Slide CIV 308 DinamikaStr PengRekGempa CIV 308 P6 7

a home base to excellence
Mata Kuliah
Kode
SKS

: Dinamika Struktur & Pengantar Rekayasa Kegempaan
: CIV – 308
: 3 SKS

Multi Degree of Freedom System
Free Vibration
Pertemuan – 6, 7

a home base to excellence


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

a home base to excellence


Sub Pokok Bahasan :
 Penentuan derajat kebebasan
 Properti matrik kekakuan, massa dan redaman

a home base to excellence



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 home base to excellence


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

a home base to excellence
u3

m3

F3(t)
k3

F3(t)

k3
m2

F2(t)
k2

m1

F1(t)

u2

m 3ü 3

F2(t)

k3(u3 u2)
k3(u3 u2)
m 2ü 2

F3(t)

k2(u2 u1)

m 1ü 1

k2 u1

k2(u2 u1)

k1u1

k1

k1

Shear Building w/ 3 DoF

Free Body Diagram

a home base to excellence


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

mass :
m1u1  k1u1  k2 u 2  u1   F1 t   0

m2u2  k2 u 2  u1   k3 u3  u 2   F2 t   0

m3u3  k3 u3  u 2   F3 t   0


M u K u  F 

(1)

Eq. (1) may conveniently be written in matrix notation as :
(2)

a home base to excellence




Where :
m1
M    0
0


And :

0
m2
0

u1 
u  u 2 
u3 

0
0
m3 

k1  k2
K     k2
 0


u1 
u  u2 
u3 

 k2
k2  k3
 k3

0 
 k3 
k3 

 F1 t 
F   F2 t 
 F3 t 

a home base to excellence
Natural Frequencies and Normal Modes






M u K u  0

EoM for MDoF system in free vibration is :

u  n qn t   n  An cos nt  Bn sin nt 

(3)

Solution of Eq. (3) is :

K   

M n qn t   0

(4)

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

n

2

(5)

a home base to excellence



K  

n





M n   0

The formulation of Eq (5) is known as an eigenvalue problem.
To indicate the formal solution to Eq. (5), it is rewritten as :
2

(6)

Eq (6) has non trivial solution if :

K  n 2 M   0

(7)

Eq.(7) gives a polynomial equation of degree n in n2, known
as characteristic equation of the system
A vibrating system with n DoF has n natural vibration frequencies n, arranged in
sequence from smallest to largest

a home base to excellence
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.


a home base to excellence
Example 1
 Determine :
1. The natural frequencies and
corresponding modal shapes

E = 2.106 kg/cm2

W1 = 11.600 kgf

u2
W10x21

W10x45 I = 10,300 cm4

W10x21

W10x21  I = 4,900 cm4

W2 = 24.000 kgf

3m

   11


21

 2  38,28

12  0,6809  3,0385

1 
22   1

W10x45

1  10,76

W10x45

u1
4,5 m

a home base to excellence

W1 = 11.600 kgf

W10x21

E = 2.106 kg/cm2

W2 = 24.000 kgf

u3
3m

W10x21

W10x45 I = 10,300 cm4

W10x21

W10x21  I = 4,900 cm4

W3 = 20.000 kgf

W10x21

Tugas 6
 Determine :
1. The natural frequencies and
corresponding modal shapes

3m

u2

 2  26,65

3  39,53

11 12 13  0 ,5546  1,004 5,3431 
   21 22 23   0,8526  0,662  2,6574

 
1
1 
 31 32 33   1

W10x45

1  7,934

W10x45

u1
4,5 m

a home base to excellence
mode 2

mode 1
3

3

1

2

-0,662

0,8526

1

-1,004

0,5546

0

0

0

0,5

1

1,5

-2

-1

mode 3
31

1

2

-2,6574

2

1

1

0

0
0

1

2

-5

5,3431

0

5

10

a home base to excellence
Modal & Spectral Matrices


The N eigenvalues, N natural frequencies, and n natural
modes can be assembled compactly into matrices
11 12
   ...21 ...22

 n1 n 2

Mode 1

2

... 1n 
... 2 n 
... ... 
... nn 

Modal matrix

n

12



2
2

2


 



2

 n 


 

DoF 1
2
…..
n

Spectral matrix

a home base to excellence
Orthogonality of Modes


The natural modes corresponding to different natural
frequencies can be shown to satisfy the following
orthogonality conditions :

n T K r   0

n  M r   0
T

When n ≠ r

a home base to excellence
Free Vibration Response : Undamped System







M u K u  0

EoM for MDoF Undamped Free Vibration is :
(8)

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).

u  1q1  2 q2  3q3  ... n qn

Or

u   q
N

n 1

n

n

u   q

(9)
(10)

a home base to excellence


Premultiply Eq. (10) with {n}T[M] :

n T M u  n T M  qn 

 n  M 1q1  n  M 2 q2  ....
T

 n  M N q N

(11)

T

T



Regarding the orthogonality conditions :
T
T
n  M u  n  M n qn
 From which :

  M u

  M u
also
q 
q 
  M  
  M  
T

T

n

n

n

T

n

T

n

n

n

n

(11)

a home base to excellence


If Eq. (8) is premultiplied by the transpose of the nth modeshape vector {n}T, and using Eq.(10) with its second
derivative it becomes :

  M  q    K  q
T

T

n

n

n

n



n

0

(11)

Kn

Mn
Generalized mass

n

Generalized stiffness

Eq.(11) is an EoM from SDoF Undamped System for mode n,
which has solution :
qn t   qn 0cos nt 

qn 0

n

sin nt

(12)

a home base to excellence


Solution for Eq. (8) becomes :



qn 0
u  n qn 0cos nt 
sin nt 
n
n 1


N





(13)

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

a home base to excellence


Gaya pegas tiap lantai dari masing-masing
mode dapat dicari dengan persamaan :

 f s   K u   n 2 M u

a home base to excellence
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.
6 10 3 
u t 0   
 3 m
8
10



0 
u t 0   
0,20m/s



a home base to excellence
Tugas 7



0
u t  0   20cm/s


0

Also plot the time history of spring
force

W10x21

W1 = 11.600 kgf

W10x21

W10x21

W2 = 24.000 kgf

u3
3m

3m

u2

u1
W10x45

0 ,6
u t 0   0 ,8cm
1,0 

E = 2.106 kg/cm2

W10x21

W10x21  I = 4,900 cm4
W10x45 I = 10,300 cm4

W3 = 20.000 kgf

W10x45

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.

4,5 m

a home base to excellence
Free Vibration Response : Damped System






M u C u K u  0

EoM for MDoF Damped Free Vibration is :

(14)

M  q C  q K  q  0

Using Eq. (10) and its derivative :

(15)

  M  q   C  q   K  q  0 (16)

Premultiply Eq. (15) with {}T :
T

T

Mn

Cn

T

Kn

a home base to excellence


Eq.(16) is an EoM from SDoF Damped System for mode n,
which has solution :
qn t   e



 nn t



qn 0   nn qn 0
sin nDt 
qn 0cos nDt 

nD



(17)

The displacement response of the system is then obtained by
substituting Eq. (17) in Eq. (10)


q 0   nn qn 0
u t   n e   t qn 0cos nDt  n
sin nDt 
nD
n 1


N

n

n

(18)