modeling of a hydraulic braking system -...

Modeling of aHydraulic Braking System

CHRISTOPHER LUNDIN

Master’s Degree ProjectStockholm, Sweden 2015

IR-EE-SB 2015:000

Abstract

The objective of this thesis is to derive an analytical model representing a re-duced form of a mine hoist hydraulic braking system. Based primarily on fluidmechanical and mechanical physical modeling, along with a number of simplifyingassumptions, the analytical model will be derived and expressed in the form of asystem of differential equations including a set of static functions. The obtainedmodel will be suitable for basic simulation and analysis of system dynamics, withthe aim to capture the fundamentals regarding feedback control of the brakesystem pressure.

The thesis will mainly cover hydraulic servo valve and brake caliper modelingincluding static modeling of brake lining stress-strain and disc spring deflection-force characteristics. Nonlinearities such as servo valve hysteresis, saturation,effects of under- or overlapping spool geometry, flow forces, velocity limitationsand brake caliper frictional forces have intentionally been excluded in order not tomake the model overly complex. The hydraulic braking system will be describedin detail and basic theory that is needed regarding fluid properties and fluidmechanics will also be covered so as to facilitate the reader in his understandingof the material presented in this work.

Overall, the scope of this thesis is broad and more work remains in orderto complement the model of the system both qualitatively and quantitatively.Although not complete in its simplified form and with known nonlinearities aside,the validity of the model in the lower frequency domain is confirmed by resultsgiven in form of measurements and dynamic simulation. Static analysis of thebrake caliper model is also verified to be essentially correct when comparingcalculated characteristics against actual measurements, as is also the case for thestatic models of the brake lining and disc-spring characteristics.

Contents

1 Introduction 21.1 Mine Hoists . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 Hoist Brake System . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.2.1 Brake Control System . . . . . . . . . . . . . . . . . . . . 41.2.2 Brake Hydraulic System . . . . . . . . . . . . . . . . . . . 51.2.3 Brake Mechanical System . . . . . . . . . . . . . . . . . . 5

1.3 Project . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.3.1 Lab Environment . . . . . . . . . . . . . . . . . . . . . . . 7

2 The Hydraulic Braking System 102.1 The Hydraulic Power Unit . . . . . . . . . . . . . . . . . . . . . . 102.2 The Hydraulic Control Unit . . . . . . . . . . . . . . . . . . . . . 112.3 Transmission Lines . . . . . . . . . . . . . . . . . . . . . . . . . . 132.4 Disc Brake Calipers . . . . . . . . . . . . . . . . . . . . . . . . . . 152.5 Reduced System . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

3 Modeling of the Hydraulic Braking System 183.1 Fluid Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

3.1.1 Fluid Mass Density . . . . . . . . . . . . . . . . . . . . . . 183.1.2 Fluid Bulk Modulus . . . . . . . . . . . . . . . . . . . . . 203.1.3 Effective Fluid Bulk Modulus . . . . . . . . . . . . . . . . 253.1.4 Fluid Viscosity . . . . . . . . . . . . . . . . . . . . . . . . 29

3.2 Fluid Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 323.2.1 Navier-Stokes Equations . . . . . . . . . . . . . . . . . . . 323.2.2 Bernoulli Equation . . . . . . . . . . . . . . . . . . . . . . 353.2.3 Orifice Flow . . . . . . . . . . . . . . . . . . . . . . . . . . 353.2.4 Pressure Dynamics in a Hydraulic Volume . . . . . . . . . 39

3.3 Hydraulic Components . . . . . . . . . . . . . . . . . . . . . . . . 413.3.1 Servo Valve . . . . . . . . . . . . . . . . . . . . . . . . . . 41

3.4 Mechanical Components . . . . . . . . . . . . . . . . . . . . . . . 543.4.1 Brake Caliper . . . . . . . . . . . . . . . . . . . . . . . . . 543.4.2 Disc Springs . . . . . . . . . . . . . . . . . . . . . . . . . . 613.4.3 Brake Lining . . . . . . . . . . . . . . . . . . . . . . . . . 64

CONTENTS ii

3.4.4 Simplified Brake Caliper Model . . . . . . . . . . . . . . . 673.5 Brake Stand . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 733.6 Dynamic Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

4 Model Parameterisation, Simulation and Validation 774.1 Model Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

4.1.1 Fluid Properties . . . . . . . . . . . . . . . . . . . . . . . . 774.1.2 Brake Calipers . . . . . . . . . . . . . . . . . . . . . . . . 784.1.3 Servo Valve . . . . . . . . . . . . . . . . . . . . . . . . . . 80

4.2 Static Characteristics . . . . . . . . . . . . . . . . . . . . . . . . . 854.3 Disc Spring Force . . . . . . . . . . . . . . . . . . . . . . . . . . . 854.4 Pressure Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . 88

4.4.1 Simulation . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

5 Discussion and Conclusions 91

Bibliography 93

List of Figures

1.1 Double-skip friction hoist. . . . . . . . . . . . . . . . . . . . . . . 31.2 Brake control system. . . . . . . . . . . . . . . . . . . . . . . . . . 41.3 Brake hydraulic system. . . . . . . . . . . . . . . . . . . . . . . . 51.4 Brake caliper. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.5 Brake stand. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.6 Hydraulic power unit. . . . . . . . . . . . . . . . . . . . . . . . . . 71.7 Hardware-In-the-Loop Simulation (HILS). . . . . . . . . . . . . . 9

2.1 Hydraulic diagram. . . . . . . . . . . . . . . . . . . . . . . . . . . 112.2 Hydraulic diagram. . . . . . . . . . . . . . . . . . . . . . . . . . . 122.3 Valve manifold. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.4 Piping. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142.5 Transmission line. . . . . . . . . . . . . . . . . . . . . . . . . . . . 142.6 Modal approximation. . . . . . . . . . . . . . . . . . . . . . . . . 152.7 Hydraulic cylinder. . . . . . . . . . . . . . . . . . . . . . . . . . . 162.8 Reduced system. . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

3.1 Fluid stress and strain. . . . . . . . . . . . . . . . . . . . . . . . . 223.2 Fluid volume and pressure. . . . . . . . . . . . . . . . . . . . . . . 233.3 Fluid bulk modulus. . . . . . . . . . . . . . . . . . . . . . . . . . 243.4 Container . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253.5 Effective fluid bulk modulus. . . . . . . . . . . . . . . . . . . . . . 293.6 Shear stress and strain. . . . . . . . . . . . . . . . . . . . . . . . . 293.7 Shear flow. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303.8 Kinematic viscosity as a function of temperature. . . . . . . . . . 323.9 Flow through an orifice. . . . . . . . . . . . . . . . . . . . . . . . 363.10 Approximated discharge coefficient. . . . . . . . . . . . . . . . . . 393.11 Hydraulic volume. . . . . . . . . . . . . . . . . . . . . . . . . . . . 393.12 Servo valve symbol. . . . . . . . . . . . . . . . . . . . . . . . . . . 423.13 Servo valve. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 423.14 Torque motor. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 433.15 Spool. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 433.16 Spool displacement. . . . . . . . . . . . . . . . . . . . . . . . . . . 44

LIST OF FIGURES iv

3.17 Valve flow. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 453.18 Wheatstone bridge. . . . . . . . . . . . . . . . . . . . . . . . . . . 463.19 Flow as a function of current. . . . . . . . . . . . . . . . . . . . . 493.20 Saturation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 493.21 Blocking both actuator ports. . . . . . . . . . . . . . . . . . . . . 503.22 Pressure Sensitivity. . . . . . . . . . . . . . . . . . . . . . . . . . . 513.23 Load attachment. . . . . . . . . . . . . . . . . . . . . . . . . . . . 523.24 Internal valve leakage. . . . . . . . . . . . . . . . . . . . . . . . . 533.25 Flow forces. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 533.26 Caliper halve. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 553.27 Model of the caliper yoke. . . . . . . . . . . . . . . . . . . . . . . 553.28 Hydraulic unit. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 563.29 Model of the hydraulic unit. . . . . . . . . . . . . . . . . . . . . . 573.30 Brake shoe. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 573.31 Brake release. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 583.32 Gap adjustment. . . . . . . . . . . . . . . . . . . . . . . . . . . . 583.33 Model of the brake shoe. . . . . . . . . . . . . . . . . . . . . . . . 593.34 Brake caliper model. . . . . . . . . . . . . . . . . . . . . . . . . . 593.35 Spring stack. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 613.36 Disc spring. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 613.37 Stack deflection and force. . . . . . . . . . . . . . . . . . . . . . . 623.38 Stack deflection and spring rate. . . . . . . . . . . . . . . . . . . . 633.39 Friction during compression. . . . . . . . . . . . . . . . . . . . . . 643.40 Stress-strain characteristic. . . . . . . . . . . . . . . . . . . . . . . 653.41 Stress-strain characteristic and Young’s Modulus. . . . . . . . . . 663.42 Simplified caliper model. . . . . . . . . . . . . . . . . . . . . . . . 683.43 Brake caliper static characteristics. . . . . . . . . . . . . . . . . . 723.44 Brake calipers. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

4.1 Measurements and pressure sensitivity. . . . . . . . . . . . . . . . 844.2 Measured characteristics. . . . . . . . . . . . . . . . . . . . . . . . 864.3 Pressure response with different signal inputs. . . . . . . . . . . . 894.4 Simulation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

List of Tables

4.1 Fluid Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . 784.2 Brake Caliper Parameters . . . . . . . . . . . . . . . . . . . . . . 784.3 Brake Lining Parameters . . . . . . . . . . . . . . . . . . . . . . . 794.4 Disc Spring Parameters . . . . . . . . . . . . . . . . . . . . . . . . 794.5 Valve Rating . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 804.6 Servo Valve Parameters . . . . . . . . . . . . . . . . . . . . . . . . 834.7 Clamping Force and Air Gap Adjustment . . . . . . . . . . . . . . 85

Chapter 1

Introduction

1.1 Mine HoistsMine hoists are used for the purpose of vertical transportation of ore, personneland equipment in underground mine shafts. With the use of a mine hoist in anunderground mine, production may increase significantly as it is the most efficientway of both elevating ore to the surface and transporting personnel to and fromthe deep levels of a mine. Hoisting distances typically lie in the range between500 and 1500 meters with payloads up to about 60 tons. Mechanically, a minehoist may be configured differently depending on a number of factors, but the twomain categories of hoists are either of a winding or frictional type. Friction hoists(or Koepe hoists), see the example depicted in figure 1.1, are based on friction fordisplacement of two conveyances placed on each side of a mine shaft. Throughfrictional forces between a number of hoisting cables called head ropes (5) andfriction inlays mounted on a hoist pulley (1), which is driven by an electrical motor(2) either directly or through a gear box, the conveyances are allowed to ascendand descend in the mine shaft. The motor in turn is mechanically connectedto a shaft onto which the rotor and pulley is mounted, which is supported bytwo bearings (3) on both sides. The conveyances can be either in form of orecontaining skips (8) or personnel carrying cages, which may be combined with acounterweight if hoisting with a single skip or cage. The head ropes are connectedto the upper end of the conveyances through rope attachments (6) and a numberof balancing tail-ropes (10) are fixed to the lower end of both conveyances withthe purpose of maintaining constant rope tensional forces on both sides of thehoist pulley. The skips are typically filled through ore-bins (9) or from conveyorbelts placed on one or both sides of the shaft depending on if the hoist is used ina single- or double-skip configuration. The diameter of the pulley is decided byrope selection criteria and a number of head sheaves (7) may be installed in orderto decrease the size of the shaft if needed. Drum hoists (or winders) are a type ofhoist where the hoisting cable is wound around the drum and may be configured in

1.2 Hoist Brake System 3

a single- or double-drum configuration with either one or two rope compartmentson each drum and a single or double clutch mechanism for double-drum hoistingin order to be able to adjust the relative distance between the conveyances (thisto either adapt for rope elongation or adjustment to different hoisting distances).Regardless of the type of hoist, a disc brake system1 is installed for safety reasonsand a number of brake calipers are mounted on brake stands (4) that are placedon each side of the pulley or drum. The mine hoist brake system will be the topicof this work and is presented in more detail in the next section.

Figure 1.1: Double-skip friction hoist.

1.2 Hoist Brake SystemFrom a safety point of view, the braking system is the most critical part of amine hoist. Failure to bring the hoist to a standstill during operational distur-bances such as a power outage or an emergency situation will lead to a so calledoverwinding, possibly with serious consequences, which in a worst-case scenariocan result in fatal outcome. When also taking into consideration the economicaspects associated with such malfunction of the system, potentially causing haltsin production for long periods of time and loss of investments, the conclusion

1Shoe brakes for mine hoists also exist.


must be that safety issues concerning mine hoists are of great importance. Pastaccidents and incidents have historically lead to more strict regulations world-wide concerning the use of mine hoist systems, with current trends pointing inthe direction of an increasing demand for high safety integrity in the mining in-dustry. The hoist brake system consists of three subsystems which spans acrossthree technical domains containing electrical, hydraulic and mechanical compo-nents. The different subsystems of the brake system are briefly presented in thefollowing.

1.2.1 Brake Control SystemThe control system mainly comprises a main control unit (MCU), a HMI2 oper-ator interface and four channel control units (CCU), see figure 1.2. The controlsystem is built on a multi-channel architecture with identical, in parallel execut-ing and independent so called brake channels. Each channel of the brake systemis controlled by a CCU in form of a PLC3, which includes control logic, moni-toring and closed-loop control functions. The main advantage of a multi-channel

Figure 1.2: Brake control system.

system is the ability to both electrically and hydraulically divide the total brakingcapacity into two or more independent subsystems, thus avoiding the dependencyon a single system. In a single channel system (or mono-channel system) one issolely dependent on a single chain of both electrical and hydraulic components.A single component failure will thus affect the entire brake capacity, with thetwo extremes being complete loss of brake force and full application of the brakeswith the hoist in motion. With a multi-channel system, a degree of redundancyis achieved and only a fraction of the brake capacity is affected in case of failure

2HMI - Human Machine Interface3PLC - Programmable Logic Controller


of the equipment in one of the brake channels (for a four-channel system a fourthand for a two-channel system one half). The disadvantages of a multi-channelsystem are of course the greater complexity, increased cost of the equipment andreduced availability since the probability of component failure will be higher.

1.2.2 Brake Hydraulic SystemThe hydraulic parts of the brake system consist of a power unit and two controlunits, see figure 1.3. Each such control unit in turn contains two so called valvemanifolds each, which are the controlling hydraulic elements of the brake chan-nels. As the term indicates, the power unit maintains the system pressure andsupplies the control units with hydraulic fluid. The main task of the brake system

Figure 1.3: Brake hydraulic system.

is to safely decelerate the mine hoist by friction-based braking in emergency orsafety stop situations. During normal operation, hoist deceleration is howeverperformed by the electric motor of the drive system and the brake system onlyprovides a static holding force to securely keep the position of the mine hoistas the drive is disconnected at standstill. The power and control units togetherhydraulically maneuver the brake calipers that mechanically produce the brakingforce that is needed during emergency braking and at hoist standstill.

1.2.3 Brake Mechanical SystemThe braking system is of a disc brake type with discs mounted on the drumor pulley of the mine hoist where the clamping force from a number of brakecalipers is applied, see figure 1.4. The calipers are mounted on brake stands (see

Figure 1.4: Brake caliper.


figure 1.5) which are positioned around the drum or pulley. The number of brakestands and calipers varies and are sized differently depending on load conditionsand regulatory requirements. The number of brake discs are however at least twobecause of the risk of contamination of the disc surfaces (loss of brake force canamount to half of the total brake capacity in the case of a single disc). The brakecalipers acts by applying a clamping force directed normally against both surfacesof the brake discs, which through friction between the brake linings and the brakedisc will induce a braking force and in turn also a braking torque. The calipers areof a fail-safe design and do therefore not need any hydraulic pressure applied inorder to apply the brake linings against the brake disc surfaces. A restoring springforce instead strives towards a full application of the brakes against the discs andis counteracted by increasing the hydraulic pressure. Maximum working pressurein the system therefore implies that the brake calipers are released from the brakedisc with a fully developed gap between the linings and brake disc surface andthat the brakes are fully applied at atmospheric pressure (zero gauge pressure)with maximum clamping force. It must however be stressed that the pulley (in

Figure 1.5: Brake stand.

the case where transport of the conveyances takes place due to friction betweenthe pulley and head ropes) or drum (when the rope or ropes are wound onto orunwound from the drum) must not be exposed to excessive braking force when thehoist is set in motion as this may give rise to dangerously high deceleration, whichin turn can lead to injuries, rope slippage or slack rope conditions. The emergenceof this can lead to serious accidents which may involve the risk of endangeringthe lives of mine personnel. The number of brake calipers is oversized for thecase of braking during dynamic conditions, this because the brake system mustmeet the requirements of static holding force as specified in applicable miningregulations. For example, the braking system must be able to keep the minehoist at standstill during overload conditions, which may mean that the minehoist is subjected to unbalanced forces that exceed twice or possibly even threetimes that of the rated maximum load. Overcapacity of available braking forceis dealt with by dividing the system into several brake channels and limitation of

1.3 Project 7

applied clamping force during dynamic conditions, this because of the risk thatcould arise if the system would fail with the mine hoist in motion and provide fullapplication of the brakes. For this reason, the braking system is always providedwith at least two brake channels.

1.3 ProjectThe objective of this thesis is to obtain an analytical model of a reduced formof the full-scale hydraulic brake system that has been briefly presented in thisintroductory chapter. The analytical model will be expressed explicitly in theform of a system of differential equations along with a number of static functionsand is primarily suitable for basic simulation and analysis of system dynamics.Results will be presented in form of dynamic simulation and static analysis ofthe system, which to some extent will verify the validity of the obtained model.The components and functionality of the system will be described in more detailin the next chapter along with a number of modeling assumptions and the scopeof model inclusion.

1.3.1 Lab EnvironmentThe hydraulic system is installed in a development and test environment alongwith the brake control system. A hydraulic power unit (see figure 1.6) and asingle valve manifold mounted therein including a brake stand with a total ofthree brake calipers, two single and one double unit, are installed in the laband available for tests. The brake stand is mounted close to the power unit

Figure 1.6: Hydraulic power unit.

and hydraulic transmission lines in form of steel pipes connect between them.Additional to what was presented in section 1.2.1, the configuration of the labcontrol system also consists of both PC- and PLC-based real-time simulationhardware for Hardware-In-the-Loop-Simulation (HILS) of the brake system hy-draulics (HySim) and the mine hoist (HoistSim) respectively. Referring to figure

1.3 Project 8

1.7a, one out of the four channel control units is connected to the valve manifoldinstalled in the power unit while the other control units are connected to HySim.The valves in turn hydraulically control the brake caliper clamping force, whichis measured with a force transducer mounted between one of the calipers on thebrake stand. The measured clamping force is then fed via an analog input toHoistSim along with the force of three simulated brake channels where a sim-ple summation is made in order to obtain the total applied clamping force forall four channels. HoistSim then simulates the hoist dynamics in order to pro-duce an analog output corresponding to the angular velocity of the hoist pulleyor drum, which is then fed back to each one of the channel control units. Asshown in figure 1.7b, the deceleration of the hoist is controlled through cascadedclosed-loop control of the brake pressure and deceleration calculated from thehoist angular velocity. In an actual hoist installation rotational speed connectsback to the control units through incremental pulse encoders, which in the labenvironment is emulated through a frequency transducer as illustrated in figure1.7b. A near full-scale model of the brake hydraulics is implemented throughcomponent-based physical modeling and is simulated in HySim. The advantagesof component-based modeling is the ease at which large complex models may beconfigured and changes to existing models are more or less trivial. The disadvan-tage is however the lack of explicit analytical insight into a model of a completesystem model, which essentially motivates the objective of this work.

1.3 Project 9

(a) Hardware configuration.

(b) Signal flow and closed-loop control.

Figure 1.7: Hardware-In-the-Loop Simulation (HILS).

Chapter 2

The Hydraulic Braking System

A more detailed description of the brake system with its individual subsystemsand components will be given in the sections of this chapter that follows. Thescope of the system to be modeled only includes a subset of the complete hydraulicbrake system and will be presented as a reduced system in the last section of thechapter. This will then form the basis and outline of the analytical model.

2.1 The Hydraulic Power UnitThe main function of the hydraulic power unit is to supply the control units withhydraulic fluid and maintain a constant system pressure. The hydraulic circuitryof the power unit is found in figure 2.1 in form of a hydraulic diagram. The powerunit mainly includes a hydraulic pump (4) driven by an electric motor (5), a tank(9) which supplies the braking system with hydraulic fluid and an accumulator(21) with the task of maintaining the system pressure at high loads. Additional tothese main items the power unit also includes other components such as controlvalves (6, 20), an oil heater with thermostat (11, 17), an oil cooler including acirculation pump, electric motor and fan (22-25), shut-off valves with and withoutan indicator (1-3, 8), oil level indicators (10, 15), oil filters with indicators (13, 26)and (14, 27), a breather filter (18), a thermometer (19), a temperature transducer(16), a check valve and pressure relief valve (29, 7), measurement points (12, 30),a pressure gauge (32) and pressure transducers (28, 31). A vast majority of thesecomponents are relatively uninteresting for the modeling of the system dynamics,at least regarding control of the hydraulic brake pressure which will be the mainfocus of this work. These components will therefore be omitted in the model ofthe system. Components that are of potential interest in the modeling of thepower unit are components (4), (5), (26) and (29). The accumulator (21) hasa great impact on the system pressure, but will however be fully loaded withvalve (20) de-energized and valve (6) energized (i.e. both valves closed). Itcan therefore be disregarded when modeling the power unit as it will not have

2.2 The Hydraulic Control Unit 11

Figure 2.1: Hydraulic diagram.

any effect on the system pressure. This exclusion is justified by the fact thatthe brake calipers in the lab environment are few and that the accumulator forthat reason is not needed. The hydraulic pump (4) is of a piston type withvariable displacement driven by an induction motor (5) and the system pressureis controlled via a built-in mechanical regulator in the pump, which through aswash-plate mechanism controls the flow of oil (the pump can thus be seen as aflow source). When the set pressure is reached, no oil will flow into the system.However, a small leakage flow will still exist but is diverted through a drain lineconnected to the tank (see the dashed line in the hydraulic diagram in figure 2.1).The filter (26) ensures that the hydraulic fluid entering the system through thepump has a certain degree of cleanliness. This is important as contaminated oilotherwise could damage the control valves both in the hydraulic power unit andthe hydraulic control units. A drop in pressure occurs across the filter, but ishowever of such small magnitude that it may be neglected. A check valve (29) isalso installed to prevent the pump supply line from acting as an oil return path.

2.2 The Hydraulic Control UnitThe hydraulic power unit is connected to the two hydraulic control units of thebrake system through a pump and tank connection, see P and T in the hy-draulic diagram of the power unit in figure 2.1. The control units are in turnconnected to the brake calipers on the brake stands, which are controlled by thebrake pressure through a number of valves installed in the control units. Thehydraulic diagram for one out of the two channels in each control unit is foundin figure 2.2. The valves used for control of the brake pressure are mounted in so

2.2 The Hydraulic Control Unit 12

Figure 2.2: Hydraulic diagram.

called valve manifolds (see figure 2.3) that physically implements the hydrauliccircuitry. Four valve manifolds, one for each hydraulic channel, are placed in thetwo control units which thus are housing two manifolds each. However, in the labenvironment only a single manifold mounted inside the power unit is available fortests. A servo valve (7) is the main component used for pressure control in eachhydraulic channel and represents the actuator in the closed-loop control of thebrake pressure. The feedback of pressure to the controller in each channel controlunit PLC is carried out by the use of one of two redundant pressure transducers(20) and (21), which are mounted on the return line of each hydraulic controlunit. In addition to these components, the manifold also contains a number ofother items. Among these the relief valve (1), the shut-off valve (3), measuringpoints (4), (12) and (13), the return line accumulator (5), the pressure gauge (19)and the pressure transducers (11), (20) and (21) may be excluded as they are ofno interest when modeling the system. The pressure relief valve (9) is used tocontrol the pressure at unregulated braking of the hoist (constant brake force)

Figure 2.3: Valve manifold.

2.3 Transmission Lines 13

and is mechanically set to a certain opening pressure. At a pressure below thissetting, the valve is closed and no flow will pass through the valve. Above the setpressure level the valve will open and let oil out of the system until the pressurehas decreased to the point where the valve will close again. Valve (10) is an elec-trically controlled pressure relief valve with a large flow capacity which is usedat initial application of the brake linings against the brake disc during a safetystop of the mine hoist. In order for the brakes to apply in as short time possible,a large return flow of oil is needed during a time span of about 200 milliseconds.The functionality of the valve is basically the same as for a purely mechanicalpressure relief valve, apart from having a pressure level of relief that is set elec-trically. The check valves (2) and (18) function similar to a relief valve, howeverthe opening pressure of a check valve is much smaller and its main purpose isrectification of the oil flow rather than a limitation of the hydraulic pressure.The directional two-way valves (14), (15), (16) and (17) act as either open orclosed elements in the hydraulic circuitry and are controlled through digital sig-nals from the control system. Hence, when energized (or activated) valve (16)will block the path for flow of oil through the valve and allow the brake pressureto be increased above the pressure relief setting of valve (9). When energizingvalve (17), oil may flow through the prime application valve (10) and the brakepressure will be constrained to be equal to or below the electrically set level ofpressure relief. The normally closed valve (14) implements a similar functionalityand blocks the actuator port of the servo valve if needed. The accumulator (6)stores potential energy by compression of a volume of nitrogen in the same wayas in the power unit and will during short periods of time allow a much higherrate of flow into the system through the servo valve compared to the flow rate ofthe pump. It effectively also acts as a shock absorber, this since the mechanicalregulator in the pump will not be able to respond to the fast dynamics of theservo valve. If suddenly placing the servo valve in its centered position with thepump delivering its maximum flow rate without an accumulator attached, thesystem pressure would increase rapidly until reaching the safety pressure relieflevel which should be avoided. Valve (8) is a pressure compensated flow controlvalve and is used together with valve (15) to discharge the accumulator duringunregulated braking. Through a mechanical function, which regulates the open-ing of the valve to maintain a constant pressure drop across its ports, a constantflow is achieved.

2.3 Transmission LinesThe different parts of the brake system are interconnected through hydraulictransmission lines in form of steel pipes. The power unit connects to both con-trol units via the pump and tank connectors P and T and the control units arein turn linked to the brake calipers on the brake stands through the L and M

2.3 Transmission Lines 14

connections, corresponding to the supply and return side of the control units re-spectively (see figure 2.2). With a single manifold mounted inside the power unitin the lab environment, the connections between the pump, tank and the valvemanifold may, due to the short distance, be either neglected or considered to beof a purely resistive character from a modeling perspective. The piping betweenthe manifold and brake stand will however form a quite extensive transmissionline network as is illustrated in figure 2.4. A main line connects the two portstogether and branched connections are made from the main line to each caliperhalve mounted on the brake stand. When increasing the brake pressure, andeventually releasing the brake linings from the brake disc, valve (16) and (17) inthe control unit are energized and de-energized respectively. Thus, both valvesare closed which effectively renders the return connection completely blocked toany throughput of oil. In order to capture the high frequency and resonant behav-

Figure 2.4: Piping.

ior of the system, modeling of the transmission line dynamics is necessary. Thedynamics of a single hydraulic transmission line may in the frequency domain bedescribed by the following transfer function [16][

PuQu

]=[

cosh Γ(s) Z(s) sinh Γ(s)1

Z(s) sinh Γ(s) cosh Γ(s)

] [PdQd

], (2.1)

where Pu, Pd, Qu and Qd correspond to the pressure and flow at the upstreamand downstream connection of the transmission line as shown in the figure below.

Figure 2.5: Transmission line.

The functions Γ(s) and Z(s) are the Laplace transforms of the propagation op-erator and characteristic impedance of the transmission line. Now, time-domain

2.4 Disc Brake Calipers 15

expressions of (2.1) may not be obtained without approximation due to the cyclichyperbolic functions in the frequency domain and several ways of approximatingtransmission line dynamics exist [8]. The method of modal approximation aspresented by Yang and Tobler [16] will yield truncated transfer functions fromcausal forms of (2.1) with n modes of the transmission line dynamics preserved.An example of modal approximation is shown in figure 2.6. At steady laminarflow, the relation between pressure and flow across and through a pipe with acircular cross-section will be linear and obey the Hagen-Poiseuille law [2]

∆P = 128µLπD4 Q, (2.2)

where ∆P is the pressure drop across the ports of the pipe, µ is the dynamic vis-cosity, L is the length of the pipe, D is the inner diameter of the pipe and Q themean volumetric rate of flow through the pipe. With sinusoidal input variationsof pressure and flow, the dynamic response initially show low-pass characteris-tic behavior in the lower regions of the spectrum and cyclic anti-resonant andresonant response in both pressure and flow in the higher regions.

Figure 2.6: Modal approximation.

2.4 Disc Brake CalipersThe hydraulic pressure connects to the brake calipers that are mounted on thebrake stands through the pipe connections L and M (see figure 2.4). Thecalipers are single acting hydraulic cylinders of a spring-return type and mayschematically be represented by the symbol shown in figure 2.7. Functionalityof the brake calipers is based on a fail-safe spring-applied and pressure-releaseprinciple with a fully developed clamping force at atmospheric pressure. This

2.5 Reduced System 16

Figure 2.7: Hydraulic cylinder.

principle therefore requires pressure to be applied to the caliper pressure portin order to decrease the clamping force. At the so called release pressure, theclamping force is balanced by the hydraulic pressure and increasing the hydraulicpressure further will release the brake linings from the brake disc. A number ofdisc springs in the brake calipers generates the clamping force FC applied againstthe brake disc of the drum or pulley through the brake lining mounted on thebrake shoe of each caliper halve. Disc springs are used due to the large force thatcan be generated for small deflections and also their compactness. The clampingforce acts normal to both surfaces of the brake disc and will yield a brake forceFB with a magnitude which depends on the coefficient of friction µ between thelinings and disc surface

FB = 2× µFC . (2.3)The braking force FB that arise from the friction between both surfaces trans-fers into a braking torque which will stop the rotation of the pulley or drum ofthe hoist during a safety stop and prevent it from rotating when stopped. Atnormal service, the brakes simply provide a static balancing holding torque tomaintain the hoist at standstill. The magnitude of the brake torque depends onthe diameter of the brake disc, the applied brake force and the width of the brakelining

TB = FB ×DB − wl

2 , (2.4)

where TB is the braking torque, DB is the brake disc diameter and wl is thewidth of the brake lining. Hence, the point of application of the brake torque willeffectively be at a distance from the center of the pulley or drum correspondingto the radius of the brake disc corrected for by halve of the width of the brakelining.

2.5 Reduced SystemThe analytical model of the hydraulic brake system will in this work correspond toa reduced form of the system that has been presented. With focus on closed-loopcontrol of the brake pressure, the system may be reduced to the simplified singlechannel system shown in figure 2.8. Hence, the pump, motor and check valveneed to be included from the hydraulic power unit and only the servo valve andaccumulator in the hydraulic control unit. However, the supply pressure will beassumed constant in the modeling of the system, leaving the servo valve and brake

2.5 Reduced System 17

calipers as the remaining components. The directional two-way valve (14) mayeither be considered as being a hydraulic restriction in form of an orifice or elseneglected if the magnitude of the flow through the valve is relatively low, which isthe case for system that is modeled. For a complex network of transmission lines,the process of modeling the system dynamics based on modal approximationwill yield accurate models but is on the other hand quite involved, see e.g. [9].Transmission line dynamics will not be taken into account in this work and thevalidity of the model is therefore limited to the low region of frequency where thetransmission lines may be treated as pure hydraulic volumes. Now, as it mightseem the presented outline of the model is a highly reduced form of the actualsystem. However, the essential components of the system are still included andthe scope of dynamic and static modeling of the different parts of the system willstill be of a considerable extent, as will be presented in the following chapters.

Figure 2.8: Reduced system.

Chapter 3

Modeling of the HydraulicBraking System

In order to model a dynamic system based on physical modeling, insight into andunderstanding of the underlying fundamentals that play a role in the behaviorof the system is of great importance. The first part of this chapter is thereforedevoted to the basic concepts and theory of fluid properties and fluid mechanicsthat is needed to model the system. In the latter parts of the chapter, thedifferent hydraulic and mechanical components that are included in the modelof the system are presented and described in detail along with both static anddynamic models.

3.1 Fluid PropertiesFluid is the medium in which energy is transmitted through in a hydraulic systemand the type of fluid used will have an impact on a number of different systemproperties. The fluid itself might be synthetic or non-synthetic and importantproperties and issues that need to be considered when selecting a hydraulic fluidare lubricity, viscosity, protection against corrosion, tendency to foam, fire resis-tance and environmental impact. The properties of a fluid medium are apart fromthe type of fluid used also dependent on temperature, pressure and entrained air.Understanding how these various parameters affect the behavior and control ofa hydraulic system constitutes the fundamentals of hydraulic system theory andis therefore presented in the following.

3.1.1 Fluid Mass DensityThe density of a homogeneous fluid is defined as its mass per unit volume, denotedwith the symbol ρ

ρ = lim∆V→0

∆m∆V = m

V, (3.1)

3.1 Fluid Properties 19

where m, V , P and T correspond to fluid mass, volume, pressure and temperaturerespectively. The definition above assumes an isothermal and isobaric condition,where the Latin prefix iso used in these terms translates into equal or similar, i.e.a condition where the fluid pressure and temperature are held constant such thatP = P0 and T = T0. The mass density of a fluid is dependent on temperatureand pressure, both causing volumetric change to a fluid under varying conditions.

The relation between fluid density, temperature and pressure may be formulatedthrough the equation of state that describes the state of a matter under a givenset of physical conditions. An example of an equation of state is the ideal gas law

PV = nRT, (3.2)

where P is the absolute pressure, V is the gas volume, n is the amount of sub-stance (number of moles), R is the universal gas constant and T is the absolutetemperature of the gas. By noting that n = m

M, where m is the mass and M is the

molar mass, equation (3.2) may be reformulated to explicitly give an expressionfor the density as a function of the state variables

PV = nRT = m

MRT ⇒ ρ = m

V= M

R· PT. (3.3)

For a liquid, no such simple relation between fluid density, temperature andpressure exists. However, a linearized equation of state in form of a first-orderTaylor expansion may be used to describe the variations of density with smallchanges in pressure and temperature [10]

ρ = ρ0 + ∂ρ

∂P

∣∣∣∣∣P=P0

(P − P0) + ∂ρ

∂T

∣∣∣∣∣T=T0

(T − T0) (3.4)

where ρ0, T0 and P0 represent the density, temperature and pressure of the fluidrespectively at a known reference point.

Equation (3.4) can be expressed on the form

ρ = ρ0

1 + 1β0

(P − P0)︸︷︷︸∆P

−α0 (T − T0)︸︷︷︸∆T

, (3.5)

where β is the isothermal bulk modulus and the reciprocal 1/β is defined as beingthe compressibility of a fluid

1β

= 1ρ0

∂ρ

∂P. (3.6)

The isobaric coefficient of fluid thermal expansion α in (3.5) is defined such that

α = − 1ρ0

∂ρ

∂T, (3.7)


which quantifies the volumetric expansion of a fluid due to varying thermal con-ditions under constant pressure. With a fluid bulk modulus β that is numericallylarge in magnitude (approximately 1.7 × 109 Pa for a petroleum based mineraloil at 40° C) and a coefficient of thermal expansion α that is numerically smallin magnitude (approximately 7 × 10−4° C−1 for a petroleum based mineral oil),fluid density is more or less unaffected by changes in temperature and pressureand may therefore be regarded as constant ρ ≈ ρ0 throughout a range of normalvariation in temperature and pressure about a known working point. Effects oftemperature on the coefficient of fluid thermal expansion are moderate through-out a relatively wide range of normal working temperatures, which is also thecase for the fluid bulk modulus. The density of a hydraulic fluid typically liesbetween 850 kg/m3 and 900 kg/m3. For an incompressible fluid, considered to bean ideal condition, the density ρ is assumed to be constant.

3.1.2 Fluid Bulk ModulusThe fluid bulk modulus of elasticity, or more specifically the isothermal fluid bulkmodulus of elasticity, is a measure of the tendency of a fluid volume to deformelastically when under uniform compression. Denoted with the letter β, it isdefined as [10]

β = lim∆ε→0

∆σ∆ε

∣∣∣∣∣T=T0

= ∂σ

∂ε, (3.8)

where σ and ε are the fluid stress and strain respectively. Stress is defined asthe force causing a deformation of the fluid divided by the area of the surfaceon which the force is applied and fluid strain may be defined as the negativerelative volumetric change (commonly known as the engineering strain) causedby an increase or decrease of fluid pressure

ε = V0 − VV0

= −∆VV0

. (3.9)

With the definition of fluid strain according to above, strain will increase witha decreased volume relative to the volume of the fluid at a reference pressure,typically under atmospheric conditions. Since stress is identical to fluid pressure

σ = P, (3.10)

the bulk modulus may be expressed with fluid pressure explicitly

β = ∂P

∂ε. (3.11)

Further to the above, by differentiating strain as defined by (3.9) one will obtain

dε = − 1V0dV, (3.12)


as well as an equivalent definition of the tangential fluid bulk modulus

β = ∂P

∂ε= −V0

∂P

∂V. (3.13)

From the definition of density and the assumption on constant mass, it is con-cluded that

dm = d(ρV ) = ρ dV + V dρ = 0⇒ − 1VdV = 1

ρdρ, (3.14)

and additionally the bulk modulus may thus also be given by

β = ρ0∂P

∂ρ, (3.15)

which is the form of the bulk modulus in equation (3.6).

The secant isothermal bulk modulus is defined as the slope between two points onthe stress-strain curve

K = ∆P∆ε . (3.16)

The secant bulk modulus of liquids varies linearly with pressure over a limitedrange [10] according to

K = K0 +mP, (3.17)where K0 represents the fluid bulk modulus at a pressure P = 0 relative to theatmospheric pressure (gauge pressure). For a mineral oil, which is a fluid typicallyused in a hydraulic system, equation (3.17) is valid up to a pressure of about 800bar1. At a normal working temperature of the oil in the range of 30− 70 °C thefluid bulk modulus K0 lies in the range of about 1.8 - 1.5 GPa and the coefficientof proportionality m is considered constant throughout the whole normal workingrange of fluid temperature, and for a mineral oil m ≈ 5.6 [10].

Calculating the strain with the reference point taken at atmospheric pressureyields

∆ε = V0 − VV0

− ε0 = V0 − VV0

− 0 = V0 − VV0

. (3.18)

With the atmospheric pressure defined as the reference pressure the difference instress, or pressure, will simply equal

∆P = P − P0 = P − 0 = P. (3.19)

Calculation of the secant bulk modulus now gives

K = ∆P∆ε = P

V0−VV0

= V0 P

V0 − V. (3.20)

1 1 bar = 1× 105 Pa


An expression relating the fluid strain and stress may be obtained by equating(3.20) with (3.17)

K = ∆P∆ε = V0 P

V0 − V= K0 +mP ⇒ ε = P

K0 +mP. (3.21)

This function is depicted in figure 3.1, where the origin corresponds to atmo-spheric conditions and the bulk modulus and the coefficient of proportionalityare given the values of K0 = 1.8 GPa and m = 5.6.

0 0.5 1 1.5 20

10

20

30

40

Strain ε [%]

Stre

ssσ

[MPa

]

Figure 3.1: Fluid stress and strain.

Rearranging the above also allows for identification of the relation between fluidvolume and pressure

V = V0 (1− ε) = V0

(1− P

K0 +mP

). (3.22)

The relative fluid volume 1− ε as a function of fluid pressure is shown in figure3.2. With a relation between stress and strain at hand, the tangential isothermalbulk modulus defined as in (3.11) may be calculated through differentiation of(3.21) with respect to P [10]

1β

= ∂ε

∂P= K0

(K0 +mP )2 ⇒ β = (K0 +mP )2

K0. (3.23)

Equivalently, one may arrive at the same expression for the tangential bulk mod-ulus through (3.13) and (3.22).

∂V

∂P= − V0K0

(K0 +mP )2 ⇒ β = −V0∂P

∂V= (K0 +mP )2

K0. (3.24)


0 10 20 3098

99

100

Pressure P [MPa]

Volu

meV

[%]

Figure 3.2: Fluid volume and pressure.

If instead a logarithmic definition of the fluid strain (commonly known as thetrue strain) is used

ε = − ln(V

V0

), (3.25)

the tangential isothermal bulk modulus is calculated according to

dε = − 1VdV ⇒ β = −V ∂P

∂V. (3.26)

For values close to V/V0 = 1, a Taylor expansion may be used to show that

− ln(V

V0

)= −

[(V

V0− 1

)− 1

2

(V

V0− 1

)2+ ...

]≈ V0 − V

V0. (3.27)

Now, using equation (3.22) together with the derivative ∂V/∂P in (3.24) yields[10]

β =(

1− P

K0 +mP

) (K0 +mP )2

K0

= K0

(1 + (m− 1)P

K0

)(1 + mP

K0

). (3.28)

A comparison between the different moduli is plotted in figure 3.3, where K0 =1.8 GPa and m = 5.6. All show a linear characteristic in the given pressure rangeand as expected, the secant bulk modulus is lower than the tangential forms. Thedifference between the tangential bulk modulus based on the engineering strainor true strain is more or less insignificant in the lower range of fluid pressure as


0 10 20 30 400

0.5

1

1.5

2

2.5

Pressure P [MPa]

Bulk

Mod

ulus

[GPa

]

K

β, ε = −∆V/V0β, ε = − ln (V/V0)

Figure 3.3: Fluid bulk modulus.

was indicated by equation (3.27).

Linear forms of (3.23) and (3.28) may be obtained by noting that

(K0 +mP )2

K0= K0 + 2mP + m2

K0P 2 ≈ K0 + 2mP (3.29)

and

K0

(1 + (m− 1)P

K0

)(1 + mP

K0

)

=K0 + (2m− 1)P + m(m− 1)K0

P 2

≈K0 + (2m− 1)P. (3.30)


3.1.3 Effective Fluid Bulk ModulusThe fluid bulk modulus quantifies the elasticity of a fluid as it undergoes vol-umetric deformation. However, apart from the fluid bulk modulus, one mustalso take into consideration the influence of additional elasticity due to entrainedair and mechanical compliance in a hydraulic system, which together with thecompressibility of the fluid forms an effective fluid bulk modulus. Elasticity is asource of resonance within a hydraulic system, which from a control point of viewis not a problematic issue as long as the effective bulk modulus is of a sufficientmagnitude. However, the presence of a relatively large fraction of unsolved airwithin the hydraulic fluid and the use of flexible hydraulic hoses instead of fixedmetal pipelines will however undoubtedly affect the hydraulic system in a soften-ing manner. This will effectively lower the frequency of resonance and thus canpotentially become a concern if feedback control is employed.

Consider an uncompressed mixture of fluid and air in a container as depicted infigure 3.4a.

(a) Uncompressed (b) Compressed

Figure 3.4: Container

The total volume of the system is assumed to equal the combined volume of liquidand entrained air [10]

V = Vl + Vg. (3.31)A pressurization of the system through compression of both the fluid and aircontent inside the container as in figure 3.4b will give rise to a decrease in volume.Pressurizing the combined volume in the container will also lead to mechanicaldeformation of the container itself. The total volume of the system can thereforealso be expressed as

V = V0 − Vε + Vδ = Vl + Vg, (3.32)where V0 represents the initial volume of the system, Vε a pressurizing decrease offluid volume and Vδ an increase in system volume due to mechanical compliance


of the container.

An effective volume may be defined such that

Ve = V0 − Vε = Vl + Vg − Vδ. (3.33)

Differentiating the above yields the differential change in volume

dVe = dVl + dVg − dVδ. (3.34)

Referring to equation (3.26), the following expression for the tangential effectivebulk modulus may be obtained from (3.34) [10]

1βe

= − 1Ve

∂Ve∂P

= − 1Ve

∂Vl∂P− 1Ve

∂Vg∂P

+ 1Ve

∂Vδ∂P

. (3.35)

By complementing each term in (3.35), a relation between the effective bulk mod-ulus, the bulk modulus of the liquid, the entrained air and that of the containermay be obtained as

1βe

= VlVe

(− 1Vl

∂Vl∂P

)+ VgVe

(− 1Vg

∂Vg∂P

)+ 1Ve

∂Vδ∂P

= VlVe

1βl

+ VgVe

1βg

+ 1βδ. (3.36)

In the above, the bulk modulus of the container has been defined such that

βδ = Ve∂P

∂Vδ. (3.37)

By noting that Ve = Vl + Vg − Vδ, equation (3.36) can be arranged as follows

1βe

=(

1 + VδVe

) 1βl

+ VgVe

1βg

+ 1βδ. (3.38)

It can be shown that the above is analogous to a series connection of threemechanical springs representing the hydraulic fluid, the entrained air and thecontainer respectively [10].

The bulk modulus of the container is assumed much greater in magnitude thanboth the bulk modulus of the hydraulic fluid and entrained air and thereforehas no significant impact on the effective bulk modulus. In the absence of anyhose connections or alike, which could have the potential effect of lowering theeffective bulk modulus, no attention will thus be paid to mechanical compliancein the model of the system. Based on this assumption, (3.38) is reduced to thefollowing form

1βe

= 1βl

+ VgV

1βg, (3.39)


where the volume of the system is Ve = V since Vδ = 0.

The tangential bulk modulus for liquid βl in (3.39) was stated in section 3.1.2.It may however be justified to simplify the effective bulk modulus by assuming aconstant fluid bulk modulus βl if the region of interest lies in a range relativelyclose to the atmospheric pressure, which is the case for the system being modeled.

To obtain an explicit expression for the bulk modulus of air βg needed in equation(3.39), one may assume that the pressurization and compression of the entrainedair occurs under adiabatic conditions (i.e. implying that no heat transfer occursto or from the surroundings of the gas). For an adiabatic process, the followingholds true

(P + P0)V γ = P0 Vγ

0 , (3.40)where P is the pressure of the gas relative to the atmospheric pressure P0, Vis the volume of the gas, V0 the gas volume at atmospheric pressure and γ theadiabatic index2.

Now, differentiating (3.40) will yield

d((P + P0)V γ) = (P + P0) γ V γ−1 dV + V γdP = 0. (3.41)

The tangential adiabatic bulk modulus for air is found by arranging the abovesuch that

V γdP = −(P + P0) γ V γ−1 dV ⇒ −dVV

= dP

γ(P + P0) , (3.42)

which gives [10]βg = γ (P + P0). (3.43)

The bulk modulus of air linearly increases with pressure and the presence ofentrained air in the hydraulic fluid has a significant impact on the effective bulkmodulus. As a comparison, the adiabatic bulk modulus of air at atmosphericpressure is βg = γP0 ≈ 1.4 × 105 Pa3, which is to be compared with the bulkmodulus of oil which typically lies in the range of 1.5×109 Pa to 1.8×109 Pa. Also,from (3.39) note that the volume of the entrained air in the fluid is an importantfactor which determines the characteristics of the effective bulk modulus in thelower region of the pressure range, and for this reason thorough deaeration ofthe hydraulic fluid is often needed in hydraulic systems. When the pressure isincreased, the bulk modulus of the entrained air in the fluid also increases until theeffective bulk modulus asymptotically approaches that of the fluid bulk modulus.What has not been mentioned is also the fact that the entrained air dissolves into

2 γ = 1.4 for air3 1 atm = 1.01325 bar = 1.01325× 105 Pa


the fluid as pressure is increased. Dissolved air has no significant effect on theeffective bulk modulus and will effectively decrease the volume of entrained airin the liquid [6]. The effect of air dissolving into the fluid and its effect on thebulk modulus will however not be taken into account in the following.

In equation (3.39), the gas volume Vg may be calculated through the relationbetween gas pressure and volume as given in (3.40) and thus

Vg = Vg0

(P0

P0 + P

)− 1γ

. (3.44)

With the bulk modulus of the liquid assumed constant, the volume of the liquidVl can be obtained through integration of (3.26) as

∂P

∂Vl= −βl

Vl⇒ P = −βl ln Vl + C. (3.45)

Exponentiation of both both sides and solving for the liquid volume gives

Vl = Vl0e− Pβl . (3.46)

An explicit expression for the effective bulk modulus is found by solving for βein equation (3.39)

βe = (Vg + Vl) βl βg(Vg + Vl) βg + Vg βl

= βlVg + Vl

Vg + Vl + βlβgVg. (3.47)

Replacing Vg, Vl and βg in (3.47) with (3.44), (3.46) and (3.43) respectively yields

βe = βl

R +(1 + P

P0

) 1γ e− Pβl

R βlγ(P+P0) +

(1 + P

P0

) 1γ e− Pβl

, (3.48)

where R = Vg0Vl0

.

If βl � P , which normally is the case, then e− Pβl ≈ 1 and the above may be

simplified so that

βe = βl

R +(1 + P

P0

) 1γ

R βlγ(P+P0) +

(1 + P

P0

) 1γ

, (3.49)

which corresponds to the model of the effective bulk modulus derived by Cho etal. [4].

The effective bulk modulus as given in (3.49) is plotted in figure 3.5 for R ∈{0.1, 0.5, 1.0} % and βl = 1.8 GPa.


0 1 2 3 4 50

0.5

1

1.5

2

R = 0.1 %

R= 0.5

%

R= 1.0

%

Pressure P [MPa]

Effec

tive

Bulk

Mod

ulusβe

[GPa

]

Figure 3.5: Effective fluid bulk modulus.

3.1.4 Fluid ViscosityIn solid mechanics, shear strain is directly proportional to the shear stress con-sidering a linear elastic deformation of a solid material. This constant of propor-tionality that relates shear strain to the shear stress is called the shear modulusof elasticity. If shear stress is applied to a solid, as illustrated in figure 3.6,the material deforms and remains in this state until the shear stress is removedwhereby the solid elastically regains its original state. This is not replicated by a

Figure 3.6: Shear stress and strain.

fluid however as a constantly applied shear stress will deform a fluid continuously.Consider the situation in figure 3.7 where two plates, both having an area A, areclosely spaced apart at a certain distance h and separated by a homogeneousfluid. For a Newtonian fluid with a parallel flow between the plates (referred toas shear flow or Coutte flow), shear stress τ is proportional to the rate of shearstrain γ in the direction of the fluid flow

τ = µdγ

dt. (3.50)


With a shear stress τ applied on the upper plate at a time instant t = 0 twopoints p1 = (x1, y1) and p2 = (x2, y2) are situated in two different layers y1 6= y2of the fluid between the plates. The shear strain γ equals zero since x1 = x2.Now, at a second time instant t = ∆t, the different layers of fluid have traveleda distance ∆x relative to each other such that the points now have moved inthe x-direction and are equal to p′1 = (x′1, y′1) and p′2 = (x′2, y′2). Note that thevelocity u of each fluid layer differs and varies from u = 0 at the bottom layerto u = U at the top layer beneath the top plate. Assuming that the traveled

Figure 3.7: Shear flow.

distance ∆x is small, the change in shear strain may be expressed as

∆γ = tan(

∆x∆y

)≈ ∆x

∆y , (3.51)

since for small angles θ ≈ 0

tan θ = sin θcos θ ≈ θ. (3.52)

The change in the horizontal direction ∆x equals

∆x = x′2 − x′1 = u2∆t− u1∆t = (u2 − u1) ∆t = ∆u∆t, (3.53)

where u1 and u2 are the velocities of the two different fluid layers. Relating thisto the change in shear strain yields

∆x∆y = ∆u∆t

∆y ≈ ∆γ ⇒ ∆γ∆t ≈

∆u∆y . (3.54)

Taking the limit now givesdγ

dt= du

dy. (3.55)

Hence, the rate of shear strain γ is equal to the gradient of the fluid velocity.

From (3.50) the shear stress may now be calculated as

τ = µdu

dy. (3.56)


The velocity varies linearly from 0 to U between the plates and the velocitygradient therefore equals

du

dy= U

h. (3.57)

Replacing the velocity gradient in (3.56) with (3.57) now yields

τ = µU

h. (3.58)

The force applied on the upper plate causing shear strain to the fluid is propor-tional to the area A of the plate

F = τA = µAU

h. (3.59)

The coefficient of proportionality µ is termed absolute viscosity, or the dynamicviscosity of the fluid. Viscosity describes the internal resistance to flow of a fluidand may be thought of as a measure of fluid friction. A fluid with a low viscositywill flow more smoothly compared to a fluid with a high viscosity, which is truewhen comparing for example water with a typical hydraulic oil, where waterhaving a lower viscosity than oil at a certain temperature exhibits a more fluentcharacteristic. This is seen analytically when rearranging equation (3.56) suchthat

du

dy= 1µτ. (3.60)

An increased viscosity will decrease the velocity gradient and thus yield a flowwhere the velocities of the different fluid layers deviate less than with a lowerviscosity. Dividing the dynamic viscosity with the density of the fluid gives thekinematic viscosity ν

ν = µ

ρ. (3.61)

The properties of a fluid are dependent on conditions such as pressure and tem-perature, which is also the case for fluid viscosity. A varying fluid pressure hasa modest impact on viscosity, whereas changes due temperature variations aresignificant. Pressure and temperature affect the viscosity exponentially accordingto [8]

µ = µ0 eα(P−P0) (3.62)

andµ = µ0 e

−λ(T−T0) (3.63)where µ0 is the viscosity given at a certain temperature T0 and pressure P0. Theviscosity of the fluid is highly dependent on temperature, which is clearly seenin figure 3.8 for the case where ν0 = 46 cSt4 at T0 = 40 °C and ν = 6.8 cSt atT = 100 °C with fluid pressure held at the level of atmospheric pressure. All real

4 1 cSt (centiStoke) = 1× 10−6 m2/s

3.2 Fluid Mechanics 32

0 20 40 60 80 100 1200

50

100

150

Temperature T [◦C]

Kin

emat

icV

iscos

ityν

[cSt

]

Figure 3.8: Kinematic viscosity as a function of temperature.

fluids exhibit resistance to shear stress and an ideal fluid, which lacks any viscousforces, is known as an inviscid fluid. In the modeling of the system, isothermalconditions will be assumed and fluid viscosity µ will thus be considered as beingconstant.

3.2 Fluid MechanicsThe flow of a fluid is elementary to hydraulic systems and basic knowledge of thefundamental laws and equations which govern fluid flow is therefore necessary inorder to understand and be able to analyze and model fluid mechanical systems.This section will briefly present the essential theory and what is needed for themodeling of orifice flow and pressure dynamics within hydraulic volumes, whichwill be needed in later sections of this chapter.

3.2.1 Navier-Stokes EquationsThe governing laws of fluid mechanics are expressed in form of momentum, masscontinuity and energy equations for fluid flow. The Navier-Stokes equations arefundamental partial differential equations that describe the conservation of mo-mentum within a fluid. In differential form, the equations are

ρ

(∂u∂t

+ (u · ∇) u− f)

= −∇p+ µ∇2u, (3.64)

where ρ is the fluid density, u is the velocity vector of the fluid flow, t is time, f isa vector containing all the body forces acting on the fluid, p is the fluid pressure


and µ is the fluid viscosity. If considering the fluid as being incompressibleand assuming isothermal conditions, both the fluid density ρ and the dynamicviscosity µ are constant. A solution of the Navier-Stokes equations will be inform of a flow field describing the velocity of the fluid at a given point in spaceand time.

Reynolds Number

The Navier-Stokes equations may under certain conditions be simplified such thatmore simple analytical solutions can be obtained. Equation (3.64) can be writtenin a non-dimensional form by introducing the following non-dimensional variablesu = uU , f = fF , t = t T , p = p P , ∇ = ∇ 1

Land ∇2 = ∇2 1

L2 , where U , F , T , Pand L represent a characteristic reference of fluid velocity, force, time, pressureand length respectively. Implemented into (3.64), the non-dimensional form ofthe Navier-Stokes equations is given by [10]

ρU

T

∂u∂t

+ ρU2

L(u · ∇) u− ρF f = −P

L∇p+ µU

L2 ∇2u. (3.65)

Fluid flow may be characterized as either being laminar, with well-defined stream-lines of smooth, constant and orderly fluid motion without disruptions, or turbu-lent with a more erratic, swirling and fluctuating type of description. Laminarflow occurs when viscous (frictional) forces are dominant while turbulent flow isdominated by inertial forces. With a non-dimensional form of the Navier-Stokesequations at hand, the relation between the inertial and viscous terms may bequantified by relating the terms of (3.65) to the viscous term (containing thefactor µ) on the right side of the equation. Dividing both sides of the equationwith µU/L2 yields

ρL2

Tµ

∂u∂t

+ Re (u · ∇) u− ρFL2

µUf = −PL

µU∇p+ ∇2u, (3.66)

whereRe = ρUL

µ= UL

ν. (3.67)

In fluid mechanics the Reynolds number, denoted Re, is a dimensionless numberthat describe whether flow conditions lead to laminar or turbulent flow. A largeReynolds number indicate that the inertial terms are dominant, whereas a smallnumber indicate dominance of viscous terms. Laminar flow tends to occur at lowfluid velocities, below a threshold of the Reynolds number at which it becomesturbulent. This corresponds with what is indicated by (3.67). Hence, above acritical Reynolds number a transition from laminar to turbulent flow will occur.However, in practice a transition may take place across a wider range of values.


In the case of a steady flow, where ∂u/∂t = 0, the Navier-Stokes equations arereduced to

ρ(u · ∇) u− µ∇2u− ρf = −∇p. (3.68)For low Reynolds numbers Re� 1, the inertial terms are much smaller than theviscous term and can therefore be ignored. Equation (3.68) then becomes

µ∇2u−∇p+ ρf = 0, (3.69)

which is known as Stokes flow. If body forces are also neglected, (3.69) simplifyto

∇p = µ∇2u. (3.70)If instead Re � 1, the viscous term is much smaller compared to the inertialterms and (3.68) then becomes

ρ(u · ∇) u +∇p− ρf = 0, (3.71)

leaving Euler’s equation of inviscid flow (µ = 0). Neglecting body forces in theabove yields

∇p = −ρ(u · ∇) u. (3.72)Apart from the fluid properties of density ρ and viscosity µ, the definition of theReynolds number also includes the fluid velocity U and a characteristic length L.This dimension is defined differently depending on the type of flow and geometry.As a common example, for a cylindrical pipeline with a circular cross-section thecharacteristic dimension is chosen to be the diameter of the pipe by convention.Hence, for a pipeline the Reynolds number is defined as

Re = ρUDh

µ= UDh

ν. (3.73)

The hydraulic diameter Dh is used for calculation of the Reynolds number incircular as well as non-circular ducts or pipes and is defined such that

Dh = 4AP, (3.74)

where A is the cross sectional area and P is the wetted perimeter (encirclingthe flow) of a duct or pipe. In the case of a circular cross-section the hydraulicdiameter is equal to the geometrical diameter since

Dh = 4 πD2

41πD

= D. (3.75)

For a rectangular duct with dimensions a and b the hydraulic diameter will equal

Dh = 4 ab

2(a+ b) = 2ab(a+ b) . (3.76)


3.2.2 Bernoulli EquationBy assuming an incompressible, inviscid and steady flow of fluid (3.64) may bereduced to the following form of the Euler’s equation

(u · ∇) u− f = −1ρ∇p. (3.77)

The vector algebraic identity

∇(A ·B) = (A · ∇)B + (B · ∇)A + A× (∇×B) + B× (∇×A) (3.78)

may be used to rewrite equation (3.77) such that

(u · ∇)u = 12∇(u · u)− u× (∇× u). (3.79)

The above may be evaluated further to give

12∇u

2 + 1ρ∇p−∇(−gz) = u× (∇× u), (3.80)

where the body forces are gravitational and assumed equal to the gradient of thepotential of gravity. With an irrotational velocity field the curl of the velocity iszero and (3.80) becomes

∇(

12u

2 + 1ρp+ gz

)= ∇

(12ρu

2 + p+ ρgz)

= 0 (3.81)

which implies that12ρu

2 + p+ ρgz = C, (3.82)

where C is a constant. The equation mathematically formulates the Bernoulliprinciple, which states that an increase in fluid velocity simultaneously implies adecrease in fluid pressure and/or potential energy, which is to be considered as astatement of the conservation of energy.

3.2.3 Orifice FlowAn orifice is a form of restriction with the purpose of controlling and restrictingflow and pressure within a fluid mechanical system. In a hydraulic system, flowthrough valves is often characterized as orifice flow and the analytical relationbetween flow and pressure through and across an orifice therefore constitute oneof the most important equations in hydraulic system theory. Flow of fluid througha sharp-edged orifice is depicted by the streamlines as seen in figure 3.9. Due to


Figure 3.9: Flow through an orifice.

the requirement on conservation of mass (with the mass flow rate calculated asρQ) the volumetric flow rate through the orifice must equal

Q = u1A1 = u2A2, (3.83)

where Q is the flow and u1, A1, u2 and A2 are the fluid velocity and stream areaon the upstream and downstream side of the orifice respectively. If an incom-pressible, inviscid and steady flow through the orifice is assumed, the Bernoulliequation (3.82) holds on both the upstream and downstream side of the orifice.Applying (3.82) to both sides of the orifice at A1 and A2 in the same plane andequating the two equations gives

p1 + 12ρu

21 + ρgz1 = p2 + 1

2ρu22 + ρgz2 = C

⇒ p1 − p2 = 12ρ(u2

2 − u21

). (3.84)

The Bernoulli equation as stated in (3.82) is not applicable at the orifice becauseof fluid acceleration, instead u2 and p2 refers to the point of contraction on thedownstream side of the orifice called the vena contracta, where the stream areaA2 < A0. Also note the indicated fluid turbulence on the downstream side of theorifice. Because of the turbulence, this type of flow is termed turbulent orificeflow (as opposed to non-turbulent laminar orifice flow). The fluid velocity u1and u2 in (3.84) may be expressed in terms of the volumetric flow and streamarea through the use of (3.83)

p1 − p2 = ∆p = 12ρ((

Q

A2

)2−(Q

A1

)2). (3.85)

Solving for the flow Q yields

Q = 1√(1A2

)2−(

1A1

)2

√2ρ

∆p = A2√1−

(A2A1

)2

√2ρ

∆p. (3.86)


Now, the jet area at the vena contracta may instead be expressed in terms of theorifice area

A2 = CcA0, (3.87)where Cc is the contraction coefficient. As was mentioned earlier in the text,the area at the vena contracta A2 < A0, and thus Cc < 1. Moreover, in orderto account for discrepancies between the theoretically predicted and the actualjet velocity u2 a velocity coefficient Cv < 1 is needed for correction. Since Q =u2A2 = u2CcA0, complementing (3.86) with the coefficients Cc and Cv gives [11]

Q = Cv CcA0√1− C2

c

(A0A1

)2

√2ρ

∆p. (3.88)

The orifice equation may now be written on the form

Q = CdA

√2ρ

∆p, (3.89)

where Cd is the discharge coefficient of the orifice and

Cd = Cv Cc√1− C2

c

(A0A1

)2. (3.90)

In practice, Cv ≈ 1 and A1 � A0 and thus consequently Cd ≈ Cc. Alternatively,the orifice equation may also be expressed on the form

Q = K√

∆p, (3.91)

where K = CdA√

2ρ.

Laminar - Turbulent Orifice Equation

Equation (3.89) describes the relation between pressure drop across and turbulentflow through an orifice. For a sufficiently low Reynolds number however, the flowwill become laminar with a linear relation between orifice flow and pressure dropand equation (3.89) will in this case therefore not describe the flow through theorifice correctly. Moreover, when the fluid flow Q and pressure drop ∆p approachzero one is likely to encounter numerical difficulties if (3.89) is implemented inits current form since a discontinuity of the derivative exist at the origin

dQ

d∆p

∣∣∣∣∣∆p=0

= CdA0√2ρ∆p

∣∣∣∣∣∆p=0

=∞. (3.92)

It is possible to combine laminar and turbulent orifice flow into a single continuousrelation by allowing the coefficient of discharge to vary with the Reynolds number


[8]. Solving for the pressure drop ∆p as a function of flowQ in (3.89) and replacingthe constant 1/C2

d with the variable ξ gives

∆p = ξ ρ1

2A2Q2. (3.93)

For high Reynolds numbers, equation (3.93) will be equal to (3.89) and thus

limRe→∞

ξ = 1C2d

. (3.94)

At low Reynolds numbers it has empirically been found that the discharge co-efficient is directly proportional to the square root of the Reynolds number [11]which implies that

1√ξ

= δ√

Re ⇒ ξ = 1δ2 Re , (3.95)

where δ is the laminar flow coefficient. With the Reynolds number defined as per(3.73) and the fluid velocity equal to U = Q/A

Re = ρUDh

µ= ρQDh

µA, (3.96)

the orifice flow at low Reynolds numbers will be linear and laminar since using(3.95) in (3.93) will yield

∆p = µ

2δ2DhAQ ⇒ Q = 2δ2DhA

µ∆p = πδ2D3

h

2µ ∆p. (3.97)

The laminar flow coefficient δ depends on both the type of orifice restriction andits geometry.

Now, by defining the coefficient ξ such that [8]

ξ = k1

Re + k2, (3.98)

a laminar-turbulent orifice equation is obtained where k1 = 1/δ2 and k2 = 1/C2d .

The approximated discharge coefficient 1/√ξ is plotted in figure 3.10 for the case

of a sharp edged orifice for which Cd = π/(π + 2) ≈ 0.611 and δ = Cd/√

(20) ≈0.137 [11].

Solving for the orifice flow Q in equation (3.93) gives

Q = πDh

8ρk2

(−k1µ+

√µ2k2

1 + 8k2ρD2h∆p

). (3.99)

The derivative of (3.99) is continuous at ∆p = 0 sincedQ

d∆p

∣∣∣∣∣∆p=0

= πD3h

2µk1= πδ2D3

h

2µ . (3.100)

Note that the tangent of (3.99) at the origin is equal to the coefficient of propor-tionality in (3.97).


0 2 4 6 8 10 12 14 160

0.2

0.4

0.6

√Re

1√ξ

Figure 3.10: Approximated discharge coefficient.

3.2.4 Pressure Dynamics in a Hydraulic VolumeThe fundamental assumption on conservation of mass within a fluid system leadsto another important equation in hydraulic system theory, namely the descriptionof the dynamics of pressure in hydraulic volumes. Consider the flow of fluid intoand out of a volume V as depicted in figure 3.11. By the principle of conservation

Figure 3.11: Hydraulic volume.

of mass, the rate at which mass is accumulated inside the volume must equalthe net difference between the rate of mass flow of incoming flow and outgoing.This is mathematically described by the mass conservation continuity equationin integral form

d

dt

ˆV

ρ dV +ˆS

ρu · dS = 0 (3.101)

⇒ d

dt

ˆV

ρ dV = −ˆS

ρu · dS, (3.102)

where V is the hydraulic volume, S is the boundary surface of the hydraulicvolume, ρ is the density of the fluid and u is the vector field of fluid velocityacross the boundary surface of the volume.

The left hand side of (3.102) equals the rate of change of the fluid mass stored


in the hydraulic volume. With a fluid that is homogeneous and a total volumeof the system known and equal to V , this may be rewritten such that

d

dt

ˆV

ρ dV = d

dt(ρV ). (3.103)

The right hand side of equation (3.102) equals the total net rate of mass flowacross the boundary of the volume V . With a single ended entry of fluid into thehydraulic volume (Qout = 0) and a mean flow Qin = Q, the rate of mass flow intothe volume is calculated asˆ

S

ρu · dS = −ρ QAA = −ρQ, (3.104)

where the mean fluid velocity is equal to Q/A.

Equating (3.103) and (3.104) according to (3.102) and applying the chain rule to(3.103) gives

d

dt(ρV ) = ρ

dV

dt+ V

dρ

dt= ρQ. (3.105)

Dividing with ρ and complementing (3.105) such that the bulk modulus and thetime rate of change of pressure may be identified yields

dV

dt+ V

1ρ

dρ

dp

dp

dt= dV

dt+ V

β

dp

dt

⇒ dV

dt+ V

β

dp

dt= Q. (3.106)

Finally, the pressure dynamics of the volume is given by rearranging the abovesuch that

dp

dt= β

V

(Q− dV

dt

). (3.107)

Now, with the volume V constant the hydraulic volume is effectively identical toa fixed capacitance Ch = V/β in the hydraulic domain since

dp

dt= β

VQ = 1

ChQ

⇒ Q = Chdp

dt, (3.108)

which is analogous with an electrical capacitance

i = Cdu

dt. (3.109)

3.3 Hydraulic Components 41

The transfer function between the volumetric flow rate Q and the pressure P ofthe hydraulic volume V is found by applying the Laplace transform to (3.108)

sP (s) = 1ChQ(s)⇒ G(s) = 1

Ch s, (3.110)

which shows that the pressure of the hydraulic volume is identical to a scaledintegration of the flow input. With a constant hydraulic capacitance Ch, a pureproportional feedback control of the pressure in the hydraulic volume gives thefollowing transfer function of the closed-loop system Gc

Gc(s) = F (s)G(s)1 + F (s)G(s) =

K 1Chs

1 +K 1Chs

⇒ Gc(s) = K

Ch s+K. (3.111)

Applying the final value theorem (which holds) with a step input to the aboveshows that control of pressure in the volume through control of the input flowproves especially simple if a pure proportional control principle is used since thesteady-state error becomes zero

limt→∞

Y (t) = lims→0

s Y (s) = lims→0

sGc(s)1sU = U. (3.112)

The assumption on a constant hydraulic capacitance Ch and fixed volume V isvalid in the mid-region of fluid pressure in the hydraulic system where the brakelinings are still applied against the brake disc and the value of the effective bulkmodulus is close to reaching that of the liquid.

In the lower pressure region the volume is constant but the hydraulic capacitancewill vary with the pressure due to its dependency on the bulk modulus, whichis identical to what is given in (3.49). Hence, the pole of the closed-loop systemwill directly depend on the effective bulk modulus and be equal to

s = −KCh

= −βeK

V. (3.113)

As seen, the bulk modulus has a great impact on system bandwidth and thefraction of entrained air within the fluid should therefore be kept at a minimum.

3.3 Hydraulic Components

3.3.1 Servo ValveThe ability to achieve continuous and stepless control of flow is made possiblethrough the use of a directional valve known as the hydraulic servo valve. This


valve forms the basis of pressure control in the hydraulic braking system andis controlled by closed-loop feedback of the brake pressure through a pressuretransducer connected to the control system. A hydraulic symbol representing theservo valve is shown below. The servo valve has four ports to which connections

Figure 3.12: Servo valve symbol.

to the hydraulic pump, reservoir tank and the actuator are made. The ports Pand T are the pump and tank connectors of the valve, where the pump connectionserves as a pressure source and fluid supply to the brake system and the tankconnection as a return exhaust path for flow of fluid out of the system. The portsA and B are actuator connectors of the valve and are connected to an actuator ofsome kind, such as a hydraulic cylinder or a hydraulic motor. In this applicationthe A port is connected to the brake calipers and the B connector is blocked,hence only the A connection of the valve is used and this due to the fact that thebrake calipers are of a spring-return type and thus have the ability to self-retractwithout any need of flow from the pump.

A cross-sectional view of the servo valve and its internals is found in figure 3.13.Electrically, the control system connects to the valve through a connector (A) with

Figure 3.13: Servo valve.


a signal in form of a current that feeds a so called torque motor (B) that is placedinternally. An illustration of the torque motor is seen in figure 3.14. The torquemotor basically consists of a permanent magnet with a north pole piece (a) and asouth pole piece (b), an armature (c) with two coils (d) attached and a lever (e).When a current is driven through the coils, an electromagnetic force is induced on

Figure 3.14: Torque motor.

the armature which will then pivot about its equilibrium position either clockwiseor counterclockwise depending on the direction of the current driven through thecoils. Hydraulically, the servo valve essentially consists of two spool valve stageswhich indirectly control the direction and magnitude of fluid flow through theservo valve. The connecting ports of the valve are located under the valve body,of which only the two actuator ports (G) and (H) are seen in the figure. Theinternal connections of the supply P and return T are referred to as (I) and (J)respectively. Through the lever (C) the torque motor connects to the pilot spoolstage (D), which is placed inside a hollow main spool stage (E). The main spoolin turn is placed inside the spool sleeve (F). By displacement of the pilot stagefrom its centered equilibrium position by the torque motor, an induced hydraulicforce will move the main stage axially until the forces on the main spool stageagain are balanced. The main stage is indirectly and hydraulically controlledby the pilot spool (hence the term pilot) which is indicated in figure 3.12 bythe triangular shaped symbol seen on the left together with an arrow indicatingelectro-hydraulic control of the valve. An illustration of the main stage spool isshown in figure 3.15. The spool consists of a number of so called control landsthat block oil flow through the valve sleeve, which internally connects to each

Figure 3.15: Spool.


of the four ports P , T , A and B as seen in the figure. In its centered position,flow through the valve is blocked with all the connections to the valve portsin the spool sleeve completely covered, but not overlapped, by the spool lands.However, with displacement of the main spool from its center position, flow offluid is allowed through the valve and can either be in a direction from P to Aand B to T or P to B and A to T . These different spool positions are reflectedin the symbol of the component and are illustrated more clearly in figure 3.16together with actual spool position in the valve. Initially the valve prevents any

Figure 3.16: Spool displacement.

flow in its centered position, which is indicated by the symbol A to the leftwhere all paths are blocked by the spool lands. When a current is driven throughthe coils of the torque motor the pilot spool, and thus also the main spool, willbe shifted axially and allow fluid to flow pass the control lands. Depending onthe direction of the current through the coils, displacement of the main spoolwill be in either direction B or C as shown in the figure and symbols on theleft reflect the allowed path of flow through the valve. With no current presentin the coils of the torque motor the pilot and the main spool will return to thecenter position, which symbolically is indicated by springs on both sides of thevalve symbol in figure 3.12. The term critical center is used to describe the typeof spool that is illustrated above. The lands of the spool may however in somecases also overlap or underlap the ports of the sleeve in which case the spool is


referred to as closed center and open center respectively.

A model of the servo valve may in a simplified form be divided into a static anddynamic part, where the static part relates spool displacement and fluid pressurewith flow of fluid through the valve and the dynamic part corresponds to thedynamic relation between input current and displacement of the actuating mainspool of the valve. Models of both the static and dynamic part of the servo valveare presented in the following.

Static Model

It is important to understand that the servo valve itself does not directly controlpressure or flow, it merely prevents or allows flow in either of the directionsindicated in figure 3.16. Flow through the valve is determined by the pressuredifferential that exists across the valve ports and the opening area between thespool and sleeve of the valve, which is controlled by the supplied electrical currentthrough the torque motor. Flow within the valve is depicted in figure 3.17 witha displacement x of the main spool according to B in figure 3.16. Note thatflow of the fluid through the valve is distributed over a number of different pathslocated along the perimeter of the sleeve and that the flow as designated thereforerepresents the total flow to simplify the analysis. Now, as expected fluid will flow

Figure 3.17: Valve flow.

from P to A and from B to T . However, due to radial clearance between thespool and the sleeve a leakage flow inevitably also exists between the path P toB and A to T . Hence, in practice no path is completely blocked regardless ofspool position. Although small in magnitude, the leakage flow is dominant atsmall displacements of the main spool (typically in the case of feedback controlof pressure when the valve is balancing about a set point at steady state) andmay not be neglected in the null region of the valve (i.e. about the centeredposition of the main spool) and must be accounted for in the analysis of thesystem. The restriction will be in form of an annulus that form between the


spool and sleeve and the internal leakage flow of the valve may be described by theorifice equation. Through fundamental circuit analysis, the following equationsare obtained describing the relations between the supply, return and actuatorport flow

QS = QSA +QSB (3.114a)QR = QRA +QRB (3.114b)QA = QSA −QRA (3.114c)QB = QRB −QSB . (3.114d)

Equating the flow through the actuator ports gives

QA = QB ⇒ QSA +QSB = QRA +QRB ⇒ QS = QR. (3.115)

With a displacement of the main spool according to C in figure 3.16, the actuatorflow QA and QB change directions such that{

QA = QRA −QSA (3.116a)QB = QSB −QRB . (3.116b)

Further to the above, the relation between the pressure drop across the valvechamber and the actuator ports may be formulated as

PS − PR = PSA + PRA = PSB + PRB . (3.117)

The orifices that effectively form between the spool and sleeve of the valve to-gether with what has been stated above may analogously be described as a Wheat-stone bridge, see figure 3.18, to facilitate the understanding and analysis of theservo valve.

Figure 3.18: Wheatstone bridge.

The flow through and pressure drop across the actuator load may be defined as{PL = PA − PB (3.118a)QL = QA = QB, (3.118b)


where PA and PB is the pressure at the A and B actuator ports of the valverespectively.

Flow through the valve will be in form of orifice flow and with an assumption onvalve symmetry, modeling of both the actuator and leakage flow may be combinedto yield the following [5]

QSA = K√PS − PA

x0 + x x ≥ 0x2

0 (x0 − k x )−1 x < 0(3.119a)

QRA = K√PA − PR

x20 (x0 + k x )−1 x ≥ 0x0 − x x < 0

(3.119b)

QSB = K√PS − PB

x20 (x0 + k x )−1 x ≥ 0x0 − x x < 0

(3.119c)

QRB = K√PB − PR

x0 + x x ≥ 0x2

0 (x0 − k x )−1 x < 0.(3.119d)

The coefficient K is defined such that K = Cd√

2ρw, where w is equal to the

area gradient of the valve (for a fixed orifice K is defined as given in section3.2.3). Multiplying the valve area gradient with the valve spool position yieldsthe opening area A of the orifice. With positive displacement of the valve spoolas in B of figure 3.16 and 3.17, direction of flow through the valve and attachedload will be according to what is depicted in figure 3.18. The internal leakagearea of the valve, due to radial clearance between the spool and sleeve, maybe interpreted as a minimum opening x0 of the spool and for an ideal valvex0 = 0 (i.e. free from internal leakage). With a displacement of the spool fromits centered position, the internal leakage will decrease which is reflected in themodel by the inversely proportional terms found in (3.119) that will effectivelydecrease the area of leakage in the valve. The four piecewise defined functionsmay be rewritten to explicitly include the valve area

QSA = 1wK√PS − PA

A0 + (Amax − A0) x x ≥ 0A2

0 (A0 − k (Amax − A0) x )−1 x < 0(3.120a)

QRA = 1wK√PA − PR

A20 (A0 + k (Amax − A0) x )−1 x ≥ 0

A0 − (Amax − A0) x x < 0(3.120b)

QSB = 1wK√PS − PB

A20 (A0 + k (Amax − A0) x )−1 x ≥ 0

A0 − (Amax − A0) x x < 0(3.120c)

QRB = 1wK√PB − PR

A0 + (Amax − A0) x x ≥ 0A2

0 (A0 − k (Amax − A0) x )−1 x < 0,(3.120d)


where the area has been calculated using the area gradient w and the spoolposition has been normalized. With the area gradient w factored out, the valvecoefficient will be 1

wK = Cd

√2ρ. By showing the area gradient equal to

w (x0 + xmax) = w x0 + w xmax = A0

x0x0 + w xmax = Amax

⇒ w = A0

x0= Amax − A0

xmax, (3.121)

equation (3.119a) is modified such that

w (x0 + x) = A0 + Amax − A0

xmaxx =

A0 + (Amax − A0) x

xmax= A0 + (Amax − A0) x (3.122)

and

wx2

0x0 − kx

= w2

w

x20

x0 − kx= A2

0A0 − k (Amax − A0) x , (3.123)

which gives (3.120a). Equations (3.119b) - (3.119d) are modified in the same wayto yield (3.120b) - (3.120d). The position of the spool may thus be defined aseither relative or absolute and may alternatively also be indirectly given as thevalve coil control current. As was discussed in section 3.2.3, the orifice equationshould for numerical reasons be implemented in its laminar-turbulent form asgiven in equation (3.99). In order to obtain the flow Q through each orifice, thehydraulic diameter Dh is needed and may be calculated from the orifice area asgiven in (3.120). Due to the lack of knowledge regarding the exact valve geometry,the hydraulic diameter will be calculated based on circular orifice geometry.

If the pressure ∆p across the valve supply and return port is held constant andneglecting leakage, flow will linearly depend on the spool position since

Q = Cd

√2ρw√

∆p x = K√

∆pw︸︷︷︸constant

x, (3.124)

where a no-load (short-circuit) connection across the actuator ports of the valveis assumed. For an ideal valve, one would thus expect a linear relation betweencurrent and flow with pressure across the valve held constant. Due to saturation,the servo valve characteristic however deviate from the ideal as is shown in figure3.19. The internals of the valve will constitute a restriction in the higher regionsof flow and effectively saturate the flow. If simply adding two orifices on thesupply and return connection of the valve in series with the bridge in figure 3.18,


Figure 3.19: Flow as a function of current.

the graph in figure 3.20 may be plotted to illustrate the deviation of the actualvalve current-flow characteristic from the ideal. Modeling of the servo valve willdespite the nonlinearity of the valve assume a linear current-flow characteristicwith a constant area gradient proportional to the slope of the tangent w ∝ ∂Q/∂ithrough the origin. This is motivated by the fact that the control signal will bekept close to the null-region in the simulation of the system. Another important

−100 100

−100

100

Current I [%]

Flow Q [%]

LinearSaturated

Figure 3.20: Saturation.

characteristic of the servo valve may be obtained by blocking the actuator portsas illustrated in figure 3.21. With the ports blocked, no load is attached to thevalve and QSA = QRA as well as QSB = QRB . Hence, two identical and parallellegs are now connected between the supply and return port of the valve. Byequating (3.119a) with (3.119b) and (3.119c) with (3.119d) one may solve for the


Figure 3.21: Blocking both actuator ports.

actuator port pressure PA and PB respectively such that [5]

PA =fPS+PR

1+f x ≥ 0PS+fPR

1+f x < 0(3.125a)

PB =PS+fPR

1+f x ≥ 0fPS+PR

1+f x < 0(3.125b)

where f is a calculation coefficient and is equal to

f(x) =(

1 + |x|x0

)2 (1 + k

|x|x0

)2

. (3.126)

Plotting actuator port pressure against input current for both port A and B willgive the pressure sensitivity of the servo valve, see figure 3.22a. As expected, thepressure is divided to half of the supply pressure at the centered position x = 0for both ports. When displacing the spool from the centered position the pressurewill either increase or decrease depending on if the displacement is positive ornegative. If the spool is displaced such that x > 0, pressure will increase atport A and decrease at port B and vice versa if x < 0. Notice that in a rangeof displacement less than about 1 % of the rated current, the pressure at theactuator ports shifts from half of the supply pressure to about maximum. Theactual pressure sensitivity curve of the valve is found in figure 3.22b. Also note thehysteretic behavior of the valve which will form a loop shaped current-pressurecurve. The characteristics of the valve might also be affected by a slight overlapor underlap at the centered position. The model of the system will however notimplement any of these nonlinearities.

When connecting a load to the A port of the valve, in this case the brake calipers,the valve and load will be connected according to the configuration illustrated


−2 −1 0 1 20

50

100

Current I [%]

Pres

sureP

[%]

(a) Modeled.

(b) Actual.

Figure 3.22: Pressure Sensitivity.


in figure 3.23. As was discussed in section 3.2.4, the brake calipers may, whenthey are applied against the brake disc, be interpreted as being identical to ahydraulic capacitance. Note well that what is shown in 3.23 is used to illustratean analogy intended to simplify the understanding of the hydraulic circuit. Whencontrolling the hydraulic pressure in the brake system one may with anotheranalogy of an electrical system interpret the servo valve as a being a variableresistance in which flow enters or leaves the hydraulic volume. If the actuatorport pressure PA initially is zero and x = 0, flow to the load QA will be nonzeroand at its maximum when t = 0. As the pressure transiently increase in thehydraulic volume, flow to the load will decrease until an equilibrium is reachedwhere PA = (PS − PR)/2 and QA = 0. At this point the flow through the valvemust be QSA = QRA , and the system will be in a steady state. If displacing thespool from the centered position, the hydraulic pressure will increase or decreaseuntil an equilibrium anew is reached with pressure at the actuator port as givenin figure 3.22a.

With the actuator ports blocked as shown in 3.21, the valve internal leakagewill be equal to the supply flow QS. By substituting the expressions of (3.119)into (3.114a) - (3.114d) and calculating the supply flow as QS = QSA +QSB thefollowing is obtained [5]

QS = 2K√PS − PR

x0 + |x|√1 + f(x)

. (3.127)

The valve internal leakage is plotted against the valve displacement x and shownin figure 3.24. Leakage will be at its maximum in the spool centered positionwhere x = 0 and decrease when the spool is displaced such that x 6= 0 in eitherdirection.

Flow through the valve will create forces on both spool stages, see figure 3.25.

Figure 3.23: Load attachment.


−4 −2 0 2 40

0.1

0.2

0.3

0.4

Current I [%]

Flow

Q[l/

min

]

Figure 3.24: Internal valve leakage.

The reactive force due to the acceleration of fluid through the orifice between themain spool and sleeve will exert a force F on the spool in both the radial andaxial direction at steady-state flow

F = Fx i + Fy j. (3.128)

The magnitude of the components of the force depends on the jet angle θ, whichwill vary with displacement of the spool. The jet angle is θ = 21° for smalldisplacements and will asymptotically approach θ = 69° for large displacements[11]. Due to the symmetry of the valve, the lateral component is compensated and

Figure 3.25: Flow forces.

will not affect the spool significantly. The axial forces on the spool will howevertend to close the valve and effectively function as a return spring on the spool.Displacement of the main spool stage is carried out indirectly and hydraulicallythrough the pilot stage, which is controlled by the torque motor. The magnitudeof the flow forces on the pilot stage is small and will not be included in the modelof the valve.

3.4 Mechanical Components 54

Dynamic Model

In order to accurately model the dynamic characteristics of the valve, modelingof the torque motor and spool dynamics is needed (see e.g. [11]). Detailedmodeling will require knowledge of parameters not readily available, and becauseof the difficulties a simplified and general black-box model approach is preferred.A second-order model approximation of the servo-vale dynamics of the followingform is suggested [8]

1ω2

0x+ 2 ζ

ω0x+ x = Ku, (3.129)

where x is the spool position, ω0 is the valve natural frequency, ζ is the valvedamping coefficient and K the valve gain. The model can be modified to alsoinclude valve hysteresis and spool velocity limitation due to viscous friction, thiswill however be left out. The current input u will control the position x of thespool and, as was mentioned earlier in the text, the static part of the model thenrelates the spool position to the flow of hydraulic fluid through the valve.

3.4 Mechanical Components

3.4.1 Brake CaliperThe disc brake calipers consist of a number of different mechanical parts that froma modeling point of view together form a quite complex object. Simplificationis therefore needed in order to obtain a model with a reasonable amount ofdynamic states that captures the essential characteristics of the brake calipers.However, before a more simple model may be formulated, insight into the detailedfunctionality of the brake caliper and its individual parts is nevertheless neededin order to first gain fundamental understanding. A cross-sectional top view andan exploded isometric view of a caliper half is shown in figure 3.26a and 3.26brespectively. Each caliper half essentially consists of five different main parts andthese are the (A) yoke, (B) hydraulic unit, (C) disc springs, (D) brake shoe and(E) brake lining. Each individual part is described in more detail in the following.

Yoke

The yoke of the caliper is split into two halves and is fitted onto the brake standwith one half on each side of the stand. The two pieces are then bolted togetherto form a complete brake caliper. The yoke serves as a caliper housing and alsoneed to withstand axial and radial forces of considerable magnitude. Clampingforce applied on both surfaces of the brake disc will yield an axial deflection ofthe yoke. Deflection will decrease the maximum clamping force, and thus also theresultant brake force, and must therefore be limited. Apart from axial clamping


(a) Cross-sectional view.

(b) Exploded view.

Figure 3.26: Caliper halve.

forces on both sides of the brake disc, the yoke is also subjected to radial forcesfrom the brake disc and gravitational forces due to the own weight of the caliper.From a modeling point a view, where clamping force of the brake caliper is ofinterest first and foremost, only the yoke deflection due to axial forces needs to betaken into account. In a model of the complete caliper, which will be presentedlater in the section, the yoke axial deflection is modeled with a mass and springelement as shown in figure 3.27.

Figure 3.27: Model of the caliper yoke.


Hydraulic Unit

Hydraulic pressure is connected to the brake caliper through ports on the con-nector plate (A) of the hydraulic unit, see figure 3.28a. Three out of the four

(a) Front view. (b) Cross-sectional view.

Figure 3.28: Hydraulic unit.

connector ports P , A, and B function as pressure ports (B) and all connecthydraulically to the piston (E) through the cylinder chamber (D) inside the hy-draulic unit. The fourth connector port R is a drain port (C) which removeshydraulic fluid from the caliper in case of any internal leakage between the cylin-der chamber and the piston seals (F) and (G) (internal paths between the cylinderchamber and drain port not shown). The connector plate is mounted onto thedisc spring support piece (H) through a set of screws (I). The hydraulic unitis then fixed to the yoke of the caliper through an adjustment screw (J) whichis placed between the connector plate and the support piece. The adjustmentscrew enables an axial positioning of the hydraulic unit relative to the caliperyoke through manual adjustment to both set the gap between the brake liningand brake disc initially and to compensate for wear of the brake lining whenneeded. In order to accurately model the hydraulic unit a total of four massesis needed, see figure 3.29. First, the piston is allowed to move freely inside thehydraulic unit and is therefore represented by a single isolated mass. However, ascan be seen in figure 3.28b, the connector plate acts as a rear stop for the pistonand a gap and spring must therefore be added in the model to account for the endof travel of the piston inside the hydraulic unit. The connector plate, the supportpiece and the adjustment screw are each represented by individual masses in themodel. A flexible interconnection is added between the connector plate and thesupport piece in form of a spring element, this due to the forces that are exertedon the connector plate by the piston at the rear end of the hydraulic unit at fullrelease of the brake lining. Backlash between the adjustment screw, the support


Figure 3.29: Model of the hydraulic unit.

piece and connector plate exists and needs to be incorporated into the model.This is added through a gap and spring on both sides of the mass representingthe adjustment screw. Finally, two forces are indicated to act on both the pistonand the support piece (with differing area of fluid contact) due to the appliedhydraulic pressure.

Brake Shoe

The brake shoe (A) (see figure 3.30) holds the brake lining (D) in place, which isclamped and secured by a set of screws (C) on both sides. A connection of thepiston (E) to the center rod (B) transfers the force generated by the hydraulicpressure to the brake shoe. This force in turn acts on the disc springs (also

Figure 3.30: Brake shoe.

called cup springs or Belleville springs), which are placed between the brakeshoe and the hydraulic unit (see figure 3.26). The disc springs are compressedto an initial deflection by the nut (F), which essentially holds the brake shoe,


center rod, disc spring and piston assembly together. The force generated by thehydraulic pressure counteracts the clamping force generated by the disc springsand when the force on the piston equals the force on the brake shoe the pressureis said to balance the clamping force, in which case no force is applied on thebrake disc. If the hydraulic pressure is increased beyond the release pressure thebrakes will release and a gap between the brake disc and the brake lining willform. This is illustrated in figure 3.31. Maximum displacement of the brake

Figure 3.31: Brake release.

shoe is achieved when the piston reaches its end of travel at the rear stop in thehydraulic unit, i.e. when the piston has reached the connector plate. The end oftravel of the piston occurs below the maximum working pressure, but increasingthe hydraulic pressure further will only displace the piston insignificantly. Withthe brake linings fully released, gap adjustment is performed if needed duringcommissioning or maintenance. This is needed due to the fact that wear ofthe brake linings will eventually affect the clamping force. Since brake force isindirectly controlled by the hydraulic pressure it is important that the gap liesin a close range of what is nominally intended. An inductive proximity sensor istherefore installed in each caliper half on the back cover of the connector plate(see figure 3.26b) to monitor the gap by measuring the distance between theback cover and the center rod. By manually turning the adjustment screw thegap between the brake lining and brake disc is adjusted to normally lie in therange of 1-3 mm, see figure 3.32. Now, although it may not be obvious, the gap

Figure 3.32: Gap adjustment.

adjustment effectively controls the maximum clamping force achieved with fullapplication of the brakes onto the brake disc at atmospheric pressure. In figure3.32, the maximum clamping force for each of the three cases of gap adjustmentwill be increasing from the left example of adjustment of the gap to the example


on the right. This since the spring deflection at full application, and thus themaximum clamping force, depends on the adjustment of the gap with the brakesfully released. From a modeling point of view, the brake shoe and piston assemblymay be seen as a single mass and thus identical to the rightmost mass seen inthe model depiction of the hydraulic unit in figure 3.29. The model of the brakeshoe is shown in the figure below. The brake lining attached to the brake shoe is

Figure 3.33: Model of the brake shoe.

compressible and therefore need to be modeled with a spring element. Hydraulicpressure applied to the surface of the piston in the hydraulic unit generates aforce on the piston and brake shoe assembly and therefore acts on the massrepresenting the whole assembly in the model.

Brake Caliper Model

From what has been discussed and presented earlier in the section, a completemodel of the brake caliper may now be formed. Interconnecting the model of thecaliper yoke, brake shoe and the hydraulic unit will give the following mechanicalmodel

Figure 3.34: Brake caliper model.

where the five different masses m1,..,5 correspond to the brake shoe, hydraulicunit, connector plate, adjustment screw and yoke respectively. The springs k1and k2 represent the brake lining and the disc springs and k3,..,9 flexible intercon-nections and hard stops of high stiffness. Two rigid points of reference (a) and


(h) are placed at the brake disc and brake stand. The gap (b) is the gap betweenthe brake lining and brake disc that form when the brake lining is released fromthe brake disc and (c) corresponds to the distance between the piston and theconnector plate. Backlash between the adjustment screw and hydraulic unit andbetween the threads of the yoke and the adjustment screw is represented by thegaps (d), (e), (f) and (g). Two forces F1 and F2 are applied on the piston andthe hydraulic unit by the hydraulic pressure that acts in both directions. Nodissipative restrictions in form of frictional elements (e.g. between the piston andthe seals in the hydraulic unit) or dampers have been added.

The brake caliper operates in one of three different discrete states which dependon the applied hydraulic pressure. Without any force applied on the piston thebrake lining is applied against the brake disc with a fully developed clampingforce and the gap (b) is closed. With the brake in its applied state, the gap (c)is open and the piston is positioned at a maximum distance from the connectorplate. The disc springs are pushing the hydraulic unit against the adjustmentscrew such that the gap (e) and (f) are closed and the gap (d) and (g) are open.The force exerted on the adjustment screw is transferred to the caliper yoke whichwill be deflected due to the torque that is created by the shear force from thedisc springs. When hydraulic pressure is applied on the piston and the supportpiece, the disc springs will compress and the brake lining will decompress. Sincethe force applied on the piston will counteract the force that is applied on thebrake shoe by the disc springs, the clamping force on the brake disc will decrease.Increasing the hydraulic pressure will also decrease the deflection of the caliperyoke due to the supporting hydraulic force on the hydraulic unit. If the hydraulicpressure is increased to the caliper release pressure, the force of the disc springswill be balanced by the force on the piston and brake shoe assembly. At thispoint no clamping force is applied on the brake disc by the brake lining. If thehydraulic pressure is increased beyond the release pressure, the brake lining willbe released from the brake disc and the gap (b) between the brake disc and thebrake lining will be nonzero. Due to the backlash, the hydraulic unit will bepushed inward by the hydraulic pressure whereby the gap (e) and (f) will openand the gap (d) and (g) will close. The force on the hydraulic unit will thustransfer to the connector plate which is pressed onto the adjustment screw andthe caliper yoke. If the pressure is increased until the piston reaches the connec-tor plate, the gap (c) will close and the gap (b) will obtain its maximum. At thispoint the brake is fully released from the brake disc. At the rear stop of pistontravel the force on the piston will counteract the force applied on the hydraulicunit, adjustment screw, connector plate and caliper yoke.

The current form of the brake caliper model is overly complex for dynamic mod-eling and therefore needs to be simplified. However, the model is suitable fordetailed static modeling which will offer great insight into the fundamental func-


tionality of the caliper. The static analysis of the complete brake caliper modelwill not be presented here.

3.4.2 Disc SpringsDisc springs are stacked back-to-back in series as shown in figure 3.35 to achievethe caliper clamping force that is needed. Through the use of disc springs large

Figure 3.35: Spring stack.

forces are obtained for small deflections. Force is exerted on both the brakeshoe and the hydraulic unit as shown in the figure. The relation between springdeflection and disc spring force is nonlinear and is described by the followingequation for a single disc [1]

F = 4E t4(1− µ2)αD2

o

s

t

((h0

t− s

t

)(h0

t− 1

2s

t

)+ 1

), (3.130)

where s is the spring deflection, E is the Young’s modulus and µ the Poisson’sratio for the disc spring material used, Di and Do the inner and outer diameterof the disc, t is the material thickness and L0 and h0 is the unloaded inner andouter height of the disc respectively. Definitions are given in figure 3.36.

Figure 3.36: Disc spring.

The calculation coefficient α in equation (3.130) is defined in the following way

α = 1π

(δ − 1δ

)2 (δ + 1δ − 1 −

2ln δ

)−1

, (3.131)


where δ is the ratio between the inner and outer diameter of the disc

δ = Do

Di

. (3.132)

Equation (3.130) gives the force generated by a single disc. To obtain the forcegenerated by a number of identical discs stacked in series as shown in 3.35, thetotal deflection of the stack is divided by the number of discs N to give thedeflection of each individual disc

s = 1N

N∑i=1

si. (3.133)

Hence, calculation of the total disc spring force may be carried out with thedeflection of each individual disc spring. The total deflection of the stack ofsprings will then simply be equal to the individual disc spring deflection multipliedwith the number of disc springs in the spring stack. As indicated in (3.130), therelation between disc spring force F and deflection s is nonlinear and thereforemay generally not be linearly approximated. A plot of the characteristic for thedisc springs used in the calipers however reveals a close to linear relation betweenspring force and deflection as seen in figure 3.37. One may linearize about a

0 5 10 15 200

20

40

60

80

100

Deflection s [mm]

Forc

eF

[kN

]

Figure 3.37: Stack deflection and force.

working point by calculating the slope of a tangent to (3.130) or use a secantbetween two points on the curve of 3.37.

The derivative of (3.130) with respect to spring deflection is

k = dF

ds= 4E t4

2 t3 (1− µ2)αD2o

(3s2 − 6h0s+ 2(t2 + h2

0)). (3.134)


The spring rate of a spring stack with a deflection-force characteristic as per 3.37is calculated with (3.134) to yield the spring rate shown in figure 3.38. Despite

0 5 10 15 20

4.6

4.8

5

5.2

5.4

Deflection s [mm]

Sprin

gR

atek

[MN/m

]

Figure 3.38: Stack deflection and spring rate.

the possible linearization of the force characteristic, (3.130) will be used to modelthe force generated by the disc springs.

Friction

Frictional forces arise during compression and decompression of the spring stackbecause of internal friction due to elastic deformation of the material, contactbetween the discs and the guide rod and at the edges of the spring stack [12], seefigure 3.39. Thus, when compressing the spring stack additional hydraulic forceis needed in order to also overcome frictional forces. At decompression of the discsprings the frictional forces will act in the opposite direction, and the force on thespring stack must be lowered to a level below that of the spring force includingthe added resisting frictional forces in order to decompress the discs. Friction willthus have a hysteretic effect on the control of the brake calipers. With frictionacting against the applied hydraulic pressure, this will imply that release of thebrakes will occur at a pressure higher than expected and that the level of pressurewhen applying the brakes will be lower than without the contribution of frictionalforces. Calculation of the total spring force, including edge and surface frictionalforces between individual discs when loading or unloading the springs, may becarried out as per DIN5 2092, where a percentile addition of frictional force tothe calculated spring force is made

F ± Ff = n

1∓ wM(n− 1)∓ wRF, (3.135)

5DIN - Deutsches Institut für Normung


Figure 3.39: Friction during compression.

where n is the number of discs in parallel, wR is the coefficient of edge frictionand wM is the coefficient of disc surface friction. For series stacking of the discsn = 1 and the friction force may therefore be calculated as

±Ff =( 1

1∓ wR− 1

)F = ± wR

1∓ wRF. (3.136)

A friction factor µ can be defined such that

µ = wR1∓ wR

, (3.137)

with the disc spring force F as a normal force acting perpendicular to the surfaceof contact between the springs and the abutment structure. The standard DIN2093 divides disc springs into three different categories depending on the ratiobetween the outer diameter Do and the thickness t of the disc. For a series C disc,which is the category of springs used in the brake calipers, the coefficient of edgefriction according to the standard is stated to lie in the range wR ∈ [0.03, 0.05].Frictional forces due to edge friction is thus estimated to equal about 3-5 % ofthe spring force and is the dominant of the resistant frictional forces in the brakecaliper.

3.4.3 Brake LiningWhen clamping force is applied on the brake disc by the calipers, the brake liningsare compressed. The resulting deformation, due to the applied compressive force,is not negligible and thus must be accounted for in the modeling of the mechanicalparts of the system. A characteristic is available in form of a stress-strain diagramfor the linings in use [3] and the relationship between stress and strain is shownin figure 3.40. The stress-strain diagram is initially nonlinear and progressive inthe elastic region before reaching a linear relation between stress and strain. For


Figure 3.40: Stress-strain characteristic.

a sufficiently large stress the material will reach its yield strength, which impliesa transition from the elastic region to the region of plastic deformation (notmodeled). From the data available, the Young’s modulus E may be determinedas the slope of the tangent to the stress-strain curve of the brake linings

E = lim∆ε→0

∆σ∆ε = dσ

dε. (3.138)

In the model of the system, the brake lining is represented by a spring elementwith stiffness k (see figure 3.34). Now, the stress-strain characteristic is materialspecific for the brake linings used whereas the stiffness depends on the dimensionsof the linings. However, by assuming that strain is defined as the negative relativechange in lining height L (engineering strain) and noting that

E = dσ

dε= d(F/A)d(−∆L/L0) = − dF

d∆L ·L0

A= k

L0

A, (3.139)

where A and L0 are the area and the uncompressed height of the brake liningsrespectively, the spring stiffness k can be calculated through the following relation

k = EA

L0. (3.140)

This may intuitively be more well understood if one adopts the view of the brakelining as consisting of distributed infinitesimal spring elements connected both inseries and in parallel. From the empirical data available, spline curve fitting maybe employed to obtain numerical expressions of the stress-strain characteristic infigure 3.40. A plot of the sampled data points from the stress-strain diagram to-gether with the fitted spline curve is shown in figure 3.41a. The Young’s modulus


0.00 1.00 2.00 3.00 4.000

1000

2000

3000

4000

5000

Strain ε [%]

Stre

ssσ

[kPa

]

Empirical DataCurve Fit

(a) Empirical stress-strain and curve fit.

0.00 1.00 2.00 3.00 4.000.00

0.05

0.10

0.15

0.20

Strain ε [%]

Youn

g’s

Mod

ulusE

[GPa

] ApproximationModel

(b) Young’s Modulus.

0.00 1.00 2.00 3.00 4.000

1000

2000

3000

4000

5000

Strain ε [%]

Stre

ssσ

[kPa

]

Empirical DataCurve FitModel

(c) Empirical, curve fit and modeled stress-strain.

Figure 3.41: Stress-strain characteristic and Young’s Modulus.


E of the brake lining may be approximated through numerical derivation of thefitted characteristic and is shown in figure 3.41b. The Young’s modulus showsexponential increase in the lower region of strain before asymptotically reachinga maximum and constant value in the elastic region. An empirical model of theYoung’s modulus may be represented by a piecewise defined function

E =0 ε ≤ 0E0 + (Emax − E0)

(1− e−ε/τ

)ε > 0,

(3.141)

where E0 is the Young’s modulus at zero strain, Emax is the maximum valueof the Young’s modulus in the elastic region and τ is a factor of exponentialdecay. Since no tensile forces are present, the Young’s modulus is non-zero onlyfor positive strain. A plot of the model is found in figure 3.41b along with theestimated Young’s modulus. Through integration of (3.141) the relation betweenstress and strain is obtained

σ =0 ε ≤ 0E0 ε+ (Emax − E0)

(ε− τ(1− e−ε/τ )

)ε > 0.

(3.142)

A plot of the available empirical data together with the spline fitted curve andmodeled data is shown in figure 3.41c. Using (3.140), the spring stiffness k mayfinally be expressed as

k = A

L0·

0 ε ≤ 0E0 + (Emax − E0)

(1− e−ε/τ

)ε > 0.

(3.143)

3.4.4 Simplified Brake Caliper ModelA simplification of the caliper model is needed in order to minimize the numberof dynamic states and reduce the overall complexity. By neglecting backlashand assuming rigidity of the hydraulic unit, the adjustment screw, the connectorplate and the caliper yoke assembly the model of the caliper will reduce to thesingle mass mechanical model shown in figure 3.42. The three distinct states ofoperation of the caliper described in section 3.4.1 are clearly illustrated in thefigure. A frictional element has been added because of the significant frictionthat arises due to the use of disc springs in the caliper as was discussed in section3.4.2. Applying Newton’s second law of motion yields the following second orderdifferential equation

mx = −F ± Ff − F1 + F2 + F3, (3.144)

where m is the mass of the brake shoe and piston assembly, F is the force appliedon the piston by the hydraulic pressure p, F1 is the force from the brake shoeand brake lining (see section 3.4.3), F2 is the force generated by the disc springs


(a) Brake applied.

(b) Brake released.

(c) Brake fully released.

Figure 3.42: Simplified caliper model.

(see section 3.4.2), F3 the force applied on the piston at the rear stop by theconnector plate and Ff represents the frictional forces. The force applied on thecylinder piston by the hydraulic pressure is F = pA, where A is the piston area.The forces F1 and F2 are calculated through the nonlinear functions given insection 3.4.2 and 3.4.3 and the force F3 generated by the caliper rear stop maybe represented by a linear relation between spring deflection and force. Brakecaliper frictional force Ff , as discussed in section 3.4.2, is added or subtracteddepending on if the brakes are released or applied (i.e. if the brake pressure isincreased or decreased).

Linear Analysis

To simplify an analysis of the caliper model, a linear model of (3.144) may beobtained by neglecting friction and assuming linear models of the springs

mx = −F − k1(x10 + x) + k2(x20 − x), (3.145)

where x = 0 corresponds to the position where the brake lining is fully appliedagainst the brake disc as per 3.42a and x10 and x20 are the initial deflections of


the springs. A static analysis of the caliper is of interest, therefore mx = 0

−F − k1(x10 + x) + k2(x20 − x) = 0. (3.146)

With the brake applied and x = 0 when p = 0 equilibrium is required and thus

−k1x10 + k2x20 = 0⇒ k1x10 = k2x20 = F0, (3.147)

where F0 corresponds to the maximum applied clamping force produced by thecaliper disc springs. With the brake applied onto the brake disc, i.e. when−x10 < x ≤ 0, equation (3.146) will reduce to

−F − k1x− k2x = 0⇒ F = −(k1 + k2)x. (3.148)

Replacing the force applied by the hydraulic pressure on the piston as equal toF = pA and solving for the position x yields

x = − A

k1 + k2p. (3.149)

When releasing the brake as shown in figure 3.42b, no clamping force is appliedby the brake lining on the brake disc. Removing the term which represents theforce from the brake lining in (3.146) will give

−F + F0 − k2x = 0⇒ F = F0 − k2x. (3.150)

Solving for the position −L < x ≤ −x10 gives

x = −F − F0

k2= −A

k2(p− p0), (3.151)

where L−x10 is the length of travel of the caliper piston equal to the distance ofthe gap between the brake disc and lining at the rear stop in the hydraulic unitand p0 is the hydraulic pressure level corresponding to the maximum clampingforce F0. The intersection of (3.149) and (3.151) is obtained by equating andsolving for the pressure p

− A

k1 + k2p = −A

k2(p− p0)⇒ p =

(1 + k2

k1

)p0, (3.152)

which is referred to as the brake release pressure or balancing pressure sinceat the release of the brake lining from the brake disc the force applied by thehydraulic pressure is balancing the force generated by the disc springs. For an


incompressible brake lining limk1→∞ p = p0 and in practice p ≈ p0 since k1 � k2.Substituting the pressure p in both (3.149) and (3.151) with (3.152) gives x =−F0/k1 = −x10 as defined in (3.147).

At full release of the brake, i.e. where x ≤ −L, a force F3 is added to accountfor the rear stop of the caliper piston

−F + F0 − k2x+ k3(−L− x) = 0. (3.153)

Solving for x in the above now yields

x = − 1k2 + k3

(F − F0 + k3L) = − A

k2 + k3

(p− p0 + k3L

A

). (3.154)

If k3 � k2 and F − F0 � k2 + k3 the above may be rewritten to show that

x = −F − F0

k2 + k3− k3L

k2 + k3≈ − k3L

k2 + k3≈ −L, (3.155)

which indicates a hard stop at x = −L.

The linear static model of the brake caliper may be summarized in a piecewisedefined function

x =

− Ak1+k2

p 0 ≤ p ≤ pr

− Ak2

(p− p0) pr < p ≤ pf

− Ak2+k3

(p− p0 + k3AL) p > pf ,

(3.156)

where pr = p0(1 + k2/k1) is the release pressure and pf = p0 + k2A/L is thepressure at full release of the brakes. A plot of the function is shown in figure3.43a for the case where k1 = 30 MN/m, k2 = 5 MN/m, k3 = 10 GN/m andF0 = 80 kN. As seen in the figure, below the release pressure pr at about 10 MPathe brake lining and disc springs will compress and decompress respectively. Oncethe brake lining is released from the brake disc, the rate of change in positionwill increase. It is important to understand that potential energy is stored in thecaliper disc springs when increasing the pressure in the system and that the powerrequired to open the calipers at a certain rate will increase throughout the wholerange above release of the brake lining from the brake disc. This since poweris calculated as P = ∆pQ, where ∆p is the pressure drop across the hydraulicpump (which will increase as the caliper pressure is increasing) and Q is the flowof fluid into the calipers.

Nonlinearity

A nonlinear static analysis of the calipers can be carried out in much the same wayas before, but with the nonlinear relations between disc spring deflection and force


according to (3.130) as well as between stress and strain given in (3.142) insteadof the linear spring models used previously. Leaving out friction in (3.144), thecondition at equilibrium is given by

−F − F1 + F2 + F3 = 0. (3.157)

By replacing the terms above the equation may be rewritten

− pA− Al σ1

(x10 + x

tl

)+ F2(x20 − x) + k3 max(0,−(x+ L)) = 0, (3.158)

where Al and tl is the brake surface area and thickness of the brake lining. Theinitial strain x10 and deflection x20 of the brake lining and disc springs are solvedfor by letting x = 0 and p = 0 which gives

−Al σ1(x01) + F2(x02) = 0⇒ Al σ1(x01) = F2(x02) = F0. (3.159)

Equation (3.158) is solved numerically to qualitatively compare the linear andnonlinear static characteristic of the caliper. The position as a function of thehydraulic pressure is found in figure 3.43b. The essential difference between thelinear and nonlinear characteristic in the static model of the caliper lies in thelower region below the release pressure where the brake lining characteristic isnonlinear. Friction may also be added in a static analysis such that

− pA− Al σ1

(x10 + x

tl

)+ F2(x20 − x)

±min( pA, µF2(x20 − x)) + k3 max(0,−(x+ L) = 0. (3.160)

When increasing the brake pressure with initial full application of the brakelinings, this will imply that the hydraulic pressure must overcome the frictionforce of the discs springs before any movement of the caliper brake shoe is possible.Plotting the full characteristic for both release and application of the brakes showsthe hysteretic effect of disc spring friction and that the actual release pressure prof the brakes due to friction is increased above what was stated earlier. A pressurelevel of brake application when applying the brakes from a released state mayalso be defined and is denoted pa, where pa < pr.


0 2 4 6 8 10 12 14

−3

−2

−1

0

Pressure p [MPa]

Posit

ionx

[mm

]

(a) Linear characteristic.

0 2 4 6 8 10 12 14

−3

−2

−1

0

Pressure p [MPa]

Posit

ionx

[mm

]

(b) Nonlinear characteristic.

0 2 4 6 8 10 12 14

−3

−2

−1

0

Pressure p [MPa]

Posit

ionx

[mm

]

Without FrictionReleaseApplication

(c) Nonlinear characteristics with friction.

Figure 3.43: Brake caliper static characteristics.

3.5 Brake Stand 73

3.5 Brake StandThe calipers of the hydraulic brake system are mounted on a brake stand andconnected in parallel. Since the hydraulic transmission lines are left out in themodel of the system, a simplification is justified as the calipers may be seen aslumped into a single equivalent caliper unit. Referring to figure 3.44, the disc

Figure 3.44: Brake calipers.

springs are effectively connected in parallel and the total clamping force Ftotalapplied on the brake disc thus equals

Ftotal =2N∑i=1

FCi = 2N × FC , (3.161)

where N is the number of calipers and FC is the clamping force generated by asingle caliper half. The hydraulic pressure is applied on each piston and the totalpiston area Atotal may thus in the same fashion be calculated as

Atotal =2N∑i=1

Ai = 2N × A, (3.162)

where A is the piston area in each caliper half. From a modeling point, thiswill have no significance regarding calculation of caliper disc spring force sincethere is a constant proportion between clamping force and piston area. However,calculation of the total volume of the hydraulic system must take into accountthe number of calipers as this will affect the pressure dynamics and release time.

The total volume of fluid in the system will increase when the brake pressure isincreased and the change in volume will equal ∆V = −2NA∆x where N is thenumber of brake calipers, A is the piston area and x is the position of the pistonin each caliper as defined in figure 3.42. Initially the position of the piston will bex = 0 with all brake linings fully applied onto the brake disc and the fluid volumeof the system will be equal to V = V0. With change of position, the volume maybe calculated as

V = V0 − 2NAx. (3.163)The time rate of change in system volume will equal

dV

dt= −2NA dx

dt. (3.164)

3.6 Dynamic Model 74

Substituting (3.163) and (3.164) into (3.107) will yield the pressure dynamics inthe system as

dp

dt= βeV0 − 2NAx

(Q+ 2NA dx

dt

). (3.165)

At a pressure level p below the release pressure pr of the brakes, and with a flowQ into the volume equal to the fluid displacement of the hydraulic pump, thechange in position will be small and pressure will rise quite rapidly. Once thebrake linings are released from the brake disc, increase in the system volume willhowever lower the time rate of change of the brake pressure as is seen in equation(3.165).

An equivalent lumped single brake caliper must also take into account the totalmoving mass mtotal, which simply will equal the sum of the mass m of the movingparts of each caliper half

mtotal =2N∑i=1

mi = 2N ×m. (3.166)

3.6 Dynamic ModelIn this section a complete dynamic model is presented along with a number ofstatic functions that has been presented throughout the text. The dynamic modelof the system consists of a system of nonlinear differential equations including anumber of different static functions, see equations (3.167a) - (3.167e) and equa-tions (3.167f) - (3.167r). The model has five different dynamic states xi wherex1 represents the spool position, x2 spool velocity, x3 brake pressure, x4 brakecaliper piston position and x5 piston velocity. Equation (3.167a) integrates thespool velocity x2 to obtain the spool position x1 and equation (3.167b) describesthe dynamics between the valve input current u and the main spool position x1,as presented in section 3.3.1. The servo valve has both a mechanical and electri-cal offset which effectively will shift the pressure sensitivity curve with respect tothe input current depending on the sign and magnitude of the offset. An offsetu0 has therefore been added to the input to account for both mechanical andelectrical offset of the spool position at zero input u = 0. Without any offset, thepressure will ideally be equal to PS/2 (which may adjusted mechanically by shift-ing the valve sleeve). The dynamics of the brake pressure is given by equation(3.167c), which was presented in section 3.2.4 and section 3.5. Equation (3.167d)integrates piston velocity to obtain piston displacement and equation (3.167e) cal-culates piston acceleration as was presented in section 3.4.4. The force from thebrake lining, the disc springs, the rear hard stop and the hydraulic pressure aresummed to calculate the resultant accelerating force acting on the mass as seenin equation (3.167e). Disc spring friction is not included in the dynamic model


of the system. Equation (3.167f) calculates the effective bulk modulus as givenin section 3.1.3. Flow through the valve is calculated with equations (3.167g) -(3.167o), which were given in section 3.2.3, 3.3.1 and 3.2.1. Calculation of theforce from the brake lining is carried out through the empiric stress-strain func-tion in (3.167p) as given in section 3.4.3 multiplied with the lining width andheight, which yields the lining area in contact with the brake disc and thus theresultant force. Equations (3.167q) - (3.167r) calculates the force from the discsprings and were presented in section 3.4.2. Dampers are not included, but aresimple to add in equation (3.167e) if needed.


x1 = x2 (3.167a)x2 = −ω2

0 x1 − 2 ζ ω0 x2 + ω20 (u+ u0) (3.167b)

x3 = βe(x3)V0 − 2NAx4

(Q(x1, x3) + 2NAx5 ) (3.167c)

x4 = x5 (3.167d)

x5 = 12Nm

(−Ax3 − F1

(x10 + 1

tlx4

)(3.167e)

+ F2(x20 − x4) + max(

0,−K (x4 + L)))

βe(P ) = βl

R +(1 + P

P0

) 1γ

R βlγ(P+P0) +

(1 + P

P0

) 1γ

(3.167f)

Q(x, p) = QS(x, p)−QR(x, p) (3.167g)

QS(x, p) = πDhS(x)8ρk2

(−k1µ+

√µ2k2

1 + 8k2ρD2hS

(x)(PS − p))

(3.167h)

QR(x, p) = πDhR(x)8ρk2

(−k1µ+

√µ2k2

1 + 8k2ρD2hR

(x)(p− PR))

(3.167i)

k1 = 1/δ2 (3.167j)k2 = 1/C2

d (3.167k)

DhS(x) =√

4AS(x)π

(3.167l)

DhR(x) =√

4AR(x)π

(3.167m)

AS(x) = w

x0 + x x ≥ 0x2

0 (x0 − kx)−1 x < 0(3.167n)

AR(x) = w

x0 − x x ≥ 0x2

0 (x0 + kx)−1 x < 0(3.167o)

F1(ε) = wl hl

0 ε < 0E0 ε+ (Emax − E0)

(ε− τ(1− e−ε/τ )

)ε ≥ 0

(3.167p)

F2(s) = 4E t4(1− µ2)αD2

o

s

t

((h0

t− s

t

)(h0

t− 1

2s

t

)+ 1

)(3.167q)

α = 1π

(δ − 1δ

)2 (δ + 1δ − 1 −

2ln δ

)−1

(3.167r)

Chapter 4

Model Parameterisation,Simulation and Validation

In the previous chapter, models of the different components of the system wereobtained through physical modeling and a complete model of the system wasgiven in form of a system of differential equations along with a number of staticfunctions. Parameterisation of the model is required in order to be able to simu-late the system and will be presented in this chapter. Parameters may either begathered directly from component specifications and other known data or indi-rectly through calculations and measurements. Comparisons between measureddata against static and dynamic model computations and simulation will also bemade to verify the correctness of the model.

4.1 Model Parameters

4.1.1 Fluid PropertiesMost of the information needed regarding the properties of the hydraulic fluidis either stated by the supplier of the fluid or given as tabulated values. Bothdensity ρ and kinematic viscosity ν are found in the product data sheet of the oiland are given in table 4.1. However, data regarding the liquid bulk modulus βlis not readily available and will be assumed equal to 1.8 GPa (see section 3.1.2).The total volume of fluid in the system V0 may be estimated by calculationof volume in the pipelines connected between the valve manifold and the brakestand. Pockets of oil also exist in the piston chamber of each caliper, the manifoldand ball valves which gives a small discrepancy between the actual oil volumeand the calculated. The atmospheric pressure P0 and the adiabatic index γ areboth given as known constants and the tank return pressure PR is assumed equalto zero (gauge pressure). Regarding supply pressure PS, this will be taken fromavailable measurements of the accumulator pressure. The fraction of entrained air

4.1 Model Parameters 78

Table 4.1: Fluid Properties

Parameter Symbol Unit ValueFluid Density ρ kg/m3 876 at 15 °C

Fluid Bulk Modulus βl GPa 1.8Kinematic Viscosity ν cSt 46 at 40 °C

cSt 6.8 at 100 °CFraction of Air R %Adiabatic Index γ 1.4

Atmospheric Pressure P0 bar 1.01325Supply Pressure PS barReturn Pressure PR bar 0.0

Volume V0 liter 4.44

R may be identified in different ways, but will in this case be roughly estimatedby curve fitting to measured data.

4.1.2 Brake CalipersData needed for the overall parameterisation of the brake caliper model is listedin table 4.2

Table 4.2: Brake Caliper Parameters

Parameter Symbol Unit ValueMass m kg 43.0

Stroke L mm 2.0Piston Area A cm2 74.6

Maximum Clamping Force F0 kN 75.6Number of Calipers N 4

Rear Stop Spring Rate K GN/m 10.0

Two single unit calipers and a double unit caliper is installed on the brake stand,which is equivalent to a total of N = 4 calipers. Mass of the individual parts ineach brake caliper halve is specified in the installation and maintenance manualavailable from the manufacturer, and the total moving mass m is obtained bysummation. Through direct measurements of the inner and outer diameter ofthe caliper piston, the area A is calculated as

A = π

4(D2o −D2

i

)= π

4(122 − 72

)= 74.6 cm2, (4.1)

which is confirmed by the manufacturer data where piston area is specified to be75 cm2 in each brake caliper halve. Force transducer measurements of the clamp-ing force F0 gives a measured maximum force of about 76 kN at full application


of the brakes. The spring rate of the caliper rear stop is set to a large value, inthis case 10 GN/m.

Brake Linings

Dimensions of the width wl, height hl and thickness tl of the brake linings aregiven in table 4.3. As mentioned in the previous section, brake lining force iscalculated from the lining area wl × hl applied against the brake disc and strainof the lining is obtained using the thickness of the lining according to what isgiven in (3.158). Through curve fitting of the static stress-strain function givenin section 3.4.3 against the available empirical data, values of the maximum andminimum Young’s modulus Emax and E0 and the exponential decay τ may beestimated.

Table 4.3: Brake Lining Parameters

Parameter Symbol Unit ValueYoung’s Modulus E0 GPa 0.01

Emax GPa 0.15Decay τ 0.08Width wl mm 300.0Height hl mm 202.5

Thickness tl mm 25.0

Disc Springs

All information that is needed for model parameterisation of the disc springsare either provided directly by the manufacturer or found available in form oftabulated material specific data. The dimensions of the individual disc springsas defined in figure 3.36 are given in table 4.4 below. The number of disc springs

Table 4.4: Disc Spring Parameters

Parameter Symbol Unit ValueYoung’s Modulus E GPa 206Poisson’s Ratio ν 0.3Outer Diameter Do mm 125Inner Diameter Di mm 64

Inner Height h0 mm 2.6Thickness t mm 8.0

Number of Discs N 9

N in each spring stack is specified by the manufacturer and confirmed through


visual inspection in figure 3.26. The Young’s modulus E and Poisson’s ratio νare both given through readily available spring material data.

4.1.3 Servo ValveData needed for parameterisation of the servo valve model can to some extentbe derived from what is found available in the valve datasheet. Other values areeither indirectly obtained from measurements or approximated. Rated values ofvalve pressure, flow and current are given in table 4.5. Rated flow is measuredunder a no-load condition with a short-circuit connection between the actuatorports at maximum spool actuation, which corresponds to rated current input tothe valve. Pressure ∆P = PS − PL − PR = PS − PR is held constant across thesupply and return line connections (see figure 3.17 and 3.18), thus the pressuredrop across each orifice will ideally equal (PS−PR)/2. By convention the pressureacross the supply and return line is chosen to be 70 bar (about 1000 psi1) whichyields a pressure drop ∆P = 35 bar across the valve orifices. Referring to figure

Table 4.5: Valve Rating

Parameter Symbol Unit ValueRated Pressure Prated bar 210

Rated Flow Qrated l/min 40Rated Current Irated mA 300

3.19, a linearization of the current-flow graph is desired. By extrapolating thetangent of the curve at the origin to the point on the graph where Q = Qrated,a linearized rated current I∗rated is obtained (i.e. the current at which rated flowis obtained without saturation, see the example depicted in figure 3.20). Thelinearized rated current I∗rated < Irated is estimated from the current-flow graphand I∗rated ≈ 249 mA. A linearized rated flow Q∗rated may also be defined such that

Q∗ratedIrated

= Qrated

I∗rated. (4.2)

Calculation of Q∗rated now yields

Q∗rated = 0.3000.249 × 40 ≈ 48.2 l/min. (4.3)

From the above, and by using the orifice equation, the maximum orifice openingarea Amax may be calculated as

Amax =√

2× 0.300/0.249× 40/600000.63×

√2

876 ×√

70 · 105≈ 14.3 mm2. (4.4)

1 1 bar = 14.5037738 pounds/inch2


Since Amax � A0, and thus w ≈ Amax/xmax, the area gradient w may be calcu-lated from the slope of the tangent through the origin of the current-flow char-acteristics

w ≈√

2Qrated

Cd√

2ρ

√∆p I∗rated

, (4.5)

which yields

w ≈√

2× 40/600000.63×

√2

876 ×√

70 · 105 × 0.249≈ 47.5 mm2/A. (4.6)

However, a normalized spool position is used in the model which implies thatthe gradient must be scaled with the rated current to obtain the normalized areagradient such that

w ≈√

2× 40/600000.63×

√2

876 ×√

70 · 105 × 0.249/0.300≈ 14.3 mm2. (4.7)

Thus, the maximum opening area of each orifice is equal to about 14.3 mm2, aswas calculated in (4.4).

The minimum spool opening x0 may be calculated by solving for x0 with x = 0in equation (3.127) [5]

QS0 = 2K√PS − PR

x0√1 + f(0)

⇒ x0 = QS0√2K√PS − PR

, (4.8)

where the flow gain K is equal to

K =√

2Qrated√70 · 105 I∗rated

. (4.9)

The total maximum valve leakage flow (quiescent flow), including both the flowthrough the pilot and main spool stage, is stated by the manufacturer to be lessthan 5 % of the rated flow at rated pressure. The valve leakage flow is thusdistributed between the pilot and main spool stage. With lack of data regardingthe main stage spool leakage, QS0 represents a degree of freedom in the modelof the servo valve. Other ways of obtaining x0 exists however. If x = x0 onemay solve for the leakage coefficient k in equation (3.125). Evaluating (3.126) forx = x0 gives

f(x0) =(

1 + |x0|x0

)2 (1 + k

|x0|x0

)2

= 4 (1 + k)2. (4.10)


Solving for k with the above yields

k = 12

√√√√ PA(x0)PS − PA(x0) − 1 = 1

2

√√√√ PA(x0)/PS1− PA(x0)/PS

− 1, (4.11)

where PA(x0)/PS is obtained from the valve pressure sensitivity graph, see figure3.22b. Note that in order to obtain a feasible solution k > 0, it is required that

PA(x0)/PS1− PA(x0)/PS

> 4⇒ PA(x0)/PS >45 = 0.80, (4.12)

which numerically will constrain the minimum spool opening x0 and in turn alsothe main spool leakage flow QS0 . From the pressure sensitivity graph one seesthat x0 > 0.006. Calculating the corresponding null main spool leakage flow gives

QS0 = 2√PratedQrated√

70 · 105 I∗ratedx0

= 2×√

210 · 105 × 40/60000√70 · 105 × 0.249

× 0.006× 0.300 ≈ 1.00 l/min, (4.13)

which is half of the maximum total quiescent flow. With the actuator portsblocked as illustrated in figure 3.21, the orifice leakage area may be calculated as

A0 ≈√

2× 0.5× 1.00/600000.63×

√2

876 ×√

210 · 105≈ 0.086 mm2. (4.14)

Referring to equation (3.121), the minimum orifice area A0 may also be calculatedfrom the area gradient and the minimum spool opening since

w = A0

x0⇒ A0 = wx0, (4.15)

which givesA0 ≈ 14.3× 0.006 ≈ 0.086 mm2. (4.16)

The maximum attained value of the minimum spool opening x0 may be calculatedfor the case where the main spool stage alone consumes the whole quiescent flowat rated pressure

x0 =√

70 · 105 × 0.2492×√

210 · 105 × 40/60000× 0.05× 40/60000 ≈ 3.59 mA. (4.17)

Calculating the relative spool opening gives 3.59/300 ≈ 0.012. The nominalworking pressure of the system is 145 bar and thus well below that of the ratedpressure. A lower working pressure will decrease the quiescent leakage flow in the


main spool stage and calculating the flow QS0 for the case where PS = 145 barand x0 = 0.006 yields

QS0 = 2×√

145 · 105 × 40/60000√70 · 105 × 0.249

× 0.006× 0.300 ≈ 0.83 l/min. (4.18)

The calculations presented correspond to parameters valid for a newly manufac-tured valve. The control lands of the valve spool will however wear with time andincrease the null leakage and thus also the minimum spool opening x0. Withoutavailable measurements of the center flow one may instead estimate x0 throughcurve fitting of the pressure sensitivity graph against measurements of equilib-rium pressure obtained with open-loop current input to the servo valve, see figure4.1a and 4.1b. By plotting the relative input current and pressure (the rated cur-rent and supply pressure is used to normalize the current and brake pressure),adjustments of x0, PA(x0) and u0 are made to fit the pressure sensitivity graphagainst the measured data as seen in figure 4.1c. By adjustment of the param-eters, the minimum spool opening is found to be approximately x0 ≈ 0.0165,which is higher than what was calculated as the expected maximum accordingto the data sheet. This may however be explained by the inevitable wear of thevalve spool which will increase x0 with time. Valve port pressure PA(x0) withthe actuator ports blocked or under equilibrium conditions (which determines theleakage coefficient k), is set to PA(x0)/PS = PA(x0) ≈ 0.82. The offset u0 simplyshifts the curve to adapt for mechanical and electrical offset, hence u0 = u0m+u0e ,and is found to be u0 ≈ −0.02. The valve frequency response from which thenatural frequency ω0 and damping ζ may be estimated is provided by the valvemanufacturer. The valve coefficient of discharge Cd and laminar flow coefficientδ are given typical values. It should be noted that the rated values providedby the manufacturer are rounded and therefore approximate. Nevertheless, thecalculated values serve an indicative purpose.

Table 4.6: Servo Valve Parameters

Parameter Symbol Unit ValueNatural Frequency ω0 rad/s 160

Damping ζ 0.6Laminar Flow Coefficient δ 0.10

Discharge Coefficient Cd 0.63Area Gradient w mm2 14.3

Minimum Spool Opening x0 0.0165Normalized Pressure Sensitivity PA(x0) 0.82

at Minimum Spool OpeningSpool offset u0 -0.02


0 20 40 60 80 100 120−15

−10

−5

0

5

Time t [s]

Cur

renti

[%]

(a) Signal input.

0 20 40 60 80 100 120 1400

20

40

60

80

100

Time t [s]

Pres

surep

[bar

]

(b) Step response.

−6 −4 −2 0 2 4 6 8 100

20

40

60

80

100

Current i [%]

Pres

surep

[%]

(c) Pressure sensitivity.

Figure 4.1: Measurements and pressure sensitivity.

4.2 Static Characteristics 85

4.2 Static CharacteristicsThe static characteristic of the brake caliper in figure 3.43b is obtained by simul-taneously measuring both the brake pressure and displacement of the center rodas a function of time. By slowly varying the pressure linearly with the brakesapplied and increasing the pressure until the brakes are fully released, the char-acteristics can be plotted in a xy-graph as shown in figure 4.2a. The positionof the center rod is measured using an inductive proximity sensor, which is po-sitioned a few millimeters from the rod in the fully applied state of the brakecaliper. Hence, when the brake pressure is at its minimum level the measureddistance between the sensor and the center rod will be at its maximum (whichin the model of the system is used as a position of reference for which x = 0).The brake pressure varies from about 20 bar2 to the maximum working pressureof about 145 bar. When increasing the hydraulic pressure the brake lining willdecompress and the distance between the center rod and the sensor will decrease.Note that nonlinearities exist in the range above the release pressure which is notreflected by the linear model of the brake caliper due to the simplification of themodel as was discussed in section 3.4.1 and 3.4.4. Hysteretic effects of frictionon the static characteristics of the brake caliper may also be measured by slowlydecreasing the brake pressure from the fully released state and maximum workingpressure to the pressure at the minimum level. This will yield the hysteresis loopof the static characteristics as seen in figure 4.2b.

4.3 Disc Spring ForceThe calculated disc spring force is confirmed correct when compared to availablemanufacturer clamping force data as tabulated in table 4.7

Table 4.7: Clamping Force and Air Gap Adjustment

Air Gap Unit Force Unit1 mm 80100 N2 mm 75500 N3 mm 70900 N

The air gap corresponds to the gap between the brake lining and the brake discat full release of the brakes (see figure 3.32). Manufacturer data in form of aninspection certificate specifies a force of 80100 N at a stack length of 79.1 mm.With an unloaded spring height of L0 = h0 + t = 2.6 mm + 8.0 mm = 10.6 mmper disc, the total stack length equals L = N × L0 = 9 × 10.6 mm = 95.4 mm.

2 Brake pressure and center rod displacement measured using a different system with alower pressure limitation set to 20 bar.

4.3 Disc Spring Force 86

0 20 40 60 80 100 120 140

4

4.5

5

5.5

Pressure p [bar]

Posit

ionx

[mm

]

(a) Static characteristics.

0 20 40 60 80 100 120 140

4

4.5

5

5.5

Pressure p [bar]

Posit

ionx

[mm

]

(b) Hysteresis.

Figure 4.2: Measured characteristics.

4.3 Disc Spring Force 87

Thus, at a stack deflection of s = 95.4 mm − 79.1 mm = 16.3 mm the calculatedclamping force should be equal to about 80100 N. This is confirmed by calculatingthe disc spring force with equation (3.130), where the deflection of each singledisc in the spring stack is equal to 16.3/9 ≈ 1.81111 mm. Calculation of δ and αgives

δ = 0.1250.064 = 1.953125 (4.19)

and

α = 1π

(δ−1δ

)2

δ+1δ−1 −

2ln δ≈ 0.684474. (4.20)

The coefficient is now calculated as4× 2.06 · 1011 × (8 · 10−3)4

(1− 0.32)× α× (125 · 10−3)2 ≈ 3.467919 · 105, (4.21)

which yields a spring force equal to

F ≈ 3.467919 · 105× (4.22)

1.8111118.0

(( 2.68.0 −

1.8111118.0

)( 2.68.0 −

12 ·

1.8111118.0

)+ 1

)≈ 3.467919 · 105 × 0.231117 ≈ 80150 N.

For a gap of 2 mm between the lining and brake disc at full release of the brakes,the deflection of the spring stack will be equal to 15.3 mm with maximum caliperclamping force. The spring force is calculated in the same way with a springdeflection of each disc equal to 15.3/9 = 1.7 mm to give

F ≈ 3.467919 · 105× (4.23)

1.7000008.0

(( 2.68.0 −

1.7000008.0

)( 2.68.0 −

12 ·

1.7000008.0

)+ 1

)≈ 3.467919 · 105 × 0.217729 ≈ 75506 N.

At a gap of 3 mm, the deflection of the spring stack will equal 14.3 mm with max-imum clamping force. Spring deflection of each spring will now equal 14.3/9 =1.588889 mm, which gives

F ≈ 3.467919 · 105× (4.24)

1.5888898.0

(( 2.68.0 −

1.5888898.0

)( 2.68.0 −

12 ·

1.5888898.0

)+ 1

)≈ 3.467919 · 105 × 0.204277 ≈ 70842 N.

4.4 Pressure Dynamics 88

4.4 Pressure DynamicsIt was concluded in section 3.2.4 that the calipers, when applied against the brakedisc, were identical to a hydraulic capacitance and that the time rate of change ofthe system pressure therefore should be proportional to the input of flow to thesystem (i.e. the system is integrating, see equation (3.108)). This is confirmedwhen applying a sinusoidal input current signal to the servo valve as is shown infigure 4.3a. The working point is placed in the upper region of the pressure rangebelow the release pressure pr (at about 100 bar) where the effective bulk modulusis close to that of the fluid and the amplitude of the current signal is chosen tolimit the peak-to-peak pressure response such that the changes in the pressuredrop ∆p is moderate. As seen in figure 4.3a, the pressure lags the input signalby 90° (the input current signal is scaled to the same amplitude and centered atthe peak-to-peak midpoint of the pressure response for phase comparison). Theintegrating nature of the system is also confirmed when measuring the responseto a triangular and square wave input as in figure 4.3b and 4.3c.

4.4.1 SimulationThe cyclic input signals confirms the integrating character of the system in steadystate. By input of a step-shaped open-loop signal to the servo valve, the transientbehavior of the system is measured, as was shown in figure 4.1b. The dynamicmodel given in (3.167a) - (3.167e) together with the static functions (3.167f) -(3.167r) is simulated with an input signal identical to the one shown in figure 4.4a,and the simulated pressure response along with actual measured data and supplypressure (dashed) are plotted in figure 4.4b. It must be stressed that this is farfrom a normal operating mode of the valve and used here only for the purpose ofvalidating the dynamic model. Apart from the hysteresis, the qualitative behaviorof the actual system is captured quite well by the simulation. Quantitatively thepressure sensitivity function has not been optimally fitted to the measured data,which also explains deviations between the actual and simulated system. Theeffects of entrained air is shown close to t = 0 s, which is also matched by themodel.


0 0.2 0.4 0.6 0.8

60

70

80

90

100

Time t [s]

Pres

surep

[bar

]

SignalPressure

(a) Sinusoidal response.

0 0.2 0.4 0.6 0.850

60

70

80

Time t [s]

Pres

surep

[bar

]

SignalPressure

(b) Triangular response.

0 0.2 0.4 0.6 0.8

10

20

30

40

50

60

Time t [s]

Pres

surep

[bar

]

SignalPressure

(c) Square response.

Figure 4.3: Pressure response with different signal inputs.


0 20 40 60 80 100 120 140−15

−10

−5

0

5

Time t [s]

Cur

renti

[%]

(a) Signal input.

0 20 40 60 80 100 120 1400

50

100

150

Time t [s]

Pres

surep

[bar

]

MeasurementSimulation

(b) Actual and simulated pressure response.

Figure 4.4: Simulation.

Chapter 5

Discussion and Conclusions

The work presented in this report covers a quite broad range of fluid mechanicaland mechanical physical modeling. Although the scope of the modeling of thesystem was limited to essentially only include the servo valve and brake calipermodels, the complete model yet attains an appreciable number of differentialequations and static functions. From a perspective of model complexity, this ofcourse speaks in favor of computer aided component based modeling where initialconfiguration and changes in the model, if needed, are more or less trivial. Thiswill especially facilitate the modeling of large complex systems such as the com-plete configuration of the hydraulic brake system. One may of course also claimthis way of modeling to be less prone to error. However, a trade-off betweensimplifying the process of system modeling and insight to and understandingof the underlying analytical relations exists. In the case of component basedphysical modeling, thorough documentation of each individual model of courseis necessary. Yet, one lacks analytical completeness of an interconnected systemand is therefore dependent on numerical analysis in order to analyze system dy-namics. The same argument may be used when comparing so called white-boxphysical modeling with black-box modeling. Another advantage of a physicallybased model is of course also the ability to extrapolate and make predictions ofthe effects of changes to the model parameters (e.g. pipe length, volume, massetc). This is not be replicated by a black-box model due to its numerical nature.However, even if an accurate, more detailed and complex physically based modelis readily available, model parameterisation will prove to be difficult if data isnot found a priori. Even if parameters are at hand, the resulting model may notqualitatively or quantitatively outweigh the benefits of a more simple model, as isthe case of the model of the servo valve dynamics presented in section 3.3.1 or thecomplete brake caliper model in section 3.4.1 compared to the simplified modelthat was presented in section 3.4.4. This is also the case for the calculation of discspring force given in section 3.4.2, where a secant based simple linear model maybe used instead. Hydraulic systems are challenging in the sense of fluid depen-dency on temperature, pressure and entrained air but also from the perspective

92

of mathematical complexity due to a vast number of nonlinear relations betweenfluid pressure and flow, fluid pressure and effective bulk modulus etc. A lackof knowledge regarding exact valve geometry may also be problematic. For thisreason numerous simplifications and assumptions are needed in order to be ableto handle the modeling of a hydraulic system rationally and examples of this hasbeen shown throughout the text. The model of the system presented in this workis by no means complete and much work still remains regarding validation of thedynamic model. The aim has been to capture the fundamentals of the systemregarding feedback control of the brake pressure, this in order to obtain an ana-lytical model that can be used for e.g. closed-loop control analysis. Many of thenonlinearities such as valve hysteresis, saturation, effects of under- or overlappingspool geometry, flow forces, velocity limitations and brake caliper frictional forceshave been intentionally left out in order not to make the model overly complex. Inits current form, the model is limited to the lower region of the frequency domainonly. The inclusion of transmission lines is therefore required to complementthe model and modeling of the piping network may be carried out through theprinciples of modal approximation techniques for hydraulic transmission lines.In practice, dynamics will differ and vary depending on actual installations. Itis therefore of interest to predict resonance frequencies, bandwidth and also therequired duration of initial brake application at emergency stops of a mine hoistto be able to give guidelines and recommendations regarding pipe installationlength and pipe dimensions. Assumption on an ideal pressure source in form ofa constant supply pressure is not correct, and modeling of the accumulator andvariable displacement pump connected to the servo valve is needed to includethe effects of a varying system supply pressure. A more complex and near fullscale model with inclusion of auxiliary components into the model should prefer-ably be implemented in a component based modeling environment due to thereasons discussed earlier. Static modeling of the mechanical parts of the systemin form of disc spring force and brake lining stress-strain characteristics wereverified quantitatively correct against available data. More detailed insight intoand understanding of the disc spring frictional forces, such as the frictional forcebetween the discs and the guide rod, are however needed. A complete model ofthe brake caliper was presented in section 3.4.1, and as mentioned before a staticanalysis of the caliper may be carried out in order to gain detail understanding ofthe caliper functionality. Static analysis of the simplified caliper model was alsoverified to be essentially correct against actual measurements of brake calipercenter rod displacement and brake pressure. Measurements taken with cyclicinputs in form of sine, triangular and square shaped signals clearly showed theintegrating character of the system in the brake applied state. Regarding param-eterisation of the model, most fluid data is normally available through supplierdata or given as known constants. The fraction of fluid entrained air is howevera common unknown which needs to be identified. Other mechanical data is alsoto a large extent available or easily measured.

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