Modeling and improving
of an Hydraulic Test Bench
for car seats
D.J.W. Belleter - 0639031
CST 2010.049
Bachelor End Project
Project supervisor: dr. Ir. W.J.A.E.M. Post
Eindhoven University of technology
Department of Mechanical Engineering
Control Systems Technology Group
Eindhoven, July, 2010
Table of contents
1. Introduction……………………………………………………………………………………………….1
2. The test bench……………………………………………………………………………………………2
3. The model………………………………………………………………………………………………….3
3.1 The hydraulic power unit………………………………………………………………………………………….3
3.2 The hydraulic control valves……………………….…………………………………………………………… 5
3.2.1 The physical model………………………………………………………………………………………………..5
3.2.2 The simulink controlled model………………………………………………………………………………6
3.2.3 The orifice settings………………………………………………………………………………………………..7
3.3 The hydraulic cylinders and the load…………………………………………………………………………9
3.4 Model expansions………………………………………………………………………………………………….. 11
3.4.1 Advanced stick-slip friction model……………………………………………………………………….11
3.4.2 Pipeline dynamics………………………………………………………………………………………………. 13
4. Model validation and simulations…….………………………………………………………14
4.1 Model validation and single cylinder simulations……………………………………………………14
4.2 Multiple cylinder simulations………………………………………………………………………………….16
5. The solution of the problem..……………………………………………………………………20
6. Conclusions………………………………………………………………………………………………22
References…………………………………………………………………………………………………….23
Symbol list…………………………………………………………………………………………………….24
Appendix A: Data sheets……………………………………………………………………………….25
Appendix B: Orifice calculations…………………………………………………………………...35
Appendix C: Single cylinder model………………………………………………………………..36
Appendix D: Pipeline and Stick-slip simulations…………………………………………….37
Appendix E: Test results on actual actuators provided by B-style………………….41
Appendix F: Simulations with an identical input signal………………………………….42
Appendix G: Accumulator simulations..………………………………………………………..43
1
1. Introduction
Assessment
The firm B-style automotive located in Eindhoven has an RDW (Rijksdienst voor het wegverkeer)
certified test bench to tests car seats and wheel chair fastening points. The tests on car seats and wheel
chair fastening points are designed to simulate the conditions of a car crash. The seats and fastening
points have to meet the criteria set by the RDW before collapsing.
However this hydraulic test bench shows some problems:
- The speed of the force build up in case of a maximum of six actuators is too low.
- Under some circumstances a sagging effect is present in the force build up of several actuators.
This effect slows down the test.
The goal of this research is to find the cause of the problems and to come up with a solution. In order to
do so the following steps have to be taken:
- Investigate the behavior of one or more actuators with a Matlab/Simulink model. The emphasis
should be on the hydraulic and load part of the system.
- Investigate if the problems stated above can be reconstructed using the model.
- Generate one or more solutions for these problems based on the insights gained from the
model. Test the feasibility of these solutions.
Report
The report contains the following process steps. First the system and its problems are briefly described.
After this the steps in the modeling of single components is described and their simulation model is
presented. A few model expansions including an advanced friction model are introduced too. Then the
complete simulation model was validated and some simulations with a single cylinder were done. Some
multiple cylinder simulations were done to check for capacity or interference problems. The simulations
showed that the model was valid and it had the same aberrations as the actual system. After these
simulations a solution was developed and an adapted simulation model was made. Finally the adapted
simulation model was tested to check if the solution is a success.
This solution will be presented and then some conclusions and further recommendations to improve the
system will be given.
2
2. The test bench
In this chapter a description of the system that was modeled will be presented. First the purpose of the
system will be described, followed by the description of the system components and finally the
problems with this system.
The systems purpose
The system is a hydraulic test bench used to test if (modified) car seats comply with legislated safety
demands. In these test the car seats have to be aggravated with defined force levels during defined
time- intervals. The purpose of these tests is to see how the car seats perform in a crash. The seat has to
withstand the defined load in order to comply with the safety demands. The car seats can be loaded on
6 possible locations on the seats. Hydraulic cylinders are applied to introduce these loads. After the test
the damage at each connection point of the seat to the car structure is assessed to ensure safety
legislation.
The system components
The system consists of a hydraulic power unit connected to a valve block consisting of 6 proportional
pressure reducing/relieving valves. A hydraulic cylinder is connected to each of these valves. All of the
valves are connected to a controller. The control electronics for the valves get their input from a
computer. A calibrated force sensor is connected between the cylinder rod and the connection point of
the seat.
The problems of the system
The problem arises when the following test procedure is executed. First the cylinders are pre-stressed.
Then the force in all cylinders is increased to a defined force level and held at that level for a defined
time-interval. After this the cylinders are made pressure-less again.
The problems occur when the force of every cylinder is increased to the prescribed level. Some cylinders
will respond to slow and the force build up will start to late. Another common problem is a sagging
effect during the force build up.
These effects make the system to slow and prevent the system from following the proper pre-described
load cycle. The results of a test with the actual actuators are shown in Appendix E.
3
3. The model
The basic model consists of four parts; the hydraulic power unit, the valves with their electronic
controller, the hydraulic cylinders and a load part. In this chapter the four basic parts will be shown and
the decisions in their modeling will be discussed. After this an advanced friction model and some
pipeline dynamics will be discussed to expand the model. All the model parts are made in the Simscap
toolbox of MatLab/Simulink. The settings in the Simscape blocks are standard unless suggested
otherwise in the data tables.
First the data for a vital system component is shown, namely the hydraulic fluid. The system uses Shell
Donax TM oil, the data for the hydraulic fluid can be found in table 3.1.
Table 3.1 Hydraulic fluid data
Viscosity at 40 °C 40.0 �����
Density at 15 °C 884 ��
���
Bulk modulus 8 · 10� ��
System Temprature 40 °C
3.1 The hydraulic power unit
The hydraulic power unit was manufactured by Bosch and the data sheet can be found in Appendix A.
The hydraulic power unit consists of a reservoir of hydraulic fluid, a pressure relief valve and an electric
motor that drives a fixed displacement pump and a return filter. In table 3.2 the most important data for
the hydraulic power unit can be found.
The pump of the power unit is modeled as a fixed displacement pump driven by a constant angular
velocity source. The pressure safety valve is modeled as a pressure relief valve. The return filter and the
reservoir of hydraulic fluid were left out of the model.
Table 3.2 Hydraulic power unit data
Effective capacity 50 L
Maximum pressure 200 bar
Electric motor speed 1500 rev/min = 50π rad/s
Pump displacement 21.5 L/min = 3.583·10-4
m3/s
Relief valve orifice maximum area 2.41·10-6
m2
Additional information about the power unit can be found in the data sheets in Appendix A. All the data
in table 3.2 is from the data sheets in Appendix A, except for the maximum are of the orifice of the relief
valve. The maximum area of the orifice of the relief valve was determined using the orifice equation:
max maxmax max max
max
2
2DD
p qq C A A
C p
ρρ
∆= → =∆
(3.1)
4
With this equation, the fluid density, maximum flow and maximum pressure for the hydraulic power
unit given in the data sheets of appendix A and the standard value �� � 0.7. The maximum area of the
orifice can be calculated. Rearranging the equation then leads to the maximum area of the orifice of the
relief valve. Results of these calculations can be found in Appendix B.
The data from table 3.2 is incorporated in a simulation model, created with the Simscape toolbox of
MatLab, the model can be seen in figure 3.1.
figure 3.1. the model for the hydraulic power unit
The electric motor is represented by an angular velocity source driving the fixed-displacement pump.
The pressure relief valve is the pressure safety valve and the hydraulic reference represents the
hydraulic reservoir.
5
3.2 The hydraulic control valves The pressure reducing/pressure relief valves were also manufactured by Bosch and the data sheets can
also be found in Appendix A. The valves are proportional pressure control valves; the pressure in the
valve is controlled by force balance on the valve spool. Forces result from the pressures acting on the
left- and right-end faces of the spool and the spring acting on the right side of the spool. A cross section
of the valve can be seen in figure 3.2.
Figure 3.2. cross section of the hydraulic valve
The forces acting on the spool, determine the movement of the spool and thus the flow through the
valve. The flow through the valve is either from pressure port P to user port A (pressure reducing) or
from the user port A to tank port T (pressure relieving). The chamber at the right side of the spool is
connected to user port A by means of a narrow channel through the spool, delaying pressure change of
that chamber. The spring ensures that the spool is in a pre-defined position in case the valve is not
actuated. The valve starts thus in pressure relief mode, i.e. user port A is connected to tank by means of
the orifice in this flow passage. The chamber at the left side of the spool is pressurized by means of the
pilot stage of the valve. The pilot stage is a direct acting pressure relief valve, controlled by means of the
position controlled proportional magnet. Electronic position control is integrated in the valve. The pilot
stage is connected to pressure port P by means of a flow control valve, setting the constant pilot flow.
The spool defines two orifices: an orifice in the flow path from pressure port P to user port A; and an
orifice in the flow path from user port A to tank port T. The first orifice represents the pressure reducing
part, the second the pressure relief part of the valve. Initially the valve is in pressure reducing mode.
When the orifice from port P to port A opens the pressure at port A will increase. In case the orifice from
port A to port T opens the pressure at port A will decrease.
For the modeling of this valve two options are considered. One is making a true and complete physical
model of the valve using force balance for the valve control. The other option is to model the control
signal for the valve opening in MatLab/Simulink signal builder. Both models will be analyzed here.
3.2.1 The physical model
In order to create a physical model of the system, the components mentioned above have to be
modeled. So the model will consist of two orifices, two cylinders to represent the chambers of the spool,
the spring on the right side of the spool, the mass of the spool and the flow control valve on the left side
6
of the spool and the pilot stage. The valve control electronics are left out of the model, because there is
not enough data available. The Simscape model for this configuration can be seen in figure 3.3.
figure 3.3. physical model of the hydraulic valve
The hydraulic damping in the model, represented by the segmented pipeline, represents the narrow
channel through the spool and introduces a small delay between the pressure in port A and the pressure
of the right chamber, represented by the cylinder. The hydraulic damping has also a numerical
advantage since it smoothens the signal which accelerates calculations.
The advantage of creating a physical model is that this represents the actual system best. Because it
allows to truly model the force balance on the spool. The conditions in the actual system determine
whether or not an orifice opens or closes. The major drawback of the physical model is that a lot of data
from the valve is needed that is not available. The only data available are the data sheets found in
Appendix A. The data that was not made available thus needs to be estimated; which makes the model
more inaccurate. Another drawback to this type of modeling is that the electronic part of the valve used
for control hardly can be modeled, because there is no data available for the electronics. This means
that the only way to control this valve is by pressure and since the pump delivers the pressure the only
way to close the orifice from port P to port A and open the orifice from port A to port T is to turn the
pump off. This may result in an inaccurate and unrealistic control of the valve.
3.2.2 The Simulink controlled model
In this model the two orifices are controlled using a Simulink signal made in the signal builder. A positive
signal will open the orifice form port P to port A while the orifice from port A to port T is closed and a
negative signal will close the orifice form port P to port A while opening the orifice form port A to port T.
7
This makes the model much simpler; all what need to be modeled are the two orifices, a Simulink
control signal and the pressure relieving action. The model for this configuration is shown in figure 3.4.
Figure 3.4. the Simulink controlled model
The model of figure 3.4 is much simpler than that of figure 3.3. There are far less parameters necessary;
the only data that is needed is the data for the orifices. Another advantage over the physical model is
that the valve can be easily opened and closed using the signal from the signal builder. A possible
disadvantage of the Simulink controlled valve is generating an unrealistic signal. Another possible
disadvantage of the Simulink controlled valve is that there is no feedback control loop in the system. It
just follows the pattern set in the signal builder.
For this specific case the Simulink controlled valve model is the best option, because of the limited
available data for the physical valve model. In a situation where all data, including the control
electronics, is available the physical model would probably be a better option, since this model can
handle feedback control and is more realistic.
3.2.3 The orifice settings
The most important settings for the valve model are the orifice settings. The data that is necessary to
model the orifices are the initial opening, the maximum opening and the maximal passage area through
the orifice. The initial opening and the maximum opening have to be estimated from the figures in the
data sheets of Appendix A. The maximum passage area can be calculated using eqn. (3.1).
Using figure 3.2 and the technical drawing with measurements from the data sheet it was estimated
that the stroke of the spool is about five millimeters. The orifice from port P to port A seems to be a
closed center valve judging from figure 3.2 with an initial opening of -0.65 millimeters. The orifice from
port A to port T appears to be more like a critically centered or open center valve.
The orifice equation, equation (3.1), can be used to calculate the maximum passage area. To do this a
pressure difference over the orifice is needed. The pressure difference over the orifice can be obtained
by using the pressure-flow relations from the data sheets shown in figure 3.5.
8
figure 3.5. pressure-flow lines
The 175 bar valve variant is present in the system. This means that the maximal pressure in port A is 175
bar and the maximum pressure in port P is the pressure of port A plus another 5 bar. The five bar
pressure difference is a result from the pressure loss to the flow control valve. To calculate the
maximum passage area the maximum flow and the minimal pressure difference have to be used. The
minimal pressure in the orifice from port A to port T at the maximum flow of 40 minL is 13% of 175 bar,
making 22.75 bar the minimal pressure difference. The minimal pressure difference between port P and
port A is 5 bar according to the data sheet. Using equation (3.1) the maximum orifice area for the orifice
from port A to port T is -5 21.328 10 m⋅ and the maximum orifice area for the orifice from port P to port
A is5 22.833 10 m−⋅ . Calculations for the orifice areas can be found in Appendix B.
9
3.3 The hydraulic cylinders and the load In this part of the model the hydraulic cylinders and the load are modeled. This part of the model
contains the hose form the valve to the cylinder, an elastic load and a force sensor. The plastic part of
the load is not modeled, because there is not enough information about the material behavior.
There are two types of hydraulic cylinders applied in the system; one has a larger piston area and the
other has larger stroke. Both types of hydraulic cylinders are modeled as single acting hydraulic cylinders
that are used to pull the load. The parameters for both the types of cylinders can be found in table 3.3.
Table 3.3 Hydraulic cylinder data
Piston area DW-60-35-1000 2.83 ��
Piston stroke DW-60-35-1000 1 �
Cylinder dead volume DW-60-35-1000 3.67 10-5
��
Piston area DW-70-40-400 3.85 ��
Piston stroke DW-70-40-400 0.7 �
Cylinder dead volume DW-70-40-400 4.22 10-5
��
The piston area was determined by calculating the circular area of the piston using its diameter, which is
the first number in the cylinders identification number. The piston stroke can be found too in the
cylinder identification number as the final number. Because data about the dead volume of the cylinders
was not made available by the manufacturer, this data had to be estimated using cross sections of
different makes of cylinders, see [2]. The data sheet for these cylinders can be found at the end of
Appendix A.
These cylinders are connected to a load which is modeled as a spring. The stiffness of this spring was
determined by the test data made available by B-style in which they specify a force and an estimated
displacement. The data for each load can be found in table 3.4.
Table 3.4 Elastic load data
Load name Force Deformation Stiffness
shoulder seatbelt 13500 N 300 mm 45000 N/m
hip seatbelt 17032 N 100 mm 170320 N/m
chair leg 5298 N 50 mm 105960 N/m
There is a pipeline in this model section as well. This is the hose between a valve and the hydraulic
cylinder. It is modeled as a segmented pipeline to account for friction losses, fluid inertia and fluid
compressibility. The pipes are 3 m long and have an internal diameter of a ½”. The pipe wall type is set
to flexible, because the hoses to the cylinder are made of rubber and will expand when internal pressure
is applied.
10
The pipeline, cylinders and load combined result in a complete model for this part of the system. The
Simscape model is shown in figure 3.6.
figure 3.6. model of the cylinders and load
3.4 Model expansion
To expand the overall system model an advanced friction model and some theory of pipeline dynamics
will be presented here.
3.4.1 Advanced Stick-slip friction
In this subsection the phenomenon of stick
theory of [1]. Furthermore a simulation
with the Simscape toolbox of MatLab.
The theory behind stick-slip friction:
Stick-slip friction is a typical behavior for a system with friction and a c
caused by the fact that friction at rest is larger than
given in figure 3.7. When this system is actuated a typical
The mass in figure 3.7 is initially at res
counteracts the spring force, and there wil
the break-away force the mass starts to slide and the friction decreases rapidly due to the Stribeck
effect. The spring will relax, and the spring force decreas
increases because of the Stribeck effect and the motion stops.
This phenomenon will repeat and causes the typical stick
figure shows both a position-time diagram and a diagram with f
To expand the overall system model an advanced friction model and some theory of pipeline dynamics
slip friction model
section the phenomenon of stick-slip friction will be described and exp
. Furthermore a simulation model of stick-slip friction will be presented. This mod
MatLab.
slip friction:
slip friction is a typical behavior for a system with friction and a certain amount of elasticity. It i
caused by the fact that friction at rest is larger than in motion. A typical model for stick
system is actuated a typical stick-slip motion will occur
is initially at rest and the spring force builds up. The static friction force
ts the spring force, and there will be a small displacement only. When the spring force reaches
away force the mass starts to slide and the friction decreases rapidly due to the Stribeck
, and the spring force decreases. The mass slows down and the friction force
increases because of the Stribeck effect and the motion stops.
figure 3.7. model of the stick-slip system
This phenomenon will repeat and causes the typical stick-slip motion that is shown
time diagram and a diagram with force and velocity plotted versus
11
To expand the overall system model an advanced friction model and some theory of pipeline dynamics
slip friction will be described and explained based on the
slip friction will be presented. This model is made
ertain amount of elasticity. It is
in motion. A typical model for stick-slip friction is
occur.
. The static friction force
. When the spring force reaches
away force the mass starts to slide and the friction decreases rapidly due to the Stribeck
es. The mass slows down and the friction force
is shown in figure 3.8. This
orce and velocity plotted versus time.
figure 3.8
Different friction forces contribute to the total friction force the main types of friction for
phenomenon are static friction, Coulomb friction, Stribeck friction and
Simulation model of the stick-slip friction:
The simulation model was created in
the same basic components as the theoretical model of
translational spring and friction. The translational friction block requires input for the break
friction, coulomb friction and the viscous friction. In
mass (Mass in figure 3.9) which is set to the minimal
problems. The system also contains
figure 3
The model is integrated as a subsystem
cylinder while the output is connected to
shown in figure 3.8. A validation of the model with arbitrary numbers for the friction
3.10.
figure 3.8. simulation of stick-slip motion
Different friction forces contribute to the total friction force the main types of friction for
tatic friction, Coulomb friction, Stribeck friction and viscous friction.
slip friction:
model was created in the MatLab Simscape toolbox, see figure 3.9
ents as the theoretical model of figure 3.7, a mass (Mass1 in
translational spring and friction. The translational friction block requires input for the break
friction, coulomb friction and the viscous friction. In addition to Mass1 this model also contains a second
) which is set to the minimal non-zero value of MatLab, to suppress numerical
. The system also contains two sensors; one for velocity and position and one sensor for force
figure 3.9. The simulation model for stick-slip friction
The model is integrated as a subsystem into the system model and gets input from the hydraulic
connected to the load. The model results in the theoretical behavior
validation of the model with arbitrary numbers for the friction
12
Different friction forces contribute to the total friction force the main types of friction for the stick-slip
friction.
.9. The model consists of
, a mass (Mass1 in figure 3.9), a
translational spring and friction. The translational friction block requires input for the break-away
this model also contains a second
MatLab, to suppress numerical
one for velocity and position and one sensor for force.
input from the hydraulic
the load. The model results in the theoretical behavior as
validation of the model with arbitrary numbers for the friction is shown in figure
13
figure 3.10. force-time diagram created with the stick-slip model
This figure clearly shows the sticking part until about 8 seconds and the slipping part from about 8 to 10
seconds into the test.
This results in a modular stick-slip model that can be easily build into a larger model.
3.4.1 Pipeline dynamics
Pipelines are an important part of the system, because their properties can seriously influence the
systems behavior. Pipelines introduce a hydraulic resistance in the system and can have an accumulating
effect. The magnitude of the resistance that results from the pipelines depends on the length and
diameter of the pipes and for the hoses in this system on the amount of expansion with pressure
variation.
Longer pipes and pipes with a smaller diameter have a larger hydraulic resistance, because more friction
is introduced. However in combination with accumulating effects making the pipes longer of smaller in
diameter does not necessarily result in a negative effect on the reached force.
When pressure is applied to the hose it expands. When the hose expands hydraulic fluid is accumulated,
this takes energy away from the system. This does not necessarily have to be a bad thing, because when
the system is used a second time the accumulated hydraulic fluid generates a larger force.
0 10 20 30 400
0.5
1
1.5
Time [s]
For
ce [
N]
14
4. Model validation and simulations
The model described in chapter 3 is used to perform simulations and hopefully finding the cause of the
systems problems. First some simulations were done with a single cylinder for model validation and to
check if some of the problems can already be identified. After this, simulations were done with multiple
cylinders to find out if there are capacity problems or if there is interference between the cylinders.
4.1 Model validation and Single cylinder simulations
The first simulations were done with a very basic model; without the hoses from the valve block to the
hydraulic power unit and without an advanced friction model. The model that was used for these
simulations can be seen in Appendix C.
The input signal for the valve is shown in figure 4.1 and the resulting force verses time; and piston
displacement versus time are shown in figure 4.2 and figure 4.3 respectively.
Figure 4.1 the valve input signal
figure 4.2 time-force graph
0 5 10 15 20 25 30-5
0
5x 10
-3
Time [s]
spoo
l pos
ition
[m
]
0 5 10 15 20 25 30-0.5
0
0.5
1
1.5
2
2.5x 10
4
Time [s]
For
ce [
N]
15
figure 4.3 time-displacement graph
As can be seen from figure 4.1 to 4.3 the model appears to be valid. When the valve opens the piston
moves and force builds up. When the input signal of the valve is negative the piston moves back and the
force reduces. When the input signal of the valve is set to zero the force is maintained.
Problems cannot be identified from these simulations. There is no sagging effect in the force and the
force increases linearly and without fluctuations. So in order to find the systems problems multiple
cylinders, an advanced friction model and pipeline dynamics should be considered.
Next some simulations were done to investigate the effects of stick-slip friction and pipeline dynamics.
In advance some simulations were done to find out suitable settings for the stick-slip friction model and
pipeline model. The models for and data from these simulations can be found in Appendix D. The
pipeline model has an additional pipeline between the power unit and the valve.
The results from stick-slip simulation 4 resemble the pattern seen in the tests performed on the actual
actuators the best. The test on the actual actuators can be found in Appendix E. So the stick-slip friction
in the model was set to the values in table 4.1. These settings result in the characteristic that is shown in
figure 4.4.
Table 4.1 Stick-slip friction settings
Elasticity 1400 N/m
Static friction 2000 N
Coulomb friction 600 N
Viscous friction 100 N
0 5 10 15 20 25 30-0.1
0
0.1
0.2
0.3
0.4
0.5
0.6
Time [s]
Pos
ition
[m
]
16
figure 4.4 The model including stick-slip
From the pipeline simulations it can be concluded that it is better to have shorter and stiffer pipelines in
the system. These characteristics seem to have the best effect on the first force peak in the simulation
and this is the most important peak for this system.
The additional signal cylinder simulations with the advanced stick-slip friction model and pipeline
dynamics do not reveal any problems with the system. There is no sagging behavior in the force
characteristic. So the problem probably involves interference between cylinders or capacity problems.
This was investigated in the multiple cylinder simulations.
4.2 Multiple cylinder simulations
To find out if there are capacity problems for the power unit or if there is interference between the
cylinders some multiple cylinder simulations have to be done. For these simulations a model with 3
cylinders and valves was used. These three cylinders represent the cylinders pulling the shoulder
seatbelt, the hip seatbelt and the chair leg. These cylinders are chosen because the simulations are
compared to the data gathered by B-style from the actual set-up. This data can be found in Appendix E.
The simulation with the standard model, with multiple cylinders does not show any capacity problems
or interference between cylinders when each valve has an identical input signal. The results of this
simulation can be seen in Appendix F.
In the next simulation the objective is trying to replicate the first test of the tests done by B-style, which
can be found in Appendix E. To do this the valve input signal for the chair leg valve is adapted, this valve
now opens one second later than then the other two valves. The input signals are shown in figure 4.5.
The input signals are chosen to be ideal; the opening and closing of the valves is controlled with a step
signal. In reality this would be impossible because the system cannot react that fast.
0 5 10 15 20 25 30-5000
0
5000
10000
15000
20000
Time [s]
For
ce [
N]
17
figure 4.5 valve input signals above: shoulder and hip seatbelt, below: chair leg
This simulation seems to reveal the problem with the system. As can be seen from figure 4.6, which is a
graph of the force on the load versus time. A sagging effect can clearly be seen in the graph for the
shoulder and hip seatbelt at the moment the valve for the chair leg opens. This means that there is
clearly interference between the cylinders.
figure 4.6 force-time diagram
0 1 2 3 4 5 6-5
0
5x 10
-3
Time [s]
Pos
ition
[m
]
shoulder and hip seatbelt
0 1 2 3 4 5 6-5
0
5x 10
-3
Time [s]
Pos
ition
[m
]
chair leg
0 2 4 6 8 10-2000
0
2000
4000
6000
8000
10000
time [s]
forc
e [N
]
shoulder seatbelt
hip seatbeltchair leg
18
To identify the cause of the problem it is necessary to look at the flow and pressure characteristics for
the flow to all the valves and the flow to the pump. The characteristics for flow and pressure are shown
in figure 4.7 and 4.8 respectively.
figure 4.7 the flow to the valves and from the pump
figure 4.8 the pressure in the system
From figure 4.7 it becomes clear that, when the valve for the chair leg opens an outflow of hydraulic
fluid from the shoulder and hip seatbelt valves arises. From figure 4.8 it can be seen that this is a flow
problem because the pressure in the system is virtually unaffected by the opening of the chair leg valve.
0 2 4 6 8 10-4
-2
0
2
4
6
8x 10
-4
time [s]
flow
[m
3 /s]
shoulder seatbelt
hip seatbeltchair leg
pump flow
0 2 4 6 8 100
0.5
1
1.5
2x 10
7
time [s]
pres
sure
[P
a]
shoulder seatbelt
hip seatbeltchair leg
pump pressure
19
As can be seen from the figures 4.6, 4.7 and 4.8 the first second of the simulation all appears to be as
expected. The force build up in the cylinders develops linearly as expected, but after one second when
the chair leg valve opens the force delivered by the shoulder and hip seatbelt starts to sag. This is caused
by a large outflow of hydraulic fluid from these cylinders. That is caused by the pressure difference
between the cylinders. The chair leg cylinder is still pressure less at the moment its valve opens while
the other two cylinders are already under pressure.
The three cylinders now act as communicating barrels. The pressure in the chair leg cylinder is raised
with pressure from the other two cylinders. To do this flow is directed back from the other two cylinders
into the chair leg cylinder. This is possible because the orifice from port P to port A in the shoulder and
hip seatbelt valves is open at the moment the orifice from port P to port A opens in the chair leg valve.
The outflow of hydraulic fluid in the shoulder and hip seatbelt cylinders causes a pressure drop in these
cylinders. This causes the force delivered by these cylinders to drop as well.
20
5. The solution of the problem
In this chapter a solution will be presented for the problem of the hydraulic test bench. This solution has
been tested with a simulation model.
The problem is that there is not enough pressurized fluid in the system to satisfy the sudden demand of
the chair leg cylinder. So to solve the systems problems a pressurized buffer of hydraulic fluids needs to
be introduced into the system. This can be done by adding an accumulator to the system. A gas charged
accumulator seems to be suited best for the job because this way the pre load pressure for the
accumulator can be adapted the easiest.
To size an accumulator three parameters need to be considered: the capacity, the pre load pressure and
the initial volume. Some simulations have been done with different settings for these parameters to
investigate their effect on the system. Graphic representations of these test results can be found in
Appendix G.
The capacity influences the speed of the system, more specifically the response of the force build up. A
relatively lower capacity speeds the force build up more than a relatively larger accumulator capacity.
However the accumulator capacity should be large enough to provide the necessary flow to the system.
So for this system a relatively small accumulator should be chosen that still can meet the demanded
flow.
The pre load pressure is the pressure at which the system starts to interact with the accumulator. If this
pressure is higher the system is faster, because it can deliver fluid faster because of the higher pressure.
However the pressure should not be raised too high, because there needs to be enough fluid in the
accumulator when the chair leg valve opens otherwise the accumulator cannot satisfy the demand long
enough and the sagging effect will still take place.
The initial volume is important for this system as well. There need to be a volume of hydraulic fluid
present when the simulation starts in order to raise the force delivered by the shoulder and hip seatbelt
cylinder quickly. Enlarging the initial volume makes the system faster, however this should be traded off
by the time to fill the accumulator. If the accumulator can be filled while the system is in steady state
before a test this will not add up to extra time for a test.
An impression of the systems behavior with an accumulator can be seen in figures 5.1 to 5.3. The system
reacts faster and there is no sagging effect. The accumulator used in the simulation of figure 5.1 to 5.3
has a capacity of 8 L, a pre load pressure of 35 bar and an initial capacity of 1 L. So a small accumulator
whit a modest pressure can be enough to fix the systems problems.
21
figure 5.1 force-time diagram with accumulator
figure 5.2 flow-time diagram with accumulator
figure 5.3 pressure-time diagram with accumulator
0 1 2 3 4 5 6-2000
0
2000
4000
6000
8000
10000
12000
14000
16000
Time [s]
For
ce [
N]
shoulder seatbelt
hip seatbelt
chairleg
0 1 2 3 4 5 6-3
-2
-1
0
1
2x 10
-3
Time [s]
Flo
w [
m3 /s
]
shoulder seatbelt
hip seatbelt
chairlegaccumulator
pump
0 1 2 3 4 5 63.5
4
4.5
5
5.5x 10
6
Time [s]
Pre
ssur
e [P
a]
22
6. Conclusions
The system was successfully modeled in Matlab/Simulink. It was used to do simulations of tests with the
hydraulic test bench. These simulations showed that the main problem which is a sagging effect in the
force of some cylinders is not caused by friction, pipeline dynamics or capacity problems. It was
demonstrated that the problem was interference between cylinders. The interference occurs when one
of the valves opens later than the others. This causes an outflow of hydraulic fluid from the pressurized
cylinders, resulting in a drop in the force that the cylinder delivers.
To solve this problem the system needs a pressurized buffer of hydraulic fluid, to have enough
pressurized flow when a valve opens later. An accumulator was integrated into the model and it was
shown that this solved the problem.
Simulations showed that a relatively small accumulator with a modest pressure can satisfy the systems
demands. Furthermore if the accumulator is filled before a test while the system is in steady state, filling
of the accumulator will not extend the time for a test.
23
References [1] C. Canudas de Wit, H. Olsson, K. J. Astrom, and P. Lischinksy, “A new model for control of
systems with friction”, IEEE Trans Automat. Contr., vol. 40, pp. 429-425, Mar. 1995
[2] www: FlexHydro, http://www.flexhydro.com/dubbelwerkendstandaardcilinders/
[3] dr. Ir. W.J.A.E.M. Post, slides lecture 4N630, Fluid Power Transmissions and Servo Systems
24
Symbol list Symbol Quantity Unit Unit abbreviation
A Area Square meter 2m q Flow Cubic meter per second 3m
s
DC Flow discharge
coefficient
- -
ρ Density Kilogram per cubic
meter 3kg
m
p∆ Pressure difference Pascal Pa
25
Appendix A: data sheets
26
27
28
29
30
31
32
33
34
35
Appendix B: Orifice calculations In this appendix the orifice area for the pressure relief valve of the hydraulic power unit and valve are
calculated. The areas for the two orifices that make up the valve are calculated too. All parameter values
are from the data sheets of Appendix A.
Relief valve for the hydraulic power unit
The calculations are done with the orifice equation:
max maxmax max max
max
2
2DD
p qq C A A
C p
ρρ
∆= → =∆
46 2max
max 5max
3.583 10 8842.41 10
2 0.7 2 200 10D
qA m
C p
ρ −−⋅= = = ⋅
∆ ⋅ ⋅
Relief valve for the valve 4
6 2maxmax 5
max
6.67 10 8844.789 10
2 0.7 2 175 10D
qA m
C p
ρ −−⋅= = = ⋅
∆ ⋅ ⋅
The valve orifices
The orifice from pump to cylinder:
45 2max
max 5max
6.67 10 8842.833 10
2 0.7 2 5 10D
qA m
C p
ρ −−⋅= = = ⋅
∆ ⋅ ⋅
The orifice from cylinder to tank:
45 2max
max 5max
6.67 10 8841.328 10
2 0.7 2 0.13 175 10D
qA m
C p
ρ −−⋅= = = ⋅
∆ ⋅⋅ ⋅ ⋅
36
Appendix C: Single cylinder model In this appendix the simulation model used in the single cylinder simulations is presented.
C A
R
sch
ou
der
go
rde
l
S AB
Va
riab
le O
rific
e P
-A
S AB
Var
iab
le O
rific
e A
-T
RC
f(x)
=0
So
lve
rC
on
figu
ratio
n
PS
S
Sim
uli
nk-
PS
Co
nve
rte
r1
PS
S
Sim
ulin
k-P
SC
on
vert
er
Sig
nal 2
Sig
na
l B
uil
der
A B
Pre
ssu
re R
elie
fV
alv
e1
A B
Pre
ssu
re R
elie
fV
alve
Po
siti
on
6
Po
siti
on
5
Po
siti
on
4
Po
siti
on
3
Pos
itio
n2
Po
siti
on
1
PS
S
PS
S
PS
S
PS
S
PS
S
PS
S
Me
cha
nica
lR
ota
tio
nal
Re
fere
nce
RC F
Ide
al F
orc
e S
en
sor
S
CR
Ide
al
An
gu
lar
Vel
oci
ty S
ou
rce
Hyd
rau
lic
Re
fere
nce
QA
BH
ydra
uli
c F
low
Ra
teS
en
sor3
QA
B
Hyd
rau
lic F
low
Ra
teS
en
sor2
QA
B
Hyd
raul
ic F
low
Ra
teS
en
sor1
QA
B
Hyd
rau
lic
Flo
w R
ate
Se
nso
r
S
P T
Fix
ed-
Dis
pla
cem
en
tP
um
p
Cu
sto
m H
ydra
ulic
Flu
id
145
0/6
0*2
*pi
Co
nst
an
t
AB
1/2
" p
ipe
3
37
Appendix D: Pipeline and stick-slip simulations This appendix shows the pipeline and stick-slip model for the single cylinder simulations and the
simulation data. First the pipeline model and simulations will be presented followed by the stick-slip
simulations.
C A
R
sch
ou
de
rgo
rde
l
S AB
Va
ria
ble
Ori
fice
P-A
S AB
Va
ria
ble
Ori
fice
A-T
RC
f(x)
=0
So
lve
rC
on
fig
ura
tio
n
PS
S
Sim
uli
nk-
PS
Co
nve
rte
r1
PS
S
Sim
uli
nk-
PS
Co
nve
rte
r
Sig
nal 2
Sig
na
l B
uil
de
r
A B
Pre
ssu
re R
eli
ef
Va
lve
1
A B
Pre
ssu
re R
eli
ef
Va
lve
Po
siti
on
6
Po
siti
on
5
Po
siti
on
4
Po
siti
on
3
Po
siti
on
2
Po
siti
on
1
PS
S
PS
S
PS
S
PS
S
PS
S
PS
S
Me
cha
nic
al
Ro
tati
on
al
Re
fere
nce
RC F
Ide
al
Fo
rce
Se
nso
r
S
CR
Ide
al
An
gu
lar
Ve
loci
ty S
ou
rce
Hyd
rau
lic
Re
fere
nce
QA
B
Hyd
rau
lic
Flo
w R
ate
Se
nso
r3
QA
B
Hyd
rau
lic
Flo
w R
ate
Se
nso
r2
QA
B
Hyd
rau
lic
Flo
w R
ate
Se
nso
r1
QA
B
Hyd
rau
lic
Flo
w R
ate
Se
nso
r
S
P T
Fix
ed
-Dis
pla
cem
en
tP
um
p
Cu
sto
m H
ydra
uli
cF
luid
14
50
/60
*2*p
i
Co
nst
an
t
AB
1/2
" p
ipe
3
AB
1/2
" p
ipe
2
38
The pipeline properties that are varied during the simulations are the length, the diameter, the Static
pressure-diameter coefficient and the number of pipeline segments.
0 5 10 15 20 25 30-0.5
0
0.5
1
1.5
2
2.5x 10
4
Time [s]
For
ce [N]
Test 1
0 5 10 15 20 25 30-0.5
0
0.5
1
1.5
2
2.5x 10
4
Time [s]
For
ce [N]
Test 2
0 5 10 15 20 25 30-0.5
0
0.5
1
1.5
2
2.5x 10
4
Time [s]
For
ce [N]
Test 2
0 5 10 15 20 25 30-0.5
0
0.5
1
1.5
2
2.5x 10
4
Time [s]
For
ce [N]
Test 4
Table A.D.1 pipeline dynamics simulation
Simulation 1 Simulation 2 Simulation 3 Simulation 4
Length [m] 3 3 6 6
Diameter [m] 1.27e-2 1.27e-2 1.27e-2 1.27e-2
Static pressure-
diameter
coefficient
[Pa/m]
2e-10 2e-10 2e-10 6e-10
Nr. of segments 20 50 20 20
Simulation 5 Simulation 6 Simulation 7 Simulation 8 Simulation 9
Length [m] 3 3 6 3 3
Diameter [m] 1.27e-2 1.27e-2 1.27e-2 2.54e-2 6.35e-3
Static pressure-
diameter
coefficient
[Pa/m]
6e-10 7e-11 7e-11 2e-10 2e-10
Nr. of segments 20 50 20 20
39
From these simulations it can be concluded that if the accumulative properties of the hoses are
increased (by making the hoses longer, bigger in diameter and weaker) the first force peak will be
lowered and the second increased. When the hoses are pressurized for the first time, they will expand
and this consumes energy. When the hoses are pressurized for the second time the accumulated
hydraulic fluid will help to reach a higher force.
For this case the first force peak is more important so hoses should be chosen short and stiff, rather
than longer and weaker.
0 5 10 15 20 25 30-0.5
0
0.5
1
1.5
2
2.5x 10
4
Time [s]
For
ce [
N]
Test 5
0 5 10 15 20 25 30-0.5
0
0.5
1
1.5
2
2.5x 10
4
Time [s]
For
ce [
N]
Test 6
0 5 10 15 20 25 30-0.5
0
0.5
1
1.5
2
2.5x 10
4
Time [s]
For
ce [
N]
Test 7
0 5 10 15 20 25 30-0.5
0
0.5
1
1.5
2
2.5x 10
4
Time [s]
For
ce [
N]
Test 8
0 5 10 15 20 25 300
0.5
1
1.5
2
2.5x 10
4
Time [s]
For
ce [
N]
Test 9
40
The stick-sip properties that are varied during the simulations are the spring rate (see figure 3.9), the
static friction force, the coulomb friction force and the viscous friction force.
Table A.D.2 Stick-slip friction settings
Simulation 1 Simulation
2
Simulation
3
Simulation
4
Simulation
5
Simulation
6
Elasticity 1400 N/m 1400 N/m 2500 N/m 1400 N/m 1400 N/m 1400 N/m
Static friction 700 N 2000 N 2000 N 2000 N 2000 N 2000 N
Coulomb friction 600 N 1800 N 1800 N 600 N 600 N 1800 N
Viscous friction 100 N 800 N 800 N 100 N 800 N 100 N
0 5 10 15 20 25 30-5000
0
5000
10000
15000
20000
Time [s]
For
ce [
N]
Test 1
0 5 10 15 20 25 30-5000
0
5000
10000
15000
20000
Time [s]
For
ce [
N]
Test 2
0 5 10 15 20 25 30-5000
0
5000
10000
15000
20000
Time [s]
For
ce [
N]
Test 3
0 5 10 15 20 25 30-5000
0
5000
10000
15000
20000
Time [s]
For
ce [
N]
Test 4
0 5 10 15 20 25 30-5000
0
5000
10000
15000
20000
Time [s]
For
ce [
N]
Test 5
0 5 10 15 20 25 30-5000
0
5000
10000
15000
20000
Time [s]
For
ce [
N]
Test 6
41
Appendix E: Test results on actual actuators
provided by B-style
42
Appendix F: Simulations with an identical input
signal In this appendix the data will be shown from a simulation where each valve has an identical input signal.
This input signal is shown in figure A.F.1. The resulting simulation data for the force, flow and pressure is
shown in figure A.F.2.
figure A.F.1 valves steering signal
figure A.F.2 test results
From the simulation data it becomes clear that when all the valves have identical input signals there is
no problem. The force builds up linearly without sagging, as can be expected from the system.
0 1 2 3 4 5 6-5
0
5x 10
-3
Time [s]
Pos
ition
[m
]
shoulder and hip seatbelt
0 2 4 6-5000
0
5000
10000
Time [s]
For
ce [
N]
0 2 4 6-2
0
2
4x 10
-4
Time [s]
Flo
w [
m3 /s
]
0 2 4 6-1
0
1
2x 10
7
Time [s]
Pre
ssur
e [P
a]
shoulder seatbelt
hip seatbeltchair leg
pump
43
Appendix G: Accumulator simulations In this appendix the data gathered in the accumulator simulations is presented. The flow and pressure in
the graphs are measured at inlet port of each valve. Important is to watch the flow. If the flow gets
negative a drop in the force will be present.
capacity [L] pre load pressure [bar] initial volume [L]
Simulation 1 8 20 1
Simulation 2 8 20 0.5
Simulation 3 8 25 0.5
Simulation 4 20 20 1
Simulation 5 8 35 1
Simulation 6 8 50 1
Simulation 1
0 1 2 3 4 5 6-2000
0
2000
4000
6000
8000
10000
Time [s]
For
ce [
N]
0 1 2 3 4 5 6-2
-1
0
1
2x 10
-3
Time [s]
Flo
w [
m3 /s
]
0 1 2 3 4 5 62.2
2.4
2.6
2.8
3
3.2
3.4x 10
6
Time [s]
Pre
ssur
e [P
a]
shoulder seatbelt
hip seatbelt
chair legaccumulator
pump
44
Simulation 2
Simulation 3
0 1 2 3 4 5 6-2000
0
2000
4000
6000
8000
10000
Time [s]
For
ce [
N]
0 1 2 3 4 5 6-2
-1
0
1
2x 10
-3
Time [s]
Flo
w [
m3 /s
]
0 1 2 3 4 5 62
2.2
2.4
2.6
2.8
3x 10
6
Time [s]
Pre
ssur
e [P
a]
shoulder seatbelt
hip seatbelt
chair legaccumulator
pump
0 1 2 3 4 5 6-5000
0
5000
10000
15000
Time [s]
For
ce [
N]
0 1 2 3 4 5 6-2
-1
0
1
2x 10
-3
Time [s]
Flo
w [
m3 /s
]
0 1 2 3 4 5 62.4
2.6
2.8
3
3.2
3.4x 10
6
Time [s]
Pre
ssur
e [P
a]
shoulder seatbelt
hip seatbelt
chair legaccumulator
pump
45
Simulation 4
Simulation 5
0 1 2 3 4 5 6-2000
0
2000
4000
6000
8000
10000
Time [s]
For
ce [
N]
0 1 2 3 4 5 6-2
-1
0
1
2x 10
-3
Time [s]
Flo
w [
m3 /s
]
0 1 2 3 4 5 62.1
2.2
2.3
2.4
2.5x 10
6
Time [s]
Pre
ssur
e [P
a]
shoulder seatbelt
hip seatbelt
chair legaccumulator
pump
0 1 2 3 4 5 6-5000
0
5000
10000
15000
Time [s]
For
ce [
N]
0 1 2 3 4 5 6-3
-2
-1
0
1
2x 10
-3
Time [s]
Flo
w [
m3 /s
]
0 1 2 3 4 5 63.5
4
4.5
5
5.5x 10
6
Time [s]
Pre
ssur
e [P
a]
shoulder seatbelt
hip seatbelt
chair legaccumulator
pump
46
Simulation 6
Comparing Simulation 1 and Simulation 4 it can be seen that the system when enlarging only the
capacity will react slower. Comparing simulation 2 to simulation 3 and simulation 1 to simulation 5,
shows that raising the pre load pressure makes the system faster. Simulation 1 and simulation 2 show
that in case of a larger capacity the system will also react faster. Finally simulation 6 shows that if the
pressure is set too high for the initial capacity the accumulator will be empty before the other valve
opens and the force will still sag.
0 1 2 3 4 5 60
0.5
1
1.5
2x 10
4
Time [s]
For
ce [
N]
0 1 2 3 4 5 6-4
-3
-2
-1
0
1
2x 10
-3
Time [s]
Flo
w [
m3 /s
]
0 1 2 3 4 5 62
3
4
5
6
7x 10
6
Time [s]
Pre
ssur
e [P
a]
shoulder seatbelt
hip seatbelt
chair legaccumulator
pump
