Solution Manual for Kinematics and Dynamics of Mechanical Systems (3rd Edition) by Kevin Russell, Qiong Shen, and Raj S. Sodhi

All 11 Chapter s Cover ed




SOLUTIONS

, CONTENTS

Preface …………………………………………...……………………………………….. 1

Chapter 2 Mathematical Concepts in Kinematics ……………………………………….. 2

Chapter 3 Fundamental Concepts in Kinematics ……………………………………….. 8

Chapter 4 Kinematic Analysis of Planar Mechanisms …………………………………… 19

Chapter 5 Dimensional Synthesis ………………………………………………………... 81

Chapter 6 Static Force Analysis of Planar Mechanisms …………………………………. 159

Chapter 7 Dynamic Force Analysis of Planar Mechanisms ……………………………… 210

Chapter 8 Design & Kinematic Analysis of Gears ……………………………………….. 288

Chapter 9 Design & Kinematic Analysis of Disk Cams …………………………………. 327

Chapter 10 Kinematic Analysis of Spatial Mechanisms ………………………………….. 364

Chapter 11 Introduction to Robotic Manipulators ………………………………………… 409




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, CHAPTER 2

Problem 2.1 Statem ent:

Formulate an equation for the vector loop illustrated in Figure P.2.1. Consider that vector V j

always lies along the real axis.




Figu re P.2.1 Vector loop (3 vectors where V j changes length) in 2-D complex space

Problem 2.1 Solu tion:

Taking the clockwise sum of the vector loop in Figure P.2.1 produces the equation

V1ei1 V2ei V j 0 .
2

When expanded and separated into real and imaginary terms, the vector loop equation becomes

V1 cos 1 V2 cos 2 V j 0
.
V1 sin 1 V2 sin 2 0

Problem 2.2 Statem ent:

Formulate an equation for the vector loop illustrated in Figure P.2.2. Consider that vector V j

always lies along the real axis and vector V 3 is always perpendicular to the real axis.




2



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, Figu re P.2.2 Vector loop (4 vectors where V j changes length) in 2-D complex space

Problem 2.2 Solu tion:

Taking the clockwise sum of the vector loop in Figure P.2.2 produces the equation

i i V 3 V j 0 .
V1e 1 V 2e 2

When expanded and separated into real and imaginary terms, the vector loop equation becomes

V1 cos 1 V2 cos 2 V j 0
.
V1 sin 1 V2 sin 2 V 3 0

Problem 2.3 Statem ent:

Calculate the first derivative of the vector loop equation solution from Problem 2.2. Consider

only angles 1 , 2 and vector V j from Problem 2 to be time-dependent.

Problem 2.3 Solution:

Differentiating the vector loop equation solution from Problem 2.2 produces the equation

i1V1ei1  i2V2e i V j 0 .
2

When expanded and separated into real and imaginary terms, the vector loop equation becomes


1V1 sin 1 2V2 sin 2 Vj 0
.
1V1 cos 1 2V2 cos 2 0


3

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,Problem 2.4 Statem ent:

Calculate the second derivative of the vector loop equation solution from problem 2.2. Consider

only angles 1 , 2 and vector V j from Problem 2 to be time-dependent.

Problem 2.4 Solu tion:

Differentiating the vector loop equation solution from Problem 2.3 produces the equation
2 i1 i1 2 i2
i2
1V1e i1V 2V i2V V j 0 .
1e 2e 2e

When expanded and separated into real and imaginary terms, the vector loop equation becomes
2
 V cos2  V sin cos   sin  0
  V
 V
21 1 1 1 1 1 2 2 2 2 2 2
 V sin 2 V sin cos  V 0.
cos   V  V
1 1 1 1 1 1 2 2 2 2 2 2 j

Problem 2.5 Statem ent:

Formulate an equation for the vector loop illustrated in Figure P.2.3.




Figu re P.2.3 Vector loop (4 vectors) in 2-D complex space

Problem 2.5 Solu tion:

Taking the clockwise sum of the vector loop in Figure P.2.3 produces the equation

i i i i 0 .
V1e 1 V 2e 2 V3e 3 V0e 0

When expanded and separated into real and imaginary terms, the vector loop equation becomes



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, V1 cos 1 V2 cos 2 V3 cos 3 V0 cos 0 0
.
V1 sin 1 V2 sin 2 V3 sin 3 V0 sin 0 0

Problem 2.6 Statem ent:

Calculate the first derivative of the vector loop equation solution from Problem 2.5. Consider

only angles 1 , 2 and 3 from Problem 5 to be time-dependent.

Problem 2.6 Solu tion:

Differentiating the vector loop equation solution from Problem 2.5 produces the equation

i iV iV 0 .
i1V1 e 1
i i3
e2 2 2 e 3 3

When expanded and separated into real and imaginary terms, the vector loop equation becomes


1V1 sin 1 2V2 sin 2 3V3 sin 3 0
.
1V1 cos 1 2V2 cos 2 3V3 cos 3 0

Problem 2.7 Statem ent:

Calculate the second derivative of the vector loop equation solution from Problem 2.5. Consider

only angles 1 , 2 and 3 from Problem 5 to be time-dependent.

Problem 2.7 Solution:

Differentiating the vector loop equation solution from Problem 2.6 produces the equation
2 i1 i1 2 i2 2 i3
i2 i3
1V1e i1V  2V2e i2V 3V i3V 0 .
1e 2e 3e 3e

When expanded and separated into real and imaginary terms, the vector loop equation becomes
2 2
 V cos2  V sin cos   sin   V cos  V sin  0
  V
 V
21 1 1 1 1 1 2 2 2 2 2 2 23 3 3 3 3 3
 V sin 2 V sin cos   V sin  V cos  0 .
cos   V  V
1 1 1 1 1 1 2 2 2 2 2 2 3 3 3 3 3 3




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,Problem 2.8 Statem ent:

Formulate an equation for the vector loop illustrated in Figure P.2.4.




Figu re P.2.4 Vector loop (5 vectors) in 2-D complex space

Problem 2.8 Solu tion:

Taking the clockwise sum of the vector loop in Figure P.2.4 produces the equation

  3  V0e 0 .
V1 e1 V 2e2 V3e V4e4

When expanded and separated into real and imaginary terms, the vector loop equation becomes

V1 cos 1 V2 cos 2 V3 cos 3 V4 cos 4 V0 cos 0 0
.
V1 sin 1 V2 sin 2 V3 sin 3 V4 cos 4 V0 sin 0 0

Problem 2.9 Statem ent:

Calculate the first derivative of the vector loop equation solution from Problem 2.8. Consider

only angles 1 , 2 , 3 and  4 from Problem 8 to be time-dependent.

Problem 2.9 Solu tion:

Differentiating the vector loop equation solution from Problem 2.8 produces the equation

i iV iV iV 0 .
i1V1e 1
i i i
e2 2 2 e33 3 e4 4 4

When expanded and separated into real and imaginary terms, the vector loop equation becomes




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, 1V1 sin 1 2V2 sin 2 3V3 sin 3 4V4 sin 4 0
.
1V1 cos 1 2V2 cos 2 3V3 cos 3 4V4 cos 4 0
Problem 2.10 Statem ent:

Calculate the second derivative of the vector loop equation solution from Problem 2.8. Consider

only angles 1 , 2 , 3 and  4 from Problem 8 to be time-dependent.

Problem 2.10 Solu tion:

Differentiating the vector loop equation solution from Problem 2.9 produces the equation

21V1ei1 iV
1 1
ei1 222V ei2 iV
2 2
ei2 3
3
2V ei3 iV ei3 2V ei4 iV ei4 0 .
3 3 4 4 4 4


When expanded and separated into real and imaginary terms, the vector loop equation becomes
2 2 2
 V cos2  V sin cos   sin   V cos  V sin   V cos 
  V
 V
1 1 1 1 1 1 2 2 2 2 2 2 3 3 3 3 3 3 4 4 4
4V4 sin 4 0
2 2 2 .
 V sin 2 V sin cos   V sin cos   V sin 
cos   V  V  V

1 1 1 1 1 1 2 2 2 2 2 2 3 3 3 3 3 3 4 4 4
4V4 cos 4 0




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, CHAPTER 3

Problem 3.1 Statem ent:

Planar four-bar linkages have many everyday applications (some are illustrated Figure 3.1). Identify and

describe four additional everyday applications for the planar four-bar linkage. Problem 3.1 Solu tion:

As illustrated in Solution 3.1 Figure 1, the (a) elliptical trainer, (b) fan-style exercise bicycle, (c) model car

steering linkage and (d) excavator bucket all include the planar four-bar linkage.




Solu tion 3.1 Figur e 1




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