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TMHP51
IEI / Fluid and Mechanical Engineering Systems
____________________________________________________________________________________
Hydraulic Systems with Load Dynamics
Karl-Erik Rydberg
2008-10-15
J1
Dm
qm
xv
KL
BL
J2
qL
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Hydraulic systems with dynamic loads
The design of hydraulic systems is often based on a simple load model, represented by single
lumped parameters. This type of load model can be used if the connection between the
hydraulic system and the mechanical load is stiff. However, in many applications, the
mechanical system, which the hydraulic power elements are connected to, is weak compared
to the stiffness of the hydraulic system. Such weak mechanical structures cause resonance’s
which can be lower than the hydraulic natural frequency. If the structural resonance’s
dominate the frequency response of the servo system, it is extremely important to take this
fact into account in system design. The main reasons are that the stability of the systems and
the bandwidths are limited by the lowest natural frequency in the control loop.
A simple valve-controlled hydraulic cylinder with a load represented by a mass Mt and an
arbitrary load force FL is shown in Figure 1. The four-way valve is assumed to be a
servovalve with constant flow gain Kq.
Figure 1: Valve-controlled hydraulic cylinder with a mass load
For the special case of centered piston, the oil volumes between the piston and the valve are
V1 = V2 = Vt/2. If then the load pressure is defined as pL = p1 - p2 the following linearized
and laplace transformed equations can be derived:
Vt ⎞
⎛
KqXv = Ap sXp + ⎜ Kce +
s⎟ P
⎝
4β e ⎠ L
(1)
Mt s 2 Xp = Ap PL − FL
(2)
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These equations resulted in the block diagram in Figure 2.
Hydraulic system
Mechanical system - Load
Kce
Xv
Kq
+
1
_____
Vt
___
s
4βe
-
PL
QL
FL
Ap
1
____
Mt s
+
.
Xp
1
_
s
Xp
Ap
Figure 2: Block diagram of a valve-controlled hydraulic cylinder with a mass load
The question is now to define a load model, which is simple to connect to the hydraulic
system. As can be seen from Figure 2 the transfer function of the mechanical load can be
expressed as
Gm(s) =
1.
QL
PL
QL = Ap sXp
;
(3)
Loads with one degree of freedom
An hydraulic cylinder acting on a simple mass load is illustrated in Figure 3. This is a load
with one degree of freedom.
Ap
p1 C1
xp
Ap
Mt
C2 p2
FL
Figure 3: Symmetric hydraulic cylinder with a mass load
For a double acting symmetric cylinder loaded by a mass Mt and an external force FL
according to Figure 3, the piston position Xp is given by equation (2) as
Xp =
Ap PL − F L
Mt s
2
2
;
FL = 0 ⇒ Gm(s) =
Ap s
Mt s
2
=
1
Mt
2
Ap
(4)
s
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The load dynamics, from force to piston velocity, is in this case represented by a pure
integrator. Introducing the total capacitance of the cylinder CL the transfer function Gm(s) can
be rewritten as
Gm(s) =
QL
CL
=
PL
CLMt
A2p
=
s
CL
s
(5)
ω2h
The capacitance of the cylinder CL is definded as
V1
1
1
1
=
+
; C1 =
och
CL C1 C2
βe
C2 =
V2
βe
(6)
Centered cylinder piston gives the Capacitance CL and the hydraulic natural frequency ωh as
Vt
V1 = V2 =
2
Vt
CL =
4β e
⇒
⇒
4β eA2p
Vt Mt
ωh =
(7)
Figure 4 shows a more general load situation where the mass load is completed with a spring
gradient KL and a viscous damping coefficient BL. The cylinder is also defective with viscous
friction, the coefficient Bp.
Ap
KL
Ap
Mt
p1 C1
C2 p2
xp Bp
FL
BL
Figure 4: Symmetric hydraulic cylinder with a mass, spring and damping load
The load transfer function will be expressed in the same way as in equation (5) which gives
Gm(s) =
QL
=
PL
CLMt
A2p
2
s +
CL s
CLBe
A2p
s+
CLKL
A2p
A2p
s
KL
=
Mt 2 Be
s +
s+1
KL
KL
where the total viscous friction coefficient is Be = BL + Bp.
(8)
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In real systems, we always have a certain elasticity in the mechanical structure. For a cylinder
this weakness can be related to spring systems in the rear mounting end (K1) and in the piston
rod (KL) as shown in Figure 5.
K1
Ap
x1
p1 C1
xp
Ap
KL
C2 p2
Bp
x2
Mt
FL
xL
Figure 5: Hydraulic cylinder with elastic mountings
Suppose that the piston and rod have no mass and the external force FL = 0. The resulting
force equation will now be expressed as
Ap P L = Mt s 2 XL = −K1 X1 = KL(X2 − XL)
(9)
The load flow to the cylinder (QL) is
QL = Ap s (X2 − X1 ) = Ap sXp
(10)
The transfer function of the mechanical load can be written as
⎛
A 2p ⎡
QL
1 ⎞⎟ ⎤
2⎜ 1
+1
Gm =
=
M
s
+
t
PL M ts ⎢⎣
⎝ K1 K L ⎠ ⎦⎥
(11)
Consider the system shown in Figure 5 and let the cylinder be controlled by a 4-port servo
valve with the coefficients Kq and Kc. This system is shown in Figure 6.
x1
xp
K1
V1
Ap V2
p1
p2
x2
xL
KL
Mt
xv
Figure 6: Valve controlled cylinder with elastic mountings
According to Figure 2 the block diagram for the system will be developed as in Figure 7.
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Kce
Xv
Kq
+
1
_____
Vt
___
s
4βe
-
PL
QL
.
⎛ 1
1 ⎞ ⎤ Xp
1 ⎡
2
⎟ + 1⎥
+
⎢M t s ⎜
⎝ K1 KL ⎠ ⎦
M ts ⎣
Ap
1_
s
Xp
Ap
Figure 7: Block diagram for a valve controlled cylinder with elastic mountings
Reduction of the block diagram in Figure 7 and completing with the transfer function from Xp
to XL gives the following diagram.
⎛ 1
1 ⎞
M ts 2 ⎜
+
⎟ +1
⎝ K1 K L ⎠
s2
δ 'h
+
2
+1
ω 'h2
ω 'h
Xv K q
Ap
.
Xp
1_
s
Xp
1
⎛ 1
1 ⎞
M ts ⎜
+
⎟ +1
⎝ K1 K L ⎠
XL
2
Figure 8: Complete block diagram for a valve controlled cylinder with elastic mountings
where ω 'h =
Ke
Mt
;
1
Vt
1
1
K ce M t ω 'h
'
=
+
+
and
δ
=
⋅
2
h
Ap
2
Ke 4β e A 2p K 1 K L
If can be noted that the effective spring gradient Ke is derived from the series connection
2
4 βe A p
and the two mechanical springs K1, KL.
between the hydraulic spring gradient
Vt
2.
Loads with two degrees of freedom
a) Mechanical flexibility between two masses on a piston rod
Assume that the load consists of two masses (M1 and M2), of which the first one is fixed
mounted to the piston and the second mass is connected by a spring (KL) and a viscous
damper (BL) to the first mass.
Ap
p1 C1
xp
Ap
C2 p2
KL
M2
M1
BL
xL
Figure 9: Hydraulic cylinder with a load with two degrees of freedom
FL
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The force balance equations for the masses M1 and M2 are for FL = 0
M1 s 2 Xp + M2 s 2 XL = Ap PL
(12)
M2 s 2 XL = −BLs (XL − Xp ) − KL(XL − Xp )
(13)
The piston position Xp can be expressed as
⎛⎜ M2 2 BL
⎞⎟ A P
⎝ KL s + KL s + 1⎠ p L
Xp =
M1 M2
BL
⎤
2
(M1 + M2 ) s2 ⎡⎢
+
s + 1⎥⎥
s
⎢⎣ KL(M1 + M2 )
KL
⎦
(14)
With the load flow QL = ApsXp we obtain the transfer function
BL
⎞
⎛ M2
CL ⎜⎝ K s 2 +
s + 1⎟⎠
QL
KL
L
Gm(s) =
=
PL
CL
M1 M2
BL
⎤
2
(M1 + M2 ) s ⎡⎢⎢
s
+
s + 1⎥⎥
2
(
)
K
L
⎦
⎣ KL M1 + M2
Ap
(15)
Equation (15) can also be written in the following form
2δa
⎛ s2
⎞
⎜⎜
⎟⎟
+
s
+
1
ωa
⎝ ω2a
⎠
A2p
Gm(s) =
(M1 + M2 ) s ⎛ s2
2δ1
⎞
⎟⎟
⎜⎜
+
s
+
1
ω1
⎠
⎝ ω21
KL
,
M2
ωa =
where
δa =
BL
2
1
,
KLM2
KL(M1 + M2 )
= ωa
M1 M2
ω1 =
δ1 =
BL
2
M1 + M2
= δa
KLM1 M2
(16)
1+
1+
M2
M1
M2
M1
From equation (16) it can be seen that the two coupled masses cause one natural frequency of
the fundamental mode ω1 and one natural frequency of nodes of vibration ωa. For these
frequencies, it is always evident that ωa < ω1 and the damping ratio δa < δ1. A Bode diagram for
equation (16) is shown in Figure 10.
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Amplitude
101
100
10-1
10-2
100
101
102
Frequency [rad/s]
Phase
50
0
-50
-100
100
101
Frequency [rad/s]
102
Figure 10: Frequency response according to equation (16). Solid line is
valid for M1 = M2 and dashed line for M1 = M2/2
Figure 10 explains how the amplitude and the phase curve change when the mass M1 is
reduced in proportion to M2. If M1 instead increases in proportion to M2 the frequency
response will be influenced in the way as expressed in Figure 11.
Amplitude
100
10-1
10-2
100
101
102
Frequency [rad/s]
Phase
50
0
-50
-100
100
101
Frequency [rad/s]
102
Figure 11: Frequency response according to equation (16). Solid line is
valid for M1 = M2 and dashed line for M1 = 2M2
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It is convenient to combine equation (16) with the hydraulic part of the system. If the external
force FL = 0, the valve cylinder combination (compare with Figure 1) and the load with two
masses give the block diagram illustrated in Figure 12.
Kq
Xv ___
Ap +
-
Ap
__________
Vt
Kce + ___ s
4βe
PL
Ap
s2
GLX(s ) =
where
1
__________
GLX(s)
(M1 + M2) s
ω2a
s2
ω21
+
2 δa
s+1
ωa
+
2 δ1
s+1
ω1
.
Xp
Figure 12: Block diagram for a valve-controlled cylinder with a load of two masses (FL = 0)
From Figure 12 the transfer function for the hole system, with valve opening Xv as input
signal and piston velocity dXp/dt as output signal, can be derived as
GHL(s ) =
GLX(s)
sXp Kq
=
Xv
Ap s 2
2
where
ωh =
(17)
2 δh
+
s + GLX(s )
ωh
ω2h
4β eAp
Vt (M1 + M2 )
;
δh =
Kce
Ap
β e(M1 + M2 )
Vt
(18)
The frequency response for equation (17), when the hydraulic natural frequency ωh is lower
than the structural resonance’s (ωa and ω1) is shown in Figure 13 (see next page). The dashed
lines in the diagrams illustrated the response when the mass M2 is increased 5 times compared
to the situation of the solid lines. The hydraulic spring rate Kh (=4βeAp2/Vt) has the same
value for both cases.
In this system, the frequency response will be dominated by the hydraulic system. In the low
frequency range is GLX(s) ≈ 1 and the system can be treated as a one-mass system with the
load mass Mt = M1 + M2.
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101
Amplitude
100
10-1
10-2
10-3
100
ωa ω1
ωh
101
Frequency [rad/s]
102
101
Frequency [rad/s]
102
0
Phase
-50
-100
-150
-200
100
Figure 13: Frequency response for GHL(s) according to equation (17). Solid line: M1 = M2 = 1.
Dashed line: M1 = 1 and M2 =5. Kh = 0,1.KL for both curves
If we now consider that the structural resonance’s are lower than the hydraulic natural
frequency, the system response will change drastically, which can be seen in Figure 14.
Amplitude
101
100
ωa
10-1
100
ω1
ω'h
101
Frequency [rad/s]
102
101
Frequency [rad/s]
102
Phase
100
0
-100
-200
100
Figure 14: Frequency response for GHL(s) according to equation (17). Solid line: M1 = M2 = M0.
Dashed line: M1 = M0 and M2 =5M0. Kh = 10.KL for both curves
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In Figure 14 the hydraulic spring rate Kh is 10 times greater than the mechanical spring rate
KL which means that ωh will be the highest frequency. If we look at the amplitude of the
transfer function GLX (see Figure 13) with frequencies greater than ω1, this equation will
reach a constant value
2
M2
⎛ ω1 ⎞⎟
⏐GLX(s)⏐ω>ω1 = ⎜⎝
=
1
+
ωa ⎠
M1
(19)
Since the hydraulic resonance occur at a frequency higher than the structural resonance’s, the
value of ωh have to be influenced by the function GLX written as equation (19). For this case,
the hydraulic frequency and damping, here named ω´h and δ´h, may be changed from the
original expression (equation 18) to
ω,h
= ωh
δ,h = δh /
M2
1+
=
M1
1+
M2
Kce
=
M1
Ap
Kh
=
M1
4β eA2p
Vt M1
(20)
β eM1
Vt
This equation shows that the dynamics of the hydraulic system only is dependent on the mass
M1. Increasing of the mass M2, of course, will lower the mechanical resonance’s but the
hydraulic frequency ω´h is not influenced by change of this mass. The reason is that the
movement of the mass M2 is approximately zero at such high frequencies as ω´h. The
hydraulic natural frequency is solely determined by the mass M1 which is stretched between
the hydraulic spring in the cylinder Kh and the load spring KL. However, the low value of KL
compared to Kh means that KL cannot be seen in equation (20).
If the system is changed so that only the hydraulic spring constant Kh (see eq 20) increases,
the variation in amplitude and phase shift according to the mechanical structure will be
reduced. This situation is shown in Figure 15. From equation (17) it can also be seen that, if
the structural resonance’s (GLX(s)) are dominant, the transfer function from valve opening to
piston speed approaches
GHL(s) = sXp/Xv ≈ Kq/Ap
(21)
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Amplitude
101
100
10-1
10-2
100
101
102
Frequency [rad/s]
Phase
100
0
-100
-200
100
101
102
Frequency [rad/s]
Figure 15: Frequency response for GHL(s) according to equation (17). Solid line: Kh = Kh0.
Dashed line: Kh = 5.Kh0. M1 = M2/2 for both curves
Another situation where ωh increases and makes the hydraulic system stiffer will arise if the
mass M1 is reduced. Figure 16 shows how the frequency response develops when M1 is
reduced from M1 = M10 to M1 = 0,2.M10. Since M1 also influences the mechanical
resonance’s this rise of ωh will not cause a reduction of the structural resonance’s as in the
case of increased hydraulic stiffness.
Amplitude
101
100
10-1
10-2
100
10 1
102
Frequency [rad/s]
Phase
100
0
-100
-200
100
10 1
102
Frequency [rad/s]
Figure 16: Frequency response for GHL(s) according to equation (17). Solid line: M1 = M10.
Dashed line: M1 = 0.2.M10. Kh and KL are the same for both curves
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The conclusion to be drawn from these examples is that the structural resonance’s are reduced
by a stiff hydraulic system if ωh is always higher than the mechanical frequencies. This
stiffness can be achieved by low hydraulic capacitance for the valve cylinder combination
and/or feedback control.
b) Two-mass system with mechanical flexibility in the rear mounting end
A further example of a hydraulic cylinder loaded by a two-mass system is shown in Figure 17.
K1
M1
Ap
Ap
p1 C1
xp
B1
x1
M2
C2 p2
FL
x2
Figure 17: Hydraulic cylinder with flexible rear flange and a load with two degrees of freedom
The transfer function for this mechanical system is
⎛ s2
2δa
⎞
⎜⎜
⎟⎟
+
s
+
1
ωa
⎝ ω2a
⎠
QL
A2p
Gm(s) =
=
PL M2 s ⎛ s2
2δ1
⎞
⎜⎜
+
s + 1⎟⎟
2
ω1
⎝ ω1
⎠
where
K1
,
M1 + M2
ωa =
δa =
B1
2
K1
= ωa
M1
ω1 =
1
K1 (M1 + M2 )
, δ1 =
(22)
B1
2
1+
1
= δa
K1 M1
M2
M1
1+
M2
M1
According to equation (17) a valve controlled cylinder yields
GHL(s ) =
sXp Kq
=
Xv
Ap s 2
G LX(s ) =
(23)
2 δh
+
s + GLX(s )
ωh
ω2h
s2
where
GLX(s)
ω2a
2
s
ω21
+
+
2 δa
s+1
ωa
2 δ1
s+1
ω1
; ωh =
4β eA2p
Kce
; δh =
Vt M2
Ap
β eM2
Vt
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The Bode diagram for equation (23) is shown in figure 18. In this figure the hydraulic
resonance ωh is the dominant frequency, lower than ωa. If the hydraulic spring rate Kh is
changed, the value of ωh will also be changed which is illustrated in the figure.
Amplitude
10 1
10 -3
10 0
10 1
10 2
Frequency [rad/s]
Phase
0
-100
-200
10 0
10 1
Frequency [rad/s]
10 2
Figure 18: Frequency response for GHL(s) according to equation (23). Solid line: Kh = 0,04 KL.
Dashed line: Kh = 0,1 KL. M1 = 1, M2 = 2 and KL = 1000 (same for both curves)
If the hydraulic cylinder is stiffer than the mechanical structure (Kh > KL), we will have a
system with a behaviour similar to that described in Figure 15. An increase of Kh is shown in
Figure 19 and we can see that the structural resonance’s are reduced by the large value of ωh.
Amplitude
10 1
10 0
10 -1
10 -2
10 0
10 1
10 2
Frequency [rad/s]
Phase
100
0
-100
-200
10 0
10 1
10 2
Frequency [rad/s]
Figure 19: Frequency response for GHL(s) according to equation (23). Solid line: Kh = 4 KL.
Dashed line: Kh = 10 KL. M1 = 1, M2 = 2 and KL = 100 (same for both curves)
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With this stiff hydraulic system, the real hydraulic frequency and damping, ω´h and δ´h , are
changed from the original expression (compare with eq 20) to
,
ωh
M2
1+
=
M1
= ωh
M2
Kce
1+
=
M1
Ap
,
δh = δh /
2
4β eAp (M1 + M2 )
Vt M1 M2
(24)
β eM1 M2
Vt (M1 + M2 )
Figure 20 shows the frequency response when the mass M2 is increased five times. Note that
the reduction of ωh increases the variation in amplitude and phase shift for the mechanical
structure which can also be seen from the expression of ωa and δa (eq 22).
Amplitude
101
100
10-1
10-2
100
101
102
Frequency [rad/s]
Phase
100
0
-100
-200
100
101
Frequency [rad/s]
102
Figure 20: Frequency response for GHL(s) according to equation (23). Solid line: M1 = M2 = 1.
Dashed line: M1 = 1, M2 = 5. Kh = 4 KL (= 100) for both curves.
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c) Hydraulic motor with mechanical flexibility between two inertia loads
Bm
KL
J1
Dm
BL
θm
TL
J2
θL
Figure 21: Hydraulic motor with a load with two degrees of freedom
This hydraulic motor system with two inertia loads is identical with the cylinder application
illustrated in Figure 9.
Consider the system shown in Figure 21 and let the motor be controlled by a 4-port servo
valve with the coefficients Kq and Kc. The block diagram of the overall system will be
developed as in Figure 22.
TL
Kq
Xv ___
Dm +
Dm
__________
Vt
Kce + ___ s
4βe
-
PL
G LT(s)
Dm
+
.
θm
1
________
G Lθ(s)
(J1 + J2) s
Figure 22: Block diagram for a valve-controlled motor with two inertia loads and a load torque (TL)
From Figure 22 it can be seen that the mechanical structure has influence on the torque
disturbance (TL) by the transfer function GLT(s). The transfer function GLθ(s) is similar to
GLX(s) in Figure 9. These transfer functions can be expressed as
1+
GLT(s ) =
s
2
ω2a
where
2δ1 ω1
2δa
s+1
ωa
KL
,
J2
ωa =
δa =
+
BL
2
s2
s
1
,
KLJ 2
;
GLθ(s ) =
s
2
ω21
+
2δa
s+1
ωa
BL
2
(25)
2δ1
s+1
ω1
KL(J1 + J 2 )
= ωa
J 1 J2
ω1 =
δ1 =
ω2a
+
J1 + J2
= δa
KLJ 1 J 2
1+
1+
J2
J1
J2
J1
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Appendix
Transfer functions for loads with 2 DOF
16
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17