
4 4
100100
0 100
100
100
1000
100 100
100100 100T
x c s x c
y s c y s
x c x
y s y
c s s
s c c
x c xs c
y
− −− = + =
−
− = −
−
− + = =
− −
− −= −
B
d
n
n d1 1 4100s y
− −
( )
( ) ( )
41
41
4 1 4 11
1 1 4 4 1 1 4 44
100
100
100 100 100
100 100 100 100 0
sx
cy
c c s s
x c x c y s y s
= +
= − = − − +
= − − + − − =
( )( ) ( )( )1 1 4 4 4 1 1 4100 100 100 100 − − + − − −x c x s c y s y = +
Translation Joint1
1Bd
c
n
158
1 1
2 1
3 1
0.0
0.0
0.0
x
y
=
=
=
Figure Kinematic modeling of a slider-crank mechanism.
Ground
revolute joints4 4 3 3
5 4 3 3
6 3 3 2 2
7 3 3 2 2
8 2 2 1
9 2 2 1
200cos 0
200sin 0
300cos 100cos 0
300sin 100sin 0
100cos 0
100sin 0
x x
y y
x x
y y
x x
y y
− + =
− + =
+ − + =
+ − + =
+ − =
+ − =
translational joint
10 1 1 4 4 4 1 1 4
11 4 1
( 100cos )( 100sin ) ( 100cos )( 100sin )
0
0
+ − − + + − − −
=
− =
x x y y
driving constraint
12 2 5.76 1.2 0t − + =
Kinematic Modeling
159
Figure The Jacobian matrix.
Jacobian Matrix
160
Cont’d
161
Example of a slider-crank mechanism
mm200
mm300
mm200
=
=
=
BO
GB
AG
( ) ( )
1
1
1
4 3 3
4 3 3
3 3 2 2
3 3 2 2
2 2 1
2 2 1
4 1 4 4 4 1 4 4
4
11 constraint equations:
0
0
0
200 0
200 0
300 100 0
300 100 0
100 0
100 0
100 100 100 100 0
=
=
=
− + =
− + =
+ − + =
+ − + =
+ − =
+ − =
− − − − − =
−
x
y
x x
y y
x x
y y
x x
y y
y y x x
cos
sin
cos cos
sin sin
cos
cos
cos sin sin cos
1
2
0
1 constraint from the driving link
5 76 1 2 0
=
− + =t
. .
1 1 1 2 2 2 3 3 3 4 4 4To solve the 12 equations for 12 unknowns T x y x y x y x y = ,
, , , , , , , , , ,q
162
T
1 4 1 1 4 4
1 4 1 1 4 4
cos sin cos sin100 0Note that =0
sin cos sin cos0 100
x x
y y
− − − −
Jacobian matrix and the velocity
equations
=
0000003600000
35000000003400
33323100000002928
000000272600240
000000230220020
0001918017160000
0001501413012000
0100980000000
006504000000
000000000300
000000000020
000000000001
12
11
10
9
8
7
6
5
4
3
2
1
J
=
2.1
0
0
0
0
0
0
0
0
0
0
0
2
3
2
3
3
cos10027 ,17
sin30015
sin10023 ,13
cos2009
sin2005
135 ,24 ,20 ,16 ,12 ,8 ,4
136 ,34 ,26 ,22 ,18 ,14 ,10 ,6 ,3 ,2 ,1
=
−=
−=
=
−=
−=
=
( ) ( ) 144144
4
4
3
sincos10033
2932
2831
cos10029
sin10028
cos30019
yyxx −+−=
−=
−=
=
−=
=
163
Acceleration equations
( )
2
3 3
2
3 3
2 2
3 3 2 2
2 2
3 3 2 2
2
2 2
2
2 2
0
0
0
2 100cos
2 100sin
300cos 100cos
300sin 100sin
100cos
100sin
10
0
0
γ
+ =
+
γ
( ) ( ) ( ) ( ) ( ) 414414
2
444144414cos100sin100sin200cos20010 where yyxxyyxx −−−−−+−=
164
Animation
165
Time response of displacement
2x
2
2y
3x
3
3y
4x
4
4y
t166
Time response of velocity
2x
2
2y
3x
3
3y
4x
4
4y
t167
Time response of acceleration
2x
2
2y
3x
3
3y
t
4x
4
4y
168
( )
( ) 0
( , ) 0d t
=
=
q
q
kinematic constraints
driving link
3.2 Solution Technique
velocity equations
( )( ) ( )
0 or 0
0 or 0
q
dd d
q tt
=
+ =
q = qq
q + = qq
( ) ( )
0d d
−
q
q t
q =
acceleration
( ) ( ) ( ) ( )
0 or ( ) 0
( ) 2 0
q q q
d d d d
q q q qt tt
+ =
+ + + =
q + q = q q qq q
q q q q
169
Solution Technique
At any given instant
(1) Solve
n equations for n unknowns
(2) Solve
n equations for n unknowns
(3) Solve
n equations for n unknowns
( )
( , )
q
d
q t( )
q q = 0
q q = (the right hand side)
( )tq
( )
( , )
q
d
q t( )
q q = 0
q q = (the right hand side)
( )tq
( )
( )
q
d
q
q q =
q q =( )
(the right hand side)
( )tq
170
4.1 Planar Rigid Body Dynamics
i i xi
i i yi
i i i
m x f
m y f
n
=
=
=
Miqi= g
i
0 0
0 0
0 0
x
y
i
m x f
m y f
n
=
171
: all forces on the rigid body ig
Illustration of Constraint Force
21,
2
3 eqs. for 5 unknowns , , , ,
02 geometric constraint equations
0
2 2 1, 0, , ,
3 3 3
c c
mx mgs f
my mgc N
I f R I mR
x y f N
x R
y R
gx gs y s N mgc f mgs
R
= − +
= − +
= =
+ =
− =
= − = = = =
Pure rolling of a disk down the slope
x
y
f
N
mg
172
0c
mx f mgs
my N mgc
I fR
− = −
− = − − =
Constraint Force
1
1
( )
( )
( , ) 0,
,
n
m
T T
q
c
c
t
=
= +
=
Φ
q
q
Mq g g
g
There exists a Lagrange Multiplier
such that is that constraint force.
The equations of motion can be written as
T
q
T
q
−
+ =
Mq g
173
Kinematic constraint equation
:external applied force
:constraint (reaction) force
representing constraint force by the
Lagrange multiplier
g
( )cg
2
Generalized coordinate , y,
Constraint eq. 0, 0
0, + 0
0
0 0
0
0 0 1 0
0 0 0 1
10 0 0
2
1 0 0 0
0 1 0 0
T
x
x R y R
x R y R
mgs
mgc
m
m
mR R
R
+ = − =
− − = − =
−
− =
−
−
−
− −
−
Mqq
q
1
2
1 2
0
0
00
2 2 15 eqs. for 5 unknowns , 0, , ,
3 3 3
1 0
The constraint force 0 1
0
T
x mgs
y mgc
gx gs y s mgs mgc
R
R
− − =
= − = = = = −
−
= −−
λq
3 1
1
3
1
3
is the friction force for pure rolling.
mgs
mgc
mgRs
=
λ
174
mifi= n
i- (y
i
P - yi)l
1+ (x
i
P - xi)l
2
In Revolute Joint
4.2 Physical meaning of constraint force
1
2
0 0 1 0
0 0 0 1
0 0 ( ) ( )
x
y
P P
i i i ii i
m x f
m y f
y y x x n
+ = − − −
175
2i i yim y f = −
1i i xim x f = −
Constraint Force in Translational Joint
176
1( )P Q
i i xi i im x f y y = + −
1( )P Q
i i yi i im y f x x = + −
For a translational joint between i and j,
the equation of motion for body i can be written as
1 2[( )( ) ( )( )]P P Q P P Q
i i i j i i i j i i in x x x x y y y y = − − − + − − +
qq =
T
q+ =Mq g
linear algebraic equations in unknowns
for and . n m+ n m+
4.3 Formulation of Multi-body Dynamic Systems
177
q
Dynamics of a four-bar linkage
m18.0
m18.0
m26.0
m08.0
=
=
=
=
OC
BC
AB
OA
A
B
CO
mN1.0 =T
24
23
25
mkg1086.4
mkg1046.1
mkg1027.4
=
=
=
−
−
−
BO
AB
OA
I
I
I
kg18.0
kg26.0
kg08.0
=
=
=
BC
AB
OA
M
M
M
= 2sm9.8g
178
A Matlab Program for
Mass matrix and external force vector
=
−
−
−
4
3
5
1086.400000000
00.180000000
000.18000000
0001046.100000
00000.260000
000000.26000
0000001027.400
000000008.00
0000000008.0
M
= 2sm9.8g
kg18.0
kg26.0
kg08.0
=
=
=
BC
AB
OA
M
M
M
24
23
25
mkg1086.4
mkg1046.1
mkg1027.4
=
=
=
−
−
−
BO
AB
OA
I
I
I
mN1.0 =T
=
0
8.918.0
0
0
8.926.0
0
1.0
8.908.0
0
g
179
Jacobian matrix and γ
m18.0
m18.0
m26.0
m08.0
=
=
=
=
OC
BC
AB
OA
180
1
1
1 2
1 2
8 9
2 3
2 3
3
3
1 0 0.04sin 0 0 0 0 0 0
0 1 0.04cos 0 0 0 0 0 0
1 0 0.04sin 1 0 0.13sin 0 0 0
0 1 0.04cos 0 1 0.13cos 0 0 0
0 0 0 1 0 0.13sin 1 0 0.09sin
0 0 0 0 1 0.13cos 0 1 0.09cos
0 0 0 0 0 0 1 0 0.09sin
0 0 0 0 0 0 0 1 0.09cos
− −
− − − −
−= − − −
− −
J
2
1 1
2
1 1
2 2
1 1 2 2
2 2
1 1 2 2
2 2
2 2 3 3
2 2
2 2 3 3
2
3 3
2
3 3
0.04cos
0.04sin
0.05cos 0.13cos
0.05sin 0.13sin
0.13cos 0.09cos
0.13sin 0.09sin
0.09cos
0.09sin
+
+ = +
+
γ
Computation
( ) ( ) ( ) ( )
9 8
8 9 8 8
9 1 8 1
In numerial computing, the first step is to solve
for 0 and 0 with initial position 0 and velocity 0 .
T
=
q gM J
λ γJ O
q λ q q
( )9 9 9 8
8 9 8 8
The objective is to solve the differential equation for .T
t
=
q gM Jq
λ γJ O
( )
( )
( )
( )( ) ( )
( ) ( )
0Then integrate , and use and
0
to repeat the above step for and .
ΔtΔt
Δt ΔtΔt
Δt Δt
⎯⎯→
q qq q
q q
q λ
Repeat until.t Δt+
181
A convenient way with Matlab solver
Solve initial value problems for ordinary differential equations with
ode45(commended), ode23, ode113…
The equations are described in the form of z‘=f(t,z)
1
1
3
1
1
3
X
x
y
Let
y
q q,
q q
= = =
z z
182
The syntax for calling solver in Matlab
[T,Z] = ode45(@Func4Bar,[0:0.005:2],Z0);
column vector of
time points
Solution array
A vector specifying the interval
of integration
A vector of initial
conditions
A function that evaluates the right side of the
differential equations
function dz=Func4Bar(t,z)
global L1 L2 L3 L4 torque gravity
phi1=z(3); phi2=z(6); phi3=z(9);
dphi1=z(12); dphi2=z(15); dphi3=z(18);
M=diag([L1 L1 L1^3/12 L2 L2 L2^3/12 L3 L3 L3^3/12]);
J=[ -1 0 -0.5*L1*sin(phi1) 0 0 0 0 0 0;
0 -1 0.5*L1*cos(phi1) 0 0 0 0 0 0;
1 0 -0.5*L1*sin(phi1) -1 0 -0.5*L2*sin(phi2) 0 0 0;
0 1 0.5*L1*cos(phi1) 0 -1 0.5*L2*cos(phi2) 0 0 0;
0 0 0 1 0 -0.5*L2*sin(phi2) -1 0 -0.5*L3*sin(phi3);
0 0 0 0 1 0.5*L2*cos(phi2) 0 -1 0.5*L3*cos(phi3);
0 0 0 0 0 0 1 0 -0.5*L3*sin(phi3);
0 0 0 0 0 0 0 1 0.5*L3*cos(phi3)];
183
The syntax for calling solver in Matlab
J=[ -1 0 -0.5*L1*sin(phi1) 0 0 0 0 0 0;
0 -1 0.5*L1*cos(phi1) 0 0 0 0 0 0;
1 0 -0.5*L1*sin(phi1) -1 0 -0.5*L2*sin(phi2) 0 0 0;
0 1 0.5*L1*cos(phi1) 0 -1 0.5*L2*cos(phi2) 0 0 0;
0 0 0 1 0 -0.5*L2*sin(phi2) -1 0 -0.5*L3*sin(phi3);
0 0 0 0 1 0.5*L2*cos(phi2) 0 -1 0.5*L3*cos(phi3);
0 0 0 0 0 0 1 0 -0.5*L3*sin(phi3);
0 0 0 0 0 0 0 1 0.5*L3*cos(phi3)];
gamma=[ 0.5*L1*cos(phi1)*dphi1^2;
0.5*L1*sin(phi1)*dphi1^2;
0.5*L1*cos(phi1)*dphi1^2+0.5*L2*cos(phi2)*dphi2^2;
0.5*L1*sin(phi1)*dphi1^2+0.5*L2*sin(phi2)*dphi2^2;
0.5*L2*cos(phi2)*dphi2^2+0.5*L3*cos(phi3)*dphi3^2;
0.5*L2*sin(phi2)*dphi2^2+0.5*L3*sin(phi3)*dphi3^2;
0.5*L3*cos(phi3)*dphi3^2;
0.5*L3*sin(phi3)*dphi3^2];
g=[0 gravity*L1 torque 0 gravity*L2 0 0 gravity*L3 0]';
Matrix=[M J';
J zeros(size(J,1),size(J,1))];
d2q=Matrix\[g;gamma];
dz=[z(10:18,:); d2q(1:9,:)];
184
Time response of displacement
1x
1
1y
2x
2
2y
3x
3
3y
t
185
Time response of velocity
1x
1
1y
2x
2
2y
3x
3
3y
t
186
Time response of acceleration
1x
1
1y
2x
2
2y
3x
3
3y
t
310
187
Time response of λ
1
t
310
2
3
4
5
6
7
8
188
A slider-crank mechanism
m2.0
m3.0
m2.0
m1.0
=
=
=
=
BO
GB
AG
AC
A
B
C
O
G
24
22
24
mkg1067.6
mkg1004.1
mkg1067.6
=
=
=
−
−
−
BO
AB
slider
I
I
I
kg2.0
kg5.0
kg1.0
=
=
=
BO
AB
AC
M
M
M
mN1.0 =T
= 2sm9.8g
189
Time response of displacement
2x
2
2y
3x
3
3y
4x
4
4y
t190
Time response of velocity
2x
2
2y
3x
3
3y
4x
4
4y
t191
Time response of acceleration
2x
2
2y
3x
3
3y
t
4x
4
4y
310
192
Time response of λ
1
t
310
2
3
4
5
6
7
8
9
10
11
193
194
For the four-bar linkage, the mass center of each link is located at the midpoint. Let
the mass (kg) and mass moment of inertia (kg-m2) of link OA, AB, and BC are m1 =
2 , I1 = 0.2 , m2 = 3, I2 =0.3, m3 = 4, I3 =0.4.
(1) Consider the mechanism has 4 rigid bodies (including the ground). Define
the inertia (global) and body-fixed (local) coordinates and determine the degree-of-
freedom.
(2) List the kinematic constraint equation of the revolute joints at point O, A, B,
and C.
(3) For kinematic analysis with the driving link BC rotating at 2 rad/s, list the
equations to solve for displacement, velocity and acceleration of all links at all time.
Also state the procedure/process in the numerical solution.
(4) For dynamic analysis with the driving link torque 100 (N-m), list the equations
to solve for the acceleration of all links at all time. Note that BC is no longer in
constant speed. Explain the physical meaning of the constraint force(s) and sketch
where they are located.
Example
Coordinates
195
( ) ( )
1 2
2 3
3 4
4 1
0 50revolute joint O
0 0
50 130revolute joint A
0 0
130 90revolute joint B
0 0
90 180revolute joint C
0 0
= + − +
− = =
− = =
− = =
= =
i i i j j j
s s
s s
s s
s s
A Ar s r s
1
1 2 3 4
1000, 127 , sin , 270
260 −
= = = =
1 2
2 3
3 4
4 1
1
1 2 3 4
0 0
0 50
0 0
50 130
0 0
130 90
0 180
90 0
1000, 37 , 90 sin , 180
260 −
= =
−
= =
= =
− −
= =
= = = + =
s s
s s
s s
s s
1
1
2
2
33
4 4
X
X
Y
Y3
Constraint Equation
196
( ) ( )
1 2O
1 2
1 2
1 2 2
1 2 2
32A
2 3
32
Revolute joint
0 50At O A A
0 0
50cos 0
50sin
50 130At A A A
0 0
= + − +
− = + − −
− − =
− −
− = + − −
i i i j j j
x x
y y
x x
y y
xx
yy
A Ar s r s
2 2 3 3
2 2 3 3
3 3 4 4B
3 3 4 4
4 4C
4 4
50cos 130cos 0
50sin 130sin
130cos 90cosAt B 0
130sin 90sin
90cos 180At C 0
90sin
+ − + = =
+ − +
+ − + = =
+ − +
+ − = =
+
x x
y y
x x
y y
x
y
Driving Constraint and Time
Response
197
1 1 1 2 2 2 3 3 3 4 4 4
1
G
1
1
O A B C
2
Ground Constraint 0
Revolute joint at point O, A, B, C. 0 0 0 0
Driving Constraint 2 127 0
12 eqs. for 12 unknowns of displ
=
= =
= = = =
= − − =
Tx y x y x y x y
x
y
t
q
acement
Velocity
198
2
2
2 3
2 3
3 4
3 4
4
4
1 0 0
1 0
1
1 0 0 1 0 50sin
1 0 0 1 50cos
1 0 50sin 1 0 130sin
0 1 50cos 0 1 130cos
1 0 130cos 1 0 90sin
0 1 130cos 0 1 90cos
1 0 90sin
0 1 90cos
0 0 0 0 0 1 0 0 0 0 0 1
=
− − −
− − − −
− − − − −
gq
0
0
0
0
0
0
0
0
0
0
0
2
12 eqs. for 12 unknowns in
=
q
q
Acceleration
199
2 2
2 2 3 3
2 2
2 2 3 3
2 2
3 3 4 4
2 2
3 3 4 4
2
4 4
2
4 4
0
0
0
50cos 130cos
50sin 130sin
130cos 90cos
130sin 90sin
90cos
90sin
0
12 eqs. for 12 unknowns in
+ +
+ +=
+ + + +
+ +
q γq =
q
Dynamic Analysis
200
1 1 1 2 2 2 3 3 3 4 4 4
1 2 3 4
0
=diag I I I I
0 - g 0 0 - g 0 - g 0 0 - g 0
is the torque at the driving link.
initial conditions (0), (0)
The initial conditio
=
=
T
q
q
m m m m m m m m
m m m m
g
λ γ
g
M q
M
q q
ns have to satify the constraints.