§ 7 － 1 Introduction § 7 － 2 Motion Equation of a Mechanical System § 7 － 5 Introduction to Aperiodic speed Fluctuation and Its Regulation § 7 － 4 Periodic.

§7 － 1 Introduction

§7 － 2 Motion Equation of a Mechanical System

§7 － 5 Introduction to Aperiodic speed Fluctuation and Its Regulation

§7 － 4 Periodic Speed Fluctuation and Its

Regulation

§7 － 3 Solution of the Motion Equation of a

Mechanical System

Chapter 7 Chapter 7 Motion of Mechanical Systems and Its RegulationMotion of Mechanical Systems and Its Regulation

一、 Introduction

In the kinematic analysis of a mechanism, the motion of

the input link should be given. In most cases, the input lin

k is supposed to run at a constant speed. This is only an ap

proximation to reality. The actual motion of the input link

depends on the mass distribution of the mechanism and th

e external forces acting. on the. mechanism. In this chapte

r, we will study the actual motion of the mechanism accor

ding to the mass distribution of the mechanism and the ex

ternal forces acting on it.

§§77 －－ 1 1 Introduction

Start

二、 Three operating phases of a machineω

tStop Running

T T

ωm

1. Starting phase

2. Steady working phase

ω →ωm

Wd = Wc + E

2) Angular velocity remains constant ωm ＝ constant Wd = Wc

1) Cyclical fluctuation

ω(t)=ω(t+Tp) ωm ＝ constant Wd = Wc

3. Stopping phase

E = -Wc

三、 Driving and resistance forces act on machines

1. Driving force A different driving motor will provide a different driving force which is a function of different kinematic parameters.

Function relations between the driving force and the kinematic parameters are called mechanical characteristic of driving motors.

2. Resistance force

On studying the movement of a machine under external forces with analytical method, the driving force provided by the driving motor must be given as the analytical formula.

B

A

3

21

y

O xφ1

As slider crank mechanism is shown in Fig..

Let crank act as the input member.The

angular velocities , masses, mass center

locations, velocities of the mass center, and

moments of inertia of the all links are given.

The driving moment is M1 and working resistance force is F3.

一、 General Expression of the Equation of Motion

According to the principle of work and energy, we have dE=dW=Ndt

S2

S3

S1

J2 m2

m3J1

ω1M1

F3

2 2 2 21 1 2 2 S2 2 3 3d d( / 2 / 2 / 2 / 2)SE J m v J m v

21 1 3 3d d( dW M F v P t

Equation of motion ：2 2 2 2

1 1 2 3 32 S2 2 3 3 1 1d( / 2 / 2 / 2 / 2) ( )dS SJ m v J m v M F v t

§§77 －－ 2 2 Motion Equation of a Mechanical System

Choose the crank as an equivalent link, the above Eq. Can be

rewritten as.

2 2 2 21 1 2 3 32 S2 2 3 3 1 1d( / 2 / 2 / 2 / 2) ( )dS SJ m v J m v M F v t

2 2 222 3 31 2

1 2 3 1 1 31 1 1 1

d d2

Sv v vJ m m M F t

2 2 21 S2 2 1 2 2 1 3 3 1( / ) ( / ) ( / )e SJ J J m v m v

e 1 3 3 1( / )M M F v

Ｏ

A

1

φ1

Je

ω1Me

21 1 e 1 1 1[ ( ) / 2] ( , , ) ded J M t t

Je —equivalent moment of inertia ， Me—equivalent moment of force

Equivalent model

S2

S3

S1

J2 m2

m3J1

B

A

3

21

y

O xφ1 F3

ω1Me

For a planar mechanism consisting of n moving links, the general equation of motion is ：

where αi is the angle between Fi and vi.

If the directions of Mi and ωi are identical, then “ ＋” is use

d,otherwise, “ －” is used.

2 2

=1 1

d ( / 2 / 2 ( cos )n n

i Si Si i i i i i ii i

m v J Fv M dt

二、 Dynamically Equivalent Model of a Mechanical System

A mechanical system with one degree of freedom can be assumed as an imaginary link. Such an imaginary link is called the equivalent link, or the dynamically equivalent model, of a mechanical system.

Choose link 3 as an equivalent link, then

2 223 22 1

1 2 3 3 1 33 3 3

d d2

Sv vJ m m v M F t

v v v

2 2 21 1 3 S2 2 3 2 2 3 3( / ) ( / ) ( / )e Sm J v J v m v v m

e 1 1 3 3( / )F M v F

S2

S3

S1

J2 m2

m3J1

B

A

3

21

y

O xφ1 F3

ω1Me 3

v ３

s3

me

Fe

me —Equivalent mass ，Ｆ e—Equivalent force

23 3 e 3 3 3d[ ( ) / 2] ( , , ) dem s v F s v t v t

Equivalent model

The rotate link is the equivalent one.

The sliding link is the equivalent one.

Equivalent mass

Equivalent force

Equivalent moment of force

Equivalent moment of inertia2 2

=1

nSi i

e i Sii

vJ m J

=1

cosn

Si ie i i i

i

vM F M

2 2

=1

nSi i

e i Sii

vm m J

v v

=1

cosn

i ie i i i

i

vF F M

v v

三、 Other Forms of Motion Equation1. The equations of motion in differential form

23 3 e 3 3 3d[ ( ) / 2] ( , , ) dem s v F s v t v t

21 1 e 1 1 1d[ ( ) / 2] ( , , ) deJ M t t

2. The equations of motion in moment of force form

2

e

ddv

dt 2 de

e

mvm F

s 22

e

dd( /2)v

d 2 de

e

JJ M

3. The equations of motion in energy integral form

0

2 20 0 e

1 1d

2 2

s

e e sm v m v F s 0

2 20 0 e

1 1d

2 2e eJ J M

一、 The equivalent moment of inertia is a function of position and the equivalent moment of force is a function of position.

二、 The equivalent moment of inertia is a constant and the equivalent moment of force is a function of velocity.

三、 The equivalent moment of inertia is a function of posi

tion and the equivalent moment of force is a function of po

sition and velocity.

§§77 －－ 3 3 Solution of the Motion Equation of a Mechanical System

一、 Reasons for Periodic Speed Fluctuation The driving moment Md and the resistant moment M

r acting on the machinery are a periodic function of rotating angle φ of driving motor. Thus, the equivalent moment of force is a periodic function of equivalent rotating angle φ.

Md Mr

a b c d e a'φ

Me

The kinetic energy increases:The work done by the equivalent link:

ed er[ ( ) - ( )] da

W M M

2 21 1

( ) ( )2 2e ea aE J J

§§77 －－ 4 4 Periodic Speed Fluctuation and Its Regulation

Md Mr

a b c d e a'

φ

φ

Me

E

a b c d e a'

ab: Md < Mr ， △ E < 0 deficiency of work“ －” , ω↓

bc: Md > Mr ， △ E > 0 excess work“+” , ω↑ cd: Md < Mr ， △ E < 0 deficiency of work“ －” , ω ↓

de: Md > Mr ， △ E > 0 excess work“+” , ω↑

within a period ，Wd=Wr ， △ E=0 ； then

This means that the work done by driving forces is equal to the work done by resistant forces within a period. This is the condition for a periodic steady working state.

' 2 2ed er ' '

1 1[ ( ) - ( )] d 0

2 2

a

aea a ea aM M J J

二、 Regulation of Periodic Speed Fluctuation

1 . Coefficient of Speed Fluctuation

T

ω

φＯ

ωmin

ωmax

Average angular speed

m max min( ) / 2

The speed fluctuation of a machine may cause an extra dynamic

load and vibration of the system and therefore should be controlled

within some limits to ensure good working quality. The coefficient o

f speed fluctuation is then

max min m( ) /

δ≤[δ]

The regulation of periodic speed fluctuation means li

mitation of the coefficient of speed fluctuation so that

2.Design method for flywheel

（ 1 ） Principle

A flywheel serves as a mechanical reservoir for storing mech

anical energy. Its function is to store the extra energy when the avai

lable energy is in excess of the load requirements and to give away t

he same when the available energy is less than the required load.

Point b ： Emin ， min ， Wmin ；

φ

E

a b c d e a'

Emax

EminPoint c ： Emax ， max ， Wmax ；

The maximum increment of work Wmax

max max min ed er[ ( ) ( )] dc

a

W E E M M

If Je =constant, then2 2 2

max max min max min( ) / 2e e mW E E J J

We can add a flywheel with enough moment of inertia to reduce the value of δ

2max max min ( )e F mW E E J J

The coefficient of speed fluctuation is2

max /( )e F mW J J

（ 2 ） Calculation of moment of inertia of a Flywheel

If Je << JF , Je can be neglected. The equivalent moment of in

ertia of the flywheel J F can be calculated by

2max /( [ ])F mJ W

Rim type flywheel ： Rim1 、 Hub2 、 Web(spokes)3.

(3) Determination of flywheel size

The moment of inertia:2 2

1 2

2

( ) /(8 )

/(4 )

F A A

A

J J G D D g

G D g

GAD2 ＝ 4gJF

where GAD2 is flywheel moment, D is the

mean diameter of the rimThickness of rim is b and density isγ(N/m3), then

GA=πDHbγ HB ＝ GA / (πDγ)

The aperiodic speed fluctuation is needed to regulate with a speed regulator.

One kind of Centrifugal speed regulator is sketched in Fig.

§§77 －－ 5 5 Introduction to Aperiodic speed Fluctuation and Its Regulation