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.
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 r has 3 degrees of freedom
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
This has 6 degrees of freedom y 3 translation P2 P1 3 rotation P3 . 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
T θ
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.
P P1 2
b a
b
P2 bδθ P1 aδθ 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 (acceleration) M
From Newton’s 2nd law of motion A F = Ma = M&x&
or
A F − M&x& = 0
Modelling of Automotive Systems 8 I Now, the term ( − M&x& ) can be regarded as a force – we call it an inertial force, and denote it F –thus
we can then write:
I F = −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 δ Total _ force : −m&x&− kx W = ()− m&x&− kx δx = 0 Hence − m&x&− kx = 0 or m&x&+ kx = 0
R (reactive force)
m inertial force − m&x& m 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:
δW = Q1δ q1
we call δq1 – generalised displacement Q1 – generalised force. From principle of virtual work δW = Q1δ q1 = 0
∴Q1 = 0 Modelling of Automotive Systems 11 Example Referring to the mass/spring system again
x= q1
k
m x
δW = (− mq&&1 − kq1 )δq1 = 0
∴ Q1 = −mq&&1 − kq1 = 0
mq&&1 + kq1 = 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& − mgsin θ) lδθ = 0
δW = 0 (PVW)
∴ml&θ& + mg sin θ = 0 or g &θ& + sin θ = 0 l
Modelling of Automotive Systems 13 Example 3 – two degrees of freedom system
x1 x2 k k
m m
free-body diagrams.
kx1 -mx 1 -mx2 m m k(x2-x1)
Modelling of Automotive Systems 14 For LH mass: δ δ δ δ δ []− kx1 − m&x&1 +δk()x2 − x1 δ x1 = W1
[]− k()x2 − x1 − m&x&2 x2 =δ W2
W = W1 + W2 = ()Q1 x1 + Q2 ()δx2
δ For δW = 0 for all x1,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 − kx1 0 + = 0 m&x&2 − k 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 q2 ,...,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 δW = Qiδ qi where Qi is an expression relating directly to the forces acting on the system. Qi is the generalised force associated with qi .
Modelling of Automotive Systems 16 From the principle of virtual work δ
W = Qiδ qi = 0
Since, δ qi is finite, we get
Qi = 0
This must be true for i = 1,2,...,n.
Q1 = 0 Q = 0 I.e. 2 these are the equations M of motion of the system
Qn = 0
Modelling of Automotive Systems 17 The generalised forces have component parts
1) inertial forces (mass x acceleration) 2) elastic or restrainingδ forces 3) damping forces (energy δdissipation) 4) external forces δ 5) constraint forces δ I E δ D A W=Qi qi = W + W + W +δW
inertial elastic damping external
I E D A = ()Qi + Qi + Qi + Qi δ qi (Noting, as before, that the constraint forces do no virtual work)
Then the equations of motion are:
I E D A Qi + Qi + Qi + Qi = 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 co- ordinates. 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) δ We now give the system an arbitrary virtual displacement δ– this can be represented in terms of generalised co-ordinates by q1, q2 ,...,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 &r&i ). δri (4) (note that this is a scalar product).δ From this result we get the total virtual work as
I r r W = ∑(−mi&r&i ). δri (5) i
Modelling of Automotive Systems 19 Using equation (1) we have δ n r r ∂ri ri = ⋅ δq j ∑ (6) δ i=1 ∂q j Hence
n ∂rr W I = (−m &r&) ⋅ i ⋅ δq ∑ i i ∑ j (7) i j =1 ∂q j and re-arranging δ n ∂r I &r& i W = ∑δq j ∑()− miri ⋅ (8) j =1 i ∂rj
However, the generalised inertialδ forces, Qj, are effectively defined by
n I r I W = ∑(−mi&r&i ) ⋅ ∑Q jδq j (9) i j =1 Comparing (8) and (9) we have r I r ∂ri Q j = ∑(−mi&r&i ) ⋅ (10) i ∂q j
It is shown in the handout notes that
I ∂ ∂T ∂T Q j = − ( )+ (11) ∂t ∂q& j ∂q j
Modelling of Automotive Systems 20 where T is the total KE
1 r r T = ∑ m i r&i ⋅ r&i (12) i 2
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
FE κχ
extension of χ spring
1 2 external work done = kx = V 2 ∴ V= f(x)=V(x)− a function of x
1 2 Here V = kx 2
The work done by the internal spring force, W, is equal and opposite to V
1 W = −V = − kx2 2
Modelling of Automotive Systems 22 Now consider a small, virtual displacement, δx . Corresponding changes to W and V are as follows: δ δ δV δ V = ⋅ x = −δW x Comparing with standard form δ 1 W = Q E δq1 (q1 = x, here )
E ∂V ∂V then Q1 = − = − ∂q1 ∂x ⋅
Generally, δ V = V (qδ1 , q2 ,...., qn ) n ∂V V = − W = ∑ δq j j =1 ∂q j Compare with δ
E W = ∑ Q j δ q j
E ∂ V ∴ Q j = − ∂ 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
D A Qi =Qi = 0 (i =1,2,...,n)
We have found that
∂ ∂T ∂T I Qi = − + (i =1,2,...,n) ∂t ∂q&i ∂qi
E ∂V Qi = − ∂qi
If we collect these results we get
∂ ∂T ∂T ∂V − + = 0 (i = 1,2,..., n) ∂t ∂q&i ∂qi ∂qi
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 q&i ’s, then Lagrange’s equation can be written as:
∂ ∂L ∂L − = 0 ∂t ∂q&i ∂qi
Example 1- mass/spring system
x k
m ∂ ∂T ∂ Here = ()Mq&1 = Mq&&1 ∂t ∂q&1 ∂t
This is a single degree of freedom system. Here ∂T = 0 ∂q1 q 1 = x
1 2 ∂V and T = m q& 1 2 = kq1 ∂q1 1 2 V = kq 1 2 Hence
The Lagrange equation is (n=1 so only one equation) mq&&1 + kq1 = 0
∂ ∂T ∂T ∂V − + = 0 ∂t ∂q&1 ∂q1 ∂q1
Modelling of Automotive Systems 25 Example 2-simple pendulum
1 Here q1 = θ
θ In terms of q1 we have l
1 2 T= m ()lq& 1 2
m V= mgl()1− cosq1
Hence
∂ ∂T ∂ 2 2 = ()ml q&1 = ml q&&1 ∂t ∂q&1 ∂t ∂T = 0 ∂q1 ∂V = mgl sin q1 ∂q1
2 ml q&&1 + mgl sin q1 = 0
g q&&1 + sin q1 = 0 or l Modelling of Automotive Systems 26