Time History Total Base Shear dengan Metode Superposisi Modal Analysis

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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 Universitas Sumatera Utara 128 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 Universitas Sumatera Utara 129

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 � Universitas Sumatera Utara 130 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� ��� �+� = �� � ��� = �� Δ� = �� ���� −1 �Δ� 1 Δ� 2 � = � 0� � 32608880 −1270000 −1270000 27314330� −1 �Δ� 1 Δ� 2 � = � 0� Universitas Sumatera Utara 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} Universitas Sumatera Utara 132 �� � = �� � + ��̇ � + ��̈ � �� 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� Universitas Sumatera Utara 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 Universitas Sumatera Utara 134 ��̇ � = � ��� �� � − � � �̇ � + �� − � �� � �̈ � �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 Universitas Sumatera Utara 135 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 . Universitas Sumatera Utara 136 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 Universitas Sumatera Utara 137

a. Time History Perpindahan Lantai 1 dengan Metode Integrasi langsung Newmark- �