MM326 SYSTEM DYNAMICSHomework 4 Solution

Prepared by Nurdan BilginSOLUTIONS:

Question 1: (From System Dynamics: An Introduction, Rowell and Wormley,1997 Problems 5.21 page 168) It is common to mount machines and rotating equipment on shock isolation pads to reduce the transmission of time varying forces to the ground. In the system shown in figure 1 a machine represented as a mass m, has a vibrational force acting at its center of mass in addition to the gravitational force. The pad that supports the machine is made of a damping material with nonlinear stiffness given by the constitutive equation F=c x2, where c=25N /m2. The material also has damping properties that are approximately linear, that is, the damping force is proportional to the velocity across the pad.

Figure 1a) Construct a linear graph for the system.b) Derive the nonlinear differential equation for the system in terms of x. (Hint: it is useful to

differentiate the constitutive equation to obtain an elemental equation in terms of power variables.)

c) If the mass of machine tool is 1000 kg, and the vibrational force has a zero average value, what is the nominal equilibrium condition for the system?

d) Derive the linearized equation for small excursions from the nominal equilibrium condition.

Solution 1:a.)

b)Determine the primary, secondary and state variables. Primary Variables: V m , FBSecondary Variables: f m ,V B

State Variables: V m

Write the elemantal equations for passive elements as B-S. Be careful about that Primary variable must be on the left side alone.B-S=4-2=2

V m=f mm

(1 )

FB=C x2(2)

Write the continuity equations as N-1-SA. Be careful about that Secondary through-variable must be on the left side alone. You have to choose appropriate nodes because there must be only one secondary through-variable and primary variables in each equation. N-1-SA=2-1-0=1

f m=f m (t )+F (t )−FB (3)Write the compatibility equations as B-N+1-ST. Be careful about that Secondary across-variable must be on the left side alone. You have to choose appropriate loops because there must be only one secondary across-variable and primary variables in each equation. B-N+1-ST=4-2+1-2=1V B=V m(4)Eliminate the secondary variables in elemental equations using continuity and compatibility equations

V m=f mm

=f m (t )+F (t )−FB

m⇒ x=

mg+F (t )−c x2

m=g+

F (t )m

− cmx2

x+ cmx2=g+

F ( t )m

c)For the equilibrium condition x= x=0 and in the question, F (t)averageis given as 0Then

x+ cmx2=g+

F ( t )m

⇒ cmx2=g⇒ x2=mg

c⟹ xnom=√mgc =√ 1000∗9.81

25≅ 19.8

d)

x+ cmx2−g−

F ( t )m

=0⇒m x+c x2−mg−F ( t )=ϕ ( x , x ,F (t ))

¿¿¿¿mδ x+990δx−δF=0⇒ 1000δ x+990δx=δF

Question 2: (Adapted From Mühendislik Sistemlerinin Modellenmesi ve Dinamiği, Extended 2. Ed. Yücel Ercan page:306 and System Dynamics, 3. Ed., Ogata, page 197) Consider the liquid-level system shown in figure 2. At steady state the inflow rate Qi=Q, the outflow rate is Qo=Q, and head is H=H . Assume that the flow is turbulent. Then we have Qo=K √H , For this system,

Figure 2a) Construct a linear graph for the system.b) Derive the nonlinear differential equation for the system in terms of position of M.c) Derive a set of linearized state equations for small excursions from the nominal

equilibrium condition.Solution 2 a)

b)Determine the primary, secondary and state variables. (Note that; all pressure is converted to height with their physical relations)Primary Variables: ( ρg )HC f

, ( ρg )H ,vm ,Q1 , F , FKSecondary Variables: QC f

,Q o , Fm , ( ρg )~H ,v , v KState Variables: ( ρg )HC f

, vm ,FKWrite the elemantal equations for passive elements as B-S. Be careful about that Primary variable must be on the left side alone.B-S=8-2=6

(ρg ) HC f=QC f

C f

(1 )

Qo=κ√H⇒H=Qo

2

κ2 (2 )

vm=Fmm

(3)

Q1=−Av (4 )F=A ( ρg )~H (5)FK=K vK(6)

Write the continuity equations as N-1-SA. Be careful about that Secondary through-variable must be on the left side alone. You have to choose appropriate nodes because there must be only one secondary through-variable and primary variables in each equation. N-1-SA=5-1-1=3Fm=−F−FK (7)Q0=Q 1 (8 )

QC f=Qi(t )−Q1(9)

Write the compatibility equations as B-N+1-ST. Be careful about that Secondary across-variable must be on the left side alone. You have to choose appropriate loops because there must be only one secondary across-variable and primary variables in each equation. B-N+1-ST=8-5+1-1=3v=vm (10)vK=vm−v (t )(11)( ρg )~H=( ρg )HC f

−( ρg )H⇒~H=H C f

−H (12)Eliminate the secondary variables in elemental equations using continuity and compatibility equations

( ρg ) HC f=QC f

C f

=Qi(t)−Q1

C f

=Qi(t)C f

+ AvC f

=Qi(t)C f

+AvmC f

vm=Fmm

=−F−FK

m=

−A ( ρg )~Hm

−FKm

=−A ( ρg ) (HC f

−H )m

−F K

m=AH ( ρg )m

−A HC f

( ρg )m

−F K

m

H=Qo

2

κ2 ∧Q0=Q1∧Q1=−Av then H= A2 v2

κ2 ∧v=vm so H=A2 vm

2

κ2

Then we can write vm again,

vm=A3 vm

2 ρg

mκ2 −A HC f

( ρg )m

−F K

mFK=K vK=K (vm−v (t))Differentiate one time vm term

vm=2 A3 vm vm ρg

mκ2 −A HC f

( ρg )m

−FKm

Rearrange the last eq. using HC f and FK

HC f= 1

( ρg ) (Q i( t)C f

+AvmC f

) and FK=K vm−Kv (t )

vm=2 A3 vm vm ρg

mκ2 −AQi(t )mC f

−A2 vmmC f

−K vmm

+Kv (t)m

xm=2 A3 xm xm ρg

mκ2 −AQi(t)mC f

−A2 xmmC f

−K xmm

+Kv( t)m

mC f κ2 xm−2 A3C f ρg xm xm+( A2κ2+K κ2C f ) xm+A κ2Qi (t )−K κ2C f v (t)=0

ϕ ( x , x , x ,Qi (t ) , v (t))=0

For the equilibrium condition x= x= x=0 and in the question, xnom=0Then ¿¿¿¿¿¿

mC f κ2 δ x+(A2κ2+K κ2C f )δ x+Aκ2 δQi (t )−K κ2Cf δv ( t )=0

Question 3: (From Mechanical Vibrations 4. Ed., Rao,2004 Problems 13.6 page 966)

A uniform bar of length l and mass m is hinged at one end (x=0), supported by a spring at x=2l3

,

and acted by a force at x=l, as shown in figure 3. (Hint: don’t use small angle assumption.)

Figure 3a) Construct a linear graph for the system and identify the state variables.b) Derive a set of nonlinear state equations for the system.c) Derive a set of linearized state equations for small excursions from the nominal

equilibrium condition.Solution 3.

b)Determine the primary, secondary and state variables. Primary Variables: ω jeq

, v2 , Fk , T 1

Secondary Variables: T jeq,F2 , vk ,ω1

State Variables: ω jeq,Fk

Before the other steps, we have to find the some relations

x1=lsin(θ)⇒ x1=l θ cos(θ)

x2=2l3

sin (θ)⇒ x2=2 l3θ cos (θ)

j eq=ml2

3Write the elemantal equations for passive elements as B-S. Be careful about that Primary variable must be on the left side alone.B-S=5-1=4

ω jeq=T jeq

jeq(1 )

v2=−2 l

3θ cos (θ)(2)

lsin(θ)(2l/3)sin(θ)

Fk=k vk (3)

T 1=2 l3F2(4)

Write the continuity equations as N-1-SA. Be careful about that Secondary through-variable must be on the left side alone. You have to choose appropriate nodes because there must be only one secondary through-variable and primary variables in each equation. N-1-SA=3-1-0=2F2=−Fk (5)T jeq

=−T 1+Teq (t )(6)Write the compatibility equations as B-N+1-ST. Be careful about that Secondary across-variable must be on the left side alone. You have to choose appropriate loops because there must be only one secondary across-variable and primary variables in each equation. B-N+1-ST=4-2+1-2=1vk=v2(7)ω1=ω jeq

(8)Eliminate the secondary variables in elemental equations using continuity and compatibility equations

ω jeq=T jeq

jeq=

−T 1+T eq ( t )jeq

=

−2 l3F2+T eq ( t )

j eq=

2l3Fk+T eq (t )

jeq

F k=k vk=k v2=−k 2l3θ cos (θ)

Integrate this term one times, F k=−2l

3sin(θ)

θ=

2 l3Fk+T eq ( t )

ml2

3

=−4k3m

sin(θ)+3

ml2T eq ( t )

θ+ 4k3m

sin (θ )− 3

m l2T eq ( t )=0

Taylor expansion of sin (θ )

sin (θ )=θ− θ3

3 !+ θ

5

5 !− θ7

7 !We can take account only the first term. Then,

θ+ 4k3m

θ− 3

m l2Teq (t )=0