machines and mechanisms: applied kinematic analysis, 4/e

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


David Myszka



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


David Myszka



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.


David Myszka



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.


David Myszka



Dynamics


David Myszka



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


David Myszka



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


David Myszka



◼ 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


David Myszka



1.5 Kinematic Diagram

Simple Link

Simple Link

(with point of

interest)

Complex Link

Revolute Joint


David Myszka



Kinematic Diagram

Cam Joint

Gear Joint

Translation Joint

1.6 Kinematic Model


David Myszka




David Myszka



1.7 MOBILITY

15

◼ Constrained mechanism: one degree of freedom

◼ Locked mechanism: zero degree of freedom

◼ Unconstrained mechanism: more than one degree of freedom


David Myszka




David Myszka



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


David Myszka




David Myszka



1.9 SPECLAL CASES OF THE MOBILITY EQUATION

Coincident Joints

• One degree of freedom actually if pivoted links are the same size


David Myszka




David Myszka



1.10 FOUR-BAR MECHANISM, 4-BAR LINKAGE


David Myszka




David Myszka



1.11 SLIDER-CRANK MECHANISM


David Myszka



1.12.3 Quick-Return Mechanisms

27

1.12.4 Scotch Yoke Mechanism

Machines and Mechanisms: Applied Kinematic Analysis,

4/eChapter 4 Displacement Analysis


David Myszka



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


David Myszka




David Myszka



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


David Myszka




David Myszka



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


David Myszka



Analysis of Mechanism Position


David Myszka



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


David Myszka




David Myszka



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


David Myszka



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


David Myszka



4.7 Limiting Positions and Stroke


David Myszka



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


David Myszka



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


David Myszka



6.2 LINEAR AND ANGULAR VELOCITY

6.2.2 Linear Velocity of a General Point


David Myszka



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


David Myszka



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


David Myszka



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


David Myszka



1 2 3

1 1 2 2 2

2

1 2

0

0

0

50

r r r

r r r

r


+ + =

+ + =

=

r2

r1

r3

X

Y


David Myszka



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


+ + =

− = = =

−

+ + =

=

r1

r3

r2

Y

X


David Myszka



6.10 INSTANTANEOUS CENTER OF ROTATION


David Myszka



6.12 GRAPHICAL VELOCITY ANALYSIS:

INSTANT CENTER METHOD


David Myszka



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


David Myszka



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


David Myszka



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


Chapter 7 Acceleration Analysis


David Myszka



7.2 LINEAR ACCELERATION

7.2.1 Linear Acceleration of Rectilinear Points


David Myszka



( ) ( ) ( )

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


David Myszka



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


David Myszka



( ) ( )

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


David Myszka



( ) ( )

( )

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


Chapter 9 Cams


David Myszka



9.1 INTRODUCTION


David Myszka



Plate cam

Cylindrical cam

Linear cam


David Myszka



9.2 TYPES OF CAMS

Follower motion


David Myszka



Follower position


David Myszka



9.3 TYPES OF FOLLOWERS


David Myszka



9.11 THE 4-STATION GENEVA MECHANISM

Constant rotation producing index motion


Chapter 13 STATIC FORCE ANALYSIS


David Myszka



13.3 MOMENTS AND TORQUES

X

Y

r


David Myszka



13.5 FREE-BODY DIAGRAMS

13.5.1 Drawing a Free-Body Diagram

13.5.2 Characterizing Contact Forces


David Myszka



13.6 STATIC EQUILIBRIUM


David Myszka



13.7 ANALYSIS OF A TWO-FORCE MEMBER


Chapter 14 DYNAMIC FORCE ANALYSIS


David Myszka



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

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,:))


52.end


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


66.end


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,:))


42.end

43.% Reset the bottom subplot to have xticks


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


2x

2

2y

3x

3

3y

4x

4

4y

t166


2x

2

2y

3x

3

3y

4x

4

4y

t167


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

+ =

+ + + =

qq

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


(3) Solve


( )

( , )

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


1x

1

1y

2x

2

2y

3x

3

3y

t

185


1x

1

1y

2x

2

2y

3x

3

3y

t

186


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


2x

2

2y

3x

3

3y

4x

4

4y

t190


2x

2

2y

3x

3

3y

4x

4

4y

t191


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.

machines and mechanisms: applied kinematic analysis, 4/e

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