167
g. Time History Total Base Shear dengan Metode Integrasi langsung Wilson- �
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 Total Base Shear dengan Metode Integrasi Langsung
Wilson- θ
Vb kgf
Universitas Sumatera Utara
168
4.1.5.2.2 Metode Integrasi Langsung dengan � = �, �� Wilson-�
Untuk � = 1,42
• Hitung � ��� � ∆� = 0,01
� = �
� � ��
� + ���
� =
6 � Δ�
� 2,91
2,52� 10
3
+ 3 �
10,0468 8,7003�
10
3
� = � 1258205
1089804 �
� = ��� +
� �� �
��
� = 3 � 2,91
2,52� 10
3
+
θ Δt 2
� 10,0468
8,7003� 10
3
� = � 8790,59
761406 �
Universitas Sumatera Utara
169
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�
� = � 2,54
−1,27 −1,27
1,27 � 10
6
���
��
�
= �
�
+
� � ��
� +
� � ��
�
� �� = �
2,54 −1,27
−1,27 1,27 �
10
6
+
� � ��
� 10,0468
8,7003� 10
3
Universitas Sumatera Utara
170
�� = � 91145999,21
−1270000 −1270000
78016740,63� �� ��
�
= ���
�
δ� =
��� ����
−1
�δ�
1
δ�
2
� =
� 0�
� 91145999,21
−1270000 −1270000
78016740,63�
−1
�δ�
1
δ�
2
� =
� 0�
��̈
�
=
� ���
�
��
�
−
� � ��
�̇
�
− ��̈
�
��� ̈
1
��̈
2
� =
6 ΘΔ �
2
�δ�
1
δ�
2
� −
6 θ Δt
��̇
1
�̇
2
� − 3 ��̈
1
�̈
2
�
��� ̈
1
��̈
2
� =
6 ΘΔ �
2
� 0�
−
6 θ Δt
− 30
��� ̈
1
��̈
2
� =
� 0�
��̈
�
=
� �
��̈
�
�Δ�̈
1
Δ�̈
2
� =
1 �
��� ̈
1
��̈
2
� �Δ�̈
1
Δ�̈
2
� =
1 �
� 0�
�Δ�̈
1
Δ�̈
2
� =
� 0�
��̇
�
= ���̈
�
+
�� �
��̈
�
�Δ�̇
1
Δ�̇
2
� =
Δ� ��̈
1
�̈
2
� +
Δt 2
�Δ�̈
1
Δ�̈
2
� �Δ�̇
1
Δ�̇
2
� =
Δ�0 +
Δt 2
� 0�
�Δ�̇
1
Δ�̇
2
� =
� 0�
Universitas Sumatera Utara
171
��
�
= ���̇
�
+
��
�
�
�̈
�
+
��
�
�
��̈
�
�Δ�
1
Δ�
2
� =
Δ� �Δ�̇
1
Δ�̇
2
� +
Δ�
2
2
��̈
1
�̈
2
� +
Δ�
2
6
�Δ�̈
1
Δ�̈
2
� �Δ�
1
Δ�
2
� =
Δ� � 0�
� 0�
+
Δ�
2
2
0 +
Δ�
2
6
� 0�
�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} ���
�
= ���
�
+ ��̇
�
+ ��̈
�
���
1
= Δ�
1
+ ��̇
1
+ ��̈
1
���
1
= �
� + �
1258205 1089804
� �
� + �
8790,59 761406
� �
�
���
1
= �
0�
Universitas Sumatera Utara
172
� = � 2,54
−1,27 −1,27
1,27 � 10
6
���
��
�
= �
�
+
� � ��
� +
� � ��
�
� �� = �
2,54 −1,27
−1,27 1,27 �
10
6
+
� � ��
� 10,0468
8,7003� 10
3
�� = � 91145999,21
−1270000 −1270000
78016740,63� �� ��
�
= ���
�
�� ��
�
= ���
�
δ�
1
= ���
1
����
−1
�δ�
1
δ�
2
�
1
= �
0� � 91145999,21
−1270000 −1270000
78016740,63�
−1
�δ�
1
δ�
2
�
1
= �
0�
��̈
�
=
� ���
�
��
�
−
� � ��
�̇
�
− ��̈
�
��� ̈
1
��̈
2
�
1
=
6 ΘΔ �
2
�δ�
1
δ�
2
�
1
−
6 θ Δt
��̇
1
�̇
2
�
1
− 3 ��̈
1
�̈
2
�
1
��� ̈
1
��̈
2
�
1
=
6 ΘΔ �
2
� 0� −
6 θ Δt
� 0� −
3 �
0�
��� ̈
1
��̈
2
�
1
= �
0�
��̈
�
=
� �
��̈
�
�Δ�̈
1
Δ�̈
2
�
1
=
1 �
��� ̈
1
��̈
2
�
1
�Δ�̈
1
Δ�̈
2
�
1
=
1 �
� 0�
�Δ�̈
1
Δ�̈
2
�
1
= �
0�
Universitas Sumatera Utara
173
��̇
�
= ���̈
�
+
�� �
��̈
�
�Δ�̇
1
Δ�̇
2
�
1
= Δ� ��̈
1
�̈
2
�
1
+
Δt 2
�Δ�̈
1
Δ�̈
2
�
1
�Δ�̇
1
Δ�̇
2
�
1
= Δ� �
0� +
Δt 2
� 0�
�Δ�̇
1
Δ�̇
2
�
1
= �
0�
��
�
= ���̇
�
+
��
�
�
�̈
�
+
��
�
�
��̈
�
�Δ�
1
Δ�
2
�
1
= Δ� �Δ�̇
1
Δ�̇
2
�
1
+
Δ�
2
2
��̈
1
�̈
2
�
1
+
Δ�
2
6
�Δ�̈
1
Δ�̈
2
�
1
�Δ�
1
Δ�
2
�
1
= Δ� �
0� +
Δ�
2
2
� 0�
+
Δ�
2
6
� 0�
�Δ�
1
Δ�
2
�
1
= �
0�
� �
1
�
2
�
2
= �
�
1
�
2
�
1
+ �Δ�
1
Δ�
2
�
1
= �
0� ��̇
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 4.5.
Universitas Sumatera Utara
174
• 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
� +
�
1258205 1089804
� � � + �
8790,59 761406
� �
�
���
3
= �
0,28512 0,24696�
� = � 2,54
−1,27 −1,27
1,27 � 10
6
���
��
�
= �
�
+
� � ��
� +
� � ��
�
� �� = �
2,54 −1,27
−1,27 1,27 �
10
6
+
� � ��
� 10,0468
8,7003� 10
3
�� = � 91145999,21
−1270000 −1270000
78016740,63� �� ��
�
= ���
�
�� ��
�
= ���
�
δ�
3
= ���
3
����
−1
�δ�
1
δ�
2
�
3
= �
0,28512 0,24696� �
91145999,21 −1270000
−1270000 78016740,63�
−1
�δ�
1
δ�
2
�
3
= �
4,51 � 10
−9
4,56 � 10
−9
�
Universitas Sumatera Utara
175
��̈
�
=
� ���
�
��
�
−
� � ��
�̇
�
− ��̈
�
��� ̈
1
��̈
2
�
3
=
6 ΘΔ �
2
�δ�
1
δ�
2
�
3
−
6 θ Δt
��̇
1
�̇
2
�
3
− 3 ��̈
1
�̈
2
�
3
��� ̈
1
��̈
2
�
3
=
6 ΘΔ �
2
� 4,51
� 10
−9
4,56 � 10
−9
� −
6 θ Δt
� 0� −
3 �
0�
��� ̈
1
��̈
2
�
3
= �
0,000134 0,000136�
��̈
�
=
� �
��̈
�
�Δ�̈
1
Δ�̈
2
�
3
=
1 �
��� ̈
1
��̈
2
�
3
�Δ�̈
1
Δ�̈
2
�
3
=
1 �
� 0,000134
0,000136�
�Δ�̈
1
Δ�̈
2
�
3
= �
9,44 � 10
−5
9,57 � 10
−5
�
��̇
�
= ���̈
�
+
�� �
��̈
�
�Δ�̇
1
Δ�̇
2
�
3
= Δ� ��̈
1
�̈
2
�
3
+
Δt 2
�Δ�̈
1
Δ�̈
2
�
3
�Δ�̇
1
Δ�̇
2
�
3
= Δ� �
0� +
Δt 2
� 9,44
� 10
−5
9,57 � 10
−5
� �Δ�̇
1
Δ�̇
2
�
3
= �
4,72 � 10
−7
4,79 � 10
−7
�
��
�
= ���̇
�
+
��
�
�
�̈
�
+
��
�
�
��̈
�
�Δ�
1
Δ�
2
�
3
= Δ� ��̇
1
�̇
2
�
3
+
Δ�
2
2
��̈
1
�̈
2
�
3
+
Δ�
2
6
�Δ�̈
1
Δ�̈
2
�
3
�Δ�
1
Δ�
2
�
3
= Δ� �
0� +
Δ�
2
2
� 0�
+
Δ�
2
6
� 9,44
� 10
−5
9,57 � 10
−5
� �Δ�
1
Δ�
2
�
3
= �
1,57 � 10
−9
1,59 � 10
−9
�
Universitas Sumatera Utara
176
� �
1
�
2
�
4
= �
�
1
�
2
�
3
+ �Δ�
1
Δ�
2
�
3
= �
1,57 � 10
−9
1,59 � 10
−9
� ��̇
1
�̇
2
�
4
= ��̇
1
�̇
2
�
3
+ �Δ�̇
1
Δ�̇
2
�
3
= �
4,72 � 10
−7
4,79 � 10
−7
� ��̈
1
�̈
2
�
4
= ��̈
1
�̈
2
�
3
+ �Δ�̈
1
Δ�̈
2
�
3
= �
9,44 � 10
−5
9,57 � 10
−5
�
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 4.6 .
Universitas Sumatera Utara
177
Tabel 4.6.Respon Struktur MDOF dengan Metode Integrasi Langsung Wilson- � dengan � = �, ��
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
1,57359E-09 1,59548E-09
4,72078E-07 4,78644E-07
9,44E-05 9,57E-05
1,57E-09 2,19E-11
0,00618 0,006266
0,012447
5 0,05
0,00002 0,0001962 1,24585E-08
1,27075E-08 1,84924E-06
1,89767E-06 0,000181
0,000188 1,25E-08
2,49E-10 0,048932
0,04991 0,098842
6
0,06 -0,00001
-9,81E-05 3,49706E-08
3,61846E-08 2,15006E-06
2,3074E-06 -0,00012
-0,00011 3,5E-08
1,21E-09 0,137351
0,142119 0,27947
7 0,07
-0,00005 -0,000491
4,41296E-08 4,76076E-08
-9,48156E-07 -6,57243E-07
-0,0005 -0,00049
4,41E-08 3,48E-09
0,173324 0,186984
0,360308
8
0,08 -0,00008
-0,000785 5,60005E-09
1,21577E-08 -7,16859E-06
-6,8865E-06 -0,00075
-0,00076 5,6E-09
6,56E-09 0,021995
0,047751 0,069745
9 0,09
-0,00007 -0,000687
-1,0049E-07 -9,26386E-08
-1,37633E-05 -1,38706E-05
-0,00057 -0,00064
-1E-07 7,85E-09
-0,39469 -0,36385
-0,75853
10
0,1 -0,00009
-0,000883 -2,6844E-07
-2,66002E-07 -1,99901E-05
-2,10788E-05 -0,00067
-0,0008 -2,7E-07
2,44E-09 -1,05433
-1,04475 -2,09908
Universitas Sumatera Utara
178
a. Time History Perpindahan pada lantai 1 dengan Metode Integrasi langsung Wilson- �