7/26/2019 Kuliah 12-13 - Pengantar Dinamika Struktur MDOF B-1
1/35
BAYZONI emester Genap 2015 - 2016
PENGANTAR
nam kaStruktur
7/26/2019 Kuliah 12-13 - Pengantar Dinamika Struktur MDOF B-1
2/35
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.
7/26/2019 Kuliah 12-13 - Pengantar Dinamika Struktur MDOF B-1
3/35
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):
7/26/2019 Kuliah 12-13 - Pengantar Dinamika Struktur MDOF B-1
4/35
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
7/26/2019 Kuliah 12-13 - Pengantar Dinamika Struktur MDOF B-1
5/35
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:
7/26/2019 Kuliah 12-13 - Pengantar Dinamika Struktur MDOF B-1
6/35
Keterangan:
7/26/2019 Kuliah 12-13 - Pengantar Dinamika Struktur MDOF B-1
7/35
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
7/26/2019 Kuliah 12-13 - Pengantar Dinamika Struktur MDOF B-1
8/35
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
7/26/2019 Kuliah 12-13 - Pengantar Dinamika Struktur MDOF B-1
9/35
DETERMINANT
7/26/2019 Kuliah 12-13 - Pengantar Dinamika Struktur MDOF B-1
10/35
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
7/26/2019 Kuliah 12-13 - Pengantar Dinamika Struktur MDOF B-1
11/35
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
7/26/2019 Kuliah 12-13 - Pengantar Dinamika Struktur MDOF B-1
12/35
(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
7/26/2019 Kuliah 12-13 - Pengantar Dinamika Struktur MDOF B-1
13/35
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
7/26/2019 Kuliah 12-13 - Pengantar Dinamika Struktur MDOF B-1
14/35
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
7/26/2019 Kuliah 12-13 - Pengantar Dinamika Struktur MDOF B-1
15/35
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
7/26/2019 Kuliah 12-13 - Pengantar Dinamika Struktur MDOF B-1
16/35
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.
7/26/2019 Kuliah 12-13 - Pengantar Dinamika Struktur MDOF B-1
17/35
Solutions for System in
Undamped Free Vibration
(Natural Mode Shapes and Frequencies)
7/26/2019 Kuliah 12-13 - Pengantar Dinamika Struktur MDOF B-1
18/35
Solutions for System in
Undamped Free Vibration (continued)
7/26/2019 Kuliah 12-13 - Pengantar Dinamika Struktur MDOF B-1
19/35
Mode Shapes for
Idealized 3-Story Frame
7/26/2019 Kuliah 12-13 - Pengantar Dinamika Struktur MDOF B-1
20/35
Concept of Linear Combination of
Mode Shapes
(Transformation of Coordinates)U=Y
7/26/2019 Kuliah 12-13 - Pengantar Dinamika Struktur MDOF B-1
21/35
Orthogonality Conditions
7/26/2019 Kuliah 12-13 - Pengantar Dinamika Struktur MDOF B-1
22/35
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
7/26/2019 Kuliah 12-13 - Pengantar Dinamika Struktur MDOF B-1
23/35
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
7/26/2019 Kuliah 12-13 - Pengantar Dinamika Struktur MDOF B-1
24/35
Development of
Uncoupled Equations of Motion
7/26/2019 Kuliah 12-13 - Pengantar Dinamika Struktur MDOF B-1
25/35
Development of
Uncoupled Equations of Motion
(Explicit Form)
7/26/2019 Kuliah 12-13 - Pengantar Dinamika Struktur MDOF B-1
26/35
Development of
Uncoupled Equations of Motion
(Explicit Form)
7/26/2019 Kuliah 12-13 - Pengantar Dinamika Struktur MDOF B-1
27/35
Earthquake Loading for
MDOF System
7/26/2019 Kuliah 12-13 - Pengantar Dinamika Struktur MDOF B-1
28/35
7/26/2019 Kuliah 12-13 - Pengantar Dinamika Struktur MDOF B-1
29/35
Vibration Analysis by Matrix Iterations
7/26/2019 Kuliah 12-13 - Pengantar Dinamika Struktur MDOF B-1
30/35
7/26/2019 Kuliah 12-13 - Pengantar Dinamika Struktur MDOF B-1
31/35
7/26/2019 Kuliah 12-13 - Pengantar Dinamika Struktur MDOF B-1
32/35
7/26/2019 Kuliah 12-13 - Pengantar Dinamika Struktur MDOF B-1
33/35
7/26/2019 Kuliah 12-13 - Pengantar Dinamika Struktur MDOF B-1
34/35
LANGKAH
7/26/2019 Kuliah 12-13 - Pengantar Dinamika Struktur MDOF B-1
35/35
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
Top Related