Time History Total Base Shear dengan Metode Integrasi langsung Wilson- �

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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- �