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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.


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


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.


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.


θT

mg

Free body diagram


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


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


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 &&


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.


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


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


( )

00

0

11

111

111

=+=−−=∴=−−=

kqqmkqqmQ

qkqqmW

&&

&&

&& δδ

k

m x

ExampleReferring to the mass/spring system again

x= q1


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)


m

Example 3 – two degrees of freedom system

free-body diagrams.

kx1

m

kx2

m

m m

kx1 -mx1

k(x2-x1)

-mx2


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

&&

&&


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 .


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

QQ

Mthese are the equationsof motion of the system


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


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)


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)


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


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


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


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


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 &&


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&&

principle of virtual work - university of...

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