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32
Advanced Vibrations /home/djsegal/UNM/VibCourse/slides/Lecture10.frm 11/30/98 Copyright Dan Segalman, 1998 1 Slides of Lecture 10 Today’s Class: Review of Homework from Lecture 8 A short quiz on linearization. Linearization of Lagrange Equations Properties of Resulting Matrix Equations

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Ad

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ced

Vib

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s

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e/dj

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M/V

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ture

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op

yrig

ht

Dan

Seg

alm

an, 1

998

1

Slid

es o

f Lec

ture

10

Toda

y’s

Cla

ss:

Rev

iew

of H

ome

wor

k fr

om L

ectu

re 8

A s

hor

t qui

z on

line

ariz

atio

n.Li

near

izat

ion

of L

agr

ang

e E

quat

ions

Pro

pert

ies

of R

esul

ting

Mat

rix E

quat

ions

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ht

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alm

an, 1

998

2

Hom

ew

ork

from

Lec

ture

8A

num

eric

al e

xper

imen

t with

line

ariz

atio

n

Man

y tim

es w

e ha

ve d

eriv

ed th

e eq

uatio

ns f

or a

spr

ing

rein

forc

ed p

endu

lum

:.

The

line

ariz

ed f

orm

is w

here

.

We

use

the

linea

rized

freq

uenc

y to

dim

ensi

onle

ss th

e tim

e pa

ram

eter

.

Defi

ne, d

efine

, and

defi

ne

m

R

θ

κθ

κm

R2

--------

--θ

g R---θ

sin

+

+

0=

θω

L2θ

+0

=

ωL2

κm

R2

--------

--g R---

+=

τω

Lt

τ()

θτ

ωL

⁄(

)=

αg

R⁄ ωL2

--------

--=

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ced

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ht

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alm

an, 1

998

3

Hom

ew

ork

for

Lect

ure

8co

ntin

ued

The

n a

nd

1.S

olve

num

eric

all

y th

e di

men

sion

less

go

vern

ing

equa

tion

for

the

initi

al c

ondi

tions

: a

nd o

ver

the

perio

d

for

the

thre

e ca

ses:

,, a

nd.

2.D

oth

esa

me

asab

ove

but

for

the

initi

alco

nditi

ons

and

3.C

ompa

re a

nd d

iscu

ss th

e y

our

resu

lts f

or p

arts

1 a

nd 2

.

τ22

ddφ

t22

ddθ

1 ωL2

------

=τ22

ddφ

–(

)φα

φsi

n+

[]

+0

=

φ0(

ddφ0

0=

06

π,

()

α0

12⁄

1=

φ0(

)π 6---

=

τddφ

00

=

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op

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ht

Dan

Seg

alm

an, 1

998

4

Sol

utio

nU

sing

Mat

lab

Usi

ng s

tate

spa

ce f

orm

ulat

ion:

, the

n

Thi

s ap

pear

s as

a fu

nctio

n ‘p

endu

1.m

’ pr

ovid

ed to

mat

lab:

func

tion

yprim

e =

pend

u1(t

,x)

glob

al a

lpha

;yp

rime

= [ -

alph

a.*s

in(x

(2))

- (

1-al

pha)

.*x(

2);

x(1)

];re

turn

;

xφ φ

=x

αx 2(

)1

α–

()x

2–

sin

x 1=

Advanced Vibrations

/home/djsegal/UNM/VibCour se/slides/11/30/98 Copyright Dan Segalman, 1998

5

ResultsCalls to Matlab

T = [0:299]*(6*pi)/200; %define time arra y% phi_max = pi; % Initial Condition φ=π% alpha = 1; [t,x] = ode23(’pendu1’, T , [0 phi_max]); %Rung e-Kuta Integration x1 = x;% alpha = 0.5; [t,x] = ode23(’pendu1’, T , [0 phi_max]); x2 = x;% alpha = 0.0; [t,x] = ode23(’pendu1’, T , [0 phi_max]); x3 = x;% plot(t,x1(:,2), t,x2(:,2), t,x3(:,2)); print pi -depsc%% phi_max = pi/6;% alpha = 1; [t,x] = ode23(’pendu1’, T , [0 phi_max]); x1 = x;% alpha = 0.5; [t,x] = ode23(’pendu1’, T , [0 phi_max]); x2 = x;% alpha = 0.0; [t,x] = ode23(’pendu1’, T , [0 phi_max]); x3 = x;% plot(t,x1(:,2), t,x2(:,2), t,x3(:,2)); print pi6 -depsc

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998

6

Sol

utio

n f

orφ 0

Whe

n th

e pe

ndul

um s

tar

tsdi

spla

ced

near

ly

vert

ical

ly,

the

rest

orin

g f

orce

due

togr

avi

ty is

als

o ne

ar z

ero.

The

mor

e of

the

rest

orin

g f

orce

that

is d

ue to

the

tor

sion

al s

trin

g, th

em

ore

that

the

resp

onse

will

app

ear

to b

e ha

rmon

ic.

α=0.5

α=0.

00

510

1520

2530

−4

−3

−2

−101234

α=1

κ θ

R

m

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998

7

Sol

utio

n f

orφ 0

=π/6

Whe

n th

epe

ndul

um s

tar

tson

ly a

littl

edi

stan

ce fr

om th

est

atic

equ

ilibr

ium

,th

e ‘s

mal

l ang

le’

appr

oxim

atio

n is

ver

y go

od.

In th

at c

ase

, the

line

arity

of

the

resu

lt is

nea

rly

inde

pend

ent o

f whe

ther

the

rest

orin

g f

orce

is d

ue to

the

sprin

g or

to g

ravi

ty.

Legi

timat

e lin

eariz

atio

n re

quire

s th

at th

e lin

eariz

ed te

rms

be n

earl

yeq

ual t

o th

e co

rres

pond

ing

nonl

inea

r te

rms

thr

ough

out t

hede

form

atio

n.

α=0.5

α=0.

0α=

10

510

1520

2530

−0.

8

−0.

6

−0.

4

−0.

20

0.2

0.4

0.6

m

R

θ

κ

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998

8

Sho

rt Q

uiz

Thi

s qu

iz s

houl

d ta

ke a

bout

15

min

utes

.

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ced

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Dan

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alm

an, 1

998

9

For

mal

Lin

eariz

atio

n of

La

gran

ge

Equ

atio

nsD

eriv

atio

n of

Sys

tem

Mat

rices

Inth

elim

iting

case

sof

smal

lde

flect

ion

(),

smal

lra

tes

,

the

Lagr

ang

e eq

uatio

ns s

impl

ify.

We

begi

n b

y lo

okin

g at

kin

etic

ene

rgy

.

Aga

in,

we

mak

eus

eof

one

ofth

eco

reob

serv

atio

nsin

the

deriv

atio

nof

the

Lagr

ang

e eq

uatio

ns:

q r1

«q r

Tm

n 2------

x˙ nx˙ n

⋅n∑

mn 2------

q r∂∂x

˙ n

q s∂∂x

˙ n⋅

q rq

ss∑

r∑n∑

==

q r∂∂x

˙ n

q r∂∂x

n=

Tm

n 2------

q r∂∂x

n

q s∂∂x

n⋅

q

rqs

s∑r∑

n∑1 2---

q rqsM

rsq{

}(

)s∑

r∑=

=

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an, 1

998

10

For

mal

Lin

eariz

atio

n of

La

gran

ge

Equ

atio

nsD

eriv

atio

n of

Sys

tem

Mat

rices

In th

e lim

it of

sm

all d

ispl

acem

ent a

bout

the

confi

gura

tion

of s

tab

le

equi

libriu

m, t

he m

ass

mat

rix b

ecom

es

.

Not

e th

at is

stil

l sym

met

ric.

Impo

rtan

t: In

the

limit

of s

mal

l dis

plac

emen

t,

&

qs{

}

Mrs

Mrs

qs{

}(

)m

n 2------

q r∂∂x

n

qs

{}

q s∂∂x

n

qs

{}

n∑=

=

Mrs

Mrs

qs{

}(

)=

T1 2---

q rq sMrs

s∑r∑

=M

rsq r

q s∂

2 ∂∂T

=

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998

11

For

mal

Lin

eariz

atio

n of

La

gran

ge

Equ

atio

nsD

eriv

atio

n of

Sys

tem

Mat

rices

Obs

erve

that

the

mas

s m

atrix

is p

ositi

ve s

emi-d

efini

te. C

onsi

der

an

arra

yof

gen

eral

ized

spee

ds.T

heco

rres

pond

ing

kine

ticen

ergy

isqt

{}

Tqt

{}

()

1 2---M

rsq rt q st

rs,∑

=

1 2---q rt q st

mn 2------

q r∂∂x

n

qs

{}

q s∂∂x

n

qs

{}

n∑r

s,∑=

mn 2------

q r∂∂x

n

qs

{}

q s∂∂x

n

qs

{}

q rt q st

s∑r∑

n∑=

mn 2------

y˙ nty˙ nt

0w

here

y˙ nt≥

⋅n∑

q r∂∂x

n

qs

{}q rt

r∑=

=

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Seg

alm

an, 1

998

12

For

mal

Lin

eariz

atio

n of

La

gran

ge

Equ

atio

nsD

eriv

atio

n of

Sys

tem

Mat

rices

The

mea

ning

of t

he m

ass

mat

rix.

Con

side

r th

e “a

ccel

erat

ion”

for

ce

.

Nex

t, co

nsid

er c

olum

n ve

ctor

of g

ener

aliz

ed a

ccel

erat

ions

that

are

all

zero

sex

cept

for

a“1

”on

the

r’th

row

:.T

he

resu

lting

“ac

cele

ratio

n f

orce

s” s

een

by

eac

h of

the

othe

r g

ener

aliz

ed

degr

ees

of fr

eedo

m a

re th

e r’t

h co

lum

n of

.

tdd

q r∂∂T

t

ddq r

∂∂1 2---

Mst

q sqt

t∑s∑

Mrs

q ss∑

==

q{} r

0.

1.

.0

T=

M

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998

13

For

mal

Lin

eariz

atio

n of

La

gran

ge

Equ

atio

nsE

xam

ple

of M

ass

Mat

rix

Lets

con

side

r a

stru

ctur

esi

mila

rto

the

disk

-sha

ftsy

stem

inM

eiro

vitc

h.

Igno

ring

the

kine

tic e

ner

gy o

fth

e sh

aft,

the

kine

tic e

ner

gy o

fth

e sy

stem

s is

.

Our

mas

s m

atrix

is

L1

L2

L4

L3

J 1,θ

1J 2

,θ2

J 3,θ

3

E, J

s

T1 2---

J 1θ 12

1 2---J 2

θ 221 2---

J 3θ 32

++

=

Mij

[]

θ iθ

j∂

2 ∂∂T

J 10

0

0J 2

0

00

J 3

==

Advanced Vibrations

/home/djsegal/UNM/VibCour se/slides/11/30/98 Copyright Dan Segalman, 1998

14

AnotherExample of Mass Matrix

Consider acantile vered beam f orwhic h we postulate adisplacement fieldwith tw o freegeneraliz edcoor dinates:

The kinetic ener gy in the beam will be

where ,

and

EI, m

L

y x t,( ) A1 t( ) f 1 x( ) A2 t( ) f 2 x( )+=

T A1 A2,( ) m2---- y( )2

xd

0

L

∫=

12--- A1 t( )

2I1 2A1 t( ) A2 t( )I2 2A2 t( )( )I3++[ ]=

I1 m f 1 x( )( )2xd

0

L

∫= I3 m f 2 x( )( )2xd

0

L

∫=

I2 m f 1 x( ) f 2 x( ) xd

0

L

∫=

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alm

an, 1

998

15

Ano

ther

Exa

mpl

e of

Mas

s M

atrix

Con

tinue

d

Our

mas

s m

atrix

isM

ij[

]A

iA

j∂

2 ∂∂T

mI 1

I 2I 2

I 3=

=

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ht

Dan

Seg

alm

an, 1

998

16

For

mal

Lin

eariz

atio

n of

La

gran

ge

Equ

atio

nsD

eriv

atio

n of

Sys

tem

Mat

rices

Con

side

rth

epo

tent

ial

ener

gyin

the

vici

nity

ofa

confi

gura

tion

ofst

able

equi

libriu

m. I

n an

equ

ilibr

ium

con

figur

atio

n, th

e su

m o

f the

pot

entia

l

forc

es in

eac

h di

rect

ion

is z

ero

.

Tayl

or s

erie

s e

xpan

sion

for

is

.

We

defin

e th

e st

iffne

ss m

atrix

to b

e.

Not

e th

at, b

y co

nstr

uctio

n, is

sym

met

ric.

0q r

∂∂V –=

V

Vq{

}(

)V

01 2---

q rq s

2 ∂∂V

qs{

}∆q r∆

q sH

.O.T

.+

s∑r∑

+=

Krs

q rq s

2 ∂∂V

qs{

}

=

K

Ad

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ced

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on

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ht

Dan

Seg

alm

an, 1

998

17

For

mal

Lin

eariz

atio

n of

La

gran

ge

Equ

atio

nsD

eriv

atio

n of

Sys

tem

Mat

rices

Pot

entia

l ene

rgy

in te

rms

of S

tiffn

ess.

In th

e lim

it of

sm

all s

trai

ns,

whe

re w

e ha

ve r

edefi

ned

the

gen

eral

ized

coo

rdin

ates

to b

e z

ero

at th

e

equi

libriu

m c

onfig

urat

ion.

Usu

all

y, w

e se

t the

dat

um s

o th

at.

Bec

ause

sta

bilit

y re

quire

s th

at p

oten

tial e

ner

gy b

e at

a lo

cal m

inim

um,

for

all

. Thi

s is

an

asse

rtio

n th

at th

e st

iffne

ss

mat

rix is

pos

itive

sem

i-defi

nite

.

VV

01 2---

Krs

q rqs

s∑r∑

+=

V0

0=

Krs

q rqs

0≥

s∑r∑

q{}

Ad

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ced

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Seg

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an, 1

998

18

For

mal

Lin

eariz

atio

n of

La

gran

ge

Equ

atio

nsD

eriv

atio

n of

Sys

tem

Mat

rices

The

mea

ning

of t

he S

tiffn

ess

mat

rix.

Con

side

r th

e g

ener

aliz

ed f

orce

.

Nex

t, co

nsid

er c

olum

n ve

ctor

of g

ener

aliz

ed d

ispl

acem

ent t

hat a

re a

ll

zero

sex

cept

for

a“1

”on

the

r’th

row

:.T

he

resu

lting

for

ce s

een

by

eac

h of

the

othe

r g

ener

aliz

ed d

egre

es o

f

free

dom

are

the

r’th

colu

mn

of.

Fr

q r∂∂

1 2---K

stq sq

tt∑

s∑

Krs

q ss∑–

==

q{} r

0.

1.

.0

T=

K

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ced

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ht

Dan

Seg

alm

an, 1

998

19

An

Exa

mpl

e P

rob

lem

of I

nvo

lvin

g th

eS

tiffn

ess

Mat

rix

Lets

con

side

r a

stru

ctur

esi

mila

rto

the

disk

-sha

ftsy

stem

inM

eiro

vitc

h.

The

stra

inen

ergy

inth

esh

aft

isco

mpu

ted

inte

rms

ofth

edi

ffer

ence

sin

the

rota

tions

at t

he d

isks

.

L1

L2

L4

L3

J 1,θ

1J 2

,θ2

J 3,θ

3

G, J

s

V1 2---

GJ s

θ 12

L1

-----

θ 2θ 1

–(

)2

L2

--------

--------

--------

θ 3θ 2

–(

)2

L3

--------

--------

--------

θ 32

L4

-----

++

+=

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20

An

Exa

mpl

e P

rob

lem

of I

nvo

lvin

g th

eS

tiffn

ess

Mat

rix

Our

stif

fnes

s m

atrix

is

Not

e th

at th

eth

ele

men

t of

is th

e to

rqu

e fe

lt on

dis

k d

ue to

a

unit

rota

tion

impo

sed

on d

isk

.

Kij

[]

θ iθ

j∂

2 ∂∂V

GJ s

1 L1

-----

1 L2

-----

+1 L2

-----

–0

1 L2

-----

–1 L2

-----

1 L3

-----

+1 L3

-----

01 L3

-----

–1 L3

-----

1 L4

-----

+

==

ij,

Ki

j

Advanced Vibrations

/home/djsegal/UNM/VibCour se/slides/11/30/98 Copyright Dan Segalman, 1998

21

Another Example In volving theStiffness Matrix

Consider acantile vered beam f orwhic h we postulate adisplacement fieldwith tw o freegeneraliz edcoor dinates:

The strain ener gy in the beam will be

,

and

EI, m

L

y x t,( ) A1 t( ) f 1 x( ) A2 t( ) f 2 x( )+=

V A1 A2,( ) EI2------ y ′ ′( )2

xd

0

L

∫=

12--- A1 t( )2

I4 2A1 t( )A2 t( )I5 A2 t( )2I6++[ ]=

I4 EI f 1′ ′ x( )( )2xd

0

L

∫= I6 EI f 2′ ′ x( )( )2xd

0

L

∫=

I5 m f 1′ ′ x( ) f 2′ ′ x( ) xd

0

L

∫=

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22

Ano

ther

Exa

mpl

e of

Mas

s M

atrix

Con

tinue

d

Our

stif

fnes

s m

atrix

isK

ij[

]A

iA

j∂

2 ∂∂V

EI

I 4I 5

I 5I 6

==

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23

Mat

rix M

ultip

licat

ion

A s

hor

t re

vie

w

Say

and

are

col

umn

vect

ors

of le

ngth

, the

n a

re

defin

ed to

be

. Thi

s is

the

inne

r pr

oduc

t of a

lgeb

raic

vec

tor

s.

Say

is a

n b

y m

atrix

, the

n th

e m

atrix

vec

tor

prod

uct

is

defin

ed s

o th

at th

eth

ele

men

t of

is

Com

bini

ng th

e ab

ove

two

defin

ition

s:

ab

NbT

aaT

b=

a rbr

r1

=N ∑

AN

NA

a

rA

aA

rsa s

s1

=N ∑î

aTA

bA

rsa rb

ss

1=N ∑

r1

=N ∑

=

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24

Mat

rix M

ultip

licat

ion

A s

hor

t re

vie

w

The

pro

duct

of t

wo

mat

rices

is d

efine

d in

a s

imila

r w

ay.

Say

is a

lso

an b

y m

atrix

als

o, th

en th

e pr

oduc

t is

defi

ned

as

The

tran

spos

e of

a m

atrix

is th

at w

hic

h is

obt

aine

d b

y re

vers

ing

the

orde

r of

the

indi

ces:

. T

he tr

ansp

ose

of p

rod

ucts

is

foun

dto

be.

In

mat

rix n

otio

n:

BN

NA

B

AB

() r

sA

rtB

tst

1=N ∑

=

AT

() r

sA(

) sr

=

AB

()T

() r

sA

B(

) sr

Ast

Btr

t1

=N ∑B

TA

T(

) rs

==

=

AB

()T

BT

AT

=

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25

Fle

xibi

lity

Mat

rix(L

ets

star

t usi

ng m

atrix

not

atio

n)

In th

ose

case

s w

here

ther

e ar

e no

rig

id b

ody

mot

ions

, the

stif

fnes

sm

atrix

is n

onsi

ngul

ar, a

nd w

e ca

n de

fine

a fle

xibi

lity

mat

rix

.

In th

is c

ase

we

obse

rve

that

sin

ce,

.

The

str

ain

ener

gy m

ay

now

be

exp

ress

ed in

term

s of

for

ce

Som

etim

es v

ibra

tion

prob

lem

s ar

e fo

rmul

ated

in te

rms

of th

e fle

xibi

lity

mat

rix, t

houg

h th

e us

e of

dis

plac

emen

t-ba

sed

finite

ele

men

ts m

akes

flexi

bilit

y f

orm

ulat

ions

dec

reas

ingl

y po

pula

r.

Dis

cuss

ion

of th

e fle

xibi

lity

mat

rix is

pre

sent

ed h

ere

for

com

plet

enes

s.W

e sh

all d

o lit

tle m

ore

with

it in

this

cla

ss.

aK

1–= F

Kq

–=

qa

F–

=

V1 2---

qTK

q1 2---

aF

–()T

Ka

F–(

)1 2---

FT

aF

==

=

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26

Mat

rix T

rans

form

atio

ns

Say

that

we

have

deriv

edth

efo

llow

ing

mat

rixre

pres

enta

tion

ofa

linea

r

syst

em:

whe

re th

e co

lum

n ve

ctor

con

tain

s al

lex

tern

ally

app

lied

load

.

No

w, s

ay w

e se

lect

ano

ther

set

of g

ener

aliz

ed c

oord

inat

es w

ith w

hic

h

we

can

spec

ify th

e fir

st s

et:

. Let

s re

-der

ive

our

gove

rnin

g

equa

tions

in te

rms

of.

Kin

etic

Ene

rgy

:

Str

ain

Ene

rgy

:

Pot

entia

l Ene

rgy

of L

oadi

ng:

Mq

Kq

+F

=F

qT

β=

β

T1 2---

qTM

q1 2---

()T

MT

β(

)1 2---

βTT

TM

T(

)β=

==

V1 2---

qTK

q1 2---

()T

KT

β(

)1 2---

βTT

TK

T(

)β=

==

Aq–

TF

()T

F–

βTT

TF

()

–=

==

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27

Mat

rix T

rans

form

atio

nLa

gran

ge

Equ

atio

n in

Ne

w S

yste

m

beco

mes

whe

re,

, and

Not

e th

at th

e m

ass

and

stiff

ness

mat

rices

rem

ain

sym

met

ric a

nd a

tle

ast p

ositi

ve s

emi-d

efini

te.

Ifis

ano

nsin

gula

r,s

quar

em

atrix

,th

etr

ansf

orm

atio

nis

calle

da

cong

ruen

ce tr

ansf

orm

atio

n of

A.

tdd

β r∂∂T

β r∂∂T–

β r∂∂V+

β r∂∂A –=

0

+F

=

MT

TM

T=

KT

TK

T=

FT

TF

=

TT

TA

T

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28

Mat

rix T

rans

form

atio

n

We

ofte

n us

e su

ch

tran

sfor

mat

ions

whe

n in

trod

ucin

g co

nstr

aint

s to

asy

stem

. Her

e w

e co

nsid

er tw

oin

depe

nden

t pen

dulu

ms.

The

kin

etic

and

pote

ntia

l ene

rgi

es a

re

and

The

pot

entia

l ene

rgy

of l

oadi

ng is

The

line

ariz

ed e

quat

ions

of m

otio

n ar

e

R1

R2

m1

m2

F1

F2

Tm

1 2------

R12θ 12

m2 2------

R22θ 22

+=

Vm

gR

1θ 1

mg

R2

θ 2()

cos

–co

s–

= AF

1R

1θ 1

F2R

2θ 2

––

=

m1R

120

0m

2R

22

θ 1 θ 2

mg

R1

0

0m

gR

2

θ 1 θ 2+

F1R

1

F2R

2

=

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29

Mat

rix T

rans

form

atio

nLe

ts a

dd

a co

nstr

aint

Con

side

r th

e ef

fec

t of c

onne

ctin

g th

etw

o m

asse

s w

ith a

mas

sles

s r

od. T

his

has

the

eff

ect o

f req

uirin

g

, per

mitt

ing

us to

writ

e

.

Defi

ne. T

hen

R1

R2

m1

m2

F1

F2

R1θ 1

R2θ 2

=

θ 1 θ 2

1

R1

R2

⁄θ 1

= T1

R1

R2

⁄=

M1

R1

R2

⁄m

1R

120

0m

2R

22

1

R1

R2

⁄m

1m

2+

()R

12=

=

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30

Mat

rix T

rans

form

atio

nW

ith a

con

stra

int

The

stif

fnes

s m

atrix

is

and

the

appl

ied

for

ce is

K1

R1

R2

⁄m

1g

R1

0

0m

2g

R2

1

R1

R2

⁄=

m1g

R1

m2g

R12

R2

------

+=

F1

R1

R2

⁄F

1R

1

F2R

2

R1

F1

F2

+(

)=

=

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31

Mat

rix T

rans

form

atio

nW

ith a

con

stra

int

The

res

ultin

g eq

uatio

n f

or is

The

ad

ditio

n of

a c

onst

rain

t red

uced

the

num

ber

of a

ctiv

e de

gree

s of

free

dom

by

one

.

θ 1

m1

m2

+(

)R12

[]θ

1m

1g

R1

m2g

R12

R2

------

+

θ 1+

R1

F1

F2

+(

)=

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32

Nex

t Tim

e

Sho

rt O

rient

atio

n

LAB

TO

UR