127
g. Time History Total Base Shear dengan Metode Superposisi Modal Analysis
Grafik a sd g Respon Struktur MDOF akibat gempa Kobe
-15000 -10000
-5000 5000
10000 15000
1, 87
3, 74
5, 61
7, 48
9, 35
11, 22
13, 09
14, 96
16, 83
18, 7
20, 57
22, 44
24, 31
26, 18
28, 05
29, 92
31, 79
33, 66
35, 53
37, 4
39, 27
Vbn k
N
Time History Base Shear dengan Metode Superposisi Modal
= Vb kgf
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4.1.5. Perhitungan Time History perpindahan, percepatan, interstory drift setiap lantai serta base shear total dengan metode Integrasi Langsung Direct
Integration 4.1.5.1 Metode Integrasi Langsung Newmark-
�
Didapat dari perhitungan sebelumnya: �
1
= 34,525 ����
�
2
= 13,565 ����
Matriks Massa: � = �
2,91 2,52�
10
3
�� ��
2
Matriks Kekakuan: � = �
2,54 −1,27
−1,27 1,27 �
10
6
���
Matriks Redaman: � = �
10,0468 8,7003�
10
3
K
2
K
2
K
1
K
1
K
1
K
1
K
1
K
2
K
2
K
2
4 x 6 m 2 x 4 m
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VI.5.1.2 Metode Integrasi Langsung dengan Constant Average Acceleration
Method Newmark- �
Untuk ConstantAverage Acceleration , � =
1 2
��� � =
1 4
• Hitung ��, � ��� � ∆� = 0,01
�� = �� +
� � ��
� +
� ���
�
��
�� = � 2,54
−1,27 −1,27
1,27 � 10
6
+
� � Δ�
� 10,0468
8,7003� 10
3
+
1 �Δ�
2
� 2,91
2,52� 10
3
�� = � 32608880
−1270000 −1270000 27314330�
� = �
� � ��
� +
� �
��
� =
1 � Δ�
� 2,91
2,52� 10
3
+
� �
� 10,0468
8,7003� 10
3
� = � 601378
520887�
� = �
� ��
� + �� �
� ��
− �� ��
� =
1 2
�
� 2,91
2,52� 10
3
+ Δ� �
� 2
�
− 1� � 10,0468
8,7003� 10
3
� = � 581284
503486 �
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Cari beban efektif
i time
sec
Accg
�̈
�
ms
2
1 0,01
2 0,02
3 0,03
4 0,04
0,00001 0,0000981
5 0,05
0,00002 0,0001962
6 0,06 -0,00001
-0,0000981 7
0,07 -0,00005 -0,0004905
8 0,08 -0,00008
-0,0007848 9
0,09 -0,00007 -0,0006867
10 0,1 -0,00009
-0,0008829
• Cycle 1 � = 0 ��
�
= �
�+�
− �
�
= [ �]��̈
�,�+�
− �̈
�,�
� Δ�
= �
1
− � = [
�]��̈
�,1
− �̈
�,0
� Δ�
= �
2,91 2,52� �
10
3
[0 − 0]
Δ� =
� 0�
Kondisi awal u = 0, u
̇ = 0, u
̈ = 0
��
�
= ��
�
+ ��̇
�
+ ��̈
�
�� =
Δ� +
��̇ +
��̈ ��
= �
0�
���
�+�
= ��
�
��� =
�� �
= ��
����
−1
�Δ�
1
Δ�
2
� =
� 0�
� 32608880
−1270000 −1270000 27314330�
−1
�Δ�
1
Δ�
2
� =
� 0�
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131
��̇
�
=
� ���
��
�
−
� �
�̇
�
+ �� −
� ��
� �̈
�
�Δ�̇
1
Δ�̇
2
� =
� �Δ�
�Δ�
1
Δ�
2
� −
γ β
��̇
1
�̇
2
� − �1 −
� 2
�
� ��̈
1
�̈
2
� �Δ�̇
1
Δ�̇
2
� =
� �Δ�
� 0� −
γ β
− �1 −
� 2
�
� 0 �Δ�̇
1
Δ�̇
2
� =
� 0�
�̈
�
=
1 �Δ�
2
Δ�
�
−
1 �Δ�
�̇
�
−
1 2
�
�̈
�
�Δ�̈
1
Δ�̈
2
� =
1 �Δ�
2
�Δ�
1
Δ�
2
� −
1 �Δ�
��̇
1
�̇
2
� −
1 2
�
��̈
1
�̈
2
� �Δ�̈
1
Δ�̈
2
� =
1 �Δ�
2
� 0� −
1 �Δ�
0 +
1 2
�
�Δ�̈
1
Δ�̈
2
� =
� 0�
� �
1
�
2
�
1
= �
�
1
�
2
� +
�Δ�
1
Δ�
2
� =
� 0�
��̇
1
�̇
2
�
1
= ��̇
1
�̇
2
� +
�Δ�̇
1
Δ�̇
2
� =
� 0�
��̈
1
�̈
2
�
1
= ��̈
1
�̈
2
� +
�Δ�̈
1
Δ�̈
2
� =
� 0�
• Cycle 2 � = 1 ��
�
= �
�+�
− �
�
= [ �]��̈
�,�+�
− �̈
�,�
� Δ�
1
= �
2
− �
1
= [ �]��̈
�,2
− �̈
�,1
� Δ�
1
= �
2,91 2,52� �
10
3
[0 − 0]
Δ�
1
= �
0�
Kondisi u
1
= {0} , u ̇
1
= {0}, u ̈
1
= {0}
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��
�
= ��
�
+ ��̇
�
+ ��̈
�
��
1
= Δ�
1
+ ��̇
1
+ ��̈
1
��
1
= {0} +
� 601378
520887� {
}
+
� 581284
503486 � {
} ��
1
= �
0�
���
�
= ��
�
���
1
= ��
1
Δ�
1
= ��
����
−1
�Δ�
1
Δ�
2
�
1
= �
0� �
32608880 −1270000
−1270000 27314330�
−1
�Δ�
1
Δ�
2
�
1
= �
0�
��̇
�
=
� ���
��
�
−
� �
�̇
�
+ �� −
� ��
� �̈
�
�Δ�̇
1
Δ�̇
2
�
1
=
� �Δ�
�Δ�
1
Δ�
2
�
1
−
γ β
��̇
1
�̇
2
�
1
+ �1 −
� 2
�
� ��̈
1
�̈
2
�
1
�Δ�̇
1
Δ�̇
2
�
1
=
� �Δ�
� 0� −
γ β
� 0�
+ �1 −
� 2
�
� � 0�
�Δ�̇
1
Δ�̇
2
�
1
= �
0�
�̈
�
=
� ���
�
��
�
−
� ���
�̇
�
−
� ��
�̈
�
�Δ�̈
1
Δ�̈
2
�
1
=
1 �Δ�
2
�Δ�
1
Δ�
2
�
1
−
1 �Δ�
��̇
1
�̇
2
�
1
−
1 2
�
��̈
1
�̈
2
�
1
�Δ�̈
1
Δ�̈
2
�
1
=
1 �Δ�
2
� 0� −
1 �Δ�
� 0�
−
1 2
�
� 0�
�Δ�̈
1
Δ�̈
2
�
1
= �
0�
� �
1
�
2
�
2
= �
�
1
�
2
�
1
+ �Δ�
1
Δ�
2
�
1
= �
0�
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133
��̇
1
�̇
2
�
2
= ��̇
1
�̇
2
�
1
+ �Δ�̇
1
Δ�̇
2
�
1
= �
0�
��̈
1
�̈
2
�
2
= ��̈
1
�̈
2
�
1
+ �Δ�̈
1
Δ�̈
2
�
1
= �
0�
• Cycle 3 � = 2, dan seterusnya proses integrasi dilanjutkan dengan cara yang sama seperti langkah-langkah diatas dan hasil nya dapat dilihat di tabel IV.3.
• Cycle 4 � = 3 ��
�
= �
�+�
− �
�
= [ �]��̈
�,�+�
− �̈
�,�
� Δ�
3
= �
4
− �
3
= [ �]��̈
�,4
− �̈
�,3
� Δ�
3
= �
2,91 2,52� �
10
3
[0,0000981 − 0]
Δ�
3
= �
0,28512 0,24696�
Kondisi u
3
= {0} , u ̇
3
= {0}, u ̈
3
= {0} ��
�
= ��
�
+ ��̇
�
+ ��̈
�
��
3
= Δ�
3
+ ��̇
3
+ ��̈
3
��
3
= �
0,28512 0,24696
�+
� 601378
520887� {
}
+
� 581284
503486 � {
} ��
3
= �
0,28512 0,24696�
���
�
= ��
�
���
3
= ��
3
Δ�
3
= ��
3
����
−1
�Δ�
1
Δ�
2
�
3
= �
0,28512 0,24696�
� 32608880
−1270000 −1270000 27314330�
−1
�Δ�
1
Δ�
2
�
3
= �
2,39 2,41� �
10
−9
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��̇
�
=
� ���
��
�
−
� �
�̇
�
+ �� −
� ��
� �̈
�
�Δ�̇
1
Δ�̇
2
�
3
=
� �Δ�
�Δ�
1
Δ�
2
�
3
−
γ β
��̇
1
�̇
2
�
3
+ �1 −
� 2
�
� ��̈
1
�̈
2
�
3
�Δ�̇
1
Δ�̇
2
�
3
=
� �Δ�
� 2,4
2,4� � 10
−9
−
γ β
� 0�
+ �1 −
� 2
�
� � 0�
�Δ�̇
1
Δ�̇
2
�
3
= �
4,77 4,82� �
10
−7
�̈
�
=
� ���
�
��
�
−
� ���
�̇
�
−
� ��
�̈
�
�Δ�̈
1
Δ�̈
2
�
3
=
1 �Δ�
2
�Δ�
1
Δ�
2
�
3
−
1 �Δ�
��̇
1
�̇
2
�
3
−
1 2
�
��̈
1
�̈
2
�
3
�Δ�̈
1
Δ�̈
2
�
3
=
1 �Δ�
2
� 2,4
2,4� � 10
−9
−
1 �Δ�
� 0�
−
1 2
�
� 0�
�Δ�̈
1
Δ�̈
2
�
3
= �
9,54 9,64�
�10
−5
� �
1
�
2
�
4
= �
�
1
�
2
�
3
+ �Δ�
1
Δ�
2
�
3
= �
2,39 2,41� �
10
−9
��̇
1
�̇
2
�
4
= ��̇
1
�̇
2
�
3
+ �Δ�̇
1
Δ�̇
2
�
3
= �
4,77 4,82� �
10
−7
��̈
1
�̈
2
�
4
= ��̈
1
�̈
2
�
3
+ �Δ�̈
1
Δ�̈
2
�
3
= �
9,54 9,64�
�10
−5
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Untuk gaya lateral statik ekivalen setiap tingkat �
�
� = �
�
�
�
� Dimana:
�
�
= Γ
�
��
�
�
�
� = �
� 2
�
�
� → �
�
� =
�̈
�,�
−��
�
−��
�
�
�
Sehingga : �
�
� = �
� 2
� Γ
�
�
�
�
�
�
Total base shear : �
��
= � �
�� �
� =1
Catatan : Hasil gaya lateral statik ekivalen dan total base shear dapat dilihat di tabel 4.3 .
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Tabel 4.3.Respon Struktur MDOF dengan Metode Integrasi Langsung Newmark- � dengan Average Acceleration Method
i time
sec Accg
�̈
�
ms
2
Perpindahan Kecepatan
Percepatan Interstorey Drift
Gaya lateral Statik Ekivalen
Total Base
Shear Vbn
�
�
�
�
�̇
�
�̇
�
ü1 ü2
Δ₁ Δ₂
1 2
1 0,01
2 0,02
3 0,03
4 0,04
0,00001 0,0000981 2,39E-09
2,41E-09 4,77E-07
4,82106E-07 9,54194E-05
9,64E-05 2,39E-09
2,5E-11 0,00954
0,00835 0,01789
5 0,05
0,00002 0,0001962 1,41E-08
1,44E-08 1,87E-06
1,91156E-06 0,000183659 0,000189
1,41E-08 2,45E-10
0,056522 0,049808 0,10633
6 0,06
-0,00001 -9,81E-05
3,44E-08 3,56E-08
2,19E-06 2,3254E-06
-0,000120231 -0,00011
3,44E-08 1,12E-09
0,137748 0,123191 0,260939
7 0,07
-0,00005 -0,000491
4,07E-08 4,39E-08
-9,3E-07 -6,57226E-07
-0,000503706 -0,00049
4,07E-08 3,16E-09
0,162934 0,152083 0,315017
8 0,08
-0,00008 -0,000785
-7,9E-11 5,99E-09
-7,2E-06 -6,92597E-06
-0,000757108 -0,00076
-7,9E-11 6,07E-09
-0,00032 0,020744 0,020428
9 0,09
-0,00007 -0,000687
-1,1E-07 -9,8E-08
-1,4E-05 -1,39572E-05
-0,000588747 -0,00064
-1,1E-07 7,64E-09
-0,42418 -0,34095
-0,76513
10 0,1
-0,00009 -0,000883
-2,8E-07 -2,7E-07
-2E-05 -2,12249E-05
-0,000689753 -0,00081
-2,8E-07 3,33E-09
-1,11043 -0,95029
-2,06072
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a. Time History Perpindahan Lantai 1 dengan Metode Integrasi langsung Newmark- �