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 coordinates, arising from geometric constraints.
Modelling of Automotive Systems
1
Degrees of Freedom of Special Systems
A particle in free motion in space has 3 degrees of freedom
z
particle in free motion in space
has 3 degrees of freedom
r
y
x
3
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
Modelling of Automotive Systems
2
Rigid Body
y
P1
P3
This has 6 degrees of
freedom
3 translation
3 rotation
P2
.
x
3
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.
Modelling of Automotive Systems
3
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.
Modelling of Automotive Systems
4
Free body diagram
θ
T
mg
Modelling of Automotive Systems
5
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.
P2
P1
b
a
b
P1 aδθ
P2 bδθ
a
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)
Modelling of Automotive Systems
6
Total work done = P1(- aδθ) + P2(bδθ) = δW
By principle of Virtual Work
δW = 0
therefore:
P1 (- aδθ) + P2(bδθ) = 0
- a P1 + b P2 = 0
P1a = P2b
Modelling of Automotive Systems
7
D’Alembert’s Principle
Consider a rigid mass, M, with force FA applied
A = &x&
FA
M
(acceleration)
From Newton’s 2nd law of motion
F A = Ma = M&x&
or
F A − M&x& = 0
Modelling of Automotive Systems
8
Now, the term ( − M&x& ) can be regarded as a force – we call it an inertial force, and denote it FI – thus
we can then write:
F I = − M&x&
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.
Modelling of Automotive Systems
9
Virtual Work and Displacements
Using the concept of virtual displacements, and virtual work, we can derive the equations of
motion of lumped parameter systems.
Example 1
Mass/Spring System
x
k
m
Here number of degrees of freedom =1
Co-ordinate to describe the motion is x
Now consider free-body diagram, at some time t
Modelling of Automotive Systems
10
Total _ force : −m&x& − kx
δW = (− m&x& − kx )δx = 0
Hence
− m&x& − kx = 0
or
m&x& + kx = 0
R (reactive force)
m
m
inertial force
− m&x&
restoring force kx
mg (gravity force)
General one degree of freedom system
If q1 is the co-ordinate used to describe the movement then the general form of δW is as follows:
δ W = Q 1 δ q1
we call δq1 – generalised displacement
Q1 – generalised force.
δW
From principle of virtual work
= Q 1 δ q1 = 0
∴ Q1 = 0
Modelling of Automotive Systems
11
Example
Referring to the mass/spring system again
x= q1
k
m
x
δ W = (− m q&&1 − kq 1 )δ q1 = 0
∴ Q1 = − m q&&1 − kq 1 = 0
m q&&1 + kq 1 = 0
Modelling of Automotive Systems
12
Example 2
Simple pendulum
− mlθ& 2
(inertial force –radial)
0
θ
− mlθ&&
(inertial force – tangential)
P
mg
This is another one degree of freedom system.
During a virtual displacement, δθ, the virtual work done is
δW =
(−ml&θ& − mg sin θ) lδθ = 0
δW = 0 (PVW)
∴ ml&θ& + mg sin θ = 0
&θ& + g sin θ = 0
l
or
Modelling of Automotive Systems
13
Example 3 – two degrees of freedom system
x1
k
x2
k
m
m
free-body diagrams.
kx1
-mx1
m
k(x2-x1)
Modelling of Automotive Systems
-mx2
m
14
For LH mass:
[− kx1 − m&x&1 + k (x2 − x1 )]δx1 = δW1
[− k (x2 − x1 ) − m&x&2 ]δx2 = δW2
δW = δW1 + δW2 = (Q1 )δx1 + (Q2 )δx2
For δW = 0 for all
δx1 , δx2
the Qi quantities must be zero.
Hence:
m&x&1 + kx1 − k ( x2 − x1 ) = 0
m&x&2 + k ( x2 − x1 ) = 0
or
m 0   &x&1   k
 0 m  &x&  + − k

 2  
− k   x1  0
=
k   x2  0
Modelling of Automotive Systems
15
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, q1 , q2 ,...., qn .
Hence a virtual displacement can be represented by small changes in these co-ordinates:-
δ q1 , δ q 2 ,..., δ qn
Suppose only one co-ordinate,
qi (1 ≤ i ≤ n ) is given a small, imaginary displacement,
δ qi . 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
acting on the system.
δW = Qiδ qi where Qi is an expression relating directly to the forces
Qi is the generalised force associated with qi .
Modelling of Automotive Systems
16
From the principle of virtual work
δW = Qiδ qi = 0
Since,
δ qi
is finite, we get
Qi = 0
This must be true for
i = 1,2,..., n.
Q1 = 0
I.e.
Q2 = 0
M
Qn = 0
these are the equations
of motion of the system
Modelling of Automotive Systems
17
The generalised forces have component parts
1)
2)
3)
4)
5)
inertial forces (mass x acceleration)
elastic or restraining forces
damping forces (energy dissipation)
external forces
constraint forces
δW = Qiδqi =δWI +δWE +δWD +δWA
inertial
elastic
damping external
(
)
= QiI + QiE + QiD + QiA δ qi
(Noting, as before, that the constraint forces do no virtual work)
Then the equations of motion are:
QiI + QiE + QiD + QiA = 0
i = 1,2,..., n
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
q1 , q2 ,...., qn .
Modelling of Automotive Systems
18
Inertial Forces (See also Handout)
The position of the ith particle of mass, in the system, is, in general, related to the
n 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 coordinates. Thus
r r
ri = ri (q1, q2 ,...,qn , t)
(1)
Now we suppose that the system is in motion and that we represent the inertial
force on the ith particle (using D’Alembert’s Principle) as
r
− mi &r&i
(2)
We now give the system an arbitrary virtual displacement – this can be
represented in terms of generalised co-ordinates by δq1 , δq2 ,..., δqn . The virtual
displacement of the Ithparticle can be represented by
r
δr
(3)
and the virtual work done by the inertia force on the Ith particle is simply
r&
r
&
(−mi ri ). δri
(4)
(note that this is a scalar product). From this result we get the total virtual work as
r
r
δW I = ∑ (−mi &r&i ). δri
i
Modelling of Automotive Systems
(5)
19
Using equation (1) we have
r
n
∂ri
r
δri = ∑
⋅ δq j
∂
q
i =1
j
(6)
Hence
r
r& n ∂ri
&
δW = ∑ (− mi ri ) ⋅ ∑
⋅ δq j
∂
q
i
j =1
j
I
(7)
and re-arranging


n
(
r
)
δW I = ∑ δq j ∑ − mi &r&i ⋅
j =1
i
∂ri 

∂rj 
(8)
However, the generalised inertial forces, Qj, are effectively defined by
r
n
δW I = ∑ (− mi &r&i ) ⋅ ∑ Q Ij δq j
j =1
i
(9)
Comparing (8) and (9) we have
r
r&& ∂ri
Q = ∑ (− mi ri ) ⋅
∂q j
i
I
j
(10)
It is shown in the handout notes that
Q Ij = −
∂ ∂T
(
∂t ∂q& j
)+
∂T
∂q j
Modelling of Automotive Systems
(11)
20
where T is the total KE
T =
∑
i
1
r r
m i r&i ⋅ r&i
2
(12)
Elastic Forces
Consider a simple spring:
←FS
•
FE
x
For static equilibrium
FE = FS
(external force)
= (internal spring force)
Modelling of Automotive Systems
21
Suppose we define the POTENTIAL ENERGY, V, as the work done by the
external force to extend the spring a distance χ
κχ
FE
χ
external work done =
1 2
kx = V
2
∴ V = f (x) =V(x)− a function of
Here
V =
extension of
spring
x
1 2
kx
2
The work done by the internal spring force, W, is equal and opposite to V
1
W = −V = − kx 2
2
Modelling of Automotive Systems
22
Now consider a small, virtual displacement,
and V are as follows:
δx . Corresponding changes to W
δV
δV = ⋅ δx = −δW
δx
Comparing with standard form
δ W = Q E1 δ q1
then
( q1 = x , here )
Q1E = −
∂V
∂V
= −
∂ q1
∂x ⋅
Generally,
V = V ( q1 , q 2 ,...., q n )
δ V = −δ W =
∂V
δq j
∑
∂
q
j =1
j
n
Compare with
δW
∴
∑
=
Q
E
j
Q
= −
E
j
δq
j
∂V
∂q j
Modelling of Automotive Systems
23
Lagrange’s equation
Suppose that no damping forces are present, and there are no externally
applied forces. Then
QiD = QiA = 0
(i = 1,2,...,n)
We have found that
QiI = −
∂  ∂T  ∂T

+
∂t  ∂q&i  ∂qi
QiE = −
∂V
∂qi
(i = 1,2,...,n)
If we collect these results we get
∂  ∂T  ∂T ∂V
−

+
=0
∂t  ∂q& i  ∂qi ∂qi
(i = 1,2,..., n )
This is LAGRANGE’s EQUATION
Modelling of Automotive Systems
24
If we define
L=T-V
And assume that V does not depend on the
be written as:
q&i ’s, then Lagrange’s equation can
∂  ∂L  ∂L

 −
=0
∂t  ∂q&i  ∂qi
Example 1- mass/spring system
x
k
m
Here
∂T
=0
∂q1
This is a single degree of freedom system. Here
q1 = x
and
∂  ∂T  ∂

 = (Mq&1 ) = Mq&&1
∂t  ∂q&1  ∂t
∂V
= kq1
∂q1
1
m q& 12
2
1
V =
kq 12
2
T =
Hence
mq&&1 + kq1 = 0
The Lagrange equation is (n=1 so only one equation)
∂  ∂T  ∂T ∂V
−

+
=0
∂t  ∂q&1  ∂q1 ∂q1
Modelling of Automotive Systems
25
Example 2-simple pendulum
1
θ
q1 = θ
Here
In terms of
l
m
q1
we have
T=
1
2
m (l q& 1 )
2
V=
mgl(1 − cos q1 )
Hence
∂  ∂T  ∂

 =
ml 2 q&1 = ml 2 q&&1
∂t  ∂q&1  ∂t
(
)
∂T
=0
∂q1
∂V
= mgl sin q1
∂q1
ml 2 q&&1 + mgl sin q1 = 0
or
q&&1 +
g
sin q1 = 0
l
Modelling of Automotive Systems
26