LAMPIRAN 1 ( PETA GEMPA INDONESIA 2010 ) Universitas Kristen Maranatha

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1 LAMPIRAN 1 ( PETA GEMPA INDONESIA 2010 ) 89

2 Gambar L1.1 Peta perepatan punak (PGA) di batuan dasar (SB) untuk probabilitas terlampaui 10% dalam 50 tahun 90

3 Gambar L1.2 Peta respon spektra perepatan 0.2 detik (SS) di batuan dasar (SB) untuk probabilitas terlampaui 10% dalam 50 tahun 91

4 Gambar L1.3 Peta respon spektra perepatan 1.0 detik (S1) di batuan dasar (SB) untuk probabilitas terlampaui 10% dalam 50 tahun 92

11 LAMPIRAN 2 ( Hasil Analisis ETABS Gedung berdasarkan Peta Gempa 2002 ) 99

12 Mode Period UX UY UZ SumUX SumUY SumUZ RX RY RZ SumRX SumRY SumRZ meter (M) Tabel L2.1 Modal Partiipating Mass Ratio Stor Point Load UX UY UZ RX RY RZ STORY4 1 E STORY3 1 E STORY2 1 E STORY1 1 E Tabel L2.2 Point Displaements 100

13 meter(m) Stor Diaphragm Load UX UY U Z X Y RZ Point X Y Z STORY4 D4 E STORY3 D3 E STORY2 D2 E STORY1 D1 E Meter(M) Tabel L2.3 Diagraph CM Displaements Stor Item Load Point X Y Z DritX DritY STORY4 Max Drit X E STORY4 Max Drit Y E STORY3 Max Drit X E STORY3 Max Drit Y E STORY2 Max Drit X E STORY2 Max Drit Y E STORY1 Max Drit X E STORY1 Max Drit Y E Tabel L2.4 Stor Drits 101

14 meter (M) Stor Point Load FX FY FZ MX MY MZ BASE 1 COMB BASE 2 COMB BASE 3 COMB BASE 4 COMB BASE 5 COMB BASE 6 COMB BASE 7 COMB BASE 8 COMB BASE 9 COMB Tabel L2.5 Support Reations 102

15 LAMPIRAN 3 ( Hasil Analisis ETABS Gedung berdasarkan Peta Gempa 2010 ) 103

16 Mode Period UX UY UZ SumUX SumUY SumUZ RX RY RZ SumRX SumRY SumRZ meter(m) Tabel L3.1 Modal Partiipating Mass Ratio Stor Point Load UX UY UZ RX RY RZ STORY 4 1 E STORY 3 1 E STORY 2 1 E STORY 1 1 E Tabel L3.2 Point Displaements 104

17 meter(m) Stor Diaphragm Load UX UY Z X Y RZ Point X Y Z STORY4 D4 E STORY3 D3 E STORY2 D2 E STORY1 D1 E Tabel L3.3 Diagram CM Displaements meter (M) Stor Item Load Point X Y Z DritX DritY STORY4 Max Drit X COMB STORY4 Max Drit Y COMB STORY3 Max Drit X COMB STORY3 Max Drit Y COMB STORY2 Max Drit X COMB STORY2 Max Drit Y COMB STORY1 Max Drit X COMB STORY1 Max Drit Y COMB Tabel 3.4 Stor Drits 105

18 meter (M) Stor Point Load FX FY FZ MX MY MZ BASE 1 COMB BASE 2 COMB BASE 3 COMB BASE 4 COMB BASE 5 COMB BASE 6 COMB BASE 7 COMB BASE 8 COMB BASE 9 COMB Tabel L3.5 Support Reations 106

19 LAMPIRAN 4 ( Perhitungan Jumlah Tulangan Balok ) 107

20 Tumpuan Tump Kiri 2002 Penampang persegi beton bertulang dengan h = 450 mm, b = 300 mm, menerima momen M u = ,779 Nmm Desain penampang tulangan ganda. Solusi: Asumsikan : d = dt = h 38 = = 412 mm, d = 38 mm b b b 1 = 25 MPa β 1 = 0,85 0, , , ,0271 A s 0,1. b. 1. b d A s1 = 0,1.0, A s1 = 334,956 mm 2 As. 1 a = 0,85.. b 334, , = 21,017 mm 108

21 1 M A.(d - a ) 2. n1 s1 M n1 = 334, ( ,017 ) M n1 = ,22 Nmm M n = M u lentur = ,779 0,9 = ,41 Nmm M n2 = M n M n1 = , ,75 = ,34 Nmm M n ,34 A s2 = A s = = = 177,293 mm 2.( d d) 400.(412 38) A s = A s1 + A s2 = 334, ,293 = 512,249 mm 2 A s,min = bw. d = = 386,25 mm 2 < ,6 mm 2 1,4 1, 4.bw.d = = A s = 512,249 mm 2 > A s,min = 432,6 mm 2,,,,,,, Ok Pakai 3D16 (603mm) ρ = A s b. d = 512,249 A = 0,00414 ρ = s 177,293 = = 0, b. d ρ - ρ 1.0,85.. d. d , , , ,0125 0,85.0, s = 600{1- s = 600{1-0, d } ( ). d 0, ,85.38 } (0,00271)

22 s = 322,117 = 400 MPa d sb = (600 ) d 38 sb = ( ) 412 sb = 507,767 MPa 400 MPa sb = = 400 MPa ρ 0,75. b +. sb 0, ,75. 0, , ,0217 0, a = A s A 0,85.. b s s = 512, , , = 21,017 mm M n = {(A s A s s )( d a/2) + A s s (d d )} M n ={(512, , )( ,017) + 177, (412 38)} M n = ,55 Nmm M u M n ,779 Nmm 0, ,55 = ,79 Nmm Ok 110

23 Lapangan 2002 Penampang persegi beton bertulang dengan h = 450 mm, b = 300 mm, mnerima momen M u = ,153 Nmm Desain penampang tulangan ganda. Solusi: Asumsikan : d = dt = h 38 = = 412 mm, d = 38 mm = 25 MPa β 1 = 0,85 b b b 1 0, , , ,0271 A s1 = 0,1. b.b.d A s1 = 0,1.0, A s1 = 334,956 mm 2 a a = As 1. 0,85.. b 334, , a = 21,017 mm M n1 = A s1. 1.( d - a ) 2 M n1 = 334, ( ,017 ) M n1 = ,75 Nmm 111

24 M n = M u lentur ,153 = = ,39 Nmm 0,9 M n2 = M n M n1 = , ,75 = ,36 Nmm M n ,36 A S2 = A S = = = 241,398 mm 2.( d d) 400.(412 38) A S = A S1 + A S2 = 334, ,398 = 576,354 mm 2 A s,min = C 25. bw. d = = 386,25 mm 2 1,4 1, 4 <.bw.d = ,6 mm = A s =576,354 mm 2 > A s,min = 432,6 mm 2 Ok Pakai 3D16 (603mm) ρ = A s b. d = 576,354 A = 0,00466 ρ = s 241,398 = = 0, b. d ,85. ρ - ρ. d C. d , , , ,0125 0,85.0, = 600{1- = 600{1-0,85. C. 1. } ( ).d d 0, ,85.38 } (0,00271 ) = 322,117 = 400 MPa = = 400 MPa d sb = (600 ) d 112

25 38 sb = ( ) 412 sb = 507,767 MPa 400 MPa sb = = 400 MPa ρ 0,75. b +. sb 0, ,75. 0, , ,029 0, a = A S A 0,85.. b S S = 576, ,398 0, = 21,205 mm M n = {(A S - A S S )( d a/2) + A S S ( d d )} M n ={(576, , )( ,205) + 241, (412 38)} M n = ,2 Nmm M u M n ,810 Nmm 0, ,2= ,08 Nmm Ok 113

26 Tumpuan Tump Kiri 2010 Penampang persegi beton bertulang dengan h = 450 mm, b = 300 mm, menerima momen M u = ,779 Nmm Desain penampang tulangan ganda. Solusi: Asumsikan : d = dt = h 38 = = 412 mm, d = 38 mm b b b 1 = 25 MPa β 1 = 0,85 0, , , ,0271 A s 0,1. b. 1. b d A s1 = 0,1.0, A s1 = 334,956 mm 2 As. 1 a 0,85.. b 114

27 a = 334, , = 21,017 mm 1 M A.(d - a ) 2. n1 s1 M n1 = 334, ( ,017 ) M n1 = ,22 Nmm M n = M u lentur = ,779 0,9 = ,41 Nmm M n2 = M n M n1 = , ,75 = ,34 Nmm M n ,34 A s2 = A s = = = 177,293 mm 2.( d d) 400.(412 38) A s = A s1 + A s2 = 334, ,293 = 512,249 mm 2 A s,min = bw. d = = 386,25 mm 2 < ,6 mm 2 1,4 1, 4.bw.d = = A s = 512,249 mm 2 > A s,min = 432,6 mm 2,,,,,,, Ok Pakai 3D16 (603mm) ρ = A s b. d = 512,249 A = 0,00414 ρ = s 177,293 = = 0, b. d ρ - ρ 1.0,85.. d. d , , , ,0125 0,85.0, s = 600{1-0, d } ( ). d 115

28 s = 600{1-0, ,85.38 } (0,00271) s = 322,117 = 400 MPa d sb = (600 ) d 38 sb = ( ) 412 sb = 507,767 MPa 400 MPa sb = = 400 MPa ρ 0,75. b +. sb 0, ,75. 0, , ,0217 0, a = A s A 0,85.. b s s = 512, , , = 21,017 mm M n = {(A s A s s )( d a/2) + A s s (d d )} M n ={(512, , )( ,017) + 177, (412 38)} M n = ,55 Nmm M u M n ,779 Nmm 0, ,55 = ,79 Nmm Ok 116

29 Lapangan 2010 Penampang persegi beton bertulang dengan h = 450 mm, b = 300 mm, mnerima momen M u = ,153 Nmm Desain penampang tulangan ganda. Solusi: Asumsikan : d = dt = h 38 = = 412 mm, d = 38 mm = 25 MPa β 1 = 0,85 b b b 1 0, , , ,0271 A s1 = 0,1. b.b.d A s1 = 0,1.0, A s1 = 334,956 mm 2 a a = As 1. 0,85.. b 334, , a = 21,017 mm M n1 = A s1. 1.( d - a ) 2 M n1 = 334, ( ,017 ) M n1 = ,75 Nmm 117

30 M n = M u lentur ,153 = = ,39 Nmm 0,9 M n2 = M n M n1 = , ,75 = ,36 Nmm M n ,36 A S2 = A S = = = 241,398 mm 2.( d d) 400.(412 38) A S = A S1 + A S2 = 334, ,398 = 576,354 mm 2 A s,min = C 25. bw. d = = 386,25 mm 2 1,4 1, 4 <.bw.d = ,6 mm = A s =576,354 mm 2 > A s,min = 432,6 mm 2 Ok Pakai 3D16 (603mm) ρ = A s b. d = 576,354 A = 0,00466 ρ = s 241,398 = = 0, b. d ,85. ρ - ρ. d C. d , , , ,0125 0,85.0, = 600{1- = 600{1-0,85. C. 1. } ( ).d d 0, ,85.38 } (0,00271 ) = 322,117 = 400 MPa = = 400 MPa d sb = (600 ) d 118

31 38 sb = ( ) 412 sb = 507,767 MPa 400 MPa sb = = 400 MPa ρ 0,75. b +. sb 0, ,75. 0, , ,029 0, a = A S A 0,85.. b S S = 576, ,398 0, = 21,205 mm M n = {(A S - A S S )( d a/2) + A S S ( d d )} M n ={(576, , )( ,205) + 241, (412 38)} M n = ,2 Nmm M u M n ,810 Nmm 0, ,2= ,08 Nmm Ok 119

32 LAMPIRAN 5 ( Perhitungan Jumlah Tulangan Kolom ) 120

33 Perhitungan Kolom Peta Gempa 2002 Gambar Nilai Pu Kolom 400 x 400 mm P u = ,01 N M u = ,7 Nmm gh = h 2 (over + d bs ) - d b = 400 2( ) 16 = 324 mm gh 324 g = = = 0,81 h 400 m = P. b. h 0,85. u = = 400 = 18,82 0, ,01 M = 0,071 u b. h = 2 Gambar Nilai Mu ,7 = 0, Kedua nilai tersebut diplotkan ke dalam graik NZS, sehingga didapatkan nilai ρ tm. dari Chart C /0,8 dan C /0,9 Untuk g = 0,8 ρ tm = 0,080 Untuk g = 0,9 ρ tm = 0,080 tm 0,18 Untuk g = 0,81 ρ tm = 0,18 ρ t = = = 0,0095 m 18, 82 A st,perlu = ρ t.b.h = 0, = 1520 mm 2 8 D 16 = 1608 mm 2 P n,max = [0,8{0,85. ( A g A st ) + A st. }] = 0,65[0,8{0,85.25.( ) }] 121

34 = ,6 N > Pu = ,01 N..OK 1608 ρ t,pakai = 2 = 0,01005 ρtm = 0, ,82 = 0, P u P P n = = = ,86 N n 0, 65. b. h = ,86 = 0, Lihat Chart C /0.8 untuk ρ tm = 0,189 Pn. b. h = 0,11. b. h M u 2 = 0,102 M n = 0,102 x x b x h 2 = 0,102 x 25 x 400 x = Nmm M n = 0,65 x M n = Nmm M n > M u Nmm > ,7 Nmm Kolom kuat Hasil jumlah tulangan ang diperoleh baik dengan menggunakan ETABS dan ara manual diperoleh jumlah tulangan 8Ø16 dan kekuatan kolom sangat kuat. 122

35 Perhitungan Kolom Peta Gempa 2010 Gambar Nilai Pu Kolom 400 x 400 mm P u = ,01 N M u = ,7 Nmm Gambar Nilai Mu gh = h 2(over + d bs ) - d b = 400 2( ) 16 = 324 mm gh 324 g = = = 0,81 h m = = =18,82 0,85. 0,85.25 Pu. b. h = ,01 M = 0,071 u ,7 = 2 2 = 0, b. h Kedua nilai tersebut diplotkan ke dalam graik NZS, sehingga didapatkan nilai ρ tm. dari Chart C /0,8 dan C /0,9 Untuk g = 0,8 ρ tm = 0,080 Untuk g = 0,9 ρ tm = 0,080 tm 0,18 Untuk g = 0,81 ρ tm = 0,18 ρ t = = = 0,0095 m 18, 82 A st,perlu = ρ t.b.h = 0, = 1520 mm 2 8 D 16 = 1608 mm 2 P n,max = [0,8{0,85. ( A g A st ) + A st. }] = 0,65[0,8{0,85.25.( ) }] 123

36 = ,6 N > Pu = ,01 N..OK 1608 ρ t,pakai = 2 = 0,01005 ρtm = 0, ,82 = 0, P u P P n = = = ,86 N n 0, 65. b. h = ,86 = 0, Lihat Chart C /0.8 untuk ρ tm = 0,189 Pn. b. h = 0,11:. b. h M u 2 = 0,102 M n = 0,102 x x b x h 2 = 0,102 x 25 x 400 x = Nmm M n = 0,65 x M n = Nmm M n > M u Nmm > ,7 Nmm Kolom kuat Hasil jumlah tulangan ang diperoleh baik dengan menggunakan ETABS dan ara manual diperoleh jumlah tulangan 8Ø16 dan kekuatan kolom sangat kuat. 124

37 LAMPIRAN 6 ( Diagram Moment dan Axial Fore) 125

38 Peta Gempa 2002 Moment 3-3 (Nmm) Pada Balok Potongan 1-1 Potongan 2-2 Potongan 3-3 Potongan A-A 126

39 Potongan B-B Potongan C-C AXIAL FORCE pada Kolom Potongan 1-1 Potongan

40 Potongan 3-3 Potongan A-A Potongan B-B Potongan C-C 128

41 MOMENT 2-2 (Nmm) pada Kolom Potongan 1-1 Potongan 2-2 Potongan 3-3 Potongan A-A 129

42 Potongan B-B Potongan C-C 130