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Transcript of § 7 - 1 Introduction § 7 - 2 Motion Equation of a Mechanical System § 7 - 5 Introduction to...
§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