Kuliah 12-13 - Pengantar Dinamika Struktur MDOF B-1

7/26/2019 Kuliah 12-13 - Pengantar Dinamika Struktur MDOF B-1

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BAYZONI emester Genap 2015 - 2016

PENGANTAR

nam kaStruktur


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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, the

mass 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 is

discussed in the following sections.


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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 this

structure 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 that

result from the motion, and theexternally applied force. The equationsof equilibrium for the two stories can bewritten as follows (using notationanalogous to the SDOF case):


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MULTI DEGREE OF FREEDOM

k1

x1

m1k2

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


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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:


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Keterangan:


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GETARAN BEBAS

Getaran Bebas Tanpa Redaman

Solusi dari persamaan di atas adalah:

() = . c o s + .sin () = . s in+ .cos()= 2 . cos 2 . sin Sehingga diperoleh persamaan:

[]{}+[]{}= 0

2[]{} + []{} = 0


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Persamaan di atas dapat ditulis:

[] 2

[]{}= 0

Dengan aturan Cramer solusi dari persamaan di atas:{}= 0[] 2[]Pemecahan non-trivial dimungkinkan[] 2[] = 0Persamaan ini disebut persamaan frekuensi sistem,

dengan memperluas determinan akan diperoleh

persaman aljabar berderajat N dalam parameter 2

untuk sistem yang mempunyai B derajat kebebasan.

2 disebut eigen-value


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DETERMINANT


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CONTOH:

Lantai kaku

Tidak ada deformasi aksial

Semua massa terkumpul

pada lantai

Asumsi:

m1=1

m2=1

m3=1

K1= 5

K2= 4

K3= 3

[]= 1 0 00 1 0

0 0 1

[]= (1+ 2) 2 02 (2+ 3) 30 3 3

[] = (5 + 4) 4 04 (4 + 3) 30 3 3=9 4 0 4 7 30 3 3


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Periode Alami |D| = 0

[] 2[]= 0

(9 2)(9 2)(3 2) (3. 3) + 44. (3 2) = 0( 9 2) 4 04 (7 2) 3

0 3 (3 2

)

= 0

6 194 + 86 2 60 = 0

(2

)3

(192

)2

+ (86 2

) 60 = 0

2 =0.8502 =0.9222 =5.52 =2.352 =12.6 =3.55


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(90.85022) 04 (7 0.85022) 30 3 (3 0.85022) 1(1)1(1)1(1) = 000Solusi untuk Ragam ke-1

8.15 04 6.15 30 3 2.15

1(1)1(1)1(1) =

000

1(1)1(1)

1(1) =

0.3510.7161


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Solusi untuk Ragam ke-2

1(2)1(2)1(2) = 1.0520.8821 Solusi untuk Ragam ke-3

1(3)1(3)1(3) = 3.623.1681


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Normalisasi Eigenvctor

Mn nT

M n n1 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


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Normalisasi Eigenvctor

Mn nT

M n n2 T

1.049 0.881 1( )

M2 n2 T

M2 2

n 2

M2 2.876( )

2 n 2

M21 1

1

2 2

0.619

0.519

0.59

Mn nT

M n n3 T

0.352 0.717 1( )

M3 n3 T

M3 3

n 3

M3 1.637( )

3 n 3

M31 1

1

2 3

0.275

0.56

0.782


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Developing a Way To Solve

the Equations of Motion

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

shapes). Because of the orthogonality property of the

natural modeshapes, the equations of motionbecome uncoupled, allowing them to besolved as SDOF equations.

After solving, we can transform back to thenormal coordinates.


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Solutions for System in

Undamped Free Vibration

(Natural Mode Shapes and Frequencies)


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Solutions for System in

Undamped Free Vibration (continued)


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Mode Shapes for

Idealized 3-Story Frame


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Concept of Linear Combination of

Mode Shapes

(Transformation of Coordinates)U=Y


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Orthogonality Conditions


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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


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Ortogonalitas : Contoh 1

T

M

1

0

0

0

1

0

0

0

1

TK

12.508

0

0

0

5.642

0

0

0

0.85

nT

M n

24.105

0

0

0

2.876

0

0

0

1.637

n

TK n

301.5

1.905 10 1 5

5.908 10 1 5

2.703 10 15

16.226

1.033 10 15

5 .98 1 0 1 5

0

1.392


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Development of

Uncoupled Equations of Motion


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Development of


(Explicit Form)


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Development of


(Explicit Form)


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Earthquake Loading for

MDOF System


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Vibration Analysis by Matrix Iterations


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LANGKAH


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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 normalisasi11) Bentuk persamaan sdof

Kuliah 12-13 - Pengantar Dinamika Struktur MDOF B-1

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