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1/27

DYNAMIC OF STRUCTURES

CHAPTER 6

RESPONSE OF MDOF SYSTEMS

SUBJ ECTED TO GROUND MOTION

Department of Civil Engineering, University of North Sumatera

Ir. DANIEL RUMBI TERUNA, MT;IP-U


2/27

RESPONSE HISTORY ANALYSIS

gu gu

ju

t

ju

j

N

Rigid-bodymotion

ju

t

ju

j

N

gu


3/27


jm

( ) ( ) ( ) )2(0=++ tkutuctum t &&&

The total displacement of the mass is

the equation of motion for free vibration MDOF system can be expressed as

( ) ( ) ( ) )1(vectorinfluence; =+= rturtutu gjt

j

Equation of motion

Subsituting equation (1) in equation (2) lead to

)3(gurmkuucum &&&&& =++

( ) )4(1

=

=N

r

rrqtu The displacement u can be expressed as superposition of the modalcontribution


4/27


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Dividing by ,noting that lenM nnn

n

M

C 2=)8(2

2

gnnnnnnn uqqq &&&&& =++ where

===

=

=

N

j

jjn

N

j

jjn

n

T

n

T

n

n

nn

m

m

m

rm

M

L

1

2

1

Modal participation factor

)9(2 2 gnnnnnn uDDD &&&&& =++ Equation (8) can be rewritten in other form

where

( ) ( ) )10(tDtq nnn =


6/27


The solution can be obtained by comparing equation (9) tequation of motion for the mode SDOF system subjected to groundacceleration . Thus

( )tDn

( )tug&&

( ) ( ) ( ) ( ) )11(sin1

0 dteutD nt

t

gnn =

&&

Or step by step integration

Modal responses

The contribution of the mode to the displacement is

( ) ( ) ( ) )13(tDtqtu nnnnnn ==The total displacement can be obtained by combination of theresponse contribution of all modes

thn

( )tu

thn ( )tu


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( ) ( ) ( ) )14(11

tDtutu nnn

N

n

n

N

n==

==

For multistory building with rigid floor diaphragms, lateral displacementfloor can be calculated as

( ) ( ) )15(tDtu njnnjn =The interstory drift is given by the diffrences of displacement of the floorabove and below:

( ) ( ) ( ) ( ) )16()( ,1,1 tDtutut nnjjnnnjjnjn == The equivalent static lateral forces at floor for mode are

( ) ( ) )17(2 tDmtf njnnnjjn =

thj

thn

thj


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The floor acceleration can be computed from

( ) ( ) ( ) )18(1

tDtutu njnn

N

n

g

t

j += =

&&&&

Influence Vector r

m1

m2

m3

u1

u2

u3

gu

11 =r

12 =r

03 =r

=

0

11

r


9/27

EXAMPLE

2w

h=3,8m

h

h

1w

3w

L=8 m

mtww /5,221==

Column 30x45


10/27

EXAMPLE

Determine the lump mass

m1=m2 =(8x2500/980)= 20,408 kg.sec2/cm

m3=(8x2000/980)=16,327 kg.sec2/cm

Determine the storey stiffness

k= 2x12 EI/h3= 19928,20 kg/cm

Compute the eigenvector

Compute natural frequencies

=

=

8020.09013.01618.2

3424.13142.07782.1

0.10.10.1

333231

232221

131211

jn

sec/

1297,57

5728,40

7184,14

3

2

1

rad

=

=


11/27

EXAMPLE

Compute the earthquake excitation factor

[ ]{ }rmL Tnn =

[ ]{ }

{ }

cmkg

rmL T

/sec.9919,91

1

1

1

408,20

8,000

00,10

000,1

1618,27782,10,1

2

11

=

=

=

[ ]{ } cmkgrmL T /sec1052,12 222

== [ ]{ } cmkgrmL T /sec1061,6 2

33==


12/27

EXAMPLE

Compute the generalized mass

[ ]{ }nT

nn mM =[ ]{ }

{ }

cmkg

mMT

/sec.2375,161

1618,2

7782,10,1

408,20

8,000

00,10000,1

1618,27782,10,1

2

111

=

=

=

[ ]{ }

cmkg

mM T

/sec.6853,352

222

=

= [ ]{ }

cmkg

mM T

/sec.6852,672

333

=

=


13/27

EXAMPLE

Compute the mode paricipation factor

n

nn

M

L=

5706,01

1

1 == M

L

3392,02

2

2 ==M

L

0902,0

3

3

3 ==M

L

Total = 1,000


14/27

EXAMPLE

Solve the diffrential equation by use Duhamel integral or step by stepintegration

gnnnnnn uDDD &&&&& =++

22

( ) ( ) ( ) ( ) dteutD nt

t

gnn =

sin1

0

&&

( ) ( ) ( )tDtutu nnnN

n

n

N

n==

== 11

[ ] { } [ ] { } [ ] { }3

33

23

13

32

32

22

12

21

31

21

11

1

3

2

1

DDD

u

u

u

+

+

==


15/27

NUMERICAL EVALUATION OF DYNAMIC RESPONSE

[ ]{ } [ ]{ } [ ]{ } ( ) )1(tPuKuCuM =++ &&&

The diffrential of motion for MDOF system subjected to dynamic forces

can be expressed as

When the system subjected to ground acceleration ,equation (1) lead

[ ]{ } [ ]{ } [ ]{ } )2(effPuKuCuM =++ &&&

( )tug&&

[ ] { }geff urMP &&=where

Coupled diffrential equation solution (Newmark-Beta Methode)

The diffrential of motion for MDOF system subjected to ground

motion in incremental form at time ican be written as


16/27


[ ]{ } [ ]{ } [ ]{ } )3(iiii PuKuCuM =++ &&&

where

( ) ( )

)4(

2

11121 iiiiii

uu

t

u

t

uuu &&&&&&&&&

==+

( ) ( )( ) )5(1

1 iiiiii utut

ut

uuu &&&&&&

+

==

+

At the beginning time step , the incremental displacement is computed from

[ ] )6(,1,1 igigiii

uuMPPP &&&& == ++

)7(

K

Pu ii

=


17/27


where )8( iiii ubuaPP &&&++=

( ) ( ))9(

12

+

+= M

tC

tKK

and

Coeficients a and b in the equation (8) can be written as

( ))10(

1

+

= CM

ta

( ) )11(1221

+= CtMb


18/27

EXAMPLE

.sec/90,33

.sec/83,11

2

1

rad

rad

=

=

cmkgk

cmkgm

/32,5487

/sec31,24

2

2

1

=

=h

h

L

cmkgk

cmkgm

/20,7918

/sec79,11

2

2

2

=

=

t

0,2 sec

1

50cm/sec

2

02.0=t


19/27

EXAMPLE

Compute stiffness and mass matrix

Compute damping matrixWe select the damping ratio for first mode =0,05 and damping isassumed mass proportional damping

Compute

[ ] [ ]

=

=

79,110

031,24,

2,79182,7918

2,79182,13405MK

183.183.1105.0220 === xxa ii[ ] [ ]

=

== 45,130

076,28

79,110

031,24

183.1MaC o

bandaK ,,


20/27

EXAMPLE

( ) ( )

+

+= M

tC

tKK

2

1

( ) ( )

+

+

=

79,110

031,241

45,130

076,28

2,79182,7918

2,79182,13405

2

tt

K

For constant average acceleration4

1

2

1 == and

=

7,1272132,7918

2,79182,259381K

( )

=

+

=

90,23850

052,4919

45,130

076,28

79,110

031,241

ta


21/27

EXAMPLE

( )

=

+

=

85,230

052,48

45,130

076,281

279,110

031,24

2

1

ta

Determined effective load

( )ondtsec 2sec/cmu g&&0,00

46,35

88,17121,35

142,665

150,00

0

0,02

0,040,06

0,08

0,10

i0

1

23

4

5


22/27

EXAMPLE

Cycle 1

[ ] igigiii uuMPPP ,1,1 &&&& == ++

( )

=

=

==

47,54677,1126035,46

79,1131,24

0,1,010 gg uuMPPP &&&&

0=i

Initial condition 0,0,0 000 === uuu &&&

From equa.(8)

=++=47,546

77,1126

0000ubuaPP &&&

From equa.(7)0

1

0 PKu =


23/27

EXAMPLE

=

02

1

u

u

=

0046,0

0045,0

47,546

77,1126

7,1272132,7918

2,79182,2593811

( ) ( )

=

+

=

46,0

45,0

2102

1

02

1

02

1

02

1

u

utu

u

u

u

tu

u

&&

&&

&

&

&

&

( ) ( )

=

=

46

45

2

111

02

1

02

1

02

1

2

02

1

u

u

u

u

tu

u

tu

u

&&

&&

&

&

&&

&&

=

+

=

0046.0

0045.0

02

1

02

1

12

1

u

u

u

u

u

u


24/27

EXAMPLE

=

+

=

0046.0

0045.0

02

1

02

1

12

1

u

u

u

u

u

u

=

+

=

46.0

45.0

02

1

02

1

12

1

u

u

u

u

u

u

&

&

&

&

&

&

=

+

=

46

45

02

1

02

1

12

1

u

u

u

u

u

u

&&

&&

&&

&&

&&

&&


25/27

EXAMPLE

Cycle 2

[ ]

( )

=

=

==

06,493

64,101635,4617,88

79,11

31,24

1,2,121 gg uuMPPP &&&&

1=i

=

+

+

=

++=

25,267582,5413

46

45

88,230

052,48

46,0

45,0

9,23850

05,4919

06,493

64,1016

1111ubuaPP &&&


26/27

EXAMPLE

=

12

1

u

u

=

02254,0

02156,0

25,2675

82,5413

7,1272132,7918

2,79182,2593811

( ) ( )

=

+

=

245,2

147,2

2112

1

12

1

12

1

12

1

u

u

tu

u

u

u

tu

u

&&

&&

&

&

&

&

( ) ( )

=

=

40,133

60,125

2

111

12

1

12

1

12

1

2

12

1

u

u

u

u

tu

u

tu

u

&&

&&

&

&

&&

&&

=

+

=

02714.0

02606.0

12

1

12

1

22

1

u

u

u

u

u

u


27/27

EXAMPLE

=

+

=

7048.2

597.2

12

1

12

1

22

1

u

u

u

u

u

u

&

&

&

&

&

&

=

+

=

4.179

60.170

12

1

12

1

22

1

uu

uu

uu

&&

&&

&&

&&

&&

&&

Cycle 3, 2=i

Follow the same procedures

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earthquake analysis [Compatibility Mode]