c9.distributed mass and elasticity2 compatibility mode

7/29/2019 C9.DIstributed Mass and Elasticity2 Compatibility Mode

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DYNAMIC OF STRUCTURES

CHAPTER 9

SYSTEMS WITH DISTRIBUTED MASS

AND ELASTICITY

Department of civil engineering, University of North Sumatera

Ir. DANIEL RUMBI TERUNA, MT;IP-U


2/12

MODAL ANALYSIS OF FORCED DYNAMIC RESPONSE

We return to the partial diffential equation of simply beamsubjected to force

The displacement is given by a linier combination of the modes:

( )( )

( )( )

( ) )1(,,,

2

2

2

2

2

2

txpt

txuxm

x

txuxEI

x=

+

Subtituting Eq.(2) in Eq.(1) gives

( )txp ,

( ) ( ) ( ) )2(,1

tqxtxu rr

r

=

=

( ) ( ) ( ) ( ) ( )[ ] ( ) ( ) )3(,11

txptqxxEItqxxm rr

rr

r

r =+

=

=

&&

Multiply each term by , integrate it over the length of the beam gives( )xn


3/12


( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ] ( ) ( ) )4(,01 01 0

dxxtxpdxxxEIxtqdxxxxmtqL

nr

r

L

nrr

r

L

nr =+

=

=

&&

This equation can be written as

nr =By use orthogonality properties of modes, all term left side vanish exceptthe one term for which . Thus, Eq.(4) becomes

( ) ( ) ( )[ ] ( ) ( ) ( ) ( )[ ] ( ) ( ) )5(,00

2

0

=+

L

nn

L

nn

L

nn dxxtxpdxxxEIxtqdxxxmtq &&

( ) ( ) ( ) )6(tptqKtqM nnnnn =+&&where

( ) ( )[ ] ( ) ( ) ( )[ ] ( ) ( ) ( )[ ] dxxxEIxdxxxEIxKdxxxmM nL

nn

L

nn

L

nn

2

00

2

0

, ===

( ) ( ) ( ) )7(,0

=L

nn dxxtxptP


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Equation (6) governs the modal coordinate , and the generalized

properties , , and depend only on the mode .Thus we have infinite number of equation like Eq.(6), one for each mode. Thepartial diffrential equation equation (1) in the unknown fungtion hasbeen transformed to an infinite set of ordinary diffrential equation (6) inunknowns .

Once the have been determined, the contribution of the mode to the

displacement is given by:

The generalized mass and generalized stiffness for modeare related

( )txu ,

nM nKthn

)8(2 nnn MK =

( )tqnthnnM nK tPn

th

n ( )xn

tqntqn

( ) ( ) ( ) )9(, tqxtxu nnn =

thn

txu ,


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( ) ( ) ( ) ( ) )10(,,11

tqxtxutxu nn

n

n

n

=

=

==

( )xu

( ) ( ) ( ) ( ) ( ) ( )[ ] )11(== xuxEIxVxuxEIxM

4. Specialize

The bending moment and shear force at any section along the length of thebeam are related to displacement as follows:

The static relationships apply at each instant of time with replaced by

which is given by Eq.(10). Thus

xu

( )txu ,

( ) ( ) ( ) ( ) ( ) ( ) ( )[ ] ( ) )12(,,11

=

=

==n

nnn

nn tqxxEItxVtqxxEItxM


6/12

EXAMPLE

( ) )(sin2

22

aL

xnx

m

EI

L

nnn

==

0p

L

EIm,

x

0p

t

SOLUTION

1. Determined the natural frequencies and modes

2. Determined the generalized mass, stiffness and force

( ) ( )[ ] )(2

sin0

2

0

bmL

dxL

xnmdxxxmM

LL

nn =

==


7/12

EXAMPLE

( ) ( ) ( )[ ] )(2

sin3

442

0

4

44

0

cL

EIndx

L

xn

L

nEIdxxxEIxK

L

n

L

nn

=

=

=

( ) ( ) ( ) )(, 00

epdxxtxpP n

L

nn ===

3. Set up modal equation

The n th modal equation is

( ) ( )( )

n

nnnn

nnnnn

M

ptqtqor

ftptqKtqM

0

2

)(

=+

=+

&&

&&

)(2

3

442 d

L

EInMKor nnn

==


8/12

EXAMPLE

( ) ( ) ( ) )(sincos 0 gK

ptBtAtq

n

nnnnnn

++=

( ) 00 =nq

4. Solved modal equation

If at , and , we get0=t00

=nq&

)(00 hBK

pA n

n

nn =

=

Subtituting Eq.(h) in Eq.(g) lead to

( ) ( ) ( ) ( ) )(2cos1 443

00 inEI

LptK

ptq nnn

nn

==

Subtituting Eq.(i) in Eq.(10) , we obtained the displacement response txu ,


9/12


( )( )

( ) )(sincos12/2

,1

44

3

0 hL

xnt

n

L

EI

Lptxu

nn

n

=

=

4. Specialiaze for

where

)(

,...11,7,31

,...9,5,11

,...6,4,20

2j

n

n

nL

n

=

=

=

=

2/L=

Subtituting Eq. (j) in Eq.(h) gives

( )

)(...........7

sin2401

cos15sin

625

cos1

3sin

81

cos1sin

1

cos12,

75

31

4

3

0

kL

xt

L

xt

L

xt

L

xt

EI

Lptxu

+

+

=


10/12


Displacement at mid span is

The displacement for the first mode contribution only can be expressed as

)(...........2401

cos1

625

cos1

3sin

81

cos1

1

cos12,

2

75

31

4

3

0

ltt

L

xtt

EI

Lpt

Lu

+

+

+

+

=

( ) )(cos16.48

,2

1

3

0 mtEI

Lpt

Lu =

dynamic displacementamplication factor

Staticdisplacement


11/12


( )

)(...........7sin49cos15sin

25cos1

3sin

9

cos1sin

1

cos12,

75

31

2

0

kL

xtL

xt

L

xt

L

xtLptxM

+

+

=

The bending moment are obtained by subtituting Eq.(h) in Eq.(12)

The bending moment at mid span is

)(...........49

cos1

25

cos1

9

cos1

1

cos12

,2

75

31

2

3

0

ott

s

ttLp

t

L

M

+

+

+

+

=


12/12


The bending moment for the first mode contribution only can be expressed as

( ) )(cos193.4

,2

10 pt

EI

Lpt

LM =

Moment dynamicamplication factor

Static bendingmoment

Solution for by use Duhamel integral

( ) ( ) ( ) ( ) ( )

dtpm

dtpm

tqn

t

nn

n

t

nn

==

sin

1sin

1

00

0

tqn

( ) ( )[ ] ( )tn

pt

m

ptq n

nt

n

n

nn

cos1

2cos

44

0

02

0 ==

c9.distributed mass and elasticity2 compatibility mode

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