This document discusses factors that influence the natural frequency of linear hydraulic actuators. It identifies three key factors: 1. The effective mass of the pipe oil volume. The mass of oil in the pipes must be accounted for as it scales up with load movement. 2. The effective bulk modulus. Structural compliances from elastic connections or piston rod anchoring can be modeled as a mechanical spring in series with the hydraulic spring, affecting the effective bulk modulus. 3. The dynamic oil volumes of the piston chambers, which depend on sealing methods. Shaft seals in particular contribute significantly to the "constructive length" of trapped oil volumes beyond the theoretical geometric volumes.

1. Int. J. Mach. Tool Des. Res. Vol. 11, pp. 199-207. Pergamon Press 1971. Printed in Great Britain
SOME FACTORS INFLUENCING THE NATURAL
FREQUENCY OF LINEAR HYDRAULIC ACTUATORS
H. J. ULRICH*
(Received 27 August 1970)
Abstraet--A discrepancy between calculated and measured values of the natural frequency ~onof hydraulic
actuators is often observed. It is shown that this discrepancy is due to three factors: 1. The effective mass of
the pipe oil volume, the pipe cross section can be optimized. 2. The effective Bulk Modulus. 3. The dynamic
oil volumes of the piston chambers, which is defined in this paper. A testing method is described and test
results of various sealing methods are shown.
INTRODUCTION
A MOSTimportant characteristic of a servo-system is its natural frequency oJn, which deter-
mines the dynamic behaviour of the system. If the system is to have a fast time response,
~on must be large. The natural frequency is also important in determining the limits of the
open loop gain and the stability of the system.
A discrepancy between calculated and measured values of o~n is often observed in
hydraulic servo-systems. This discrepancy is particularly evident in rotary motors, where it
is usual for measured natural frequencies to be some 40 per cent lower than theoretical
predictions [1]. Investigations with linear hydraulic actuators of copying systems at the
E.T.H.* Machine Tool Laboratory have brought to light certain relevant, but frequently
neglected factors, which are presented in this paper.
ANALYSIS
Consider a system actuator-load with one ]degree of freedom (Fig. 1).
'
Servo-vo/ve
÷y
v,
System Actuator-load (1Degree of Freedom)
FIG. 1.
Since the valve has no effect on the natural frequency of the actuator, the pressure lines
can be shut off after the valve ports. The compressibility of the trapped oil volumes can be
simulated by two spring-loaded pistons with displacements Sl and s2 (Fig. 2): the oil volumes
can now be considered as absolutely rigid.
* Swiss Federal Institute of Technology, Zurich.
199

2. 200 H.J. ULRICH
Let the piston be slightly displaced in direction -- y from its neutral position, and then
released. Newton's Second Law applied to the forces on the piston yields:
My + By~-= Arpz (1)
where M is the mass, and B the viscous drag coefficient of the load. The load pressure PL
is a result of the pressure changes of Apt and -- Ap2 in chambers 1 and 2 respectively:
that is
pz ----Apl -- Ap2. (2)
$1 S 2
Equivalent 2nd Order System
FIG. 2.
The pressure variations Apl and -- Ap2 result from the trapped oil volumes being com-
pressed and decompressed, and their values are given by the Isothermal State Equation for
Liquids:
/~AVI}@1 = .~ (3)
/~ AV2
where
fl ----Bulk Modulus
V1, 112----total volumes under pressure in chambers 1 and 2.
Since, from Fig. 2,
equations (3) can be replaced by:
AVi = Ar.si ]
(4)
JAV2 = Ar.s2
- ArSl }Apl = ~ ~
Ars2Ap2 = ~ ~f.
Applying continuity conditions to the movement of the pistons gives:
-- Ar~i -----Arf2 ]
JAr~2 -----Arp.
Since no initial conditions need be considered, equations (6) yield:
81 = --y /
J$2 ~y,
(5)
(6)
(7)

3. Some Factors Influencing the Natural Frequency of Linear Hydraulic Actuators 201
Finally we combine equations (1), (2), (5) and (7), and take Laplace transforms, giving:
[Ms 2 -~-Bs + A2r[3 (1 1121)]~i+ y=0. (s)
This is evidently an equation of the second order, with a hydraulic spring constant:
+ '),1 ~72" (9)
which gives a natural frequency
COn ~ ~ M
The natural frequency COndepends on V1 and //2, and varies with the position of the
piston as plotted in Fig. 3 [2]. COnhas a minimum when 1/1 = V~, and this case must be
i-l-
~=o
CMn
¢° no I
i
v,:v,:v : v,:o
i,
L_3 T I
Natural frequency versus rnm position
leourtesy of M,Guillon)
FIG. 3.
treated in the designing of servo-systems. When 1/1 = 1/2,equation (10) reduces to the well-
known relation:
V in the above equation can be replaced by half the piston stroke ls times the ram area
Ar, plus the dead volume At. la, thus:
=If (12)
wn ~/ k(ls -f- la). M!
DETERMINATION AND INFLUENCE OF THE PARAMETERS
The piston area Ar
From equation (12) a high natural frequency results from making the ram area Ar
large. However three factors limit this possibility:
1. Normally the actuator space should be as small as possible.

4. 202 H. ,]. ULRICH
2. A larger piston enhances the mass of the load.
3. In order to maintain a maximum speed, a larger supply volume and bigger valves are
necessary.
Increasing Ar is thus not the ideal means of improving the natural frequency of hydraulic
actuators.
The dynamic volume
The volumes of a hydraulic actuator consist of all volumes under pressure: normally
these are the stroke-volume and the dead volumes of the valve and piston chamber connect-
ing lines. Under this assumption the Isothermal State Equation, equation (3), does not
give correct results. Test results show a difference between the geometrical and effective
values of the volumes.
The isothermal law of liquid state gives:
Ap = fl._~l (13)
after eliminating the piston area Ar in equation (3). In each chamber l is equal to the sum of
Is + la as shown in equation (12). The linear relation between l and AI is illustrated in the
curves plotted in Fig. 4.
On the other hand, experimental results yield the curves plotted in Fig. 5 in which all
curves for various pressure drops Ap intersect in a common point A. This means that a
further linear compliance is present, in addition to the spring of the trapped oil volume.
A!
-~ = COl'/St.
> l
Lld i Is ! oilcolumn
,- =i=
0
l = Id + ls
Compression versus stroke (theoretic~z/)
F]c. 4.
,'1
o
ASVlc d_l_pc -r
,~p
- - = COOSt.
Compression versus stroke (measured)
Fzo. 5.

5. Some Factors Influencingthe Natural Frequency of Linear Hydraulic Actuators 203
It was found that this further factor was due to the compliance of the seals in the pressurized
compartments between valve ports and the piston. The piston rod seals in particular play
a large part in this spring effect.
The assumed spring length Is + la is therefore modified by the addition of a component
l,, which depends on the construction of the system, and which may be called the "con-
structive length le". This gives the following effective length of the trapped oil volumes:
leff = ls + la + lc (15)
Several methods of sealing have been investigated (Fig. 6) since the shaft seals play such
an important part in the constructive length lc.
Description of the test device
The piston, controlled by a symmetrical 4-port-open-centre valve, was moved to give an
oil column length l, and balanced so as to remain stationary. Both chambers were then
under one-half the supply pressure, i.e. pU2.
l">~" "" " 7;'i. {~i!~7I I " 7 7~.... t " ~i{ "{'7"''""'';~(~I.).,~.~s'"~.
1 2
( t
(~
/
3 4
. . . . . . l
5, S
FIG. 6.
l
.~. / i."¸,// 7:" i ~
)

6. 204 H.J. ULRICH
The shut-off valve in the pressure line of control chamber 2, containing the seals under
test, was closed. The full supply pressure ps was developed in chamber 1 by giving a full
stroke to the sleeve of the valve (Fig. 7). Consequently the piston was displaced until the
l=O
dial gage
Ps
Test device:
measuring the dynamic volume
Fio. 7.
6C
isc
4C
3C
2O
AL
7O
/+
0#: 5E 50
+
Temp: 450C
+
/ :/
Geol I
- see, 2
÷ Seal 5 alight& pre/oaded
e Seal 6 heavily prelooded
o Sea/ 7 slight/),preloaded
i I I I ] i J I
,o 20 30 40 50 60 ~o 80 ~[~m]
Oil column
Fio. 8.
oil pressure in the control chamber was equal to the supply pressure. This procedure was
repeated for various values of 1, giving the straight line plots of Fig. 8. Now the effective
length and the piston stroke are related by a coefficient of volume kvol:
/eft = kvoz.ls (16)

7. Some Factors Influencingthe Natural Frequency of Linear Hydraulic Actuators 205
and substituting the above left into equation (15) gives:
la + le
kvo~ = 1 + (17)
ls
Thus kvol is always larger than 1. The larger the actual piston stroke ls, the smaller
will be the influence of la and lc on kvol. When designing normal servo-systems, with rather
small strokes 18,it is therefore important to keep le and la as small as possible.
The bulk modulus fi
The above described test methods of finding le can also be used to determine the bulk
modulus ft. Since, by equation (14), Al is a linear function of l and Ap/fl, constant pressure
drop curves give the value of ft. Values of fl so determined are in agreement with tables
supplied by the oil manufacturers, except in cases where there are air inclusions in the oil,
|tl, , " 11",1 ~ ^ l ~ l
~ "" " ~'" " o o o * oo oo
a) b)
structural complience
FIe. 9.
,4 Ch Cm
Equivalent spring
Fla. 10.
or when very compliant cylinder walls and elastic connecting pipes are used. The spring
constant 2. A~. fl/V of equation (11) takes the compliance of the whole system into account,
which in this case is merely the compliance of the oil volume. This idealized model should
be treated with caution since a structural compliance, not covered by any of the quoted
parameters, is normally present.
The most usual structural compliances result from an elastic connection between actua-
tor and load (Fig. 9a), or from an elastic anchoring of the piston rod (Fig. 9b). In these
cases two springs in series are active (Fig. 10): they are the hydraulic spring having a constant
eh = 2. A~. fl/V, and the mechanical spring having a constant Cm. The constants for the two
springs in series can be combined:
1 1 1
. . . . ÷ (18)
CtotM Ch Cm
which can be rewritten:
(19)

8. 206 H.J. ULRICH
Eliminating the first appearance of cn we get:
)Ctotal - leff I + C~/Cm "
It seems reasonable to include the factor 1/(1 + ch/cm) into the bulk modulus fl, to
give an effective bulk modulus:
1
~e = ~ 1 + c~/cm" (21)
The load mass
In general the mass of the load is a design feature which cannot be adjusted. It is,
however, strange to find that the mass of the oil volume in the pipes is usually ignored,
forgetting the fact that this mass has to be scaled up to the load movement. A part of the
total inertia force is due to the generation of the kinetic energy of the oil mass in the pipes.
Its effective mass is obtained by comparison of its kinetic energy as seen from the output
and that of its real motion variable.
If the pipe cross section in A~ and the fluid velocity is u, the continuity equations give a
relation between the fluid velocity and the piston velocity:
u.A~ = Ar.p.
Hence the fluid velocity, u, is:
Ar .
u = Ap y" (22)
The kinetic energy of the oil mass is:
which gives the effective mass:
u2 p2
Mptpe.~ = Mett.~ (23)
~Ar~ 2 (24)Metr = Mplpe. Ap] •
As an example, a ram diameter of 3 in. and a pipe diameter of 0.5 in. produce an effec-
tive mass seen from the output 1300 times greater than the real mass. If the connecting pipes
have length l~ and contain oil of density p, then the total load mass is (ignoring the mass of
the oil in the piston chamber):
Mt ----M-J- 2"p'I~'Aer
A~ (25)
Optimization of the pipe cross section
According to equation (25), Ap should be large in order to obtain a minimum natural
frequency, but small in order to keep the dead length la of equation (12) small. Since both
influences affect the denominator of the natural frequency equation, the latter can be
maximized by minimizing the product Mr. lem

9. SomeFactorsInfluencingthe Natural Frequencyof LinearHydraulicActuators 207
Differentiating Mt.leff by A~ gives:
d(Mt.lefr) _ M. lp _ 2. p.lp.A~ (Is + lc). (26)
d(a~) Ar A~
By setting equation (26) equal to zero an optimum is obtained when
Ap= Ar,/ (2"p'A~'(MI~ + lc)) (27)
giving an optimum ratio:
A_YAr ----Jmass of the effectiVeloadfUllmassStrokeoil volume (28)
The effective natural frequency
All the above considerations show that to calculate the natural frequency of a hydraulic
actuator accurately 3 factors must be taken into account. The dynamic volume of the
pressurized chambers, equation (16); the effective bulk modulus, equation (21); and the
effective mass of the pipe oil volumes, equation (25).
Combining all these factors
=/2-Ar. e (29)
tOn ~[ Mt./elf"
REFERENCES
[1] H. E. MERRITT,,Hydraulic ControlSystems, Wiley,New York(1967).
[2] M. GUILLON,Etudeet determinationdessystdmeshydrauliques,Dunod,Paris(1961).