Stabilization of Hybrid
Lumped-Distributed Parameter Mechanical
Systems with Multiple Equilibria.
M. V. Trivedi ∗ R. N. Banavar ∗∗ B. M. Maschke ∗∗∗
∗
Indian Institute of Technology, Bombay, Mumbai 400076. (Phone:
+91 22 2576-7888, fax: +91 22 2572-0057, e-mail:
[email protected])
∗∗
Indian Institute of Technology, Bombay, Mumbai 400076. (e-mail:
[email protected])
∗∗∗
LAGEP, UMR CNRS 5007, Universit de Lyon, Universit Lyon 1,
F-60622 Villeurbanne, France (email: [email protected])
Abstract: The objective of this effort is to initiate a study in the control of hybrid lumpeddistributed parameter mechanical systems with multiple equilibria. To do so, we study an
illustrative problem in the form of a vertical flexible beam with a tip mass at one end and
fixed to a cart at the other end. The tip mass is considered as a rigid body and the control force
is the horizontal actuation on the cart. The objective is to stabilize the beam in the vertically
upright position. This paper examines the modeling aspect alone and determines the equilibrium
profiles.
Keywords: Hybrid lumped-distributed parameter mechanical systems, multiple equilibria,
infinite dimensional system, asymptotic stability, boundary conditions, equilibrium profile.
1. INTRODUCTION
Moving end
Tip-mass
m
A flexible beam with a tip mass fixed to a moving cart
is representative of a physical system that needs both
a partial differential equation (also termed as infinite
dimensional system) and an ordinary differential equation
(also termed as finite dimensional system) to describe its
dynamic behaviour. It is an interconnection of a finite
dimensional tip mass and the cart (modeled by ordinary
differential equation) with an infinite dimensional flexible
beam (modeled using partial differential equations). Such
an assembly finds application in structures as robot arms
with flexible links and tip load. For precise control of these
systems it is desirable to account for the the dynamics of
the flexible part with utmost fidelity. Our objective here is
to stabilize the system in the vertically upright position.
A preliminary attempt has been made in this direction to
stabilize a flexible beam fixed to a moving cart (See Banavar, Dey [2010]). The control objective was to asymptotically stabilize the beam in the vertically upright position
under disturbances, using the motion of the cart as the
control effort. The problem of stabilizing the displacement
of an Euler Bernoulli beam using a boundary control
approach is discussed in (Osita et al. [2001]). Here the tip
mass acts as an actuator and its dynamics are incorporated
into the beam model. The approach of control by interconnection and energy shaping of Timoshenko beam can be
found in (Macchelli, Melchiorri [2004]) and (Macchelli,
Melchiorri [2004]).
L
Fixed end
x(t)
Fcontrol
M
Fig. 1. Flexible beam with a tip-mass on a cart
1.1 System description
A schematic of the system of our interest is shown in the
Fig. 1. It comprises of the following sub systems: a moving
cart, a flexible beam is fixed to the cart at the bottom, a
rigid tip-mass attached to the tip of the flexible beam. The
cart has a mass M . The actuating force Fcontrol applied to
the cart causes a motion of the cart. The coupling between
the cart and beam causes motion of the beam and the tipmass. We move on to nomenclature of variables. The beam
has a length L, width a and thickness b. It is symmetric
with respect to the body frame(xB − yB − zB ) as shown
in Fig. 2. Let the horizontal deflection of the beam from
the vertically upright position, the axial deflection due to
zB
RL
l=L
ΓL
FL
R0
yB
b
Γ0
ρLg
FL sinθ
FL
F0
θ
Fcontrol
b/2
b
xB
F0
RL sinθ
FL cosθ
Rg
R0
Γ0
Figure (a)
a/2
Mg
RL cosθ
ΓL
Figure (b)
RL
mg
Figure (c)
l=0
Fig. 5. Free body diagrams of the flexible beam, cart ,and
the tip-mass
xB
Figure (a)
Figure (b)
Fig. 2. (a) Undeflected position and (b) cross section of
the beam
zB
zB
zB
w(l, t)
P o
b
b
u0
P
P o
b
b
o
b
b
P
≈ ζφ(l, t)
l
φ(l, t)
l=0
xB
xB
xB
Figure (b)
Figure (a)
o
ζ
Figure (d)
Figure (c)
Fig. 3. (a) Undeflected position of the beam (b) Beam with
axial compression (c) Beam under axial and tranverse
deflection (d) Magnified view of a point in the crosssection of the beam
z
zB
w(L, t)
loading effect of the tip-mass and the rotation of the crosssection of the beam due to bending at any point l ∈ [0, L]
be w(l, t), u0 (l, t) ,and φ(l, t) respectively(Refer to Fig. 3).
The tip-mass is attached to the flexible beam with the
centre at l = L. It has a radius r, mass m and momentum
of inertia J. We assume that the motion of the mass is
a combination of both translation and rotation. Let xm ,
ym ,and θm represent the horizontal displacement, vertical
displacement and rotation of the tip-mass from the vertical
axis respectively. x(t) is the displacement of the cart with
respect to fixed frame of reference as shown in Fig. 4.
From Fig. 3.c and Fig. 4, it is straightforward to note that
xm = w(L, t) + x(t), ym = u0 (L, t) and θ = φ(L, t).
1.2 Modelling issues
Fig. 5 shows the free body diagram of the subsystems
where the following notation is being followed: FL , RL
,and ΓL represent the shear force, reaction force and the
reaction torque at l = L due to the interconnection of the
beam with the tip-mass, F0 and Γ0 represent the reaction
force and torque at l = 0 due to the interconnection of the
beam with the cart , R0 is the reaction due to the weight
of the beam, Rg is the reaction at the ground due to the
weight of the cart.
1
xm (t)
The beam
We now model the first of our subsystems the beam. We use Timoshenko beam theory to do so. From
Fig. 3 it can be observed that the displacement vector
of the beam is u = [w, 0, u0 − ζφ]. The gross transverse
displacement of a point on the beam at a distance l
from the base with reference to the fixed frame x − z is
v(l, t) = w(l, t) + x(t). We note that,
m
θ
ym
1
xB
x(t)
M
Fcontrol
x
1
Fig. 4. Displaced position of the complete system
v(0, t) = x(t)
∂v
∂w
(l, t) =
(l, t) + ẋ(t)
∂t
∂t
∂w
∂v
(l, t) =
(l, t)
∂z
∂z
The gross displacement vector of the flexible beam on a
moving cart is u = [v, 0, u0 −ζφ]. The normal strain ǫ along
the length of the beam(in the z direction) at a distance
ζ ∈ [− a2 , a2 ] from the centreline and the shear strain γ in
the zx direction are given by (See Kaya [2006])
1
ǫ = u0 ′ − ζφ′ + v ′2
2
γ = v′ − φ
(1)
t0
The variational of the integral can be now expressed as
the variation of the two terms - the kinetic and potential
energy. The variational of kinetic energy is
L
Z
0
Z
L
= −ρ
L
= −ρ
Z
0
Z
L
Z
L
=
Z
L
Z
Gγ(δv ′ − δφ) dAdz
Z
GγδvdA|L
0 −
A
0
−
Gγδγ dAdz
A
0
=
Z
Z
A
L
Z
L
0
∂
∂z
Z
Gγδv dAdz
A
GAγδφdz
0
Equation (5) must be true for any arbitary value of δv , δφ,
and δu0 . Further, δv(t2 ) = δφ(t2 ) = δu0 (t2 ) = δv(t1 ) =
δφ(t1 ) = δu0 (t1 ) = 0. With these assumptions, (5) leads
to the following equations of motion,
Z
∂
∂
ṗ1 =
Eǫ(v ′ )dA + kGA γ
∂z A
∂z
Z
∂
Eǫ(−ζ)dA + kGAγ
ṗ2 =
∂z A
Z
∂
EǫdA
ṗ3 =
∂z A
(6)
A
Z
=−
ρ||ü||δu dAdz
Z
0
(2)
Note: f ′ = ∂f
∂z . The potential energy of the beam is the
sum of strain energies due to bending and shear.
Z Z
Z Z
1 L
1 L
Eǫ2 dAdz +
Gγ 2 dAdz (3)
PB =
2 0 A
2 0 A
The kinetic energy is given by (See VoB et al. [2008])
Z
1
ρ||u̇||2 dV
(4)
KB =
2 V
where ρ, E, G, A, V represent the density per unit volume,
Young’s modulus, shear modulus and area of cross-section
and volume of the beam respectively. The equations of
motion for the Timoshenko beam can be obtained using
Hamilton’s principle (See Yu [2006]),
Z t1
(KB − PB )dt = 0
(5)
δ
δKB = −
δP2 =
Z
v̈δv + (ü0 − ζ φ̈)(δu0 − ζδφ) dAdz
We now represent these equations in the form of a statespace model and as the variation of a Hamiltonian.
A
Av̈δv + (Aü0 + I0 φ̈)δu0 + (I0 ü0 + I φ̈)δφ dz
0
L
p˙1 δv + p˙2 δφ + p˙3 δu0 dz
0
R
where I0 = A ζdA =R 0 (since the beam is symmetric in
xB − zB plane), I = A ζ 2 dA and
p1 , ρAv̇
p2 , ρAu̇0
Beam equations in a Hamiltonian framework
α , (p1 , p2 , p3 , v ′ , φ′ , u′0 )T represent the state vector.
Hamiltonian of the beam HB is the sum of kinetic
potential energy,
Z L
HB (α)dz
HB = KB + PB =
Let
The
and
(7)
0
where HB (α) represents the Hamiltonian density. The
variational derivative of HB with respect to the state
variables is
p3 , ρI φ̇
represent the generalised momenta. The variational of the
potential energy is
δPB =
Z
0
L
Z
Eǫδǫ + GγδγdAdz
A
= δP1 + δP2
where,
δP1 =
Z
L
0
Z
=+
Z
+
Z
A
A
Z
A
and
Eǫ(v ′ δv ′ + δu0 ′ − ζδφ′ )dAdz
A
Eǫ(v ′ )δvdA|L
0 −
−
Z
0
L
∂
∂z
0
Eǫ(−ζ)δφdA|L
0 −
Eǫδu0 dA|L
0
L
Z
Z
∂
∂z
L
0
Z
A
Z
∂
∂z
Eǫ(v ′ )δv dAdz
A
Z
Eǫ(−ζ)δφ dAdz
A
Eǫδu0 dAdz
δHB
δp1
δHB
δp2
δHB
δp3
δHB
δv ′
δHB
δu′0
δHB
δφ′
δHB
δφ
= v̇
(8)
= φ̇
(9)
= u̇0
Z
Eǫ(v ′ )dA + kGAγ
=
A
Z
EǫdA
=
ZA
Eǫ(−ζ)dA
=
(10)
(11)
(12)
(13)
A
= −kGAγ
(14)
The equations of motion (6) can now be recast in the
Hamiltonian framework as,
∂α
δHB
=J
∂t
δα

0 0


 0 0



 0 0

where, J = 
 ∂ 0
 ∂z

 0 ∂

∂z

 0 1

0 0
(15)

∂
0 0 0

∂z

∂
0 0
−1 0 

∂z
∂ 

0 0 0 0

∂z 

0 0 0 0 0 


0 0 0 0 0 


0 0 0 0 0 

∂
0 0 0 0
∂z
is a 7 × 7 matrix of differential operators, ∂α
∂t represents
the vector of efforts and
T
δHB
δHB δHB δHB δHB δHB δHB δHB
,
,
,
,
,
=
,
δα
δp1 δp2 δp3 δv ′ δφ′
δφ δu′0
represents the vector of flows (See Macchelli et al. [2004]).
The rate of change of the Hamiltonian HB is given by
dHB
=
dt
Z
L
Z
L
0
The tip mass The differential equations governing the
translational and rotational motion of the tip-mass are
0
δHB ∂α
dz
δα ∂t
δHB δHB
=
J
dz
δα
δα
0
Z L
∂ δHB δHB
δHB δHB
δHB δHB
=
dz
+
+
δp1 δv ′
δp2 δφ′
δp3 δu′0
0 ∂z
δHB δHB
δHB δHB L
δHB δHB
|0
(16)
+
+
=
δp1 δv ′
δp2 δφ′
δp3 δu′0
It can be observed that the power balance equation is a
quadratic function of the flows evaluated at the boundary
of the spatial domain.
The Cart The cart is described by the 2nd order differential equation M ẍ = Fcontrol − F0 . Since the cart traverses
in the horizontal plane, its total energy(Hamiltonian) Hc
is just the kinetic energy given by 12 M ẋ2 . Defining the
1
momentum px = M ẋ, Hc = 12 M x2 = 2M
px 2 and
px
∂Hc
=
M
∂px
ṗx = M ẍ = Fcontrol − F0
In matrix notation, these equations can be written as


∂Hc
0
ẋ
0 1  ∂x 
+
(17)
=
Fcontrol − F0
p˙x
−1 0  ∂Hc 
∂px
The rate of change of Hamiltonian Hc is,
ẋ =
dHc ∂Hc
∂Hc
ṗx
=
ẋ +
dt
∂x
∂px
∂Hc
∂Hc
(Fcontrol − F0 )
(ẋ) +
=
∂x
∂px
= ẋ(Fcontrol − F0 )
mẍm = RL sinθ − FL cosθ
mÿm = mg − RL cosθ − FL sinθ
J θ̈ = ΓL
(19)
Let (xm , pxm = mẋm ),(ym , pym = mẏm ) ,and (θ, pθ = J θ̇)
rerpresent the canonical coordinate pairs. The Hamiltonian of the tip-mass Hm (xm , ym , θm , pxm , pym , pθm ) is
1
1
1 2
2
Hm = mẋ2m + mẏm
+ J θ̇m
− mgym
2
2
2
1
1 2
1
px 2 +
py 2 +
pθ − mgym
=
2m m
2m m
2J
(20)
In the Hamiltonian framework the equations of motion
(19) are

x˙m
0
 y˙m   0
 

 θ̇   0
=

 px˙m   −1
 p˙   0
ym
0
p˙θ



0
0
0
0
−1
0
0
0
0
0
0
−1
1
0
0
0
0
0
0
1
0
0
0
0




0 

0

1

0


0

0 



∂Hm
∂xm
∂Hm
∂ym
∂Hm
∂θ
∂Hm
∂pxm
∂Hm
∂pym
∂Hm
∂p
 θ
0


0


0


+

 RL sinθ − FL cosθ 
 −RL cosθ − FL sinθ 
ΓL


















(21)
The rate of change of Hamiltonian is
dHm ∂Hm
∂Hm
∂Hm
ẋm +
ẏm +
θ̇
=
dt
∂xm
∂ym
∂θ
∂Hm
∂Hm
∂Hm
+
ṗθ
ṗx +
ṗy +
∂pxm m ∂pym m
∂pθ
(22)
= 0(ẋm ) − mg(ẏm ) + 0(θ̇)
+(ẋm )(RL sinθ − FL cosθ)
+(ẏm )(−RL cosθ − FL sinθ + mg) + (θ̇)(ΓL )
= (ẋm )(RL sinθ − FL cosθ)
+(ẏm )(−RL cosθ − FL sinθ) + θ̇ΓL
(23)
2. POWER BALANCE AND BOUNDARY
CONDITIONS
(18)
The total Hamiltonian HT of the system is the sum of
the Hamiltonians of the beam, cart and the tip-mass
HT = HB + Hm + Hc . The rate of change of HT is
dHT
dHB
dHm
dHc
=
+
+
dt
dt
dt
dt
δHB
δHB
δHB
δHB
(L) ′ (L) +
(L)
(L)
=
δp1
δv
δp2
δφ′
δHB
δHB
δHB
δHB
+
(L) ′ (L) −
(0) ′ (0)
δp3
δu0
δp1
δv
δHB
δHB
δHB
δHB
−
(0)
(0) −
(0) ′ (0)
δp2
δφ′
δp3
δu0
+ẋm (RL sinθ − FL cosθ) + ẏm (−RL cosθ − FL sinθ)
+(θ̇)(ΓL ) + ẋ(Fcontrol − F0 )
(24)
From (14) we have,
δHB
(L) = v̇(L) = ẋm
δp1
δHB
∂φ
(L) =
(L, t) = rotational velocity at (l = L) = θ̇
δp2
∂t
δHB
(L) = u̇0 (L) = ẏm
δp3
δHB
(0) = translational velocity at (l = 0) = ẋ
δp1
∂φ
δHB
(0) =
(0, t) = rotational velocity at (l = 0) = 0
δp2
∂t
∂v
δHB
(0) =
(0, t) = u̇0 (0) = 0
δp3
∂t
stationary. The control force Fcontrol and the net force
acting on it is zero. i.e.
M ẍ = Fcontrol − F0 = 0
(29)
⇒ F0 = 0
(30)
Conditions from the tip mass At the equilibrium the net
force and the net torque on the tip mass are zero. i.e.
mẍm = RL sinθ − FL cosθ = 0
(31)
mÿm = −RL cosθ − FL sinθ + mg = 0
(32)
J θ̈ = ΓL = 0
Equations (31) and (32) lead to
Since the system is conservative, the rate of change of
total energy must be equal to the external power supplied.
T
i.e. dH
dt = ẋFcontrol . This yields the remaining boundary
conditions on the beam as
δHB
(L) = −RL sinθ + FL cosθ
(25)
δv ′
δHB
(L) = RL cosθ + FL sinθ
(26)
δu′0
δHB
(L) = −ΓL
(27)
δφ′
δHB
(0) = −F0
(28)
δv ′
Our interest now lies in examining the equilibria of the
system described by the (15), (17) ,and (21). For this
purpose we carry out the following analysis.
Equilibria of the system
Condition from the cart
At the equilibirum the cart is
(34)
FL = mgsinθ
(35)
δHB
(L) = 0
δv ′
δHB
(L) = mg
δu′0
δHB
(L) = 0
δφ′
δHB
(0) = 0
δv ′
δHB
(L) + RL sinθ − FL cosθ
δv ′
δHB
(L) − RL cosθ − FL sinθ
+ẏm
δu′0
δHB
+θ̇
(L) + ΓL
δφ′
δHB
(0) + ẋFcontrol
+ẋ −F0 −
δv ′
RL = mgcosθ
Using the conditions obtained from the cart, tip-mass ,and
the beam, the boundary conditions (25),(26), (27) and 28
become,
Using the above relations, (24) can be written as
dHT
= ẋm
dt
(33)
(36)
(37)
(38)
(39)
Conditions from the beam From the differential equations
of the beam (15), the equilibria configurations are the
solutions to
∂ δHB
=0
∂z δv ′
δHB
∂ δHB
−
=0
ṗ2 =
∂z δφ′
δφ
∂ δHB
=0
ṗ3 =
∂z δu′0
ṗ1 =
(40)
(41)
(42)
Using the set of boundary conditions, the equilibrium
configuration of the beam is the solution to the following
equations
Z
δHB
Eǫ(v ′ )dA = −kGAγ
=
0
⇒
δv ′
A
Z
δHB
=
mg
⇒
EǫdA = mg
δu′0
A
Z
∂ δHB
δHB
∂
=
⇒
Eǫ(−ζ)dA = −kGAγ
∂z δφ′
δφ
∂z A
Using (1) and (2), the above equations take the form
1
E(u0 ′ − ζφ′ + v ′2 )(v ′ )dA = −kGA(v ′ − φ)
2
A
1
′
⇒ EA(u0 + v ′2 )(v ′ ) = −kGA(v ′ − φ)
2
(43)
Z
1
E(u0 ′ − ζφ′ + v ′2 )dA = mg
2
A
1
⇒ EA(u0 ′ + v ′2 ) = mg
(44)
2
Z
1
∂
E(u0 ′ − ζφ′ + v ′2 )(−ζ)dA = −kGA(v ′ − φ)
∂z A
2
⇒ EIφ′′ = −kGA(v ′ − φ)
Z
(45)
An intutive surmise of the possible equilibria is
(1) Vertically upright position of the beam
(2) An inclination to the right
(3) An inclination to the left
Let K , kGA. From (43), (44) we have,
Kφ
v′ =
mg + K
Substitution of (46) in (44) and (45) yields,
K 2 φ2
mg
−
EA 2(mg + K)2
mgK
φ
φ′′ =
EI(mg + K)
Equation (48) has the solution of the form
u′0 =
φ = c1 (eK1 z − e−K1 z )
mgK
EI(mg+K) and c1 is constant
(46)
(47)
(48)
(49)
Where K12 ,
dependent on
the position of the beam. Substituting (49) in (46) and
(47) and subsequently integrating them with respect to z
over the spatial domain [0, L] gives,
Kc1
(eK1 z − e−K1 z )
(50)
(mg + K)K1
c2
c2
K2
mg
( 1 e2K1 z − 1 e−2K1 z
z−
u0 =
2
EA
2(mg + K) 2K1
2K1
v=
−2c21 z)
(51)
Equations (49), (50), and (51 ) define the equilibrium
profile of the beam.
In the vertically upright position of the beam c1 = 0. From
(50), (49), and (51 ) we get the equilibrium profile of the
beam as
v=0
φ=0
mg
u0 =
z, z ∈ [0, L]
EA
For the deflection of the beam towards right from the
vertically upright position φ > 0(⇒ c1 > 0) and for the
deflection towards left φ < 0(⇒ c1 < 0). For both the
cases, the symmetry of the transverse displacement v can
be observerved from (50). This leads to an intution about
of existance of 2 equilibria in the transverse direction (right
and left of veritcally upright position).
3. CONCLUSION
Our future work will focus on stabilizing the beam at the
vertically upright position using the cart motion as an
actuator. We also need to show the nature of the equilbria
that we have obtained.
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