Hydraulic Servo and Related
Systems
ME4803 Motion Control
Wayne J. Book
HUSCO/Ramirez Chair in Fluid Power and
Motion Control
G.W. Woodruff School of Mechanical
Engineering
Georgia Institute of Technology
Hydraulics is Especially critical to the
Mobile Equipment Industry
References
1.
2.
3.
4.
5.
6.
Norvelle, F.D. Fluid Power Control Systems, Prentice
Hall, 2000.
Fitch, E.C. and Hong I.T. Hydraulic Component Design
and Selection, BarDyne, Stillwater, OK, 2001.
Cundiff, J.S. Fluid Power Circuits and Controls, CRC
Press, Boca Raton, FL, 2002.
Merritt, H.E. Hydraulic Control Systems, John Wiley and
Sons, New York, 1967.
Fluid Power Design Engineers Handbook, Parker
Hannifin Company (various editions).
Cetinkunt, Sabri, Mechatronics, John Wiley & Sons,
Hoboken, NJ, 2007.
The Strengths of Fluid Power
(Hydraulic, to a lesser extent pneumatic)
• High force at moderate speed
• High power density at point of action
– Fluid removes waste heat
– Prime mover is removed from point of action
– Conditioned power can be routed in flexible a fashion
• Potentially “Stiff” position control
• Controllable either electrically or manually
– Resulting high bandwidth motion control at high forces
• NO SUBSTITUTE FOR MANY HEAVY
APPLICATIONS
Difficulties with Fluid Power
• Possible leakage
• Noise generated by pumps and transmitted by
lines
• Energy loss due to fluid flows
• Expensive in some applications
• Susceptibility of working fluid to contamination
• Lack of understanding of recently graduated
practicing engineers
– Multidisciplinary
– Cost of laboratories
– Displaced in curriculum by more recent technologies
Voltsamp
Electric
or IC
prime
mover
System Overview
Rpm- Pump
torque
Transmis
Flowpress. sion line
& valve
Flow- Motor or
press. cylinder
Rpmtorque
or
force
Coupling
mechanism
• The system consists of a series of
transformation of power variables
• Power is either converted to another useful
form or waste heat
• Impedance is modified (unit force/unit flow)
• Power is controlled
• Function is achieved
Rpmtorque
or
force
Load
Simple open-loop open-center circuit
cylinder
Actuating solenoid
Spring return
Pressure relief valve
4-way, 3 position
valve
filter
Fixed displacement pump
Fluid tank or reservoir
Simple open-loop closed-center circuit
Closed-loop (hydrostatic) system
Motor
Check valve
Variable
displacement
reversible pump
Drain or auxiliary line
Pilot operated valve
Proportional Valve
Various valve schematics
Basic Operation of the Servo Valve
(single stage)
Flow
enters
Torque motor
moves spool
left
Flow
exits
Torque motor
moves spool
right
Positive motor
Negative motor
rotation
rotation
Orifice Model
Q  Cd Ao
2

p
Cd  orifice flow discharge coef.
Ao  orifice flow area  w x
4 Way Proportional Spool Valve Model
• Spool assumptions
q

q
1
2
– No leakage, equal actuator areas
q1  C ps  p1 x, C  a constant
– Sharp edged, steady flow
q2  C p2  p0 x
– Opening area proportional to x
ps  p1  p2  p0  p2
– Symmetrical
Load pressure : p  p1  p2
– Return pressure is zero
p  p
p  p
so : p1  s
; p2  s
– Zero overlap
2
2
• Fluid assumptions
ps  p
q1  C ps  p1 x  C
x
p
p
p
2
s
0
– Incompressible 0
– Mass density 
x
p2, q2
p1, q1
Dynamic Equations (cont.)
Expand in a Taylor series to first order to linearize
 q
q1  q1   1
 x
 q


 (p  p )   (high order term s)
( x  x )   1
x x 
 p x  x 
p  p 
p  p 

Taking partial derivative s :
q1
x
x x
p  p
ps  p
C
;
2
q1
p
x x
p  p
C

x
2 2 ps  p
p0
Choose operating point, commonly
x  0; p  0, at which q1  0
q1
x
x x
p  p
ps
q1
C
 K1 ;
2
p
x x
p  p
ps
p0
x
 K2  0
p2, q2
p1, q1
Dynamic Equations: the Actuator
q1
Change in
volume
Change in
density
y
q1  K1 x  K2p  K1 x  Ay
Net area Ap
If truly incompressible:
•Specification of flow without a response in pressure
brings a causality problem
•For example, if the piston has mass, and flow can
change instantaneously, infinite force is required for
infinite acceleration
•Need to account for change of density and compliance
of walls of cylinder and tubes
Compressibility of Fluids and Elasticity of Walls
p  N 
  m 2 
1 1 d 1 d ( M / V )


  dp 
dp
Bulk modulus :   
For the pure definition, the volume is fixed.
  
q dt
dM  q dt; dp  
 V 
More useful here is an effective bulk modulus that includes expansion of
the walls and compression of entrapped gasses
1
 eff
1 d ( M / V ) 1  1 M M V 

 
 2

dp
 V p V p 
Using this to solve for the change in pressure
  eff
dp  
 V

dM  k M dt  k q dt

Choices for modeling the hydraulic actuator
(simplified actuator)
Change of variables: consider q as volumetric flow rate
m
With no compliance or compressibility we get actuator
velocity v as
q
dy/dt
q
1/A
y
Area A
With compliance and/or compressibility combined into a factor k
And with moving mass m we get the following block diagram. Steady state
(dy/dt)/q is the same. Transient is different. (Show with final value theorem.)
feedback load pressure may reduce q
q
p
dv/dt
dy/dt
k
 dt
 dt
A/m
-
A
Manufacturer’s Data: BD15 Servovalve on HAL
Manufacturer’s Data: BD15 Servovalve on HAL
Two-stage Servo Valve
Torque motor rotates
flapper, obstructs left nozzle
With flapper
centered the flow
and pressure is
balanced
Feedback spring balances
torque motor force
Pressure increases
Spool is driven right
Flow gives
negative rotation
Details of Force Feedback Design
2 Sharp
edged
orifices,
symmetrical
opening
Shown line to line; no
overlap or underlap
Another valve design with direct
feedback
Position Servo Block Diagram
Position
measurement
Proportional
control
Load torque Flow gain / motor
displacement
May be negligible
Net flow /
displacement
Design of some components
(with issues pertinent to this class)
• The conduit (tubing) is
subject to requirements
for
– flow (pressure drop)
• 2 to 4 ft/sec for suction
line bulk fluid velocity
• 7 to 20 ft/sec for pressure
line bulk fluid velocity
– pressure (stress)
• The piston-cylinder is the
most common actuator
– Must withstand pressure
– Must not buckle
• Flow
– Darcy’s formula
– Orifice flow models
• Stress
– Thin-walled tubes
(t<0.1D)
– Thick-walled tubes
(t>0.1D)
• Guidelines
– Fluid speed
– Strengths
– Factors of safety (light
service: 2.5, general:
3.15, heavy: 4-5 or
more)
Darcy’s formula from Bernoulli’s Eq.
L
p  f D
2 Dh
2
Q
 
 A
p  pressure drop along the tube
f D  friction factor (depends on N R )
L  tube length
Dh  hydraulic diameter
or
Dh4
Q
p, N R  2000
128L
  absolute viscosity
(Hagen - Poiseuille law)
 4 x (flow section diameter)/ (section perimeter)
  fluid mass density
Q  flow rate
A  flow section area
NR 
uDh

u  fluid velocity
Friction factor for smooth pipes
(empirical) from e.g. Fitch
Orifice Model
Q  Cd Ao
2

p
Cd  orifice flow discharge coef.
Ao  orifice flow area
•
•
•
•
•
Buckling in the Piston Rod (Fitch)
Rod is constrained by cylinder at two points
Constrained by load at one point
Diameter must resist buckling
Theory of composite “swaged column” applies
Composite column fully extended is A-B-E shown
below consisting of 2 segments
– A-B segment buckles as if loaded by force F on a
column A-B-C
– B-E segment buckles as if loaded by F on DBE
– Require tangency at B
Results of Composite Column Model
Equating the slope of the two
column segments at B where
they join yields:

Ia
F
tan  La
Ib
Ea I a



   tan  Lb F


Eb I b






0.25
 64 I b 
Dr  
  diameter of rod
  
The first equation may be solved iterativel y
Composite column model
then solve for Dr
matches manufacturer’s
recommendations with
factor of safety of 4
Cylinder construction (tie-rod design)
Resulting loading on cylinder walls
Stress Formulas for Cylinders or Conduits
Barlow' s formula (thin, open)
PDi
t
 wall thickness required
2 sd
Di  inside diameter
Clavarino' s formula (closed,
thick, ductile)
Di
t
2
sd  design or allowable stress
 sd  (1  2 ) P 



1
 s  (1  ) P

d


 strength/( factor of safety * stress concentrat ion)
  Poisson' s ratio
Lame' s formula (brittle)
Di
t
2
 sd  P 



1
 s P 
 d

Birnie' s formula (open, thick, ductile)
Di
t
2
 sd  (1   ) P 


 s  (1   ) P  1
 d

Pressure Specifications
•
•
•
•
Nominal pressure = expected operating
Design pressure = Nominal
Proof pressure (for test) = 2x Design
Burst pressure (expect failure) = 4x Design
Pipes versus tubes
Tubes are preferred over pipes since fewer joints mean
•Lower resistance
•Less leakage
•Easier construction
Fittings between tube and other
components require multiple seals
Flared tube design