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Principle of Virtual Work - University of...
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Modelling of Automotive Systems 1
Principle of Virtual Work
Degrees of Freedom
Associated with the concept of the lumped-mass approximation is the idea of the NUMBER OF DEGREES OF FREEDOM.
This can be defined as “the number of independent co-ordinates required to specify the configuration of the system”.
The word “independent” here implies that there is no fixed relationship between the co-ordinates, arising from geometric constraints.
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Degrees of Freedom of Special Systems
A particle in free motion in space has 3 degrees of freedom
z
r
y
3
particle in free motion in space has 3 degrees of freedom
x
If we introduce one constraint – e.g. r is fixed then the number of degrees of freedom reduces to 2.note generally:
no. of degrees of freedom = no. of co-ordinates –no. of equations of constraint
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P2P1 P3
.
y
This has 6 degrees of freedom3 translation3 rotation
Rigid Body
3
x
e.g. for partials P1, P2 and P3 we have 3 x 3 = 9 co-ordinates but the distances between these particles are fixed – for a rigid body – thus there are 3 equations of constraint.
The no. of degrees of freedom = no. of co-ordinates (9) - no. of equations of constraint (3)= 6.
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Formulation of the Equations of Motion
Two basic approaches:
1. application of Newton’s laws of motion to free-body diagrams
Disadvantages of Newton’s law approach are that we need to deal with vector quantities – force and displacement.
thus we need to resolve in two or three dimensions – choice of method of resolution needs to be made. Also need to introduce all internal forces on free-body diagrams – these usually disappear when the final equation of motion is found.
2. use of work
with work based approach we deal with scalar quantities – e.g. work – we can develop a routine method – no need to take arbitrary decisions.
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θT
mg
Free body diagram
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P2
Principle of Virtual Work
The work done by all the forces acting on a system, during a small virtual displacement is ZERO.
Definition A virtual displacement is a small displacement of the system which is compatible with the geometric constraints.
ab
a
bP2 bδθP1 aδθ
e.g.This is a one-degree of freedom system, only possible movement is a rotation.
work done by P1 = P1(- aδθ)work done by P2 = P2(bδθ)
P1
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Total work done = P1(- aδθ) + P2(bδθ) = δW
By principle of Virtual WorkδW = 0
therefore: P1 (- aδθ) + P2(bδθ) = 0
- a P1 + b P2 = 0
P1a = P2b
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xA &&=
FA
M(acceleration)
D’Alembert’s Principle
Consider a rigid mass, M, with force FA applied
From Newton’s 2nd law of motion
xMMaF A &&==
or
0=− xMF A &&
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Now, the term ( xM &&− ) can be regarded as a force – we call it an inertial force, and denote it FI – thus
xMF I &&−=
we can then write:
FA + FI = 0
In words – the sum of all forces acting on a body (including the inertial force) is zero – this is a statics principle.In fact all statics principles apply if we include inertial forces, including the Principle of Virtual Work.
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Virtual Work and Displacements
Using the concept of virtual displacements, and virtual work, we can derive the equations of motion of lumped parameter systems.
kx
m
Example 1Mass/Spring System
Here number of degrees of freedom =1Co-ordinate to describe the motion is xNow consider free-body diagram, at some time t
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mm
R (reactive force)
inertial force
mg (gravity force)
restoring force kx
xm &&−
( )
0
0
0:_
=+
=−−
=−−=−−
kxxmor
kxxmHence
xkxxmWkxxmforceTotal
&&
&&
&&
&&
δδ
11 qQW δδ =
General one degree of freedom systemIf q1 is the co-ordinate used to describe the movement then the general form of δW is as follows:
we call δq1 – generalised displacementQ1 – generalised force.
From principle of virtual work
0
0
1
11
=∴
==
Q
QW qδδ
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( )
00
0
11
111
111
=+=−−=∴=−−=
kqqmkqqmQ
qkqqmW
&&
&&
&& δδ
k
m x
ExampleReferring to the mass/spring system again
x= q1
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θ
Example 2
Simple pendulum2θ&ml−
θ&&ml−
This is another one degree of freedom system.During a virtual displacement, δθ, the virtual work done is
δW = )sin( θθ mgml −− && 0=δθl
δW = 0 (PVW)
0sin
0sin
=+
=+∴
θθ
θθ
lg
mgml
&&
&& or
0
mg
P
(inertial force – tangential)
(inertial force –radial)
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m
Example 3 – two degrees of freedom system
free-body diagrams.
kx1
m
kx2
m
m m
kx1 -mx1
k(x2-x1)
-mx2
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For LH mass:
( )[ ]( )[ ]
( ) ( ) 221121
22212
111211
xQxQWWWWxxmxxk
Wxxxkxmkx
δδδδδδδ
δδ
+=+==−−−
=−+−−&&
&&
For δW = 0 for all 21, xx δδ the Qi quantities must be zero.
Hence:
0)(
0)(
122
1211
=−+
=−−+
xxkxm
xxkkxxm
&&
&&
or
=
−
−+
00
00
2
1
2
1
xx
kkkk
xx
mm
&&
&&
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n degree of freedom systems
Having discussed single and two degree of freedom systems, and introduced the concept of generalised forces we can now consider the general case of an n degree of freedom system. A virtual displacement must be consistent with the constraints on the system. The motion can be described by n independent, generalised co-ordinates, nqqq ,....,, 21 . Hence a virtual displacement can be represented by small changes in these co-ordinates:-
qnqq δδδ ,...,, 21
Suppose only one co-ordinate, ( )niqi ≤≤1 is given a small, imaginary displacement, .qiδ As a result every particle in the system will be, in general, displaced a certain amount. The virtual work
done will be of the form qiiQW δδ = where iQ is an expression relating directly to the forces
acting on the system. iQ is the generalised force associated with iq .
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From the principle of virtual work
0== qiiQW δδ
Since, qiδ is finite, we get
0=iQ
This must be true for .,...,2,1 ni =
I.e.
0
00
2
1
=
==
nQ
Mthese are the equationsof motion of the system
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The generalised forces have component parts
1) inertial forces (mass x acceleration)2) elastic or restraining forces3) damping forces (energy dissipation)4) external forces5) constraint forces
ADEIqii WWWWQW δδδδδδ +++==
( ) qiA
iDi
Ei
Ii QQQQ δ+++=
(Noting, as before, that the constraint forces do no virtual work)
Then the equations of motion are:
0=+++ Ai
Di
Ei
Ii QQQQ ni ,...,2,1=
These are the n equations of motion.
We will examine each of these components now, in more detail. The aim is to
relate these component forces to the generalised co-ordinates .,....,, 21 nqqq
inertial elastic damping external
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Inertial Forces (See also Handout)
The position of the ith particle of mass, in the system, is, in general, related to then generalised co-ordinates, and time (if the constraints are independent of time)then the position of the ith particle depends only on the n generalised co-ordinates. Thus
( )tqqqrr nii ,,...,, 21rr
= (1)
Now we suppose that the system is in motion and that we represent the inertialforce on the ith particle (using D’Alembert’s Principle) as
iirm &&r− (2)
We now give the system an arbitrary virtual displacement – this can be
represented in terms of generalised co-ordinates by nqqq δδδ ,...,, 21 . The virtualdisplacement of the Ithparticle can be represented by
rrδ (3)and the virtual work done by the inertia force on the Ith particle is simply
iii rrm r&&r δ).(− (4)(note that this is a scalar product). From this result we get the total virtual work as
iii
iI rrmW r&&r δδ ).(∑ −= (5)
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Using equation (1) we have
j
n
i j
ii q
qrr δδ ⋅
∂∂
= ∑=1
rr
(6)
Hence
j
n
j j
ii
ii
I qqrrmW δδ ⋅
∂∂
⋅−= ∑∑=1
)(r
&&r (7)
and re-arranging
( )
∂∂
⋅−= ∑∑= i j
iii
n
jj
I
rrrmqW &&r
1δδ (8)
However, the generalised inertial forces, Qj, are effectively defined by
j
n
j
Iji
ii
I qQrmW δδ ∑∑=
⋅−=1
)( &&r (9)
Comparing (8) and (9) we have
∑ ∂∂
⋅−=i j
iii
Ij q
rrmQr
&&r)( (10)
It is shown in the handout notes that
( )jj
Ij q
TqT
tQ
∂∂
+∂∂
∂∂
−=& (11)
Modelling of Automotive Systems 21
where T is the total KE
ii
ii rrmT &r&r ⋅= ∑ 21
(12)
Elastic Forces
Consider a simple spring:
For static equilibrium
SE FF = (external force) = (internal spring force)
•
←FS
x
FE
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Suppose we define the POTENTIAL ENERGY, V, as the work done by theexternal force to extend the spring a distance χ
external work done = Vkx =2
21
−==∴ )()( xVxfV a function of x
Here 2
21 kxV =
The work done by the internal spring force, W, is equal and opposite to V
2
21 kxVW −=−=
FE κχ
χextension ofspring
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Now consider a small, virtual displacement, xδ . Corresponding changes to Wand V are as follows:
WxxVV δδδδδ −=⋅=
Comparing with standard form
⋅∂∂
−=∂∂
−=
==
xV
qVQthen
herexqqQW
E
E
11
111 ),(δδ
Generally,
∑= ∂∂
=−=
=n
jj
j
n
qqVWV
qqqVV
1
21 ),....,,(
δδδ
Compare with
j
Ej
jEj
qVQ
qQW
∂∂
−=∴
= ∑ δδ
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Lagrange’s equation
Suppose that no damping forces are present, and there are no externallyapplied forces. Then
),...,2,1(0 niQQ Ai
Di ===
We have found that
i
Ei
ii
Ii
qVQ
niqT
qT
tQ
∂∂
−=
=∂∂
+
∂∂
∂∂
−= ),...,2,1(&
If we collect these results we get
),...,2,1(0 niqV
qT
qT
t iii
==∂∂
+∂∂
−
∂∂
∂∂
&
This is LAGRANGE’s EQUATION
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If we define
L=T-V
And assume that V does not depend on the iq& ’s, then Lagrange’s equation canbe written as:
0=∂∂
−
∂∂
∂∂
ii qL
qL
t &
Example 1- mass/spring system
This is a single degree of freedom system. Here
21
21
1
2121
kqV
qmT
xq
=
=
=
&
The Lagrange equation is (n=1 so only one equation)
0111
=∂∂
+∂∂
−
∂∂
∂∂
qV
qT
qT
t &
kx
m
and
Here ( ) 111
qMqMtq
Tt
&&&&
=∂∂
=
∂∂
∂∂
01
=∂∂qT
11
kqqV
=∂∂
Hence 011 =+ kqqm &&
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θ
1
l
m
Example 2-simple pendulum
Hence
( ) 12
12
1
qmlqmltq
Tt
&&&&
=∂∂
=
∂∂
∂∂
11
1
sin
0
qmglqVqT
=∂∂
=∂∂
Here θ=1q
In terms of 1q we have
T= ( )212
1 qlm &
V= ( )1cos1 qmgl −
0sin 112 =+ qmglqml &&
or 0sin 11 =+ qlgq&&