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Lobontiu: System Dynamics for Engineering StudentsSolutions: Chapter 5 20

Problem 5.15

Figure P5.1 shows the original system of Fig. 5.35 with the relevant pressures and

volume flow rates indicated.

Figure P5.1 Liquid-level system with pressures and volume flow rates

The liquid resistances are

21 2

1 2; al l

o

p pp pR R

q q

= = (P5.1)

and the liquid capacitances are

1 2

1 2;

i o

l l

q q q q

C Cp p

= = (P5.2)

The second Eq. (P5.1) yields

2 2a l op p R q= + (P5.3)

whose time derivative is

2 2l op R q= (P5.4)

The first Eq. (P5.1) is combined with Eq. (P5.3) to generate

1 2 1a l o lp p R q R q= + + (P5.5)

The second Eq. (P5.2) is combined with Eq. (P5.4), which results in

2 2l l o oq R C q q= + (P5.6)

As a consequence, Eq. (P5.5) becomes

( )1 2 1 2 2a l o l l l o op p R q R R C q q= + + + (P5.7)

Rl1 Rl2

pa

pa

pa

qi

qo

Cl1 Cl2

q

p1 p2 p2

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Applying the time derivative to Eq. (P5.7) results in

( )1 1 2 1 2 2l l o l l l op R R q R R C q= + + (P5.8)

The flow rate q of Eq. (P5.6) together with Eq. (P5.8) are used in conjunction with the

first Eq. (P5.2) to produce

( )1 2 1 2 1 1 2 1 2l l l l o l l l l l o o iR R C C q R C R C C q q q+ + + + = (P5.9)

The differential Eq. (P5.9) is the mathematical model of this liquid system where qi is

the input and qo is the output.

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Problem 7.12

Based on Fig. 7.43, the following dynamic equations are written for the two rotary

plates which are connected by torsional springs:

1 1 2

2 2 1

( ) 2 ( ) ( ) ( )

( ) 2 ( ) ( ) 0

+ = + =

tJ t k t k t m t

J t k t k t

(P7.1)

The Laplace transform with zero initial conditions is applied to Eqs. (P7.1), which

results in:

( )

( )

2

1 2

2

1 2

2 ( ) ( ) ( )

( ) 2 ( ) 0

+ =

+ + =

tJs k s k s M s

k s Js k s(P7.2)

Equations (P7.2) can be written as:

21

2

2

( ) ( )2

( ) 02

+ = +

ts M sJs k k

sk Js k (P7.3)

Because the input-output connection for this example is:

[ ]1

2

( ) ( )( )

( ) 0

=

ts M s

G ss

(P7.4)

it follows that the transfer function is obtained from Eq. (P7.3) as:

[ ]

2

12 2 4 2 2 2 4 2 2

2 2

2 4 2 2 2 4 2 2

2

2 4 3 4 3( )2 2

4 3 4 3

Js k k

Js k k J s Jks k J s Jks k G sk Js k k Js k

J s Jks k J s Jks k

+

+ + + + + = = + +

+ + + +

(P7.5)

Figure P7.1 Two-mesh impedance-based mechanical system

By using the mechanical impedance circuit of Fig. P7.1, the following equations are

formulated based on the impedance-form Newtons second law of motion:

Ze

Zm

2(s)

1(s) 2(s)

Zm

Ze

1(s)

Mt(s)

Ze

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Problem 3.18

Figure P3.1 shows the free-body diagrams of the mechanical system. This system has

two DOF, which are the pulley rotation angle and the mass m3 translation x2. For small

motions, the coordinate x1 is:

1 2x R= (P3.1)

Figure P3.1 Free-body diagrams of pulley and translatory body

Newtons second law of motion for the pulley rotation and the body translation is

expressed as:

2 1 1 2

3 2 2

e e

e d

J f R f R

m x f f f

=

=

(P3.2)

where:

( )

( )

2 2

1 1 2 2

1 1 1 1 2

2 2 2 1

2

1

2

e

e

d

J m R m R

f k x k R

f k x R

f cx

= + = = =

=

(P3.3)

Substitution of Eqs. (P3.3) into Eqs. (P3.2) results in:

( ) ( )2 2 2 21 1 2 2 1 2 2 1 2 1 2

3 2 2 2 1 2 2

10

2m R m R k R k R k R x

m x cx k R k x f

+ + + =

+ + =

(P3.4)

Equations (P3.4) can be written with the numerical values of the problem as:

2

2 2 2

258 2963 0

64 200 3 2

x

x x x f

= + =

= + +

(P3.5)

The forcing input can be specified by means of the Signal Builder block in the

x2

f

m3

fe2 fdx1

R2

m2

R1m1

fe2

fe1

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Sources library of Simulink. Horizontal and vertical plot segments can be displaced

and values of the function and variable can be changed interactively. The input force in

this problem has been defined as shown in Fig. P3.2.

0 1 2 3 4 5 6 7 8 9 10

0

1

2

3

4

5Signal 1

Time (sec)

pr2_28/Signal Builder : Group 1

Figure P3.2 Input created by means of a Signal Builder Simulink

source

The Simulink

diagram and time response curves (t) and x2(t) are shown in Figs. P3.3and P3.4.

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Figure P3.3 Simulink

diagram of the pulley-mass mechanical system

(a)

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(b)

Figure P3.4 Simulink

time response: (a) pulley rotation angle (t); (b) linear-motion body

displacement x2(t)

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