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Pertemuan – 6, 7

Multi Degree of Freedom System Free Vibration

Mata Kuliah : Dinamika Struktur & Pengantar Rekayasa Kegempaan

Kode : CIV – 308

SKS : 3 SKS


TIU : Mahasiswa dapat menjelaskan tentang teori dinamika struktur.

Mahasiswa dapat membuat model matematik dari masalah teknis yang

ada serta mencari solusinya.

TIK : Mahasiswa mampu mendefinisikan derajat kebebasan, membangun

persamaan gerak sistem MDOF


Sub Pokok Bahasan :

Penentuan derajat kebebasan

Properti matrik kekakuan, massa dan redaman


Structures cannot always be described by a SDoF

model

In fact, structures are continuous systems and as such

possess an infinite number of DoF

There are analytical methods to describe the

dynamic behavior of continuous structures, which

are rather complex and require considerable

mathematical analysis.

For practical purposes to study Multi Degree of

Freedom System (MDoF), we shall consider the

multistory shear bulding model.


A shear building may be defined as a structure in which there

is no rotation of a horizontal section at the level of the floors

Assumption for shear building

The total mass of the structure is concentrated at the

levels of the floors

The slabs/girders on the floors are infinitely rigid as

compared to the columns

The deformation of the structure is independent of the

axial forces present in the columns


k1 k1

k2 k2

k3 k3

m2

m3

m1

u2

u3

u1

F1(t)

F2(t)

F3(t) F3(t)

k3(u3 u2)

m3ü3

F2(t)

k2(u2 u1)

m2ü2

k3(u3 u2)

F3(t)

k1u1

m1ü1

k2(u2 u1)

Shear Building w/ 3 DoF

Free Body Diagram


By equating to zero the sum of the forces acting on each

mass :

Eq. (1) may conveniently be written in matrix notation as :

0

0

0

323333

223312222

11221111

tFuukum

tFuukuukum

tFuukukum

FuKuM (2)

(1)


Where :

And :

3

2

1

00

00

00

m

m

m

M

33

3322

221

0

0

kk

kkkk

kkk

K

3

2

1

u

u

u

u

3

2

1

u

u

u

u

tF

tF

tF

F

3

2

1


Natural Frequencies and Normal Modes EoM for MDoF system in free vibration is :

Solution of Eq. (3) is :

The substitution of Eq. (4) into Eq. (3) gives :

0 uKuM (3)

tsinBtcosAtqu nnnnnnn (4)

02 tqMK nnnn (5)


The formulation of Eq (5) is known as an eigenvalue problem.

To indicate the formal solution to Eq. (5), it is rewritten as :

Eq (6) has non trivial solution if :

Eq.(7) gives a polynomial equation of degree n in n2, known

as characteristic equation of the system

02 nn MK (6)

A vibrating system with n DoF has n natural vibration frequencies n, arranged in

sequence from smallest to largest

0K2

Mn (7)


The n roots of Eq. (7) determine the n natural frequencies n

of vibration.

These roots of the characteristic equation are also known as

eigenvalues

When a natural frequency n is known, Eq.(6) can be solved

for the corresponding vector n

There are n independent vectors , which are known as

natural modes of vibration, or natural mode shape of

vibration

These vectors are also known as eigenvector.


Example 1

Determine :

1. The natural frequencies and

corresponding modal shapes

W1 = 11.600 kgf

4,5 m

W1

0x2

1

W2 = 24.000 kgf

u2

u1

3 m

W1

0x4

5

W1

0x2

1

W1

0x4

5

W10x21 I = 4,900 cm4

W10x45 I = 10,300 cm4

E = 2.106 kg/cm2

110385368090

2838 7610

2221

1211

21

,,

,,


Tugas 6

Determine :

1. The natural frequencies and

corresponding modal shapes

W10x21 I = 4,900 cm4

W10x45 I = 10,300 cm4

E = 2.106 kg/cm2 W1 = 11.600 kgf

4,5 m

W1

0x2

1

W2 = 24.000 kgf

u2

u1

3 m

W1

0x4

5

W1

0x2

1

W1

0x4

5

W3 = 20.000 kgf

W1

0x2

1

W1

0x2

1 u3

3 m

11165742662085260

34315004155460

5339 6526 7,934

333231

232221

131211

321

,,,,,,

,,


1

0,8526

0,5546

0 0

1

2

3

0 0,5 1 1,5

mode 1 1

-0,662

-1,004

0 0

1

2

3

-2 -1 0 1 2

mode 2 1

-2,6574

5,3431

0 0

1

2

3

-5 0 5 10

mode 3


Modal & Spectral Matrices The N eigenvalues, N natural frequencies, and n natural

modes can be assembled compactly into matrices

nnnn

n

n

...............

...

...

21

22221

11211

2

22

21

2

n

Mode 1 2 n

DoF 1 2 ….. n

Modal matrix

Spectral matrix


Orthogonality of Modes The natural modes corresponding to different natural

frequencies can be shown to satisfy the following

orthogonality conditions :

0

0

rT

n

rT

n

M

K

When n ≠ r


Free Vibration Response : Undamped System

EoM for MDoF Undamped Free Vibration is :

The differential equation (8) to be solved had led to the

matrix eigenvalue problem of Eq.(6).

The general solution of Eq. (8) is given by a superposition of

the response in individual modes given by Eq. (4).

Or

0 uKuM (8)

nn q...qqqu 332211

qu

(9)

(10)

N

nnn

qu1


Premultiply Eq. (10) with {n}T[M] :

Regarding the orthogonality conditions :

From which :

also

nT

nT

n qMuM (11)

nn

T

n

T

nqMuM

NN

T

n

T

n

T

n

qM

....qMqM

2211

n

T

n

T

n

nM

uMq

n

T

n

T

n

nM

uMq

(11)


If Eq. (8) is premultiplied by the transpose of the nth mode-

shape vector {n}T, and using Eq.(10) with its second

derivative it becomes :

Eq.(11) is an EoM from SDoF Undamped System for mode n,

which has solution :

0nn

T

nnn

T

nqKqM

Mn Kn

(11)

Generalized mass Generalized stiffness

tsinq

tcosqtqn

n

n

nnn

00

(12)


Solution for Eq. (8) becomes :

The procedure described above can be used to obtain an

independent SDoF equation for each mode of vibration of

the undamped structure.

This procedure is called the mode-superposition method

N

nn

n

n

nnntsin

qtcosqu

1

00

(13)


Gaya pegas tiap lantai dari masing-masing

mode dapat dicari dengan persamaan :

uMuKf ns

2


Example 2

Based on data from Example 1, and the

following initial condition, for each DoF plot

the time history of displacement, regarding

the 1st, and 2nd mode contribution.

mu t

3

3

0 108

106

m/s0,20

00

tu


Tugas 7

Based on data from Assignment 6, and the

following initial condition, for each DoF

plot the time history of displacement,

regarding the 1st, 2nd , and 3rd mode

contribution.

Also plot the time history of spring

force

W1 = 11.600 kgf

4,5 m

W1

0x2

1

W2 = 24.000 kgf

u2

u1

3 m

W1

0x4

5

W1

0x2

1

W1

0x4

5

W10x21 I = 4,900 cm4

W10x45 I = 10,300 cm4

E = 2.106 kg/cm2

W3 = 20.000 kgf

W1

0x2

1

W1

0x2

1 u3

3 m

cm,,,

u t

018060

0 cm/s0200

0

tu


Free Vibration Response : Damped System

EoM for MDoF Damped Free Vibration is :

Using Eq. (10) and its derivative :

Premultiply Eq. (15) with {}T :

0 uKuCuM (14)

(15)

(16)

0 qKqCqM

0 qKqCqMTTT

Mn Cn Kn


Eq.(16) is an EoM from SDoF Damped System for mode n,

which has solution :

The displacement response of the system is then obtained by

substituting Eq. (17) in Eq. (10)

tsin

qqtcosqetq

nD

nD

nnnn

nDn

t

n

nn

00

0

(17)

tsinqq

tcosqetunD

nD

nnnn

nDn

tN

nn

nn

00

01

(18)

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