Slide CIV308 CIV308 Slide 06 07

a home base to excellence

Pertemuan – 6, 7

Multi Degree of Freedom System Free Vibration

Mata Kuliah : Dinamika Struktur & Pengantar Rekayasa Kegempaan

Kode : CIV – 308

SKS : 3 SKS

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

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)

F₁(t)

k₁u₁ m₁ü₁ k₂(u₂ u₁)

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 :

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

   

     

   

⁰

0 0

3 2

3 3 3

3

2 2

3 3 1

2 2 2

2

1 1

2 2 1

1 1

1































t F u

u k u

m

t F u

u k u

u k u

m

t F u

u k u

k u

m







  M   u     K     u  F

⁽²⁾

(1)

a home base to excellence



Where :



And :

  











3 2 1

0 0

0 0

0 0

m m m

M  













 

 



3 3

3 3

2 2

2 2

1

0

0 k k

k k

k k

k k

k K

 













3 2 1

u u u

u

 













3 2 1

u u u u 



   

   

_











t F

t F

t F F

3 2 1

a home base to excellence

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 :

  M   u     K     u  0

⁽³⁾

       

^u 



n ^qn ^t 



n ^An ^cos



n^t  ^Bn ^sin



n^t



⁽⁴⁾

 

 

 

 



^{K n  n2 M n}



^{qn   t  0} ₍₅₎

a home base to excellence

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

   

 K  n2 M      n  0 (6)

A vibrating system with n DoF has n natural vibration frequencies _n, arranged in sequence from smallest to largest

 

^{K  n2}

 

^{M 0} (7)

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

W₁ = 11.600 kgf

4,5 m

W10x21

W₂ = 24.000 kgf

u₂

u₁

3 m

W10x45 W10x21W10x45

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

E = 2.10⁶ kg/cm²

 ^_^ _^{ }_^{ } _^





1

1 30385

6809 0

28 38

76 10

22 21

12 11

2 1

, ,

, ,



 

 





a home base to excellence

Tugas 6

Determine :

1. The natural frequencies and corresponding

modal shapesW10x21  I = 4,900 cm⁴ W10x45 I = 10,300 cm⁴

E = 2.10⁶ kg/cm² W₁ = 11.600 kgf

4,5 m

W10x21

W₂ = 24.000 kgf

u₂

u₁

3 m

W10x45 W10x21W10x45

W₃ = 20.000 kgf

W10x21 W10x21 u₃

3 m

  









  

 



















1 1

1 0662 26574

8526

05546 1004 53431 0

53 39

65 26

7,934

33 32

31

23 22

21

13 12

11

3 2

1

, ,

, , ,

,

, ,





  

  

 







a home base to excellence

0 0.2 0.4 0.6 0.8 1 1.2 0

1 2

3 Series1

mode 1

-1.5 -1 -0.5 0 0.5 1 1.5 0

1 2

3 Series1

mode 2

-4 -3 -2 -1 0 1 2 3 4 5 6 0

1 2 3

Series1

mode 3

a home base to excellence

Modal & Spectral Matrices

The N eigenvalues, N natural frequencies, and n natural modes can be assembled compactly into matrices

  















nn n

n

n n

... ...

...

...

...

...

...











  



2 1

2 22

21

1 12

11

 





















2 22

12 2

n







Mode 1 2 n

DoF 1 2 …..

n

Modal matrix

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 :

    

     ⁰

0





T r n

T r n

M K







 When _n ≠ _r

a home base to excellence

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

  M   u     K     u  0

⁽⁸⁾

   

u  ₁ q₁ 

 

₂ q₂ 

 

₃ q₃ ...

 

n qn

 

^{u }

 

^

 

^q

(9)

  

(10)



 ^N

n nqn

u

1

a home base to excellence



Premultiply Eq. (10) with { 

_n

}

^T

[M] :



Regarding the orthogonality conditions :



From which :

also

       

T

    

n T n

n M u



M



q



 ⁽¹¹⁾

           

n n T

n T

n

M u  M  q

 

           

     

N N T

n

T n T

n

q M

....

q M

q M





















2 2 1

1

     

 

^{n T}

   

ⁿ

T n

n M

u q M





 

     

 

^{n T}

   

ⁿ

T n

n M

u q M





 ^

  (11)

a home base to excellence

If Eq. (8) is premultiplied by the transpose of the n^th 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 :

     

^

     

n n ^0 T

n n

n T

n M



q



K



q



^

M_n K_n

(11)

Generalized mass

Generalized stiffness

 

_{t q}

 

_{cos t} ^q

 

_{sin t}

q _n

n n n

n

n 

 ⁰

0 



 (12)

a home base to excellence

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

n n

n n n

n

n

q sin t

t cos

q u

1

0 0 

 

 

⁽¹³⁾

a home base to excellence



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

  f

s

   K   u  

n2

  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 1^st, and 2^nd mode contribution

  m

u _t











 _^

 3

3

0 8 10

10 6

  0,200 m/s

0 



  u t

W₁ = 11.600 kgf

4,5 m

W10x21

W₂ = 24.000 kgf

u₂

u₁

3 m

W10x45 W10x21W10x45

 ^_^ _^{ }_^{ } _^





1

1 30385

6809 0

28 38

76 10

22 21

12 11

2 1

, ,

, ,



 

 





a home base to excellence

W₁ = 11.600 kgf

4,5 m

W10x21

W₂ = 24.000 kgf

u₂

u₁

3 m

W10x45 W10x21W10x45

Time History Displacement (cm)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2

Mode 1 Lantai 1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2

Mode 1 Lantai 2

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2

Mode 2 Lantai 1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2

Mode 2 Lantai 2

a home base to excellence

W₁ = 11.600 kgf

4,5 m

W10x21

W₂ = 24.000 kgf

fs₂

fs₁

3 m

W10x45 W10x21W10x45

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

-6000 -4000 -2000 0 2000 4000 6000

fs, mode 1 Lt 1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

-6000 -4000 -2000 0 2000 4000 6000

fs, mode 1 Lt 2

Time History Spring Force (kg)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

-6000 -4000 -2000 0 2000 4000 6000

fs, mode 2 Lt 1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

-6000 -4000 -2000 0 2000 4000 6000

fs, Mode 2 Lt 2

a home base to excellence

W₁ = 11.600 kgf

4,5 m

W10x21

W₂ = 24.000 kgf

u₂

u₁

3 m

W10x45 W10x21W10x45

Time History Base Shear (kg)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

-8000 -6000 -4000 -2000 0 2000 4000 6000 8000

Base Shear Mode 1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

-8000 -6000 -4000 -2000 0 2000 4000 6000 8000

Base Shear Mode 2

Base Shear (Gaya Geser

Dasar)

a home base to excellence

Assignment 7

Based on data from Assignment 6, and the following initial condition, for each DoF plot the time history of

displacement, regarding the 1^st, 2^nd , and 3^rd mode contribution.

Also plot the time history of spring force and base shear Make a conclusion !!

Submit at the first meeting after Mid Test

W₁ = 11.600 kgf

4,5 m

W10x21

W₂ = 24.000 kgf

u₂

u₁

3 m

W10x45 W10x21W10x45

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

E = 2.10⁶ kg/cm²

W₃ = 20.000 kgf

W10x21 W10x21 u₃

3 m

  cm

,,, u _t













 100086

0

  cm/s

2000

0 













u t

a home base to excellence

Free Vibration Response : Damped System

EoM for MDoF Damped Free Vibration is :

Using Eq. (10) and its derivative :

Premultiply Eq. (15) with {}^T :

  M   u     C   u     K     u  0

⁽¹⁴⁾

(15)

(16)

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

     

^T

M    q        

^T

C    q        

^T

K      q  0

M_n C_n K_n

a home base to excellence

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)

       



 



  

 ^ q q sin t

t cos

q e

t

q _nD

nD n n n n

nD n

t n

n

n 





 



 0 0

0 

(17)

       



 



  

 ^



 ^{e q cos t q q sin t}

t

u _nD

nD n n n n

nD n

N t

n n

n

n 





 

 ^{  0 0 0}

1

 (18)