Teacher: Prof. Veljko Potkonjak - suetf.orgsuetf.org/materijali/os3hps/HPS Skripta.pdf · 6 2.4....

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ME in actuator technology Course title: Principles designing hydraulic servoactuator systems Code: 521

Teacher: Prof. Veljko Potkonjak

Abstract. Principles of hydraulic systems. Actuators. Hydraulic cylinder with piston. Rotary actuator. Mathematical models of actuator dynamics. Electrohydraulic servovalves – principles and mathematics. Permanent-magnet torque motor. Single-stage electrohydraulic servovalve. Two-stage electrohydraulic servovalve with direct feedback. Two-stage electrohydraulic servovalve with force feedback. Specification, selection and use of servovalves. Mathematical modeling. Mathematical model of the complete system. Linearization of the 5-th order model. Reduction of the system (to 3-rd order form). Linearization of the 3-rd order model. Nonlinearities. Saturation. Deadband. Backlash and hysteresis. Friction. etc. Closed-loop control of electrohydraulic system. Simulation. Simulation model. Simulation in system design. Literature: H. E. Merit, Hydraulic Control Systems, John Wiley & Sons, New York

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1. INTRODUCTION

Advantages and Disadvantages of Hydraulic Systems ADVANTAGES:

- No heating problems ... the fluid carries away the heat ...

- Lubrification ... - No saturation ... - Fast response ... fast start/stop ... high torque-to-

inertia ratio => high accelerations ... - All working modes ... continuous, intermittent,

reversing, ... - High stiffness ... little drop in speed as loads are

applied ... - Open and closed loop control ... - Other aspects ...

DISADVANTAGES:

- Power not so readily available ... - High costs for small tolerances ... - Upper temperature limit ... fire danger ; messy due

to leakage - Fluid contamination ... dirt in fluid (contamination)

is chief source of hydraulic control failure ... - Complex modeling ... very often the design is not

based on a sophisticated mathematical model ... - Inappropriate for low power ...

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2. HYDRAULIC FLUIDS (LIQUIDS, OIL) NOT GAS!

2.1. Density

)(

)()(

volumeV

weightGdensityweight ,

typically 33 /......../03.0 mNinlb

)(

)()(

volumeV

massmdensitymass ,

typically 3424 /......../sec1078.0 mkginlb

g )/81.9( 2smg (2.1)

2.2. Equation of State ● Expression that relates density ρ (or volume V), pressure P , and

temperature T . Volume (and density) changes little. So, a linear approximation is justified:

)()( 000 TTT

PPP PT

(2.2)

or

))()(1

1( 000 TTPP

(2.3)

where

4

TT

V

PV

P

00 ,

PP T

V

VT

00

11

(2.4), (2.5)

β – isotermal bulk modulus (compressibility). IMPORTANT!

- It relates to the stiffness of the liquid (a kind of a sping effect). - It have in important influence to the precision of hydraulic

actuator. - It is desired to be as high as possible. - Presence of air (gas) in the liquid, even small, decreases sharply

the bulk modulus.

▪ ρ and β depend on the temperature:

2.3. Viscosity ● It expresses the internal friction of the liquid and its resistance

to shear. ▪ Necessary for lubrification. ▪ If too low leakage! ▪ If too large power loss due to friction (lower efficiency)!

ρ lnβ

T T

5

♦ Friction force is proporional to the contact area A and to the velocity

x , and inversly proportional to the film thickness rC :

xC

DL

C

xAF

rr

, μ – absolute viscosity (coeff. of visc.) (2.7)

v – kinematic viscosity (2.8)

▪ μ depends on the temperature:

μ = μ0 e – λ (T - T0) (2.9)

leakage

leakage of liquid motion x ,

velocity x

Cr – radial clearance

F

L

D

Piston in a cylinder

resistive friction force

Fig. 2.2

μ

T

6

2.4. Thermal Properties ► Specific heat is the amount of energy (heat) needed to raise the

temperature by 10. ► Thermal conductivity is the measure of the rate of heat flow

through an area for a temperature gradient in the direction of heat flow.

2.5. Effective Bulk Modulus ♦ Interaction of the spring effect of a liquid and the masses of

mechanical parts gives a resonance in nearly all hydrauilic components.

▪ The bulk modulus can be lowered by intruducing

- mechanical compliance and/or - air compliance.

▫ For instance:

- the container can be flexible (mechanical compliance), and/or

- bubbles or pocket of gas are present inside (gas compliance).

(see Fig. 2.4)

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▫ The expression for the effective (total) bulk modulus βe can be

found in the form:

)11

(111

lgt

g

lce V

V

+ (2.20)

where: βc – the bulk modulus for the container, βl – for the liquid, βg – for the gas; Vg – the volume of the gas, and Vt – the total volume. Since gl , (2.20) becomes:

)1

(111

gt

g

lce V

V

(2.21)

▫ If there is no gas (so, only mechanical compliance), one obtains:

lce

111 (2.22)

ΔVt ΔVc

liquid, volume Vl

gas pocket, volume Vg

liquid

ΔVg

gas

Fig. 2.4

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2.6. Chemical and Related Properties - Lubricity - Thermal stability - Oxidative stability - Hydrolytic stability - Compatibility - Foaming - Flash point, fire point, autogenous ignition temperature - Pour point - handling properties (toxity, color, odor, ...) 2.7. Types of Hydraulic Fluids

► Petroleum based fluids, and ► Synthetic fluids ♦ Characteristics

2.8. Selection of the Hydraulic Fluid

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3. FLUID (LIQUID) FLOW FUNDAMENTALS It is assumed that the general theory of fluid flow is elaborated in the previous courses. Among numerous problems, we highlight here the topic: 3.4. Flow Through Orifices – Turbulent Flow

0

2

A

ACc – contraction coefficient (3.28)

▪ Let: u – fluid velocity, P – pressure . We apply:

- Bernulli’s equation )(2

2121

22 PPuu

(3.29)

- Equation of incompressibility 332211 uAuAuA (3.30)

- Volumetric flow rate (the flow) 22uAQ - Contaction coefficient (3.28) 02 / AACc

- velocity coefficient 98.0vC (sometimes adopted 1vC ) (velocity is slightly smaller due to friction)

A2 , jet area is minimum jet area A0

vena contracta – the jet area is minimimum

Fig. 3.10. 1 2 3

10

and we obtain

)(2

210 PPACQ d (3.33)

where

210

2 )/(1 AAC

CCC

c

cvd

(3.34)

is the discharge coefficient. Since 1vC and 10 AA it follows that cd CC .

If 10 AA , the theoretical value for the the discharge coefficient for all sharp-edged orifices, regardless of the geometry, is 6.0611.0)2/( cC .

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4. HYDRAULIC PUMPS and MOTORS ♦ Conversion of energy:

◦ Pump: mechanical energy hydraulic energy

◦ Motor (actuator): hydraulic energy mechanical energy

our primary interest hydrodynamic machines (turbines, etc.) ♦ Hydraulic machines positive displacement mach.! limited travel machines ♦ continuous travel machines rotary machines ♦ piston machines (translation)

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♦ Piston actuator (cylinder with a piston) – limited travel mach.

cylinder

xp

piston position

fluid IN fluid OUT

fluid IN : pressure P1

fluid OUT : pressure P2

Single rod actuator

Double rod actuator

forward chamber

backward chamber

motion

motion

pressure force

load force

piston

Fig. 4.1

piston parameters:

Mt – mass of the piston plus refered masses

Ap – effective piston area

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▪ The piston moves due to the pressure force created by the different pressures on the two sides of the piston: P1 in the forward chamber and P2 in the backward chamber. When the piston moves to the right, the fluid enters the forward chamber (fluid IN), and leaves the backward chamber (fluid OUT).

▪ Mathematical description: ▫ Differential pressure PL (difference between the two

pressures):

21 PPPL ▫ Pressure force (generated force) is

Lpg PAF

▫ Load force or output force is FL

▫ There is a spring effect associated with the piston: Kxp , where K is the gradient (stiffness).

▫ There is a viscous damping effect associated with the piston:

pp xB , where Bp is the viscous damping coefficient. ▪ Dynamics of the motor (i.e. dynamics of the piston)

Newton’s law gives:

LpppptLp FKxxBxMPA (A.1)

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♦ Vane rotary actuator – limited travel mnachine

▫ Pressure torque force (generated torque) is rPA Lpg ▫ Load torque or output torque is τL

▫ There is a torsion spring effect associated with the rotor: Kφ ,

where K is the gradient (torsion stiffness).

rotor

backward chambre

forward chambre

Vane

Rotation angle φ

Pressure makes

a resultant force and

consequently a torque

fluid IN : pressure P1

fluid OUT : pressure P2

r

rotor parameters:

It – moment of inertia

Ap – effective vane area

housing (stator)

Fig. 4.2 a

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mechanic energy

OUTPUT: mechanic

energy

▫ There is a viscous damping effect associated with the piston: B , where B is the viscous damping coefficient.

▪ Dynamics of the motor (i.e. dynamics of the rotor)

Newton’s law for rotation gives:

LtLp KBIrPA (A.2) ♦ Double vane rotary actuator is shown in ♦ Spur gear rotary machine (actuator or pump) is shown in

It allows continuous rotation. ♦ show different types (examples) of hydraulic

machines.

→ In this course, we are primarily interested in actuators. The ususl example will be a piston actator or a vane rotary motor

→ The pumps are used just as a source of hydraulic energy.

hydraulic energy

Fig. 4.2 (b) .

Fig. 4.3 .

Figs. 4.4 – 4.12

PUMP (source of hydro energy): converts

mechanical energy into

hydraulic energy

HYDRO ACTUATOR: converts hydro

energy into mechanical

energy

ELECTRIC MOTOR (source of

mechanic energy): converts electric

ener. into mechanical ener.

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5. HYDRAULIC CONTROL VAVES ♦ Valves are are the interface between the the sorce of hydraulic

energy and the actuator.

▪ Actuator (motor) is e.g. a cylinder with a piston or a vane rotary motor.

▪ Energy source is a pump (of any type). ♦ Valve is a devices that uses mechanical motion to control the

delivery of power to the actuator.

control the delivery of

energy

controlled source of energy (controlled by means of mechanical motion)

source of hydrauluic

energy

Oil supply (pressure supplay)

VALVE Actuator

oil flow oil flow

Unit which creates the mechanical motion

that controls the valve

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5.1. Valve Configurations sliding type (a, b, c, d in Fig. 5.1)

♦ Config. classification seating type (e in Fig. 5.1)

flow deviding type ( f in Fig. 5.1) ♦ Sliding valves are classified according to:

- number of ways - the number of input/output oil lines;

- number of lands, - type of center when spool is in neutral position.

(a) two-land-four-way spool valve:

flow to source

return

supply

flow to actuator

mechanical motion that controls the valve – spool stroke xv

Fig. 5.1 (a)

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(b) three-land-four-way spool valve:

(c) four-land-four-way spool valve:


flow to source

return

supply

flow to actuator

Fig. 5.1 (c)


flow to source

return

supply flow to actuator

Fig. 5.1 (b)

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(d) two-land-three-way spool valve:

(e) two-jet flapper valve:

flapper

supply

pivot

Fig. 5.1 (e)

flow to actuator

motion of the flapper controls the valve

return

to source

flow to source

return

supply flow to actuator

Fig. 5.1 (d)


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(f) jet pipe valve:

♦ Spool valves:

matching tolerances are required => - expensive and - sensitive top oil contamination

♦ Flapper valves: leakage =>

- for low power or - as a first stage in a two-stage systems.

♦ Jet pipe valves: - large null flow, - characteristics are not easy to predict, - slow response.

supply

pivot

rotation of the jet controls the valve

Fig. 5.1 (f)

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For further discussion – spool valves. ♦ Number of lands:

- two , in primitive valves; - three or four , in a usual case - up to six , for special valves.

♦ Ratio between the land width and the port:

▪ If land width < port : open center or underlapped valve

▪ If land width = port : critical center or zero lapped valve

▪ If land width > port : closed center or overlapped valve

width port

width port

width port

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▪ open center valve : large power loss ion neutral position; only

for some special systems

▪ critical center valve : our choice; linear characteristics

▪ closed center valve : deadband near null causes steady state error and stability problems.

flow Q

spool stroke xv

critical center

closed center

overlap region

underlap region

flow gain doubles near null

Fig. 5.2

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5.2. General Valve Analysis ● General Flow Equations

▪ Neglecting the compressibility, continuity request yields:

- to actuator: 41 QQQL (5.1)

- from actuator: 23 QQQL (5.2) ▪ The differential pressure is

21 PPPL (5.3)

L2

L1

L2

L1

spool stroke vx

P2

P1

3

2

1

4Supply: - flow Qs - pressure Ps

Return: - flow Qs - pressure P0 ≈ 0

To actuator: - flow QL - pressure P1

From actuator: - flow QL - pressure P2

PL= P1 – P2

Force Fi

Fig. 5.3.

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▪ According to equation (3.33), the flows through the valving

orifices is:

)(2

111 PPACQ sd (5.4)

)(2

222 PPACQ sd (5.5)

233

2PACQ d

(5.6)

144

2PACQ d

(5.7)

▪ The orifices areas depend on the valve geometry and the valve

displacement (spool stroke) xv :

)(,)(,)(,)( 44332211 vvvv xAAxAAxAAxAA (5.8)

▪ The set (5.1) – (5.8) copntains 11 equations that can be combined

to give the load flow as a function of the spool stroke xv and the diffeerential pressure PL:

),( LvLL PxQQ (5.9)

The plot of (5.9) is known as as the pressure-flow curves for the valve and is a complete description of stady state valve performance. All of the performance parameters, such as valve coefficients, can be obtained from such curves.

▪ In the vast majority of cases, the valving orifices are matched and

symmetrical. Matched orifices require

4231 , AAAA (5.10), (5.11)

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and symmetrical orifices require

)()(,)()( 4321 vvvv xAxAxAxA (5.12), (5.13) Therefore, in the neutral position of the spool, all four areas are aqual:

4,3,2,1,)0( 0 jAA j So, only one orifice area need to be described. If the orifice area is linear with the valve stroke (as is usually tha case), only one defining parameter is needed:

w – the width of the slot (hole) in the valve sleeve (cover) .

w – For linear valves (like with rectangular ports), this is the area gradient for each orifice (and so for the whole valve).

▪ For matched and symmetrical orifices, it holds that

4231 , QQQQ (5.15), (5.16)

▪ Substituting (5.4), (5.5) and (5.6) into (5.15) one obtains:

21 PPPs (5.17)

Relation (5.16) may give the same result. ▪ Equations (5.3) and (5.17) can be combined to produce:

21Ls PP

P

(5.18)

22Ls PP

P

(5.19)

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▪ From Fig. 5.3, it follows that the total supply flow can be written

as

21 QQQ s (5.20) and as

21 QQQ s (5.21)

▪ In summary, for a matched and symmatrical valve, relations (5.15), (5.16) and (5.18), (5.19) applies and equations (5.1) and (5.2) both become

)(1

)(1

21 LsdLsdL PPACPPACQ (5.22)

and similar treatment yields (using (5.20) and (5.21)):

)(1

)(1

21 LsdLsds PPACPPACQ (5.23)

● Linearization – Valve Coefficients ♦ Sometimes, a nonlinear form of the matyhematical model causes

problems and linearization is needed. ▪ Equation (5.9), describing the load flow, can be expanded in the

Taylor’s series about a particular operating point 1:

111 ),( LLv QPx producing

LL

Lv

v

LLL P

P

Qx

x

QQQ

11

1

27

▪ If the working mode is such that ),( Lv Px are kept in the vicinity

of the operating point 1, i.e. close to ),( 11 Lv Px , then

),( 11 Lv Px will be small and it is jusrtified to keep only the linear terms in the Taylor’s expansion. Thus:

LL

Lv

v

LLLL P

P

Qx

x

QQQQ

11

1 (5.24)

The partial derivatives are obtained analytically or numerically. ▪ Valve coefficients (!!!)

- Flow gain : 0

v

Lq x

QK (5.25)

- Flow-pressure coef. 0

L

Lc P

QK (5.26)

- Pressure sensitivity c

q

v

Lp K

K

x

PK

(5.27), (5.28)

► Flow gain affects the open-loop gain constant and thus has

a direct influience on the system stability. ► Flow-pressure coeficient directly affects the damping

ratio of valve-motor combination. ► Pressure sensitivity of valves is quite large which shows

the ability of valve-motor combination to breakaway large friction loads with little error.

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▪ Now, (5.24) becomes

LcvqL PKxKQ (5.29) ▪ The most important operating point is the origin:

0,0,0 111 LLv QPx .

- In this case, qK is largest (thus, high system gain)

and cK is smallest (thus, low damping), and accordingly this operating point is most critical from a stability viewpoint.

- If we achieve stability for this point, the system will be stable for all other operating points.

- Valve coefficinets calculated for thgis point are called null valve coefficients.

For this operating point ( 0,0,0 111 LLv QPx ), it holds that:

,

,

,

1

1

11

LLLL

LLLL

vvvv

QQQQ

PPPP

xxxx

and accordingly, (5.29) becomes

LcvqL PKxKQ (A.3)

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5.3. Critical-Center Spool Valve Analysis ● Pressure-Flow Curves We are going to derive the exact form of the relation (5.9)

),( LvLL PxQQ

for the case of a critical-center valve. ▪ We assume the ideal valve geometry, and hence, leakage iz zero:

0,0,0 42 vxforQQ , (so, (5.1) becomes QL= Q1 ) and

0,0,0 31 vxforQQ , (so, (5.2) becomes QL= – Q2 = – Q4) ▪ Substituting (5.18), (5.4) into (5.1), one obtains

0,2

21

vLs

dL xforPP

ACQ (5.30)

▪ For negativevalve displacements, (5.18), (5.7), substituting into (5.2), yield

0,2

22

vLs

dL xforPP

ACQ (5.31)

▪ For symmetrical valve, eq. (5.12) holds and (5.30) and (5.31) can be written as a single relation:

L

v

vs

v

vdLvLL P

x

xP

x

xACPxQQ

1

),( 1 (5.32)

30

QL

▪ If rectangular ports are used with an area gradient w, one obtains

L

v

vsvdLvLL P

x

xPxwCPxQQ

1

),( (5.33)

This is the pressure-flow curve mentioned earlier as eq. (5.9). Family of curves, for different xv is shown in Fig. 5.4.

PL

Ps

– Ps

xv increasing in positive sense

xv increasing in negative sense

Fig. 5.4

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● Valve Coefficients We recall the linearized form (A.3),

LcvqL PKxKQ (A.3)

and look for the coefficients. ▪ Differentiation of (5.33) gives

► )(1

Lsdv

Lq PPwC

x

QK

(5.35)

► )(2

))(/1(

Ls

Lsvd

L

Lc PP

PPxwC

P

QK

(5.36)

► v

Ls

c

qp x

PP

K

KK

)(2 (5.37)

▪ For the null operating point (being the most important) i.e. for

0,0,0 LLv QPx , the null coefficients for the ideal critical-center valve are:

► s

dq

PwCK 0 (5.38)

► 00 cK (5.39)

► 0pK (5.40)

▪ The computed value for 0qK is close to a realistic value (obtained

by tests). However, the computed values for 0cK and 0pK are far from the values obtained by testing a realistic valve.

=> So, we have to consider leakage !!!

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● Leakage Characteristics of Practical Critical-Center Valves

– just some comments –

▪ Ideal valve ↔ ideal geometry => no leakage ▪ Real valve ↔ radial clearance => leakage ▪ Example: Realistic pressure sensitivity curve for blocked lines

(so, only leakege flolw exists)

Ps

– Ps

load pressure difference PL

valve stroke xv

the slope is not infinite, i.e., Kp ≠ ∞

Fig. 5.5

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● Stroking Forces – Dynamics of the Valve (Spool)

Analysis is based on the Figure 5.3. ▪ Mathematical description:

▫ Force Fi is imposed to control the spool motion (stroke) i.e.

to control the valve

▫ There are flow forces that oppose the spool motion. These forces are derived from eqs. (5.90) and (5.93) in Section 5.6. and from (5.48) and (5.49) in Section 5.3. The result is:

◦ There is a spring effect associated with the spool motion (like a centerung spring). It is the steady-state flow force: Kf xp, where )(cos2 Lsvdf PPwCCK is the gradient (like a stiffness). ◦ There is a viscous damping effect associated with the spool

motion. It is the transient flow force: vf xB , where

)()( 12 Lsdf PPwCLLB is the damping coefficient.

▫ Mass Ms defines the inertia: vs xM .


vfvfvsi xKxBxMF (5.50)

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5.4. Open-Center Spool Valve Analysis HOMEWORK 1a - Ramadan 5.5. Three-Way Spool Valve Analysis HOMEWORK 1b - Mohamad 5.6. Flow Forces on Spool Valves HOMEWORK 1c - Ismail 5.7. Lateral Forces on Spool Valves 5.8. Spool Valve Design NOT DISCUSSED FOR THE MOMENT 5.9. Flapper Valve Analysis and Design

Single-jet, Double-Jet, Flow Forces HOMEWORK 1d - Abdulhalim

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6. HYDRAULIC POWER ELEMENTS 6.1. Valve Controlled Rotary Motor NOT DISCUSSED FOR THE MOMENT 6.2. Valve Controlled Piston

NOTE a difference regarding previous figures. The forward flow (to the actuator: Q 1) is not equal to the return flow (from the actuator: Q 2). Previously, it was equal: Q 1= Q 2= Q L This is due to some effects that have been neglected in the previous discussions and now we take care of them. These effects are:

- Leakage, - Compression.

xp Return line: Q 2 , P2

Forward line:

P1 , Q 1

Cylinder with a Piston

VALVE

Supply Ps

Fig. A.1

36

♦ Valve controlled flow – Linear analysis

▪ Starting from relation (A.3) ( LcvqL PKxKQ ), one may write experessions for Q 1 and Q 2 :

11 2 PKxKQ cvq (6.1)

22 2 PKxKQ cvq (6.2)

- If the valve is matched and symmetrical, the pressures in the

lines will rise above and below 2/sP by equal amounts so that the pressure drops across the two valve orifices are

identical. Hance the valve coefficients qK for forward and return flows are the same.

- The flow-pressure coefficient cK is twice that for the whole

valve since qK was defined with respect to PL and the change in PL is twice that which occurs across a port.

▪ Adding tha above two equations, it follows that

LcvqL PKxKQ (6.3)

So, the same form was obtained like expression (A.3). However, here, the “load flow” is the average :

221 QQ

QL

. (6.4)

and it is not equal to the flow in each line ( 21 QQQL ).

The load pressure (diffrerencial pressure) is still 21 PPPL .

37

♦ Valve controlled flow – Non-linear analysis ▪ Instead of (6.3) , the nonlinear expression for the flow (eq.

(5.33)), can be applied (like in later Section 6.7.)

L

v

vsvdLvLL P

x

xPxwCPxQQ

1

),( (5.33)

♦ Flow through the actuator – continuity relations .

Let us turn to the actuator chambers and look at Fig. 6.6.

Force Fi and motion xv (to control the valve)

PL = P1 – P2

P2 , V 2 P1 , V 1

xp

VALVE

Fig. 6.6

Load. - Force FL

- spring effect - damping effect

Piston parameters: Mt – mass of the

piston plus refered masses

Ap – effective piston area

Forward line:

P1 , Q 1

External leakage External leakage

Return line: Q 2 , P2

Internal leakage

Supply

38

▪ Analyzing the flow, we take care of → Piston motion. The corresponding flow is the rate of volume

change: dV/dt. → Leakage (internal and external). Flow due to leakage is

proportional to the pressure drop. → Compression (effective – due to air and mech. compliance;

oil itself might be considered noncompressible or compressible). Flow due to compression is derived starting from eq. (2.4) – the definition of the bulk modulus:

(2.4) :

V

PV0 =>

dtdV

dtdPV

/

/0 =>

dt

dPV

dt

dV

0

▪ Applying the equation of continuity for chambers 1 and 2, one obtains

dt

dPV

dt

dVPCPPCQ

eepip

1111211 )(

(6.27)

dt

dPV

dt

dVQPCPPC

eepip

2222221 )(

(6.28)

where

V1 –volume of the chamber 1 of the actuator plus related volumes: connecting line, and the refered volume in the valve)

V2 – volume of the chamber 2 plus related volumes Cip – internal leakage coefficient Cep – external leakage coefficient

▪ The volumes of the chambers may be writted as

pp xAVV 011 (6.29)

pp xAVV 022 (6.30)

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where V01 and V02 are the initial volumes (for the null position of the piston, xp= 0). The piston is usually centered, and then: V01= V02 = V0 .

▪ Now, from (29) and (6.30), the derivatives are

dt

dxA

dt

dV

dt

dxA

dt

dV pp

pp 11 ; ;

dt

dV

dt

dV 21 (A.4)

▪ The sum of the two volumes is contant and independent of piston

motion:

0020121 2VVVVVVt (6.32)

Vt is the total volume of fluid under compression in both chambers.

▪ ▪ We now combine (6.29), (6.30), (A.4) and (6.27), (6.28) to obtain

dt

PPdxA

dt

PPdVPP

CC

dt

dxA

QQQ

e

pp

e

epip

pp

L

)(

2

)(

2))(

2(

2

2121021

21

▪ If 0VxA pp , the last term may be neglected

▪ ▪ So, we finaly come to

Le

tLtpppL P

VPCxAQ

4 (6.33)

where 2/epiptp CCC is the total leakage coefficient.

40

♦ Mathematical description of the piston dynamics (this has been already discussed in Ch. 4 – we repeat here): ▫ Differential pressure PL (difference between the two

pressures):

21 PPPL

▫ Pressure force (generated force) is

Lpg PAF

▫ Load force or output force is FL

▫ There is a spring effect associated with the piston: Kxp , where K is the gradient (stiffness).

▫ There is a viscous damping effect associated with the piston:

pp xB , where Bp is the viscous damping coefficient.

▪ ▪ Dynamics of the motor (i.e. dynamics of the piston)


LpppptLp FKxxBxMPA (A.1)=(6.34)

41

6.A. Mathematical Model of the Valve-Controlled Actuator

♦ Actuator controlled by the valve stroke

As mentioned several times, the velve control the actuator by the spool stroke xv .

(I) ► Dynamics of the piston motion is desribed by (6.34):

LpppptLp FKxxBxMPA (6.34)

(II) ► Load flow is described by continuity equation (6.33):

Le

tLtpppL P

VPCxAQ

4 (6.33)

(III) ► Valve control the flow by relation - (6.3) in the case of linear analysis, or - (5.33) in the case of non-linear analysis:

LcvqL PKxKQ (6.3) or

L

v

vsvdLvLL P

x

xPxwCPxQQ

1

),( (5.33)

● Eqs. (I)–(III), i.e. - (6.34), (6.33) and (6.3) (for lin. case) or ` - (6.34), (6.33) and (5.33) (for non-lin. case),

define the mathematical model.

▪ State variables are piston position, its velocity, and load

pressure: px , px , LP .

▪ Control input is the valve spool stroke, vx .

♠ Question: If the spool stroke controls the actuator, how to generate the appropriate spool stroke ?

42

♠ ANSWER: We use a force to move the spool !

=>

♦ Actuator and valve controlled by the force imposed to the spool

Figure 5.3 showed that the spool stroke is generated by the force

Fi imposed to the spool.

(IV) ► We relate the force Fi with the spool motion xv by dynamic equation (5.50):

vfvfvsi xKxBxMF (5.50) ● Eqs. (I)–(IV), i.e. - (6.34), (6.33), (6.3), (5.50) (linear case) or

` - (6.34), (6.33), (5.33), (5.50) (non-lin.), define the mathematical model.

▪ State variables are piston position and velocity, load

pressure, spool position (stroke) and velocity: px , px , LP ,

vx , vx ,

▪ Control input is the force imposed to valve spool: Fi.

♠ Question: If the force imposed to the spool controls the valve and the actuator, how to generate the appropriate force ? ?

♠ ANSWER requires a more detailed analysis of the valve. Some

kind of motor will be needed to create the force ! This will be elasborated in Chapter 7.

43

♦ Important notes about the load.

▪ The model derived (eqs. (I) – (IV)) includes the load force FL. It is not a known force but it depends on the dynamics of the load.

▪ In a general case, the load is a dynamic system (that may have its

own degrees of freedom). So, the load force FL represents the interaction between the two systems (actuator and load – see Fig. A.2). According to the law of action and reaction, the force that acts from the actuator to the load (action) is equal and oposite to the force that acts from the load to the actuator (reaction).

▪ So, the load force FL is unknown and has to be expressed from

the mathematical model of the load dynamics. Hence, in order to complete the system of equations (i.e. to make it solveble), it will be necessary to specify the load and formulate its mathematical model.

♦♦ Canonic form of the mathematical model - For the analysis of system: dynamic characteristics, control

syntehis, stability analysis, and finally simulation, it is desired to put thge mathematical model in the canonic form.

- Let ),( ,21 zzz be state vector and let u be the input control signal.

action FL

Load

reaction FL

Actuator

Fig. A.2

44

▪ The canonic form is then:

),( uzfz for nonlinear systems, (A.5)

and

uEzDz for linear systems, (A.6)

where D and E are system matrices.

▪ The model that we discuss includes the load force FL , and it may introduce additional state variables. So, with the force FL (A.5) and (A.6) become

),,( LFuzfz for nonlin. case, (A.7)

and

LHFuEzDz for linear case. (A.8) ♦ Actuator controlled by the valve stroke

The model involves (I) – (III) .

▪ The state variables and state vactor are:

pxz 1 , pxz 2 , LPz 3 , ),,( Lpp Pxxz (A.9) ▪ Control input is the valve spool stroke,

vxu . (A.10)

45

▪ Let us rewrite (I)-(III) acoording to notation (A.9) and (A.10):

(I) Lptp FKzzBzMzA 1223

(II) 332 4z

VzCzAQ

e

ttppL

(III) 3zKuKQ cqL (for linear analysis), or

3

1z

u

uPuwCQ sdL (for nonlinear analysis)

From (A.9), it follows that 21 zxz p . ◘ By combining the above relations, for the linear case one

gets:

21 zz

Ltt

p

t

p

t

FM

zM

Az

M

Bz

M

Kz

13212 (A.11)

uV

KzV

CKzV

Azt

eq

t

etpc

t

ep

44)(

4323

i.e.. in a matrix form (A.8) it is:

Lt

t

eq

t

etpc

t

ep

t

p

t

p

t

FM

u

VKz

z

z

VCK

VA

M

A

M

B

M

K

z

z

z

HED

0

10

40

0

4)(

40

010

3

2

1

3

2

1

(A.12)

46

◘ For the nonlinear case one gets the form (A.7):

21 zz

Ltt

p

t

p

t

FM

zM

Az

M

Bz

M

Kz

13212

3323

1444z

u

uPuwC

Vz

VCz

VAz sd

t

e

t

etp

t

ep

(A.13)

◘ How to handle the load ? Let us explain this by examples !

♠ EXAMPLE 1 Form the complete mathematical model for the system of Fig. A.3! The control input is the valve stroke. NOTE: The load does not introduce any new state variable.

Load

Rolling without sliding

Cylinder:mass m radius r

Fig. A.3

Actuator

47

○ The actuator is modeled by (A.11) for a linear analysis or (A.13) for a nonlinear analysis.

○ The model includes the load force FL .

● We now look for the mathematical model of the load in order to express the load force FL .

▫ Eqations of load dynamics:

frL FFam , for translation

rFI fr , for rotation (about the center)

where a is the acceleration, α is the angular acceleration, and I is the moment of inertia.

Note that there is no sliding and accordingly NFfr (thus

friction frF is unknown).

▫ Having in mind: ra / and 2

2

1rmI , the above equations

yields:

maFL 3

2 , maFfr 3

1 .

FL Load force i.e. actuator output force

FL

Friction force (dry)

Ffr

48

▫ The motion of the wheel center equals the the piston motion

xp, and so:

2zxa p => 23

2zmFL .

One can see that FL does not introduce new state variables but depends on the existing one.

▪ For a linear analysis, load force is substituted into (A.11) (or, may be it is simpler to substitute into (I)). In any case, one gets:

21 zz

3212 )3/2()3/2()3/2(z

mM

Az

mM

Bz

mM

Kz

t

p

t

p

t

uV

KzV

CKzV

Azt

eq

t

etpc

t

ep

44)(

4323

or in a matrix form

u

VKz

z

z

VCK

VA

mM

A

mM

B

mM

K

z

z

z

ED

t

eq

t

etpc

t

ep

t

p

t

p

t

40

0

4)(

40

)3/2()3/2()3/2(

010

3

2

1

3

2

1

which is the final form (A.6).

49

▪ For a nonlinear analysis, load force is substituted into (A.13) (or, into (I)), to get:

21 zz

3212 )3/2()3/2()3/2(z

mM

Az

mM

Bz

mM

Kz

t

p

t

p

t

3323

1444z

u

uPuwC

Vz

VCz

VAz sd

t

e

t

etp

t

ep

which is the final form (A.5).

♠ EXAMPLE 2 Form the linear mathematical model for the system of Fig. A.4 ! The control input is the valve stroke. NOTE: The load introduces one additional degree of freedom

(x2r) and accordingly two additional state variabls ),( 22 rr xx .

NOTE: FL is in reverse direction (negative)

FL

x2r

Load

Body: mass m2

Cylinder: mass m1

radius r

Fig. A.4

Actuator

xp= x1

50

○ The actuator is modeled by (A.11) for a linear analysis or

(A.13) for a nonlinear analysis. ○ The model includes the load force FL .

● We now look for the mathematical model of the load in order to express the load force FL .

▫ Eqations of load dynamics:

- for the wheel: translation and rotation gmFFam L 1111

rFI 11 - for the body (translation only)

1222 Fgmam

▫ Accelerations are:

pxa 1 , rpr xxaaxa 22122 , rr xa 22

x2= =xp+ x2r

m2g

x2rF1

F1

xp= x1

FL

FL

m1

m2

m1g

51

▫ Having in mind: ra r /2 and 2

11 2

1rmI , the equations of

load dynamics, after some transformations, become

Lrp Fxmxm 211 2

1

gmxmmxm rp 22122 )2

1(

▫ Besides the “old” state variables (comming from the actuator), i.e. z1, z2, z3, we have introduced two “new” state variables (due to the new degree of freedom of the load, x2r):

rxz 24 , rxz 25 . ▫ In this case the above equations of dynamics become

LFzmzm 5121 2

1

gmzmmzm 251222 )2

1(

with

54 zz , or, after additional transformation,

gmm

mmz

mm

mmmFL

12

212

12

2121

)2/1(

)2/1(

)2/1(

)2/1()2/3(

212

2

12

25 )2/1()2/1(

zmm

mg

mm

mz

(*)

54 zz

52

▪ For a linear analysis, (A.11) is combined with the above three relations. First, FL from the first relation is substituted into the second equation from (A.11) (note that the sign of FL has changed due to the oposite action of the force). Then, from this

modified second equation of (A.11), 2z is substituted into the second relation of the above set (*). Now, this modified second relation form (*), and the third relation from (*) are supplemented to the set (A.11). In this way, five state equations are obtained:

21 zz

23232221212 GzDzDzDz

uEzDzDz 33332323

54 zz

53532521515 GzDzDzDz

where

1)2/1(2

21)2/1(21)2/3(

21

mm

mmmtM

KD

;

1)2/1(2

21)2/1(21)2/3(

22

mm

mmmtM

pBD

1)2/1(2

21)2/1(21)2/3(

23

mm

mmmtM

pAD

;

gG

mm

mmmtM

mm

mm

1)2/1(2

21)2/1(21)2/3(

1)2/1(2

21)2/1(

2

t

ep V

AD4

32 ; t

etpc V

CKD4

)(33 ; t

eq V

KE4

3

53

2112

251 )2/1(

Dmm

mD

; 22

12

252 )2/1(

Dmm

mD

2312

253 )2/1(

Dmm

mD

; 2

12

2

12

25 )2/1()2/1(

Gmm

mg

mm

mG

The obtained model describes the dynamics of the entire system. The model is in a linear canonical form, like (A.6). The matrix form is

5

2

3

5

4

3

2

1

535251

3332

232221

5

4

3

2

1

0

0

0

0

0

0

0

00

10000

000

00

00010

G

G

u

E

E

z

z

z

z

z

D

DDD

DD

DDD

z

z

z

z

z

♦ Actuator and valve controlled by the force on the spool

The model involves (I) – (IV) .

▪ The state variables and state vactor are:

pxz 1 , pxz 2 , LPz 3 , vxz 4 , vxz 5 ,

),,,,( vvLpp xxPxxz (A.14) ▪ Control input is the force to valve spool,

iFu . (A.15)

54

▪ Let us rewrite (I)-(IV) acoording to notation (A.14) and (A.15):

(I) Lptp FKzzBzMzA 1223

(II) 332 4z

VzCzAQ

e

ttppL

(III) 34 zKzKQ cqL (for linear analysis), or

3

4

44

1z

z

zPzwCQ sdL (for nonlinear analysis)

(IV) 455 zKzBzMu ffs From (A.14), it follows that

21 zxz p and 54 zxz v

◘ By combining the above relations, for the linear case one gets:

21 zz

Ltt

p

t

p

t

FM

zM

Az

M

Bz

M

Kz

13212

4323

44)(

4z

VKz

VCKz

VAz

t

eq

t

etpc

t

ep

(A.16)

54 zz

uM

zM

Bz

M

Kz

ss

f

s

f 1545

55

i.e.. in a matrix form (A.8) it is

L

t

s

t

eq

t

etpc

t

ep

t

p

t

p

t

FM

u

Mz

z

z

z

z

VK

VCK

VA

M

A

M

B

M

K

z

z

z

z

z

HED

0

0

0

10

10

0

0

0

10000

044

)(4

0

00

00010

5

4

3

2

1

5

4

3

2

1

(A.17)

◘ For the nonlinear case one gets the form (A.7):

21 zz

Ltt

p

t

p

t

FM

zM

Az

M

Bz

M

Kz

13212

3

4

44323

1444z

z

zPzwC

Vz

VCz

VAz sd

t

e

t

etp

t

ep

54 zz

uM

zM

Bz

M

Kz

ss

f

s

f 1545

(A.18)

◘ How to handle the load ?

We could explain this by examples ! The examples would be done completely analogously like Examples 1 and 2, so like it was done for the spool-stroke controlled actuator.

56

6.3. Three-Way Valve Controlled Piston 6.4. Pump Controlled Motor NOT DISCUSSED FOR THE MOMENT 6.5. Valve Controlled Motor with Load Having Many

Degrees of Freedom Let the load be in the form of n masses connected by means of springs (stiffness) and dampers, as shown in Fig. 6.8. A combination of a spring and a damper will be called simply “spring” (a real spring actually involves stiffness and damping).

m1, m2, ... , mn – masses

k1, k2, ... , kn – stiffnesses

b1, b2, ... , bn – damping constants

QL

kn

xn x2 x1

b1

k2

m1

k1 xp

Fi

xv

QL

Valve

m2 b2

mn bn

Fig. 6.8

load FL

57

▪ Position coordinates (degrees of freedom) for the entire system:

- xp , xv (for the acruator and valve) plus

- x1, x2, ... , xn (for the load) ▪▪ Dynamics of the actuator and the valve is described by

eqs. (I) – (IV) . This model includes the load force FL. ▪▪ Dynamics of the load can be described by the following set of n

equations:

2springinforce)(1springinforce

)]()([)]()([ 212212111111 xxbxxkxxbxxkxm

LF

pp

3springinforce2springinforce

)]()([)]()([ 32332321221222 xxbxxkxxbxxkxm

. . .

. . .

. . .

n

nnnnnnnn xxbxxkxm

springinforce

)]()([ 11

(A.19) ▪▪ The complete mathematical model (actuator plus load) includes:

- eqs. (I) – (IV) , fot the acatuator and valve, plus - set of n equations (A.19).

Force FL in (I)–(IV) can be eliminated since it is the force in spring 1 and it is

)()( 1111 xxbxxkF ppL , as given in the first equation of the set (A.19).

58

▪▪ The load has intruduced additional degrees of freedom and

accordingly additional state variables. The entire set of state

variables (vector z) is :

z = ( px , px , LP , vx , vx , (from the actuator)

nn xxxxxx ,,,,,, 2211 (from the load)). ▪▪ The model can be put in a canonical form. 6.6. Pressure Transients in Power Elements NOT DISCUSSED FOR THE MOMENT 6.7. Non-linear Analysis of Valve Controlled Actuators

We, in our course (and this text), discussed nonlinear analysis in Section 6.2. Equation (5.33), used in Sec. 6.2., concides with (6.93) being crucial in the current section 6.7.

59

7. ELECTROHYDRAULIC SERVOVALVES As we have mentioned, the valve and the actuator were controlled by

- spoll stroke xv , or - force Fi imposed on the valve spool.

In any case, there is a question: ♠ Question: How to generate the appropriate stroke or force ? ? ♠ ANSWER: Some kind of motor is needed to create the force (or

torque) and consequently the stroke ! It is called the torque motor.

So, servovalve means the valve (one or two stages) plus the torque motor .

7.1. Types of Electrohydraulic Servovalves ♦ Single-stage servovalve

▪ The torque motor is directly connected to the spool valve.

▪ Torque motors have limited power capabilities. This

- limits the torque/force that can be generated, - limits the flow capacity of the valve, and - may lead to stability problems in some applications.

Force/torque Torque motor

Spool of the valve

60

♦ Two-stage servovalve

▫ Stage 1 is a hydraulic preamplifier. It augments the

force/torque generated by the motor to the level that can overcome all the problems: flow forces, stiction, acceleration, vibrations, etc.

▫ Stage 1 can be:

- spool valve, - jet pipe valve, and - flapper valve.

▫ Stage 2, the main spool, is alvays a spool valve.

■ Types of feedback between the two stages (most common types):

- direct feedback , - force feedback , and - spring centered spool.

▫ With direct feedback, the main spool follows the first stage in

a one-to-one relation. We talk about hydraulic follower. ▫ With force feedback, there is a deformable element, a spring,

between the two stages.

Torque motor Stage 1 Valve of

different type

Stage 2 Spool valve

force/torque amplified

force/torque

61

7.2. Permanemnt Magnet Torque Motor

62

7. ELECTROHYDRAULIC SERVOVALVES As we have mentioned, the valve and the actuator were controlled by

- spoll stroke xv , or - force Fi imposed on the valve spool.

In any case, there is a question: ♠ Question: How to generate the appropriate stroke or force ? ? ♠ ANSWER: Some kind of motor is needed to create the force (or

torque) and consequently the stroke ! It is called the torque motor.

So, servovalve means the valve (one or two stages) plus the torque motor .

7.1. Types of Electrohydraulic Servovalves ♦ Single-stage servovalve

▪ The torque motor is directly connected to the spool valve.

▪ Torque motors have limited power capabilities. This

- limits the torque/force that can be generated, - limits the flow capacity of the valve, and - may lead to stability problems in some applications.

Force/torque Torque motor

Spool of the valve

♦ Two-stage servovalve

▫ Stage 1 is a hydraulic preamplifier. It augments the

force/torque generated by the motor to the level that can overcome all the problems: flow forces, stiction, acceleration, vibrations, etc.

▫ Stage 1 can be:

- spool valve, - jet pipe valve, and - flapper valve.

▫ Stage 2, the main valve, is always a spool valve.

■ Types of feedback between the two stages (most common types):

- direct feedback , - force feedback , and - spring centered spool.

▫ With direct feedback, the main spool follows the first stage in

a one-to-one relation. We talk about hydraulic follower. ▫ With force feedback, there is a deformable element, a spring,

between the two stages.

Torque motor Stage 1 Valve of

different type

Stage 2 Spool valve

force/torque amplified

force/torque

7.2. Permanent-Magnet Torque Motor

▪▪ The torque motor generates the torque (that can be converted into force). This force will control the valve.

▪▪ The torque motor is controlled by the current i that flows through the armature (coil). We cannot impose the current

directly. We use a source of voltage u to generate the current.

So, the voltage u is finally the true control input !! ● Dynamics of the electrical circuit.

The circuit has some resistance and some inductivity. The relation between the voltage and the current is the Ohm’s law:

dtdiLKiRu cb / (A.20)

where: R is the armature resistance, Kb is the constant of counter electromotive force

(induced by the magnetic field acting on a moving coil), and

Lc is the coil inductivity.

Fig. A.5 Fig. 7.2

● Dynamics of the armature rotation.

▪ The magnetic field acts on the coil with current. Lorentz’s force appears that makes the torque. Additional torque is due to the armature angular displacement from the null position (angle ). So, the total developed torque is

mtd KiKT (A.21) = (7.29)

where Kt is the torque constant, and Km is the magnetic spring constant. ▪ The developed torque has to solve some opposing effects. In

the opposite direction, there are: - inertial effects, - viscous friction, - spring effect of the pivot, - the load.

▪ So, the Newton’s law for the armature rotation gives

Laaamtd TKBJKiKT i.e.

Lmaaat TKKBJiK )( (A.22)=(7.31)

where Ja is the moment of inertia of the armature and related masses,

Ba is the viscous friction coefficient (damping), usually negligible,

Ka is the torsion spring constant of the armature pivot, and

TL is the load torque to the armature, i.e., the output motor torque.

●● Equation (A.20) and (A.22) describe the motor dynamics.

▪▪ Torque TL is imposed to the valve in order to control it !!

7.3. Single-Stage Electrohydraulic Servovalves

0

QL

Valve (single)

Actuator

Spool Force developed by the motor and directly imposed on the spool: Fi

Rigid bar

Pivot connection

Torque motor

Fig. A.6 Fig. 7.11

r

▪ We consider small stroke and small angle ( tansin )

▪ Now, the relation between the angle and the stroke is

rxv / (A.23)

▪ The motor output torque ( TL ) develops a force which is directly imposed to the spool of the valve (force Fi ). It holds that

iL FrT (A.24)

θ

Spool

Force developed by the motor and directly imposed on the spool: Fi

Rigid bar

Pivot connection

Fig. A.7

distance

r

spool stroke xv

Torque TL

and angle θ

♦♦ Mathematical model ▪ The actuator and the valve are described by relations (I) - (IV).

- Force Fi was considered as input. - Spool stroke position xv is involved.

▪ Now, force Fi is not the input any more. The control input is the

voltage u imposed to the armature of the torque motor. ▪ So, it is necessary to join the mathematical model of the actuator

and valve with the model of the torque motor. (I)-(IV) should be combined with (A.20), (A.22).

▪ We first modify (A.20) and (A.22) by substituting (A.23), (A.24):

dt

diLx

r

KiRu cv

b (A.25)

ivma

va

vat Fx

r

KKx

r

Bx

r

Ji

r

K

222 (A.26)

▪ So, it is necessary to join (I)-(IV) and (A.25), (A.26).

This is done by substituting Fi from (A.26) into (I)-(IV)

▪ Equations (I)-(IV) inroduce 5 state variables :

- vvLpp xxPxx ,,,, (see (A.14)) equations (A.25), (A.26) introduce 1 additional state variable :

- current i.

So, after combining, there would be 6 state variables.

♦ Let us find the canonical form of the entire mathematical model, and restrict consideration to linear case. ▪ Equations (I)-(IV) have already been transformed into state-space

form (A.16).

21 zz

Ltt

p

t

p

t

FM

zM

Az

M

Bz

M

Kz

13212

4323

44)(

4z

VKz

VCKz

VAz

t

eq

t

etpc

t

ep

(A.16)

54 zz

iss

f

s

f FM

zM

Bz

M

Kz

1545

▪ Introducing new state variable,

iz 6

and combining (A.16) with (A.25), (A.26) one obtains the canonic model of the six order :

21 zz

Ltt

p

t

p

t

FM

zM

Az

M

Bz

M

Kz

13212

4323

44)(

4z

VKz

VCKz

VAz

t

eq

t

etpc

t

ep

(A.27)

54 zz

6252

2

42

2

5 )/(/

/

/

/)(z

rrJM

Kz

rJM

rBBz

rJM

rKKKz

as

t

as

af

as

maf

uL

zL

Rz

rL

Kz

ccc

b 1646

▪ This linear model can easily be put into matrix form ! ! ▪ Nonlinear case can be elaborated in the same way ! ! ▪ Note that there is still FL in the model. To solve FL one needs to

specify the load !! The procedure for handling the load is analogous to the earlier described !!

♦♦ Simplification of the mathematical model ▪ Different approximations are possible in order to simplify the

mathematical model.

- Some of approx. are reduce the model substantially (“qualitatively”) – the order of the model is reduced (the number of state variables is reduced).

- Some other approx. reduce the model just “quantitatively” – coefficients are simpler and there is smaller number of parameters needed.

▪ Besides, the approximations have to be justified. First approximation – qualitative (substantial). The inductivity of the coil (motor armature) may be neglected:

0cL (A.28)

◦ In this case, the term dtdiLc / in eq. (A.20) and (A.25) disappears. Electric dynamics is neglected.

◦ The current i stops being the state variable. ◦ So, the model order reduce from 6 to 5. ◦ The last relation from (A.27) (the 6-th one) becomes

uR

zRr

KzuzRz

r

K bb 10 4664

and it allows to eliminate iz 6 , by substituting into the 5-th relation of (A.27).

◦ The entire model becomes:

21 zz

Ltt

p

t

p

t

FM

zM

Az

M

Bz

M

Kz

13212

4323

44)(

4z

VKz

VCKz

VAz

t

eq

t

etpc

t

ep

(A.29)

54 zz

urrJM

RKz

rJM

rBBz

rJM

RrKKrKKKz

as

t

as

af

as

btmaf

)/(

/

/

/

/

//)(252

2

42

22

5

This is a 5-th order canonic model. ▪ This linear model can easily be put into matrix form ! ! ▪ Nonlinear case can be elaborated in the same way ! ! ▪ Note that there is still FL in the model. To solve FL one needs to specify

the load !! The procedure for handling the load is analogous to the earlier described !!

Second approximation - quantitative. Some torque motor parametrs are neglected (in addition to the first approx.):

0,0,0,0,0 maaab KKBJK (A.30)

◦ This means that the mechanical dynamics of the armature is neglected.

◦ In this case, equations (A.20) and (A.22) become:

iRu

uR

KT t

L (A.31)

Lt TiK

◦ Model (A.29) now acquires a simpler form, but still preserving

the 5-th order.

21 zz

Ltt

p

t

p

t

FM

zM

Az

M

Bz

M

Kz

13212

4323

44)(

4z

VKz

VCKz

VAz

t

eq

t

etpc

t

ep

(A.32)

54 zz

urM

RKz

M

Bz

M

Kz

s

t

s

f

s

f /545

This is still a 5-th order canonic model, but simpler. ▪ This linear model can easily be put into matrix form ! ! ▪ Nonlinear case can be elaborated in the same way ! ! ▪ Note that there is still FL in the model. To solve FL one needs to specify


Third approximation – qualitative (substantial). Some valve parametrs are neglected (in addition to the first and second approx.):

0,0 fs BM (A.33)

◦ This means that

- the mechanical dynamics of the spool motion is neglected )0( sM , and

- all transient effects are neglected )0( fB .

- Only steady-state effects are preserved ( Kf ). ◦ In this case, equations (5.50) (i.e. relation (IV)) becomes:

vfi xKF (A.34) ◦ Model (A.32) now reduces the order and become of the 3-rd

order .

NOTE: vxz 4 is eliminated from the 5-th relation of (A.32),

urK

RKz

f

t /4 , producing

21 zz

Ltt

p

t

p

t

FM

zM

Az

M

Bz

M

Kz

13212 (A.35)

urK

RK

VKz

VCKz

VAz

f

t

t

eq

t

etpc

t

ep

/44)(

4323

This is a 3-rd order canonic model. ▪ This linear model can easily be put into matrix form ! ! ▪ Nonlinear case can be elaborated in the same way ! ! ▪ Note that there is still FL in the model. To solve FL one needs to specify


♠ EXAMPLE 1

Formulate the (a) linear and (b) nonlinear mathematical model of the system shown in Fig. A.8. The actuator is controlled by a single-stage spool valve. In the analysis, neglect the inductivity of the torque-motor coil.

▪ According to the above discussions, the linear model that describes the actuator controlled by a single-stage valve is (A.27). If inductivity Lc is neglected, this model reduces to (A.29). Let us write this model again:

21 zz

Ltt

p

t

p

t

FM

zM

Az

M

Bz

M

Kz

13212

4323

44)(

4z

VKz

VCKz

VAz

t

eq

t

etpc

t

ep

(A.29)

54 zz

urrJM

RKz

rJM

rBBz

rJM

RrKKrKKKz

as

t

as

af

as

btmaf

)/(

/

/

/

/

//)(252

2

42

22

5

FL to load

Actuator

xp

rolling without sliding

1 3 2

Fig. A.8 Parameters of the load (wheels):

m1= m2= m3=m , r1= r2= r3=rw

Load

FL from load

where the state is

),,,,(),,,,( 54321 vvLpp xxPxxzzzzzz ▪ Note that there is FL in the model. To solve FL one needs to

describe mathematically the dynamics of the load !! ▪ The motion of each wheel equals the motion of the piston:

pxxxx 321

and so, there is no additional degree of freedom and no additional state variable.

NOTE that 1zxp

▪ Consider the wheel 1. ▫ for translation

11 frFFam ▫ for rotation

wfrw raFrI /,1

▫ For a cylinder: 2)2/1( wrmI

▫ Combining the above relations, one obtains:

12

3Fam (*)

z1

pushing force F1

Friction force Ffr1


212 frFFFam ▫ for rotation

2frw FrI



122

3FFam (**)


32 frL FFFam ▫ for rotation

3frw FrI



22

3FFam L (***)

oposing force F1

pushing force F2

Friction force Ffr2

z1

oposing force F2

pushing force FL

Friction force Ffr3

z1

▪ We now make a sum of equations (*), (**), and (***):

1)2/3( Fam (*)

12)2/3( FFam (**)

2)2/3( FFam L (***)

obtaining

LFam )2/9(

Since pxa , 1zxp , and 2zx p , it follows that LFzm 2)2/9(

▪ Now, the obtained expression for the force FL is substituted into

model (A.29) (in the second relation), to obtain:

21 zz

3212 )2/9()2/9()2/9(z

mM

Az

mM

Bz

mM

Kz

t

p

t

p

t

4323

44)(

4z

VKz

VCKz

VAz

t

eq

t

etpc

t

ep

(A.36)

54 zz

urrJM

RKz

rJM

rBBz

rJM

RrKKrKKKz

as

t

as

af

as

btmaf

)/(

/

/

/

/

//)(252

2

42

22

5

This is the final linear model in a canonic form, ready for simulation etc.

+

▪ It is possible to put the model in a matrix form.

u

E

Ez

z

z

z

z

D

DD

DDD

DDD

z

z

z

z

z

55

4

3

2

1

5554

343332

232221

5

4

3

2

1

0

0

0

0

000

10000

00

00

00010

(A.37)

where

mM

KD

t )2/9(21 ,

mM

BD

t

p

)2/9(22 ,

mM

AD

t

p

)2/9(23

t

ep V

AD4

32 , t

etpc V

CKD4

)(33 , t

eq V

KD4

34

2

22

54 /

//)(

rJM

RrKKrKKKD

as

btmaf

, 2

2

55 /

/

rJM

rBBD

as

af

,

rrJM

RKE

as

t

)/(

/25

♦ Nonlinear analysis ▪ In the case of a nonlinear analysis, the third relation of (A.29)

and (A.36), which is linearized, based on linear flow law (A.3):

34 zKzKPKxKQ cqLcvqL , is modified based on nonlinear flow law (5.33):

3

4

44

11z

z

zPzwCP

x

xPxwCQ sdL

v

vsvdL .

▪ With the modified third experssion, (A.36) becomes:

21 zz

3212 )2/9()2/9()2/9(z

mM

Az

mM

Bz

mM

Kz

t

p

t

p

t

3

4

44323

1444z

z

zPzwC

Vz

VCz

VAz sd

t

e

t

etp

t

ep

(A.38)

54 zz

urrJM

RKz

rJM

rBBz

rJM

RrKKrKKKz

as

t

as

af

as

btmaf

)/(

/

/

/

/

//)(252

2

42

22

5

This is the final nonlinear model in a canonic form, ready for simulation etc.

♠ EXAMPLE 2 Formulate the (a) linear and (b) nonlinear mathematical model of the system shown in Fig. A.9. The actuator is controlled by a single-stage spool valve. In the analysis,

- neglect the inductivity of the torque-motor coil; 0cL ; - neglect some other torque-motor parameters,

0,0,0,0,0 maaab KKBJK ;

- neglect the mechanical dynamics of the spool motion )0( sM ; and

- neglect all transient flow effects in the valve )0( fB .

▪ According to the above discussions, the linear model that

describes the actuator controlled by a single-stage valve is (A.27). If the listed parameters are neglected, the model reduces to (A.35). Let us write this model again

21 zz

Ltt

p

t

p

t

FM

zM

Az

M

Bz

M

Kz

13212 (A.35)

urK

RK

VKz

VCKz

VAz

f

t

t

eq

t

etpc

t

ep

/44)(

4323

This is a 3-rd order canonic model, where the state is

),,(),,( 321 Lpp Pxxzzzz

from load | to load FL

Actuator

xp

no friction

Fig. A.9 Paramaters of the load:

mass m , stiffness k and damping b

Load

▪ Note that there is FL in the model. To solve FL one needs to

describe mathematically the dynamics of the load !! ▪ The motion of the load body equals the motion of the piston:

pL xx

and so, there is no additional degree of freeedom and no additional state variable.

NOTE that 1zxp

▪ Consider the load body. ▫ Newton’s law

bkL FFFam

▫ Spring (stiffness) force is 1zkFk

Damping force is 21 zbzbFb

Acceleration is 21 zza ▫ Now, Newton’s law is

212 zbzkFzm L 212 zbzkzmFL ▪ Now, the obtained expression for the force FL is substituted into

actuator model (A.35) (in the second relation), to obtain:

Fb

Fk FL

z1

21 zz

3212 zmM

Az

mM

bBz

mM

kKz

t

p

t

p

t

(A.39)

urK

RK

VKz

VCKz

VAz

f

t

t

eq

t

etpc

t

ep

/44)(

4323

This is the final linear model in a canonic form, ready for simulation etc. ▪ It is possible to put the model in a matrix form.

u

E

Ez

z

z

D

DD

DDD

z

z

z

33

2

1

3332

232221

3

2

1

0

0

0

010

(A.40)

where

mM

kKD

t

21 , mM

bBD

t

p

22 ,

mM

AD

t

p

23

t

ep V

AD4

32 , t

etpc V

CKD4

)(33 , rK

RK

VKE

f

t

t

eq

/43

♦ Nonlinear analysis

▪ In the case of a nonlinear analysis, the third relation of (A.35) and (A.39), which is linearized, based on linear flow law (A.3):

34 zKzKPKxKQ cqLcvqL , is modified based on nonlinear flow law (5.33):

3

4

44

11z

z

zPzwCP

x

xPxwCQ sdL

v

vsvdL .

▪ With the modified third expression, (A.39) becomes:

21 zz

3212 zmM

Az

mM

bBz

mM

kKz

t

p

t

p

t

(A.41)

3

4

44323

1444z

z

zPzwC

Vz

VCz

VAz sd

t

e

t

etp

t

ep

▪ Variable 4zxv is eliminated from the 5-th relation of (A.32):

urK

RKz

f

t /4

producing finally

21 zz

3212 zmM

Az

mM

bBz

mM

kKz

t

p

t

p

t

(A.42)

uzu

uPwCzCzAz

rRKK

rRKK

rK

RK

VVV ft

ft

f

t

t

e

t

e

t

esdtpp

3323

)/(

)/(1/444

This is the final nonlinear model in a canonic form, ready for simulation etc.

♠ EXAMPLE 3 The preceding example is modified

- introducing a different load (see Fig. A.10), and - looking for the linear model only.

▪ According to the above discussions, the linear model that

describes the actuator controlled by a single-stage valve is (A.27). If the listed parameters are neglected, the model reduces to (A.35). Let us write this model again

21 zz

Ltt

p

t

p

t

FM

zM

Az

M

Bz

M

Kz

13212 (A.35)

urK

RK

VKz

VCKz

VAz

f

t

t

eq

t

etpc

t

ep

/44)(

4323

xL xp

from load | to load FL

Actuator

xp

no friction

Fig. A.10 Parameters of the load:

mass m , stiffness k and damping b

Load


),,(),,( 321 Lpp Pxxzzz ▪ Note that there is FL in the model. To solve FL one needs to

describe mathematically the dynamics of the load !! ▪ First, we note that FL eqauls the summ of spring force and

damper force. If we assume that the spring is compressed, then

)()( LpLpbkL xxbxxkFFF (A.43) ▪ On the right-hand side of the spring, the same force acts on the

body. ▪ It is important to notice that the load intruduces one additional

degree of freedom – motion of the body ( xL ) is not not geometrically dependent on the piston motion ( xp ). This is due to the deformable spring (and damper).

This additional degree of freedom ( xL ) introduces two new

state variables LL xzxz 54 , , with the property

54 zz (A.44)

The complete state vector is now

),,,,(),,,,( 54321 LLLpp xxPxxzzzzzz

▪ On the right-hand side of the spring, the force given by (A.43)

acts on the body. It can be rewritten in terms of state variables:

5421

5241 )()(

bzkzbzkz

zzbzzkFL

(A.45)

▫ Consider the dynamics (Newton’s law)

bk FFam i.e.

)()( 5241 zzbzzkam

▫ Acceleration is 5zxa L ▫ Now, Newton’s law becomes

)()( 52415 zzbzzkzm (A.46)

▪ The complete mathematical model of the system ios now

obtained by substituting FL from (A.45) into the second relatioon of (A.35) , and supplemented equations (A.44) and (A.46). This yields:

21 zz

543212 zM

bz

M

kz

M

Az

M

bBz

M

kKz

ttt

p

t

p

t

urK

RK

VKz

VCKz

VAz

f

t

t

eq

t

etpc

t

ep

/44)(

4323

(A.47)

54 zz

54215 zm

bz

m

kz

m

bz

m

kz

mass

m

Fk

Fb

This is the final linear model in a canonic form, ready for simulation etc. ▪ The model can be put in a matrix form:

u

E

rK

RK

VK

z

z

z

z

z

Dm

b

m

k

m

b

m

k

VCK

VA

M

b

M

k

M

A

M

bB

M

kK

z

z

z

z

z

f

t

t

eq

t

etpc

t

ep

ttt

p

t

p

t

0

0

/40

0

0

10000

004

)(4

0

00010

5

4

3

2

1

5

4

3

2

1

(A.48)

HOMEWORK – Ramadan Formulate the nonlinear mathematical model of the shown system. The actuator is controlled by a single-stage spool valve. In the analysis,



- neglect the mechanical dynamics of the spool motion )0( sM ;

- neglect all all transient flow effects in the valve )0( fB .

Paramaters of the load: mass m , radius r stiffness k and damping b

All paramaters of the actuator are known

no sliding

HOMEWORK – Mohamed Formulate the linear mathematical model (including matrix form) of the shown system. The actuator is controlled by a single-stage spool valve. In the analysis,



- neglect the mechanical dynamics of the spool motion )0( sM ;

- neglect all all transient flow effects in the valve )0( fB .

no sliding

Paramaters of the load: mass m , radius r, and damping b


HOMEWORK – Ismail

Formulate the linear mathematical model of the shown system. The actuator is controlled by a single-stage spool valve. In the analysis, neglect

- the inductivity of the torque-motor coil, 0cL ; - neglect some other torque-motor parameters,

0,0,0,0,0 maaab KKBJK

HOMEWORK – Abd alhalim

Formulate the linear mathematical model of the shown system. The actuator is controlled by a single-stage spool valve. In the analysis, neglect

- the inductivity of the torque-motor coil, 0cL ; - neglect some other torque-motor parameters,

0,0,0,0,0 maaab KKBJK

no friction

Paramaters of the load: mass m , and stiffness k


no sliding

Paramaters of the load: mass m , radius r, and damping b


7.4. and 7.5. Two-Stage Electrohydraulic Servovalve ▪ Two-stage servovalves overcome the disadvantages of limited

flow capacity and instability which are inherent in single-stage servovalves.

▪ Two-stage servovalves with position feedback are most common

and have pressure-flow curves as shown in Fig. 7.1a.

▪ Position feedback may be achieved in three basic ways:

- direct position feedback , - using a spring to convert position to a force signal

which is feedback to the torque motor, - placing stiff springs to center the spool.

We are going to discuss the second type. The first type should be read from the book (only the principles) ; We don’t have enough time to elaborate this type here. The third type is not considered in this course.

Increasing

current i

Ps 0

QL

0 PL

Fig. 7.1 a

7.4. Two-Stage Electrohydraulic Servovalve with Force Feedback

▪ Two-stage valve means the two valves. ▪ The first stage is a flapper valve (Fig.A.11).

The flapper can open or close the nozzles D1 and D2.

▪ The second step is a spool valve (Fig. A.12). ▪ The flapper is connected to the spool by means of an elastic

element (feedback spring in the form of a flexible cantilever element).

Torque motor

Flapper valve

Fig. A.11 D1 , D2 - nozzles

Fig. A.12

xv

Flexible cantilever element – spring

Horizontal position of the armature

Vertical position of the flapper

xv represents the deformation (deflection)

Stady-state position of the spool

Fig. A.13

r

► How does this system work ? ▪ In our explanation, we adopt some simplifications.

○ We neglect the entire dynamics of the torque-motor : - neglect the inductivity of the torque-motor coil; 0cL ; - neglect some other torque-motor parameters,

0,0 ab JK , where Ja includes the flapper inertia; and

0,0,0 maa KKB ○ We neglect the entire dynamics of the valve :

- neglect the mechanical dynamics of the spool motion )0( sM ; and

- neglect all transient flow effects in the valve )0( fB . ▪ The original equations of torque-motor dynamics (electrical and

mechanical) are (A.20) and (A.22). Let us write them again:

dtdiLKiRu cb / (A.20)

Lmaaat TKKBJiK )( (A.22)=(7.31)

▪ For the adopted simplifications, one obtains

iRu (A.49)

Lt TiK (A.50)

where: u is the control volatage, R is the armature resistance,

i is the current, Kt is the torque constant, TL is the output motor torque.

► Let us explain the work !

Suppose that some voltage is applied, and that the current forces the armature to turn anti clockwise, as in Fig. A.12a. The flapper moves to the right, thus closing nozzle D2.

The pressure supply line Ps 2 (the right one) is now closed and the oil from the left line, Ps 1 , flows through pipe C1 into the cylinder. So, the actuator piston moves to the right. Pipe C2 allows the oil to flow out from the cylinder to the return line R (back to the reservoir).

Since nozzle D2 is closed, the oil in the right supply line exerts strong pressure upon the right-hand side of the valve spool forcing it to move to the left.

This motion causes the deformation of the feedback spring. At some deformation, the elastic torque of the deformed spring starts to turn the armature back (clockwise) and the flapper to the left thus opening nozzle D2.

When the oil begins to flow through D2 , the pressure acting upon the right-hand side of the spool reduces, but it is still stronger then the pressure acting upon the left-hand side. Hence, the spool continues moving to the left.

The pressure on the both sides of the valve spool balances when the flows through D1 and D2 become equal. This means the vertical position of the flapper, that is, the horiozontal position of the armature (as shown in Fig. A.12b and A.13). The motion of the spool stops.

In this position, the motor torque equals the spring deformation torque. The coordinate xv defines the position of the spool. This can be seen as the cantilever deformation (deflection), as shown ion Fig. A.13. So, it holds that :

- The motor torque is iKT tL , according to (A.50).

- The deformation force is vf xKF , where Kf is the spring constant (dependent on the length of the beam, its cross-section, and the E modulus).

The balance means:

rFTL i.e.

vft xKriK irK

Kx

f

tv (A.51)

or

uRKr

Kx

f

tv (A.52)

This ballanced position of the valve spool xv corresponds to some value of oil flow and accordingly to some velocity of the piston in the cylinder. Since xv is directly controlled by the voltage u (relation (A.52)), we achieve the possibility of controlling the flow and the actuator speed.

The expression (A.52) means that there exists a direct

relation between the input control volatage u and the spool position xv. Since all the dynamics of the servovalve (valve plus torque motor) is neglected, any change of control voltage is followed by an instantaneous change of the spool position. This is a simplified situation. In reality, the change of voltage would be folowed by a transient phase before this ballanced position of the spool is reached (this is due to the dynamics of the servovalve) One should say that is rather common to look at current i as an input control. In such case, (A.51) is more important then (A.52)

♦ Mathematical modeling We start modeling by using relations (I)-(III), that described the

actuator controlled by the spool motion xv. Let us repeat these relations: (I) ► Dynamics of the piston motion:

LpppptLp FKxxBxMPA (6.34)

(II) ► Load flow:

Le

tLtpppL P

VPCxAQ

4 (6.33)

(III) ► Valve control the flow by relation - (6.3) in the case of linear analysis, or - (5.33) in the case of non-linear analysis:

LcvqL PKxKQ (6.3) or

L

v

vsvdLvLL P

x

xPxwCPxQQ

1

),( (5.33)

▪ State variables were piston position, its velocity, and load

pressure: ),,(),,( 111 Lpp Pxxzzzz .

▪ Model (I)-(III), in a liner case, was transformed into a canonic state-space form (A.11). We repeat this model :

21 zz

Ltt

p

t

p

t

FM

zM

Az

M

Bz

M

Kz

13212 (A.11)

vt

eq

t

etpc

t

ep x

VKz

VCKz

VAz

44)(

4323

▪ In a nonliner case, (I)-(III) was transformed into (A.13). We

repeat:

21 zz

Ltt

p

t

p

t

FM

zM

Az

M

Bz

M

Kz

13212 (A.13)

3323

1444z

x

xPxwC

Vz

VCz

VAz

v

vsvd

t

e

t

etp

t

ep

Note that when repeating (A.11) and (A.13), notation for spool motion is xv , and not u, as it was originally in (A.11) and (A.13). This is due to the fact that in original models, spool motion was the control input and now it is not ! Now, u is used for the voltage (new control input).

▪ Since xv directly follows from the applied voltage, according to (A.52), this expression should be substituted into (A.11) and (A.13).

▪ In this way the linar model is obtained in the form:

21 zz

Ltt

p

t

p

t

FM

zM

Az

M

Bz

M

Kz

13212 (A.53)

uRKr

K

VKz

VCKz

VAz

f

t

t

eq

t

etpc

t

ep

44)(

4323

(A.55)

▪ The model can be put in a matrix form:

LF

H

tMu

E

fRKrtK

tVe

qKz

z

z

D

tVe

tpCcKtVe

pA

tM

pA

tM

pB

tM

K

zzz

0

10

40

0

3

2

1

4)(

40

010

3

2

1

(A.54)

▪ For the nonlinar model, it is obtained that :

21 zz

Ltt

p

t

p

t

FM

zM

Az

M

Bz

M

Kz

13212

3323

1444z

u

u

PuRKr

KwC

Vz

VCz

VAz

fRKrtK

fRKrtK

sf

td

t

e

t

etp

t

ep

NOTE that there is still FL in all these models. In order to eliminate the load force, it is necessary to specify the load and write the mathematical expressions that describe the load dynamics. NOTE alsio that (A.51) is commonly used instead of (A.52) when substituting into (A.11) and (A.13). This means that the

current i is seen as a control input. u and i are related by simple expression (A.49).

♠ EXAMPLE 1 Formulate the (a) linear and (b) nonlinear mathematical model of the system shown in Fig. A.14. The actuator is controlled by a two-stage servovalve with force feedback. In the analysis, neglect the parameters according to the above discussion !

▪ According to the above discussions, the linear model that describes the actuator controlled by a two-stage valve with force feedback is (A.53). Write it again:

21 zz

Ltt

p

t

p

t

FM

zM

Az

M

Bz

M

Kz

13212 (A.53)

uRKr

K

VKz

VCKz

VAz

f

t

t

eq

t

etpc

t

ep

44)(

4323

1 2

FL to load

Actuator

xp


Fig. A.14

Parameters of the load wheels: m1= m2=m , r1= r2=rw

spring: stiffness k

Load

FL from load



describe mathematically the dynamics of the load !! ▪ It is important to notice that the load intruduces one additional

degree of freedom – motion of the wheel 1 (coordinate x1 ) is

not not geometrically dependent on the piston motion ( xp ). This is due to the deformable spring.

This additional degree of freedom ( x1 ) introduces two new

state variables 1514 , xzxz , with the property

54 zz (A.56)


),,,,(),,,,( 1154321 xxPxxzzzzzz Lpp

▪ Dynamics of the wheel 1 ▫ Translation:

11111 )( frpfr FxxkFFam ▫ Rotation:

wfr rFI 11 , where wra /11 and 2)2/1( wrmI

x1

Ffr1

F1

▫ By combination:

)(2

311 xxkma p

i.e.

)(3

211 xx

m

ka p (A.57)


11

112

)( frpL

frL

FxxkF

FFFam

▫ Rotation:

wfr rFI 22 , where wra /22 and 2)2/1( wrmI ▫ By combination:

)(2

312 xxkFma pL

i.e.

)(2

312 xxkmaF pL . (A.58)

▪ Let us rewrite (A.57) and (A.58) in the state-space notation. It

holds that:

22511 , zxazxa p

xp

Ffr2

FLF1

(A.57) and (A.58) becomes:

)(3

2415 zz

m

kz (A.59)

)(2

3412 zzkzmFL . (A.60)

▪ Now, load force FL is substituted from (A.60) into the model

(A.53) (into its second relation). After that, (A.56) and (A.59) are suppelemented to this set.

In this way, one obtains:

21 zz

43212 )2/3()2/3()2/3()2/3(z

mM

kz

mM

Az

mM

Bz

mM

kKz

tt

p

t

p

t

uRKr

K

VKz

VCKz

VAz

f

t

t

eq

t

etpc

t

ep

44)(

4323

54 zz

)(3

2415 zz

m

kz

▪ This is a final linear model of the cemplete system. ▪ We now look for a nonlinear model. (A.55) is used instead of (A.53). In fact, the difference is in the third expression only:

(A.61)

21 zz

43212 )2/3()2/3()2/3()2/3(z

mM

kz

mM

Az

mM

Bz

mM

kKz

tt

p

t

p

t

3323

1444z

u

u

PuRKr

KwC

Vz

VCz

VAz

fRKrtK

fRKrtK

sf

td

t

e

t

etp

t

ep

54 zz

)(3

2415 zz

m

kz

▪ This is a final nonlinear model of the cemplete system.

(A.62)

♠ EXAMPLE 2 Formulate the linear mathematical model in a matrix form, for the system shown in Fig. A.15. The actuator is controlled by a two-stage servovalve with force feedback. In the analysis, neglect the parameters according to the above discussion !

▪ The linear model that describes the actuator controlled by a two-stage valve with force feedback has a third order cononic form (A.53). Write it again:

21 zz

Ltt

p

t

p

t

FM

zM

Az

M

Bz

M

Kz

13212 (A.53)

uRKr

K

VKz

VCKz

VAz

f

t

t

eq

t

etpc

t

ep

44)(

4323

x1

xp

FL from load

1

2 FL to load

Actuator

Fig. A.15

masses m1= m2=m , radius rw

spring: stiffness k

Load

where the state is ),,(),,( 321 Lpp Pxxzzz ▪ In the model, the sign of the load force FL is changed since the

direction of the force in this example is oposite. ▪ Load force FL is unknown. In order to eliminate it, one needs

the mathematical description of the load dynamics !!

▪ Let us see if there is additional degee of freedom with the load. ▫ Rotation of the wheel (number 2) is directly (geoemetrically)

depending on the piston motion xp and can be expressed in

terms of xp. ▫ However, the motion of the body 1 is not geometrically

dependent on the piston motion xp . This is due to the deformable spring. So, there is an additional degree of freedom and the body 1 has its independent position coordinate x1.

▫ This additional degree of freedom ( x1 ) introduces two new


54 zz (A.63)


),,,,(),,,,( 1154321 xxPxxzzzzzz Lpp

▪ Dynamics of the body 1 ▫ Translation (Newton’s law):

11 Fmgam

▫ Spring force is: )( 11 pxxkF

▫ And hence )( 11 pxxkmgam

▫ Since 411, zxzx p and 511 zxa , the state-space form is

)( 145 zzkmgzm i.e.

)( 145 zzm

kgz (A.64)

▪ Dynamics of the wheel 2 ▫ Rotation:

wLw rFrFI 12 ,

where wra /22

and 2)2/1( wrmI

▫ By combination:

Lp FFxm 12

1

and introducing the expression for the spring force:

Lpp Fxxkxm )(2

11 ppL xmxxkF

2

1)( 1

load force FL

spring

force F1

gravity

force mg

spring

force F1

and in the state-space:

214 2

1)( zmzzkFL (A.65)

▪ Now, we combine the actuator and the load.

- Load force FL is substituted from (A.65) into the actuator model (A.53) (into its second relation).

- After that, (A.63) and (A.64) are suppelemented to this set.


21 zz

43212 )2/1()2/1()2/1()2/1(z

mM

kz

mM

Az

mM

Bz

mM

kKz

tt

p

t

p

t

uRKr

K

VKz

VCKz

VAz

f

t

t

eq

t

etpc

t

ep

44)(

4323

54 zz

415 zm

kz

m

kgz

▪ This is a final linear model of the cemplete system. ▪ The matrix form is:

(A.66)

g

uRKr

K

VK

z

z

z

z

z

m

k

m

k

VCK

VA

mM

k

mM

A

mM

B

mM

kK

z

z

z

z

z

E

D

f

t

t

eq

t

etpc

t

ep

tt

p

t

p

t

0

0

0

0

0

0

40

0

000

10000

004

)(4

0

0)2/1()2/1()2/1()2/1(

00010

5

4

3

2

1

5

4

3

2

1

(A.67)

♠ EXAMPLE 3 Formulate the nonlinear mathematical model of the system shown in Fig. A.16. The actuator is controlled by a two-stage servovalve with force feedback.

angle

θ xp

FL from load

FL to load

Fig. A.16

mass m ,

length b

slider: mass

neglectedd

(A.55)

▪ The nonlinear model that describes the actuator controlled by a two-stage valve with force feedback has a third order cononic form (A.53). Write it again:

21 zz

Ltt

p

t

p

t

FM

zM

Az

M

Bz

M

Kz

13212

3323

1444z

u

u

PuRKr

KwC

Vz

VCz

VAz

fRKrtK

fRKrtK

sf

td

t

e

t

etp

t

ep

where the state is ),,(),,( 321 Lpp Pxxzzz ▪ The system is nonlinear from two reasons:

- the actuator is considered nonlinear, and - the load is nonlinear.

▪ There is no additional degree of freedom with the load. The

rotation of the load (angle θ) is directly (i.e. geometrically)

dependent on the piston motion xp. So, no additional state variable !

Angle θ can be expressed in terms of xp :

d

xptan (A.68)

d

θ

xp

▪ We write the dynamics of the load in order to eliminate FL. ▫ Rotation

aFMI n ▫ Moment of inertia is

2

3

1bmI

▫ It holds that:

cosFFn , cos/da dFaFM Ln ▫ Rotation equation becomes

dFbm L23

1

d

bmFL 3

2

(A.69)

▫ Angular acceleration has to be expressed in terms of xp and its derivatives because the idea is to substitute (A.69) into the actuator (A.55) model and thus eliminate FL.

We use (A.68) to relate θ and xp:

tandx p (A.68)

and accordingly

2cos

1dx p (A.70)

23 cos

1

cos

sin2ddx p (A.71)

a θ

θ Fn – normal component

FL

d

▫ There exists ono-to-one correspondence between θ and xp and accordingly between angular acceleration and

piston acceleration px . However, the relations (A68), (A.70) and (A.71) are rather complex and it is not easy to

express in terms of xp and its derivatives. ▫ Let us try to do what is stated above (express in terms

of ppp xxx ,, .

- we substitute from (A.70) into (A.71):

2cos

1

cos

sin2dxx pp

- wherefrom:

pp xd

xd

cossin2cos 2

- sin and cosin functions can be expressed in terms

of tan:

2tan1

tansin

,

2tan1

1cos

- So, one obtains

pp xd

xd

)tan1(

tan2

)tan1(

122

- Using (A.68) (i.e. dxdx pp /tantan ), it is

pp

pp

p

xdxd

dxx

dxd

))/(1(

)/(2

))/(1(

122

(A.74)

▫ Now, the load force (A.69) becomes

pp

pp

p

pp

pp

pL

xdxd

bmx

dx

dx

d

bm

xdxd

dxx

dxdd

bmF

22

2

22

2

22

2

)/(1

1

3)/(1

)/(2

3

)))/(1(

)/(2

))/(1(

1(

3

(A.72) ▫ In the state-space:

221

2

2

221

12

2

)/(1

1

3)/(1

)/(2

3z

dzd

bmz

dz

dz

d

bmFL

(A.73)

▪ Let us combine the actuator and the load. Load force FL is

substituted from (A.73) into the actuator model (A.55) (into its second relation).

21 zz

3

212

22

212

2

21

1

2

2

1

212

22

)(1

1

3)(1

1

3

)(1

)(2

3

)(1

1

3

zM

d

zd

bm

Az

M

d

zd

bm

B

d

zd

z

d

bm

zM

d

zd

bmK

z

t

p

t

p

t

3323

1444z

u

u

PuRKr

KwC

Vz

VCz

VAz

fRKrtK

fRKrtK

sf

td

t

e

t

etp

t

ep

▪ This is a final model of the cemplete system. It is strongly nonlinear. It is of the third order.

♠ EXAMPLE 1 Formulate the (a) linear and (b) nonlinear mathematical model of the system shown in Fig. A.14. The actuator is controlled by a two-stage servovalve with force feedback. In the analysis, neglect the parameters according to the above discussion !

▪ According to the above discussions, the linear model that describes the actuator controlled by a two-stage valve with force feedback is (A.53). Write it again:

21 zz

Ltt

p

t

p

t

FM

zM

Az

M

Bz

M

Kz

13212 (A.53)

uRKr

K

VKz

VCKz

VAz

f

t

t

eq

t

etpc

t

ep

44)(

4323

1 2

FL to load

Actuator

xp


Fig. A.14

Parameters of the load wheels: m1= m2=m , r1= r2=rw

spring: stiffness k

Load

FL from load



describe mathematically the dynamics of the load !! ▪ It is important to notice that the load intruduces one additional

degree of freedom – motion of the wheel 1 (coordinate x1 ) is

not not geometrically dependent on the piston motion ( xp ). This is due to the deformable spring.

This additional degree of freedom ( x1 ) introduces two new


54 zz (A.56)


),,,,(),,,,( 1154321 xxPxxzzzzzz Lpp


11111 )( frpfr FxxkFFam ▫ Rotation:

wfr rFI 11 , where wra /11 and 2)2/1( wrmI

x1

Ffr1

F1

▫ By combination:

)(2

311 xxkma p

i.e.

)(3

211 xx

m

ka p (A.57)


11

112

)( frpL

frL

FxxkF

FFFam

▫ Rotation:

wfr rFI 22 , where wra /22 and 2)2/1( wrmI ▫ By combination:

)(2

312 xxkFma pL

i.e.

)(2

312 xxkmaF pL . (A.58)

▪ Let us rewrite (A.57) and (A.58) in the state-space notation. It

holds that:

22511 , zxazxa p

xp

Ffr2

FLF1

(A.57) and (A.58) becomes:

)(3

2415 zz

m

kz (A.59)

)(2

3412 zzkzmFL . (A.60)

▪ Now, load force FL is substituted from (A.60) into the model

(A.53) (into its second relation). After that, (A.56) and (A.59) are suppelemented to this set.


21 zz

43212 )2/3()2/3()2/3()2/3(z

mM

kz

mM

Az

mM

Bz

mM

kKz

tt

p

t

p

t

uRKr

K

VKz

VCKz

VAz

f

t

t

eq

t

etpc

t

ep

44)(

4323

54 zz

)(3

2415 zz

m

kz

▪ This is a final linear model of the cemplete system. ▪ We now look for a nonlinear model. (A.55) is used instead of (A.53). In fact, the difference is in the third expression only:

(A.61)

21 zz

43212 )2/3()2/3()2/3()2/3(z

mM

kz

mM

Az

mM

Bz

mM

kKz

tt

p

t

p

t

3323

1444z

u

u

PuRKr

KwC

Vz

VCz

VAz

fRKrtK

fRKrtK

sf

td

t

e

t

etp

t

ep

54 zz

)(3

2415 zz

m

kz

▪ This is a final nonlinear model of the cemplete system.

(A.62)

♠ EXAMPLE 2 Formulate the linear mathematical model in a matrix form, for the system shown in Fig. A.15. The actuator is controlled by a two-stage servovalve with force feedback. In the analysis, neglect the parameters according to the above discussion !

▪ The linear model that describes the actuator controlled by a two-stage valve with force feedback has a third order cononic form (A.53). Write it again:

21 zz

Ltt

p

t

p

t

FM

zM

Az

M

Bz

M

Kz

13212 (A.53)

uRKr

K

VKz

VCKz

VAz

f

t

t

eq

t

etpc

t

ep

44)(

4323

x1

xp

FL from load

1

2 FL to load

Actuator

Fig. A.15

masses m1= m2=m , radius rw

spring: stiffness k

Load

where the state is ),,(),,( 321 Lpp Pxxzzz ▪ In the model, the sign of the load force FL is changed since the

direction of the force in this example is oposite. ▪ Load force FL is unknown. In order to eliminate it, one needs

the mathematical description of the load dynamics !!

▪ Let us see if there is additional degee of freedom with the load. ▫ Rotation of the wheel (number 2) is directly (geoemetrically)

depending on the piston motion xp and can be expressed in

terms of xp. ▫ However, the motion of the body 1 is not geometrically

dependent on the piston motion xp . This is due to the deformable spring. So, there is an additional degree of freedom and the body 1 has its independent position coordinate x1.

▫ This additional degree of freedom ( x1 ) introduces two new


54 zz (A.63)


),,,,(),,,,( 1154321 xxPxxzzzzzz Lpp

▪ Dynamics of the body 1 ▫ Translation (Newton’s law):

11 Fmgam

▫ Spring force is: )( 11 pxxkF

▫ And hence )( 11 pxxkmgam

▫ Since 411, zxzx p and 511 zxa , the state-space form is

)( 145 zzkmgzm i.e.

)( 145 zzm

kgz (A.64)

▪ Dynamics of the wheel 2 ▫ Rotation:

wLw rFrFI 12 ,

where wra /22

and 2)2/1( wrmI

▫ By combination:

Lp FFxm 12

1

and introducing the expression for the spring force:

Lpp Fxxkxm )(2

11 ppL xmxxkF

2

1)( 1

load force FL

spring

force F1

gravity

force mg

spring

force F1

and in the state-space:

214 2

1)( zmzzkFL (A.65)

▪ Now, we combine the actuator and the load.

- Load force FL is substituted from (A.65) into the actuator model (A.53) (into its second relation).

- After that, (A.63) and (A.64) are suppelemented to this set.


21 zz

43212 )2/1()2/1()2/1()2/1(z

mM

kz

mM

Az

mM

Bz

mM

kKz

tt

p

t

p

t

uRKr

K

VKz

VCKz

VAz

f

t

t

eq

t

etpc

t

ep

44)(

4323

54 zz

415 zm

kz

m

kgz

▪ This is a final linear model of the cemplete system. ▪ The matrix form is:

(A.66)

g

uRKr

K

VK

z

z

z

z

z

m

k

m

k

VCK

VA

mM

k

mM

A

mM

B

mM

kK

z

z

z

z

z

E

D

f

t

t

eq

t

etpc

t

ep

tt

p

t

p

t

0

0

0

0

0

0

40

0

000

10000

004

)(4

0

0)2/1()2/1()2/1()2/1(

00010

5

4

3

2

1

5

4

3

2

1

(A.67)

♠ EXAMPLE 3 Formulate the nonlinear mathematical model of the system shown in Fig. A.16. The actuator is controlled by a two-stage servovalve with force feedback.

angle

θ xp

FL from load

FL to load

Fig. A.16

mass m ,

length b

slider: mass

neglectedd

(A.55)

▪ The nonlinear model that describes the actuator controlled by a two-stage valve with force feedback has a third order cononic form (A.53). Write it again:

21 zz

Ltt

p

t

p

t

FM

zM

Az

M

Bz

M

Kz

13212

3323

1444z

u

u

PuRKr

KwC

Vz

VCz

VAz

fRKrtK

fRKrtK

sf

td

t

e

t

etp

t

ep

where the state is ),,(),,( 321 Lpp Pxxzzz ▪ The system is nonlinear from two reasons:

- the actuator is considered nonlinear, and - the load is nonlinear.

▪ There is no additional degree of freedom with the load. The

rotation of the load (angle θ) is directly (i.e. geometrically)

dependent on the piston motion xp. So, no additional state variable !

Angle θ can be expressed in terms of xp :

d

xptan (A.68)

d

θ

xp

▪ We write the dynamics of the load in order to eliminate FL. ▫ Rotation

aFMI n ▫ Moment of inertia is

2

3

1bmI

▫ It holds that:

cosFFn , cos/da dFaFM Ln ▫ Rotation equation becomes

dFbm L23

1

d

bmFL 3

2

(A.69)

▫ Angular acceleration has to be expressed in terms of xp and its derivatives because the idea is to substitute (A.69) into the actuator (A.55) model and thus eliminate FL.

We use (A.68) to relate θ and xp:

tandx p (A.68)

and accordingly

2cos

1dx p (A.70)

23 cos

1

cos

sin2ddx p (A.71)

a θ

θ Fn – normal component

FL

d

▫ There exists ono-to-one correspondence between θ and xp and accordingly between angular acceleration and

piston acceleration px . However, the relations (A68), (A.70) and (A.71) are rather complex and it is not easy to

express in terms of xp and its derivatives. ▫ Let us try to do what is stated above (express in terms

of ppp xxx ,, .

- we substitute from (A.70) into (A.71):

2cos

1

cos

sin2dxx pp

- wherefrom:

pp xd

xd

cossin2cos 2

- sin and cosin functions can be expressed in terms

of tan:

2tan1

tansin

,

2tan1

1cos

- So, one obtains

pp xd

xd

)tan1(

tan2

)tan1(

122

- Using (A.68) (i.e. dxdx pp /tantan ), it is

pp

pp

p

xdxd

dxx

dxd

))/(1(

)/(2

))/(1(

122

(A.74)

▫ Now, the load force (A.69) becomes

pp

pp

p

pp

pp

pL

xdxd

bmx

dx

dx

d

bm

xdxd

dxx

dxdd

bmF

22

2

22

2

22

2

)/(1

1

3)/(1

)/(2

3

)))/(1(

)/(2

))/(1(

1(

3

(A.72) ▫ In the state-space:

221

2

2

221

12

2

)/(1

1

3)/(1

)/(2

3z

dzd

bmz

dz

dz

d

bmFL

(A.73)

▪ Let us combine the actuator and the load. Load force FL is

substituted from (A.73) into the actuator model (A.55) (into its second relation).

21 zz

3

212

22

212

2

21

1

2

2

1

212

22

)(1

1

3)(1

1

3

)(1

)(2

3

)(1

1

3

zM

d

zd

bm

Az

M

d

zd

bm

B

d

zd

z

d

bm

zM

d

zd

bmK

z

t

p

t

p

t

3323

1444z

u

u

PuRKr

KwC

Vz

VCz

VAz

fRKrtK

fRKrtK

sf

td

t

e

t

etp

t

ep

▪ This is a final model of the cemplete system. It is strongly nonlinear. It is of the third order.

DIFFERENT EXAMPLES

♠ EXAMPLE 1 Formulate the mathematical model of the system shown in Fig. A.17. The actuator is controlled by a single-stage servovalve. In the analysis, use the linear model of the actuator, but don’t use any other approximation!

angle θ

negligible mass, length l

FL to load

Actuator

xp


Fig. A.17

Load parameters.

wheel: m , rw

Load

FL from load

x1

▪ At the begining, we look for an appropriate mathematical model of the actuator. → For an actuator controled by a single-stage servovalve, with

no approximation, we use model (A.27). It is a linear model (as required in this example).

We write it again:

21 zz

Ltt

p

t

p

t

FM

zM

Az

M

Bz

M

Kz

13212

4323

44)(

4z

VKz

VCKz

VAz

t

eq

t

etpc

t

ep

(A.27)

54 zz

6252

2

42

2

5 )/(/

/

/

/)(z

rrJM

Kz

rJM

rBBz

rJM

rKKKz

as

t

as

af

as

maf

uL

zL

Rz

rL

Kz

ccc

b 1646

where the state variables (state vector) are:

),,,,,(),,,,,( 654321 ixxPxxzzzzzzz vvLpp → Force FL has to be eleiminated from the equations of load

dynamics.

▪ We note that the load does not introduce any additional degree of freedom. So, the satate vector remains as it is.

→ Position variables of the load, angle θ and coordinate x1 are

directly geometrically dependent on the piston motion xp.

▪ Let us look for the relation between θ, x1 and xp !

▫ First, it is clear that

www rxrxrx 111 ,, (A.75)

▫ Now, let us relate θ and xp

From the figure: sincoscos wwp rrllx

Next, one use the sinus theorem: lrw

)2/sin(sin

sinwr x1= rw θ

α

cosl

θ φ

coslxp

i.e.

lrw

cossin 2

2

cos1coscossin

l

r

l

r ww

(A.76)

So, the relation between θ and xp is:

22

cos1sincos

l

rlrrlx w

wwp (A.77)

It allows to express θ (and its derivatives) in terms of xp (and its derivatives):

),,(...,),(...,)(... pppppp xxxfxxfxf (A.78)

(Well, this is not simple, but generally it is possible to do!!)

▪ Dynamics of the load.

π/2-θ-φ

FL /cos φ

FL FL /cos φ

FL

φ θ

(FL /cos φ)sin(π/2-θ-φ) = = (FL /cos φ)cos(θ+φ)

Ffr

▫ Translation: frL FFxmma 1 and according to (A.75):

frLw FFrm (A.79)

▫ Rotation:

wfrwL rFrFII )cos()cos/( (A.80)

and by elimination Ffr from (A.79) and (A.80), dynamics of the load becomes:

wwLwL rrmFrFI )()cos()cos/( (A.81)

▪ FL can ber expressed from (A.81):

w

wL r

rmIF

)cos/)cos(1(

2

(A.82)

▪ Now, this FL should be substituted into actuator model (A.27). Before that, one should do the following:

- express φ in terms of θ (by using (A.76)) , and

- express , in terms of ppp xxx ,, by using (A.78).

So, FL will be a function of ppp xxx ,, .

▪ When FL is substituted into actuator model (A.27), the example will be accomplished. There will be 6 equations in canonic form, with 6 state variables.

▪ What is important to notice is that a “simple” load can produce

huge problems in modeling the system.

♠ EXAMPLE 2 Formulate the mathematical model of the system shown in Fig. A.18. The actuator is controlled by a single-stage servovalve. In the analysis,

- use the nonlinear model of the actuator, - neglect the torque-motor electrical dynamics ( 0cL ), - neglect the torque-motor mechanical dynamics

( 0,0,0,0,0 maaab KKBJK ), - neglect the dynamics of the spool and the transient effects

in the valve )0,0( fs BM . ▪ At the begining, we look for an appropriate mathematical model

of the actuator. → For an actuator controled by a single-stage servovalve, with

all mentioned approximations, we use the linear model (A.35). We did not write the nonlinear model.

→ So, we have to derive the nonlliner model starting from the liner (A.35).

θ

Beam: mass m , length l

FL to load

Actuator

xp

Fig. A.18

Load

FL from load

mass m

We write (A.35) again:

21 zz

Ltt

p

t

p

t

FM

zM

Az

M

Bz

M

Kz

13212 (A.35)

urK

RK

VKz

VCKz

VAz

f

t

t

eq

t

etpc

t

ep

/44)(

4323

with the state vector

),,(),,( 321 Lpp Pxxzzzz

▪ We recall that while formulating (A.35), the piston stroke vx was eliminated by using the 5-th relation of (A.32),

urK

RKxz

f

tv

/4 (A.83)

▪ In order to find the non lionear model, we put vx back into (A.35), by introducing (A.83):

21 zz

Ltt

p

t

p

t

FM

zM

Az

M

Bz

M

Kz

13212 (A.84)

vt

eq

t

etpc

t

ep x

VKz

VCKz

VAz

44)(

4323

▪ The linear form of (A.84) comes out from the linearized

expression for the flow (eq. (A.3)):

3zKxKPKxKQ cvqLcvqL (A.3)

Nonlinear form of the model will be obtained if the nonlinear expression for the flow is used (eq. (5.33)):

3

1z

x

xPxwCQ

v

vsvdL (5.33)

▪ When we replace (A.3) with (5.33), model (A.84) becomes first:

21 zz

Ltt

p

t

p

t

FM

zM

Az

M

Bz

M

Kz

13212

t

evqc

t

etp

t

ep V

xKzKzV

CzV

Az 4

)(44

3323

and then finally:

21 zz

Ltt

p

t

p

t

FM

zM

Az

M

Bz

M

Kz

13212

t

e

v

vsvd

t

e

t

etp

t

ep V

zx

xPxwC

Vz

VCz

VAz

414443323

▪ We note that the load does not introduce any additional degree of

freedom. So, the satate vector remains as it is.

→ Position variable of the load (angle θ), is directly

geometrically dependent on the piston motion xp.

→ Force FL has to be eleiminated from the equations of load dynamics.

(A.85)

▪ Let us look for the relation between θ and xp !

▫ It holds that

llxp cos )cos1( lxp (A.86)

wherefrom

pxl

11cos ,

21

11sin

px

l (A.87)

▫ First derivative gives:

pxl 1

sin px

pxl

l

2

111

1

(A.88)

A

B

gravity center (C) and grav. force

force FL

velocities current center of

rotation, P

xp

θ

l cosθ

l

▫ Second derivative gives:

pxl 1

sincos 2

2

22 3

2

11

111

111

1p

p

p xx

x

lp

p xxl

l l

l

(A.89)

▪ ▪ Dynamics of the load ! ▫ It will be described by using Lagrange equations.

▫ There is one degree of freedom, and accordingly one equation. ▫ Let, for the moment, the generalized coordinate be the angle θ.

Later, when we accomplish the equation, we will replace θ with xp.

▫ Lagrange equation has the form:

QEE

dt

d kk

(A.90)

where Ek is the kinetic energy and Q is the generalized force. ▫ The beam AB is rotation about the current center of rotation

(point P in the figure). So, the kinetic energy is:

22

2

1

2

1BPk ymIE (A.91)

where IP is the moment inertia of the beam, calculated for the point P.

▫ For the parallel axes, the moment of inertia is

2

CPmII CP which gives

222

3

1)

2(

12

1ml

lmmlIP (A.92)

▫ Position and velocity of the slider B are:

sinlyB (A.93)

coslyB (A.94) ▫ So, kinetic energy is:

222 )cos

3

1(

2

1 mlEk (A.95)

▪ Let us form the Lagrange equation (A.90).

▫ The elements of the left-hand side of the equation are:

)cos3

1( 22

mlEk

(*) 2222 sincos2)cos

3

1(

mlml

E

dt

d k

(*) 22 sincos

ml

Ek

▫ The right-hand side (generalized force) is obtained from the virtual work:

BCAL

gravitybslidergravitybeamforceactuator

mgdxmgdxdxF

dWdWdWdW

)cos1( lxx pA dldxA sin

sin2

lxC d

ldxC cos

2

sinlxB dldxB cos

dmgllFdW L )cos2

3sin(

▫ So, by definition, the generalized force is :

(*) cos2

3sin mgllFQ L

▫ Now, all the terms marked by (*) are substituted into Lagrange

equation (A.90):

cos2

3sin

sincossincos2)cos3

1( 222222

mgllF

mlmlml

L

i.e.

cos2

3sinsincos)cos

3

1( 2222 mgllFmlml L

(A.96)

▪ From (A.96), the load force FL can be found

sin

cos

2

3cos

sin

cos3

12

2

mgmlmlFL

(A.97)

▫ Since FL should be expressed in terms of ppp xxx ,, , we substitute (A.87), (A.88) and (A.89) into (A.97):

2

2

22

2222

2

111

11

2

31

1

)1

1(3

1

1112

112)

11(

3

1

111

111

p

p

p

p

pp

p

L

xl

xlmgx

xlml

xmxx

lmF

p

pp

xl

px

lpx

l

l

l xxll

or in terms of the state-space coordinates:

2

1

12

222

1

1

2222

1

22

1

21

111

11

2

31

1

)1

1(3

1

1112

11

12)1

11(

3

1

111

111

zl

zlmgz

z

zlml

z

z

mz

z

zlmF

l

zl

zl

l

lll

L

(A.98)

(A.99)

▪ This expression is substituted into the actuator model (A.85) (into

its second expression) to give:

21 zz

2

1

12

222

1

12

222

1

1

31

21

11

2

111

11

2

31

1

)(

1

)()()(

1112

11

12)1

11(

3

1

111 z

l

zlmgz

z

zlmlz

z

mzM

zzM

Az

zM

Bz

zM

Kz

l

zl

zl

lll

eff

eff

p

eff

p

eff

t

e

v

vsvd

t

e

t

etp

t

ep V

zx

xPxwC

Vz

VCz

VAz

414443323

where

2

1

21

11

11

)1

1(3

1

)(

z

zlmMzM

l

teff

▪ This is the final model !!

Note that it would be more simple if written in terms of θ

(instead of xp).

(A.85)

Naslov predmeta:

HIDRAULIČKI I PNEUMATIČKI SISTEMI Fond časova: 3+1 Nastavnik: prof. Veljko Potkonjak Kabinet: 105 Obaveštenja: tabla pored kab. 105 Literatura: Merit, Hydraulic Control Systems

UVODNA RAZMATRANJA Problemi u nastavi Hidraulike i pneumatike:

- nedostatak predznanja iz mehanike hidromehanike - prevazilaženje problema ....preskakanje ! .....gotove formule ! .....

Način ispitivanja

Kolokvijum plus domaći rad na simulaciji Upravljanje mehaničkim sistemom

ulaz izlaz

MEHANIČKI SISTEM

stanje

Stanje: pozicione koordinate i njihovi izvodi (brzine)

Izlaz(i): - pozicija (standardno) - brzina (standardno) - sila (komplikovano)

Ulaz: Sila/moment Može li sila biti upravljački ulaz ? Teoretski da, ali u praksi teško !! Sila se ne može menjati po želji već je proizvod jednog složenog sistema – pogonskog sistema.

Pogonski sistemi:

- električni (razne vrste elektromotora)

električna energija mehanička energija

- hidraulički hidraulička energija mehanička energija

- pneumatički

pneumatička energija mehanička energija

upravljački ulaz

sila / moment

izlaz

MEHANIČKI SISTEM

stanje

POGONSKI SISTEM

UKUPNI SISTEM

Elektromotori – nedostaci:

- samo rotacija ... sistem za “konverziju” kretanja (rot-transl., videćemo kasnije probleme)

- odnos moment/brzina ... već pri “normalnim”, a pogotovo jako velikim silama

- redukcija ... komplikovanija konstrukcija, poskupljenje, gubitak snage, zazor i elast. def.

- loš odnos masa prema momentu - izmeštanje i transmisija ... komplikovanija konstrukcija,

poskupljenje, gubitak snage, zazor i elast. def. - ...............

Elektromotori – osnovne prednosti:

- jednostavno upravljanje (manje-više) - pogodni za povezivanje sa računarom

Hidraulički pogon – osnovne prednosti

- i rotacija i translacija - bez problema pravi jako velike sile/momente !! - ......

Hidraulički pogon – osnovni nedostaci

- složenije upravljanje – precizno upravljanje skupo - problemi sa curenjem i prljanjem ulja - .......

Pneumatički pogon – osnovne prednosti

- i rotacija i translacija - velika brzina - niska cena - precizni – tamo gde mogu da se upotrebe - ..........

Pneumatički pogon – osnovni nedostaci

- kretanje samo između dva graničnika - nema upravljanja između dva graničnika - buka - .......

Princip hidrauličkog pogona

- Ulje je nestišljivo protok jednoznačno definiše pomeranje - Regulisanjem protoka reguliše se kretanje Problemi pneumatičkog pogona - Vazduh je stišljiv ...................... - Nema regulacije kretanja ............

kretanje

sila pritiska

izlaz ulja

ulaz ulja

pumpa

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