kuliah 12-13 - pengantar dinamika struktur mdof b-1.pdf

BAYZONISemester Genap 2015 - 2016

PENGANTAR Dinamika Struktur

Multidegree-of-Freedom Systems

• A structure can be modeled and its responseanalyzed using a SDOF model if the mass isessentially concentrated at a single point that canmove, translate, or rotate only in one direction, or ifthe system is constrained in such a way as to permitonly a single mode of displacement. In general, themass of a larger building or structure is distributedthroughout the structure and can move in manyways.

• A realistic description of the dynamic response ofsuch systems generally requires the use of anumber of independent displacement coordinates,and modeling of the system as a multidegree-of-freedom (MDOF) system.

• Dynamic analysis of such MDOF systems isdiscussed in the following sections.

Equations of Motion

• The MDOF analysis procedure isillustrated by examining the dynamicresponse of the idealized threestorybuilding shown in figure below. Themass of the structure is assumed to beconcentrated at the floor levels, whichare further assumed to be rigid anddisplace in one translational directiononly. Thus, the dynamic behavior of thisstructure is completely defined by thethree-story displacements u1(t), u2(t) andu3(t).

• The equation of motion of any story canbe derived from the expression ofdynamic equilibrium of all of the forcesacting on the story mass, including theinertia, damping, and elastic forces thatresult from the motion, and theexternally applied force. The equationsof equilibrium for the two stories can bewritten as follows (using notationanalogous to the SDOF case):

MULTI DEGREE OF FREEDOM

k1

x1

m1

k2

F1(t)

m3m2

x2

k3

F2(t) F3(t)

x3

0)(... 223312222 tFxxkxxkxm

0)(... 11221111 tFxxkxkxm

0)(.. 323333 tFxxkxm

• Model 3 derajat kebebasan

• Keseimbangan Gaya

• Dalam bentuk Matrik

• Dalam hal ini:

tFXKXM ..

3

2

1

00

00

00

m

m

m

M

33

3322

221

0

0k

kk

kkkk

kk

K

3

2

1

x

x

x

X

3

2

1

x

x

x

X

)t(F

)t(F

)t(F

F

3

2

1

t

tFXKXCXM ...

• Dalam hal terdapat redaman maka:

• Keterangan:

GETARAN BEBAS

• Getaran Bebas Tanpa Redaman

Solusi dari persamaan di atas adalah:

�� (�) = ��. cos �� + �� . sin ��

��̇(�) = −���. sin �� + ���. cos ��

��̈(�) = −�2��. cos �� − �2�� . sin ��

Sehingga diperoleh persamaan:

[�]{�̈} + [�]{�} = 0

−�2[�]{�} + [�]{�} = 0

Persamaan di atas dapat ditulis:

�[�] − �2[�]�{�} = 0

Dengan aturan Cramer solusi dari persamaan di atas:

{�} =0

‖[�] − �2[�]‖

Pemecahan non-trivial dimungkinkan

‖[�] − �2[�]‖ = 0

Persamaan ini disebut persamaan frekuensi sistem,dengan memperluas determinan akan diperolehpersaman aljabar berderajat N dalam parameter 2

untuk sistem yang mempunyai B derajat kebebasan.2 disebut “eigen-value”

DETERMINANT

CONTOH:

• Lantai kaku

• Tidak ada deformasi aksial

• Semua massa terkumpulpada lantai

Asumsi:

m1=1

m2=1

m3=1

K1= 5

K2= 4

K3= 3

[�] = �1 0 00 1 00 0 1

� [�] = �

(�1 + �2) −�2 0−�2 (�2 + �3 ) −�3

0 −�3 �3

�

[�] = �(5 + 4) −4 0

−4 (4 + 3 ) −30 −3 3

� = �9 −4 0

−4 7 −30 −3 3

�

Periode Alami |D| = 0

‖[�] − �2[�]‖ = 0

(9 − �2)�(9 − �2)(3 − �2) − (−3 . −3 )� + 4�−4. (3 − �2)� = 0

�

(9 − �2) −4 0

−4 (7 − �2) −3

0 −3 (3 − �2)

� = 0

�6 − 19 �4 + 86 �2 − 60 = 0

(�2)3 − (19 �2)2 + (86 �2) − 60 = 0

�2 = 0.8502 � = 0.922

�2 = 5.52 � = 2.35

�2 = 12 .6 � = 3 .55

�

(9 − 0.85022) −4 0

−4 (7 − 0.85022) −3

0 −3 (3 − 0.85022)

� �

�1(1 )

�1(1 )

�1(1 )

� = �000

�

Solusi untuk Ragam ke-1

�8.15 −4 0−4 6.15 −30 −3 2.15

� �

�1(1 )

�1(1 )

�1(1 )

� = �000

�

�

�1(1 )

�1(1 )

�1(1 )

� = �0.3510.716

1�

Solusi untuk Ragam ke-2

�

�1(2)

�1(2)

�1(2)

� = �−1 .052−0.882

1�

Solusi untuk Ragam ke-3

�

�1(3 )

�1(3 )

�1(3 )

� = �3 .62

−3 .1681

�

Normalisasi Eigenvctor

Mn nTM n n

1 T

3.614 3.169 1( )

M1 n1 T

M1 1

n1

M1 24.105( )

1 n1 M11 1

1

2 1

0.736

0.646

0.204

Normalisasi Eigenvctor

Mn nTM n n

2 T

1.049 0.881 1( )

M2 n2 T

M2 2

n2

M2 2.876( )

2 n2 M21 1

1

2 2

0.619

0.519

0.59

Mn nTM n n

3 T

0.352 0.717 1( )

M3 n3 TM3 3

n3

M3 1.637( )

3 n3 M31 1

1

2 3

0.275

0.56

0.782

Developing a Way To Solvethe Equations of Motion

• This will be done by a transformation ofcoordinates from normal coordinates(displacements at the nodes) To modalcoordinates (amplitudes of the natural Modeshapes).

• Because of the orthogonality property of thenatural mode shapes, the equations of motionbecome uncoupled, allowing them to besolved as SDOF equations.

• After solving, we can transform back to thenormal coordinates.

Solutions for System in Undamped Free Vibration(Natural Mode Shapes and Frequencies)

Solutions for System in Undamped Free Vibration (continued)

Mode Shapes for Idealized 3-Story Frame

Concept of Linear Combination of Mode Shapes(Transformation of Coordinates)

U=ΦY

Orthogonality Conditions

Ortogonalitas : Contoh 1

Matrix Kekakuan : Matrix Massa : dim :

n 3K 1

9

4

0

4

7

3

0

3

3

M

1

0

0

0

1

0

0

0

1

Eigenvalue : Eigenvectors :

2

12.508

5.642

0.85

0.736

0.646

0.204

0.619

0.519

0.59

0.275

0.56

0.782

i 2i

3.537

2.375

0.922

n

3.614

3.169

1

1.049

0.881

1

0.352

0.717

1

Ortogonalitas : Contoh 1

TM

1

0

0

0

1

0

0

0

1

TK

12.508

0

0

0

5.642

0

0

0

0.85

nTM n

24.105

0

0

0

2.876

0

0

0

1.637

nTK n

301.5

1.905 1015

5.908 1015

2.703 1015

16.226

1.033 1015

5.98 1015

0

1.392

Development of Uncoupled Equations of Motion

Development of Uncoupled Equations of Motion(Explicit Form)

Development of Uncoupled Equations of Motion(Explicit Form)

Earthquake “Loading” for MDOF System

Vibration Analysis by Matrix Iterations

LANGKAH PENYELESAIAN TAKE HOME1) Pilih bangunan

2) Tentukan ukuran balok, kolom dan pelat

3) Tentukan Beban Hidup dan Beban Mati

4) Hitung Massa tiap-tiap lantai

5) Hitung kekakuan masing-masing kolom

6) Bentuk Matrik Massa

7) Bentuk Matrik Kekakuan

8) Hitung w2

9) Hitung mode shape

10) Hitung mode shape normalisasi

11) Bentuk persamaan sdof