discrete dynamical systems : inverted double pendulum : consider the double pendulum shown:

1

(a) Discrete Dynamical Systems:1. Inverted Double Pendulum: Consider the double

pendulum shown:

1 2,

( )

k k linear torsional spring

l lengthof each rod

P external conservative load

m mass of each rod

2. Some analytical models of nonlinear physical systems

1

2

l

l

k2

k1

P

g

O

A

y

x

2

Inverted Double Pendulum – Equation of motion

Here,

• - are the generalized coordinates, • T and V- are the respective kinetic and potential

Energies for the system, • -- the generalized forces due to non-

conservative effects.

nci

i i i

d T T V( ) Q , i 1,2,3,......

dt q q q

The equations of motion can be determined by using Lagrange’s equations:

3

Inverted Double Pendulum – Equations of motion

2 2 2 21 2 G1 G1 1 G2 G2 2

2 21 1

G2 A G2/A A 1 OA

OA 1 1 A 1 1 1

G2/A 2 G2/A G2/A 2 2

G2/A 2 2 2

T T T [(mv I ) (mv I )] / 2

T (ml / 3) / 2;

v v v ; v k r

r l(cos i sin j ); v l( sin i cos j );

v k r ; r l(cos i sin j ) / 2;

v l( sin i cos j ) / 2;

2 2 2 2 22 2 1 2 1 2 2 1T (ml / 12) / 2 ml [ / 4 cos( ) / 2] / 2

For the double pendulum- kinetic energy:

4

Inverted Double Pendulum – Equations of motion

The work done by the external force P in a virtual

displacement from straight vertical position is:

1 2 1 1 2

2 21 1 2 2 1

V V V mglcos / 2 mg[lcos lcos / 2]

[k k ( ) ] / 2

1 1 2 2

1 1 1 2 2 2

1 1 2 2

1 1 2 2

B

B

B

W P i r ;

r l(cos i sin j ) l(cos i sin j )

r l( sin i cos j ) l( sin i cos j )

W P[ lsin lsin ]

The are :

Q Plsin ; Q Plsin ;

generalized forces

Potential energy:

5

Inverted Double Pendulum – Equations of motion:

Equation for 1:

21 1 2 2 1

1

2 1 2 2 1

22 1 2 1

1

1 1 2 1 2 21

2 21 2 2 1 2 2 1

1 2 1 2 2

d T( ) ml [ / 3 cos( ) / 4

dt

( )sin( ) / 4]

Tml sin( ) / 4

V3mglsin / 2 (k k ) k

ml [4 / 3 cos( ) / 4 sin( ) / 4]

(k k ) k 3mglsin 1 1/ 2 Plsin

6

Inverted Double Pendulum – Equations of motion:

Equation for 2:

22 1 2 1

2

1 1 2 2 1

22 1 2 1

2

2 2 1 2 21

2 22 1 2 1 1 2 1

2 1 2 2 2 2

d T( ) ml [ / 3 cos( ) / 4

dt

( )sin( ) / 4]

Tml sin( ) / 4

Vmglsin / 2 k k

ml [ / 3 cos( ) / 4 sin( ) / 4]

k k mglsin / 2 Plsin

7

Inverted Double Pendulum –Equations of motion:

We now consider a simplified version with k1= k2=k

Let

Then, the equations are:

Equation for 1:

Equation for 2:

22 1 2 1 1 2 1

1 2 2

[ / 3 cos( ) / 4 sin( ) / 4]

k k (P M)sin

21 2 2 1 2 2 1

1 2 1

[4 / 3 cos( ) / 4 sin( ) / 4]

2k k (P 3M)sin

2 2 2k k /ml , P Pl /ml , M mgl / 2ml

8

Discrete dynamical systems…….

2. Inverted Double Pendulum with Follower Force: Consider the same system as in last example, except that the force P changes direction depending on the orientation of the body on which

it acts.The force P now acts

at an angle to the rod AB and always maintains this

direction relative to the rod

regardless of the position in space of the system during its oscillations.

1

2

l

l

k2

k1

P

g

O

A

y

x

B

9

Inverted Double Pendulum with Follower….– Equations of motion

The work done by the external force P in a virtual

displacement from straight vertical position is:

1 2 1 1 2

2 21 1 2 2 1

V V V mglcos / 2 mg[lcos lcos / 2]

[k k ( ) ] / 2

2 2

1 1 2 2

B

B

W P[cos( ) i sin( ) j ] r ;

r l(cos i sin j ) l(cos i sin j )

Note that the only change is in the effect of the external force P. The potential and kinetic energy expressions remain the same. So,potential energy:

10

Inverted Double Pendulum with Follower– Equations of motion

The resulting equations of motion are:

1 1 1 2 2 2

1 1 2 2

Br l( sin i cos j ) l( sin i cos j )

So,the virtual work done is

W Pl[ sin( ) sin ]

1 1 2 2Q Plsin( ); Q Plsin .

2 21 2 2 1 2 2 1

1 2 1 2 2 1 1 2

ml [4 / 3 cos( ) / 4 sin( ) / 4]

(k k ) k 3mglsin / 2 Plsin( )

Thus, the generalized forces are:

The virtual displacement is:

11

Inverted Double Pendulum with Follower– Equations of motion

In the reduced case, with equal springs etc., the

equations are:

2 22 1 2 1 1 2 1

2 1 2 2 2

ml [ / 3 cos( ) / 4 sin( ) / 4]

k k mglsin / 2 Plsin

21 2 2 1 2 2 1

1 2 1 2 1

[4 / 3 cos( ) / 4 sin( ) / 4]

2k k Psin( ) 3Msin

22 1 2 1 1 2 1

1 2 2

[ / 3 cos( ) / 4 sin( ) / 4]

k k PSin Msin

and

12

Discrete dynamical systems…….

3. Dynamics of a Bouncing Ball: Consider a

ball bouncing above a horizontal table. The table

oscillates vertically in a specified manner (here

we assume harmonic oscillations).

The motion of the ball, during free flight, is governed by

Integrating once gives:

0 0y=y ( )g t t ground

table

Y(t)

ballgmg

ormy mg y g ( ) sin( )X t A t

13

Dynamics of a Bouncing Ball …….

Integrating once gives:

Integrating again, we get:

The position of the ball has to remain above the

table Finally, we have the law

of interaction between the table and the ball:

0 0y=y -g(t-t )2

0 0 0 0

1( ) ( )

2y y y t t g t t

0y(t) X(t), t t

the relative velocities

before and after impact are relate

If w

d

by c

simple law o

oefficient

e a

of

f

i

re

mpac

stit

ssume

t,

ution.ground

table

Y(t)

ballgmg

( ) sin( )X t A t

14


Thus, we have the relation:

These relations provide a complete description of

motion of the ball. Note that, given an initial condition, we have to piece together the

motion in forward time

i i i i

i

i

V(t )-X(t )=e[X(t ) U(t )]

V(t ) - velocity of the ball immediately after impact

and U(t ) - velocity of the ball just before impact

where

ground

table

Y(t)

ballgmg

15


Let us now proceed in a systematic manner. For

nonlinear analysis, it is always advisable to non-

dimensionalize equations: So we define

ground

table

Y(t)

ballgmg

2 2

2 2

2

: / , 2

: 2

: ( 2)

. : ( ) 2

, ( ) (2 ) ( )

sin( ) sin(2 )

( ) sin(2 );2

Time t t T whereT

Acceleration units g

Veloctiy units g T g

Pos units g T g

So X t g X

A t A

AX

g

16


Ball motion: z(t)= (22g/2)Z(),

2 2

2 2

2 2 2 2

2 2 2 2

dz 2 g dZ d 2 g dZ g dZ( ) ( ) ( )

dt d dt d 2 d

d z g d Z d g d Z g d Z( ) ( ) ( )

dt d dt d 2 2 d

2 .z g Z

0 0

20 0 0 0

( ) 2( ),

( ) ( ) ( )

Z Z

Z Z Z

ground

table

Y(t)

ballg

mg

Integrating,

Now,

17


Thus, we have

Ball motion:

Table motion:

Initially: ball starts at time 0 when it is in contact with the table, and just about to

leave

X( ) sin(2 ) (3)

2

0 0

20 0 0 0

2; ( ) 2( ), (1)

( ) ( ) ( ) (2)

Z Z Z

Z Z Z

0 0 0 0X( ) Z( ) Z sin(2 ) (4)2

ground

table

Y(t)

ballgmg

18


Ball velocity at =0:

Next collision at time 1 > 0 when

Now,

Using (3)-(5)

Z( ) X( ) W( ) 0

0 0 0

0

0 0 0

0

dZZ W X ( )

d

Z W cos(2 ) (5)

(H is relative velocity of the balere W 0 an unknl, own)

0 0 0

20

W( ) Z Z ( )

( ) sin(2 ) (6)2

ground

table

Y(t)

ballgmg

19


Then, the time instant 1 is defined by

(this is a relation in W’0, 1

and 0, and it depends on )

0

20 0 0 0

W( ) [sin(2 ) sin(2 )]2

[W cos(2 )]( ) ( )

1 0 1

0 0 1 0

21 0

W( ) [sin(2 ) sin(2 )]2

[W cos(2 )]( )

( ) 0 (7)

ground

table

Y(t)

ballgmg

20


or:

1 1 1 1

1 0 0 1 0

1

W W ( ) Z ( ) X ( )

dW( )cos(2 ) [W cos(2 )] 2( ) (8)

d

1 1

coefficient of restitut

W eW

(e ion)

1 0

1 0 1 0

W e[ {cos(2 )

cos(2 )} W 2( )] (9)ground

table

Y(t)

ballgmg

When just about to contact at this time instant 1 the

relative velocity is :

On impact, the ball relative velocity changes:

21


One can write the equations now in a more compact

form:

i 1 i i 1 i

i 1 i

W e[ {cos(2 ) cos(2 )} W

2( )] (B)

i 1 i i 1

2i i i 1 i i 1 i

W ( ) [sin(2 ) sin(2 )]2

[W cos(2 )]( ) ( ) 0 (A)and

ground

table

Y(t)

ballgmg

Knowing (I,W’i), equations (A) and (B) can be used to compute (I+1,W’i+1), thus generating the trajectory.

22

(b) Continuous Dynamical Systems

1. Rotating Thermosyphon: Consider a closed

circular tube in a vertical plane. The tube is filled with

a liquid of constant properties, except for variation of its density with

temperature in buoyancy and

centrifugal terms, i.e. in body forces. One part

of the loop is heated, and the other cooled.

The tube is spun about the vertical axis.

α

heating

cooling

R

r

23

Rotating Thermosyphon…

There are many ways to develop a model for the system. If the tube radius ‘r’ is much smaller than the torus radius ‘R’, one can assume that there is

negligible flow in the radial direction. Another approach is to average

the velocity and temperature over the

tube radius. Then, the equations for fluid

motion are:

α

heating

cooling

R

r

24

Rotating thermosyphon: equations…

Continuity:Here, V – average flow velocity at any section ; - density of the fluid and (V) is independent of .

Momentum:

Here, p – fluid pressure at a

section,

w- shear stress at the wall

1 (ρV)0 (1)

t R

α

heating

cooling

R

r 2

2

(ρV) 1 (ρV) 1cos

2cos sin (2)w

pg

t R R

Rr

25

Rotating thermosyphon:Equations…

Energy:

where Cp- specific heat of the fluid, T – mean fluid temperature at a section,

k – thermal conductivity,

q’ – applied heat source per unit length.

Remark: Here viscous dissipation term is neglected

22 2

2 2

( )( ) (3)

tp

T V T k Tr C r q

R R

α

heating

cooling

R

r

26

Rotating thermosyphon: Equations…

Simplification and nondimensionalization:Integrating the momentum eqn. (2) along the loop

Shear stress:

Friction factor: α

heating

cooling

R

r

2 2 2 22

0 0 0 0

2 22

0 0

(ρV) 1 (ρV) 1cos

2cos sin (4)w

pd d d g d

t R R

d R dr

2 / 2 (5)w f V

16 / Re, Re 2 /f Vr

27


Simplification:Introduce the variation of density with temperature in buoyancy and centrifugal terms, and use periodicity of variables (eqn. (4))

2

20

22

0

V 32( )cos

(2 ) 2

( )cos sin (6)2

r

r

V gT T d

t r

RT T d

where -kinematic viscosity, Tr-reference temperature

- coefficient of thermal expansion

28


Non-dimensional variables:

2

2

2

4 2

3

2

32 (2 ), , ,

(2 ) 32

2048 '( ) * 1, ( ) ( )

(2 ) 8 Pr 8192

Pr /

2 ( '

P

)*

r

2

r

p

p

p

t r T TU V

r R T

R q Ra rT Q

g r C T R

Here C k

r q CRa

k

andtl number

Modified Rayleigh number

29


Non-dimensional equations:The resulting momentum and energy equations are:

Note: - combination of geometric parameter and the Prandtl number Pr. Pr > 1 for ordinary fluids (air, water etc.) and <1 for liquid metals.

2 2

0 0

2

2

2 2

cos cos sin (7)

( ) (8)

/ , ( ) / 8Pr

UU d d

U Q

rwhere R g

R

2( / )r R

30


Solution approach: The solutions to (7) and (8) can

be expressed as:

1( , ) [ ( )sin( ) ( ) cos( )] (9)

n n

nB n C n

1( ) [ sin( ) cos( )] (10)

n nn

Q A n D n

where as, the externally imposed heat flux can be represented in a Fourier series as:

Substituting (9) and (10) in equations (7) and (8), and collecting the appropriate Fourier coefficients gives an infinite set of ordinary differential equations In the unknowns U(), Bn(), and Cn().

31


Solution approach…: It turns out that only five of these equations are independent-master equations.The remaining equations are linear equations for the remaining variable (called “slave variables” and “slave equations”). Equation (8) gives:

2

20 0

0

1

[ ( )sin( ) ( ) cos( )] [ ( )sin( ) ( ) cos( )]

[ ( )sin( ) ( ) cos( )]

[ sin( ) cos( )]

n n n n

n n

n n

n n

n

n

B n C n B n C n

B n C n

U A n D n

32


Solution approach…: Collecting terms of different n’s give:

0 0

1 1 1 1

1 1 1 1

2 2 2 2

2 2 2 2

3 3 3 3

3 3 3 3

2

2

0 :

1: (sin )

(cos )

2 : 4 2 (sin 2 )

4 2 (cos 2 )

3 : 9 3 (sin 3 )

9 3 (cos3 )

: (sin )

p p p p

p

n C C

n B B UC A

C C UB D

n B B UC A

C C UB D

n B B UC A

C C UB D

n p B p B pUC A p

C p C

(cos ) (11)

3,4,5,6,...........

p p ppUB D p

p

33


Solution approach…: Now, considering equation (7)we get:

Evaluating the two integral terms on the right-hand side, it is clear that only coefficients of will survive. Thus, we get

Equations (11) and (12) govern the dynamics.

2

0

2

0

1

1

[ ( )sin( ) ( )cos( )]cos

[ ( )sin( ) ( )cos( )]cos sin

n n

n n

n

n

UU B n C n d

B n C n d

cos( ) sin(2 )and

1 2 (12) U U C B

34


2. Buckling of Elastic Columns: Consider a thin

beam that is initially straight. O xyz is coordinate

system with x-y plane coinciding with undisturbed

neutral axis of the beam. Let EI is the bending stiffness

X

V

35

Buckling of elastic…..

V(s) – vertical displacement of the centroidal axis,

X – distance measured along the centroidal axis from left end.

Define:

Then, the strain energy of the system is:

X

V

2 2 3 3x X /L; u V /L; L P / EI; L K / EI;

2

2

2

2 2

0

0

1 u(u, , ) )dx

2 1 u

1 u )dx

1 1u (0) u ( )

2 2

(

(1

36


3. Thin rectangular plates: Consider a thin plate

that is initially flat. Oxyz is coordinate system with x-y

plane coinciding with undisturbed middle surface of

the plate. Let h – plate thickness. The equations of motion for the plate,

for moderately large displacements

– von Karman equations.

In here, we give a short review of the

derivation of these equations.

x, u

y, v

z, w

a

37

Thin rectangular plates…..

Consider a differential plate element: The equations

in the three directions are: 2

2

2

2

(1)

(2)

xyx

xy y

NN uh

x y t

N N vh

x y t

x

y

z

Nx

Ny

Nxy

Nxy

Ny

Nx

F

2 22

2 2

2

2

2

( ) ( )

( ) ( ) (3)

y xyx

x y

xy xy

M MM

x yx y

w wN N

x x y y

w w wN N h F

x y y x t

38


The constitutive equations for a linearly elastic and

isotropic material are:

In these expressions, N, Ni – forces, M - moments

2 1 2

2

2 1 2

2

(1 ) [ / 2

/ 2] (4)

(1 ) [ / 2

/ 2] (5)

x x x

iy y x

y y y

ix x y

N Eh u w

v w N

N Eh v w

u w N

x

y

z

Nx

Ny

Nxy

Nxy

Ny

Nx

F

[ ] (6)

( ) (7)

ixy y x x y xy

x xx yy

N Gh u v w w N

M D w w

39


The constitutive equations for a linearly elastic and

isotropic material are:

Also u,v,w – displacements

Substituting the force – displacement

relations in the dynamic equations give:

x

y

z

Nx

Ny

Nxy

Nxy

Ny

Nx

F

3 2

( ) (8)

(1 ) (9)

2 /(1 );

/12(1 )

y yy xx

xy xy

M D w w

M D w

G E

D Eh

40


The dynamic equations for a plate made of linearly

elastic and isotropic material are then simplified by

introducing a stress function such that:

Then, (1) and (2) are automatically satisfied if in-

plane inertia terms are neglected. Furthermore, the

expressions (7)-(9) can be substituted in (3) to get

equation for transverse displacement. Also, a

compatibility condition is (gives an equation for ):

2 2 2

2 2, , (10)

x y xyN N Nx yy x

( ) ( ) 0 (11) xyy yxx y x xyu v u v

41


4. Flow between concentric rotating cylinders:

Consider two concentric cylinders with radii a, b;

Let 1, 2 – angular velocities of inner

and outer

cylinders; let (ur,uθ,uz) – velocity

components in a cylindrical coordinate system;

p – pressure at a point;

We now define the equations of motion for the system.

a

b 12

42

Flow between concentric rotating….

Equations of motion for the system: In cylindrical

coordinate system the NS equations are

ab 1

2

2r r

r 2 2

r r2 2

zz

r z

u uDu u1 p 2( u ),

Dt r r r r

Du u u u u1 p 2( u ),

Dt r r r r

Du 1 p( u ), (1)

Dt z

where

Du u u

Dt t r r z

43


Equations of motion for the system: and

There is also the equation for mass conservation:

The basic flow is defined by:

r r zuu u u10. (2)

r r r z

ab 1

2

2 2 2

2 2 2 2r rr r z

r z

2

u 0, u 0, u V(r) r (r) and p P(r)

1 dP Vor and DD V 0 (3)

dr r

44


The basic flow…: Equations (3) have solution of the form

Let us consider perturbations to the basic flow:

We can then obtain linearized equations about the basic

flow.

ab 1

2

2

22

1 1 12 2

2 1 1 2

(r) A B / r

1where A , B R

1 1

and / , R /R (4)

r zu (u ,V u ,u ), p P p

45


Linearized equations about the basic flow:

r r r

r 2 2

rr 2 2

z zz

r r z

uu u u1 p 22 u ( u )

t r r r

u u u u1 p 2(D V)u ( u )

t r r r

u u 1 p( u )

t z

uu u u1and 0

r r r z

discrete dynamical systems : inverted double pendulum : consider the double pendulum shown:

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