140
g. Time History Total Base Shear dengan Metode Integrasi langsung Newmark- �
Grafik a sd g Respon Struktur MDOF akibat gempa Kobe
-300000 -200000
-100000 100000
200000 300000
1, 95
3, 9
5, 85
7, 8
9, 75
11, 7
13, 65
15, 6
17, 55
19, 5
21, 45
23, 4
25, 35
27, 3
29, 25
31, 2
33, 15
35, 1
37, 05
39
Vb K
g f
Time History Base Shear dengan Metode Integrasi Langsung Newmark-
β
Vb kgf
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4.1.5.1.2 Metode Integrasi Langsung dengan Linear Acceleration Method
Newmark- �
Untuk Constant Average Acceleration , � =
1 2
��� � =
1 6
• 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
�� = � 47643320
−1270000 −1270000 40336495�
� = �
� � ��
� +
� �
��
� =
1 � Δ�
� 2,91
2,52� 10
3
+
� �
� 10,0468
8,7003� 10
3
� = � 902066,4
781329,9�
� = �
� ��
� + �� �
� ��
− �� ��
� =
1 2
�
� 2,91
2,52� 10
3
+ Δ� �
� 2
�
− 1� � 10,0468
8,7003� 10
3
� = � 8819,728
7639,293�
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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�
�
+ ��̇
�
+ ��̈
�
�� =
Δ� +
��̇ +
��̈ ��
= {0}
���
�+�
= ��
�
��� =
�� �
= ��
����
−1
� Δ�
1
Δ�
2
� = {0}
� 326,4468
−12,7 −12,7
273,4003� 10
5 −1
� Δ�
1
Δ�
2
� =
� 0�
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��̇
�
=
� ���
��
�
−
� �
�̇
�
+ �� −
� ��
� �̈
�
� 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�
�̈
�
=
� ���
�
��
�
−
� ���
�̇
�
−
� ��
�̈
�
� 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 awal u
1
= �
0� , u
̇
1
= �
0� , u
̈
1
= �
0�
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��
�
= ��
�
+ ��̇
�
+ ��̈
�
��
1
= Δ�
1
+ ��̇
1
+ ��̈
1
��
1
= �
0� +
� 902066,4
781329,9� � 0�
+ �
8819,728 7639,293� �
0� ��
1
= �
0�
���
�
= ��
�
���
1
= ��
1
Δ�
1
= ��
����
−1
� Δ�
1
Δ�
2
�
1
= �
0� �
47643320 −1270000
−1270000 40336495�
−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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��̇
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.4.
• 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
�+
� 902066,4
781329,9� {
}
+
� 8819,728
7639,293� {
} ��
3
= �
0,28512 0,24696�
���
�
= ��
�
���
3
= ��
3
Δ�
3
= ��
3
����
−1
�Δ�
1
Δ�
2
�
3
= �
0,28512 0,24696�
� 47643320
−1270000 −1270000 40336495�
−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 4.4 .
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Tabel 4.4.Respon Struktur MDOF dengan Metode Integrasi Langsung Newmark- � dengan Linear 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- �