Bazzucchi Campolmi Zatti
Bazzucchi Campolmi Zatti
Bazzucchi Campolmi Zatti
01
=-51.27 mm
02
=-0.71 mm
03
=-31.81 mm
The static preloads of the system are the pre-compressions that the system generates
in the springs to get the equilibrium position. Under these compressions the system
can sustain its weight, that in this particular case is enormous, i.e. more or less 50
tons.
We have a really little preload on the horizontal actuator, that have to suffer less
weight with respect to the other two actuators.
The two vertical actuators preloads are not so small since they are in the order of cm,
since the weight is considerable, even if the oil of the pistons is quite incompressible.
Linearization of the equation of motion
The equations found using the Lagrange method are non linear and so the resolution
of them is too difficult (we have 3 differential equations with not constant
coefficients). However this formulation must be used in large displacements and
solved by means of numerical integration. If we deal with small displacements, we
can linearize the 3 equations of motion around the equilibrium position, making the
solution of the system easier. So we will linearize the 4 contributions (potential
energy, dissipative function, kinetic energy and Q) of the Lagrange equations.
Linearization of potential energy V
For a one degree of freedom system the approach is the following one: we perform
the Taylor expansion of
z
T
[K] z;
The equivalent stiffness matrix becomes:
Where:
i=1,2,3 (there are 3 actuators);
k
i
is the stiffness of each actuator;
m
i
is the mass of each body;
K
F
is a diagonal matrix containing the 3 actuator stiffnesses ;
Y
k
is a vector containing the 3 elongations
1
,
2
,
3
;
J
k
is the stiffness jacobian;
H is the hessian matrix;
0
is the static preload of the actuators.
J
k
1
0 0 1134.3
2
-1162.3 1162.3 0
3
-2477.4 -14.3 0
H
l1
= [
]
H
l2
= [
]
H
l3
= [
]
H
hg1
= 1.0e+03 * [
]
H
hg2
= [
]
H
hg3
= [
]
K_el_I = 1.0e+12 * [
] N/m
K_el_II = 1.0e+09 * [
] N/m
K_g = 1.0e+08 * [
] N/m
[K]=[ K_el_I] + [K_el_II] + [K_g] =1.0e+12*[
] N/m
Linearization of dissipative function D
The dissipative function for a multidegree of freedom system is: D =
k
T
[R
F
]
k
,
where:
k
is a vector containing the elongation velocities of the 3 actuators;
[R
F
] is a diagonal matrix containing the damping coefficients of the 3
actuators.
Since
k
= [J
k
] z
T
(J
k
is the stiffness jacobian), D becomes:
D =
T
[R] with [R] = [J
k
]
T
[R
F
] [J
k
];
[R]= 1.0e+10 * [
] (N*s)/m
Linearization of kinetic energy T
In order to perform the linearization of the kinetic energy, we have linearized the
kinematic relationships getting the following results:
V
1x
-
LG
1
cos(6.81-
o
) - NC cos(o) -
n sin(o)
V
1y
LG
1
sin(6.81-
o
) - NC sin(o) +
n cos(o)
V
2
CG
2
2
V
3
FG
3
o,
o and o are the values of the independent coordinates , , at the
equilibrium position.
(V
1
)
2
= (V
1x
)
2
+ (V
1y
)
2
We have projected the velocity V
1
on x and y direction to get an easier matrix
representation in the mass jacobian.
Mass jacobian
J
M
V
1x
-564.6 -408.8 2291.2
V
1y
369.9 -1021.2 201.2
1
0 0 623.0
V
2
0 0 467.0
2
1 0 0
V
3
0 0 1
3
0 0 1
The kinetic energy for a multidegree of freedom system is: T=
M
T
[M
F
]
M
where:
M
is a vector containing all the velocities of the bodies in physical
coordinates;
[M
F
] is a diagonal matrix containing masses and inertia properties of the 3
bodies; since we have considered V
1x
and V
1y
, M
F
becomes diag ( m
1
, m
1
, J
1
,
m
2
, J
2
, m
3
, J
3
);
Since
M
= [J
M
] , T becomes: T =
T
[M] with [M] = [J
M
]
T
[M
F
] [J
M
]; [M] is the
mass matrix.
[M] = 1.0e+11 [
] kg
Linearization of lagrangian component Q
In general an external force F(t) having non null average value can be written as the
sum of two contributions:
mean value;
*
L =
F
T
F = z
T
( [J
M
]
T
F ) = z
T
Q
F
where:
F
{
s
}
F = {
} z = {
}
[J
M
] is the force jacobian:
J
M
s
x
-335.1 2291.2 -408.8
s
y
-2169.3 -201.2 -1021.2
Natural frequencies and mode shapes
Once we have linearized T, D, V and Q, we apply the Lagrange method and we
obtain three linearized equations of motion.
[M] + [R] + [K] z = Q
These equations are coupled; this means that
from the mathematical point of view, we can not start solving one equation
independently from the other;
from the physical point of view, the motion of a certain mass will influence the
motion of the other ones.
Now we can solve the system and find the solution. The first step is to calculate the
natural frequencies. To do this, we will consider the unforced and undamped system:
[M] + [K] z = 0
Its solution will be: z = z
0
e
it
z
0
= {
}
Deriving z with respect to time, we obtain:
= iz
0
e
it
; = -
2
z
0
e
it
Substituting these terms inside the system:
( -
2
[M] + [K] ) z
0
e
it
= 0
since this expression must be satisfied for all values of time t
( -
2
[M] + [K] ) z
0
= 0
This is an algebraic system in which the unknown is z
0
. If the determinant of the
coefficients matrix is different from zero, there will be only one solution, that will be
the trivial one since this system is also homogenous.
Trivial solution means {
} = {
} no vibration !!!
In order to avoid this situation, we must impose that the determinant of the
coefficients matrix is equal to zero and so we obtain an equation called frequencies
or characteristic equation in which the unknown is .
, are the
amplitudes of the , ,
respectively
Solving this equation, we obtain the natural frequencies of our system:
1
= 27.3928 rad/s
2
= 2.7260 rad/s
3
= 1.0936 rad/s
We could have used another procedure in order to get the natural frequencies:
( -
2
[M] + [K] ) z
0
= 0
-
2
[I] + [M]
-1
[K]
[A]
So we obtain:
2
[I] z
0
= [A] z
0
[] = [
]
Observation: z = [] q
This is an eigenvalue problem in which:
are the eigenvalues;
the eigenvectors are collected in a
matrix called (mode shapes).
q is the vector containing
the modal coordinates
Frequency response function
We will start from the linearize equations of motion ( [M] + [R] + [K] z = Q ) and
we will suppose that the external force is F = F
0
e
it
(approach with complex
numbers).
Q = Q
0
e
it
z = z
0
e
it
Deriving z with respect time, we obtain:
=iz
0
e
it
; = -
2
z
0
e
it
This is the
particular
solution
Substituting these terms inside the system:
( -
2
[M] + i [R] + [K] ) z
0
e
it
= Q
0
e
it
( -
2
[M] + i [R] + [K] ) z
0
= Q
0
[A]=[A()]
So: z
0
= [A]
-1
Q
0
.
an amplitude
z
0
is a complex number and hence it has
a phase
In the following pictures, we have considered = [ 0 2*
1
].
Matrix of the
mechanical
impedence of the
system
Amplitude
Amplitude
Amplitude
When the external disturbance forces the system with a frequency close to the three
natural frequencies of the system, the resonance condition is reached and the
amplitudes increase (as we can see from these figures).
Some comments:
when is equal to
1
, the amplitude of and is very small with respect to
the ones obtained when is equal to
2
or
3
;
the order of magnitude of the amplitudes is 10
-7
10
-8
;
the amplitude of does not start from zero as it could appear.
Phase
Phase
Phase
When we are in the resonance condition, the phase is