Page 580 - Mechatronics with Experiments

JWST499-Cetinkunt
JWST499-c07
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control application case, the base motion would be zero, whereas the base motion is non
zero in mobile equipment applications due to ground height variations as the equipment
travels on the ground.
There are three inertias in the system: m , m , m , connected to each other via springs
1
2
3
and dampers, as well as the hydraulic cylinder. The motion of the hydraulic system has three
degrees of freedom, y (t), y (t), y (t), which are the positions of each inertia relative to a
3
2
1
reference ground. The ground height variation y (t) is an input variable as a function of the
0
actual ground conditions and travel speed of the equipment. In web-thickness roll control
applications, it is zero, y (t) = 0.0. The motion of each mass is defined by a second-order
0
differential equation. The coupling between the motion of the three inertias is defined by
the springs and dampers, as well as the pressure in the cylinder. The cylinder pressure is
modeled as a compressible fluid volume. The flow in and out of the cylinder is controlled
by a flow control valve. The flow control valve is modeled to have a spool displacement
versus orifice area relationship with a symmetric deadband. In other words, around the
null-position of the spool, there is a deadband in equal amounts in both directions of the
spool motion. While the spool position is in the deadband region, there is no orifice opening,
hence no flow.
The resultant dynamic equations are the three coupled second-order and two first-
order differential equations, where the flow and pressure conditions affect the motion of
the inertia through pressure in the cylinder head-end and rod-end.
m ̈ y (t) =−c ( ̇ y (t) − ̇ y (t)) − k (y (t) − y (t)) − (A p (t) − A p (t)) (7.639)
1
0
re re
0
1
1
he he
1 1
1
m ̈ y (t) =−c ( ̇ y (t) − ̇ y (t)) − k (y (t) − y (t)) + (A p (t) − A p (t)) (7.640)
2
3
re re
3
he he
3
2 2
2
3
m ̈ y (t) =−c ( ̇ y (t) − ̇ y (t)) − k (y (t) − y (t)) − F (t) (7.641)
3 3 3 3 2 3 3 2 load
where,
dp (t)
he
= ⋅ (Q (t) − A ( ̇ y (t) − ̇ y (t))) (7.642)
2
he
he
1
dt V (t)
he
dp (t)
re
= ⋅ (−Q (t) + A ( ̇ y (t) − ̇ y (t))) (7.643)
re
re
2
1
dt V (t)
re
For the spool position ranges in x ≥ x , the valve spool has positive displacement past the
db
s
deadband region and flow is from pump to head-end,
√
Q (t) = C ⋅ A (x ) ⋅ (7.644)
P
he
he d1 he1 s |p (t) − p (t)|
√
Q (t) = C ⋅ A (x ) ⋅ (7.645)
re
t
re d1 re1 s |p (t) − p (t)|
for the spool position ranges in x ≤ −x , the valve spool has negative displacement past
s db
the deadband region and the flow is from pump to rod-end,
√
Q (t) =−C d2 ⋅ A he2 (x ) ⋅ |p (t) − p (t)| (7.646)
he
he
s
t
√
Q (t) =−C d2 ⋅ A re2 (x ) ⋅ |p (t) − p (t)| (7.647)
P
s
re
re
for the spool position ranges in −x ≤ x ≤ x , the valve spool is in the deadband region
db s db
and there is no flow to either side of the cylinder,
Q (t) = 0.0 (7.648)
he
Q (t) = 0.0 (7.649)
re
and
V (t) = V (0) + A ⋅ (y (t) − y (t)) (7.650)
1
he
he
he
2
V (t) = V (0) + A ⋅ (y (t) − y (t)) (7.651)
re
2
1
re
re