Preliminary Desain Lantai Preliminary Balok

159 Universitas Kristen Maranatha LAMPIRAN 3 PRELIMINARY DESAIN

1. Preliminary Desain Lantai

Menentukan tebal pelat minimum [RSNI, 2002]. Asumsi: bentang bersih : Ln panjang = 6000 – 250 = 5750 mm : Lnpendek = 5000 – 250 = 4750 mm  = = = 1,2 2 ………pelat 2 arah Penetuan tebal pelat 2 arah : h min = = = 109 mm h max = = = 171mm jadi tebal pelat lantai dan atap = 109 mm h 171 mm h = 150 mm Pembebanan Lantai  Beban mati q dl Tebal pelat = 0,15 m berat jenis beton = 2400 kgm 3 berat sendiri pelat = 0,15 x 2400 = 360 kgm 2 160 Universitas Kristen Maranatha  Beban mati tambahan q sdl Adukan 3 cm = 3 x 21 kgm 2 = 63 kgm 2 Keramik 2cm = 2 x 24 kgm 2 = 48 kgm 2 Berat plafond + penggantung = 14 kgm 2 Mekanikal elektrikal = 20 kgm 2 + q sdl = 145 kgm 2  Beban hidup LL Beban hidup = 250 kgm 2

2. Preliminary Balok

1. Balok lantai 3 sumbu C antara as 4 dan 3  Beban mati q dl = 360 kgm 2  Beban mati tambahan q sdl Adukan 1 cm = 1 x 21 kgm 2 = 21 kgm 2 Berat plafond + penggantung = 14 kgm 2 Mekanikal elektrikal = 20 kgm 2 + q sdl = 55 kgm 2  Beban hidup LL = 250 kgm 2  q ult = 1,2 q dl + q sdl + 1,6 LL q ult = 1,2 360 + 55 + 1,6 250 = 898 kgm 2  q ek = q ult x l q ek = 898 x 5 = 4490 kgm  Mult = x q ek x l 2 = x 4490 x 5 2 = 9354,166667 kgm = 93541666,67 Nmm  M ult =  x Zx x f y = 0,9 x Zx x 410 Zx = 253500,4517 mm 3 161 Universitas Kristen Maranatha Diambil profil : IWF 250.250.9.14  Zx = x Luas = 125 – 20,8 x 9218 = 960515,6 mm 3 253500,4517 mm 3 ………OK 2. Balok lantai 2 sumbu C antara as 4 dan 3  Beban mati q dl Tebal pelat = 0,15 m berat jenis beton = 2400 kgm 3 berat sendiri pelat = 0,15 x 2400 = 360 kgm 2  Beban mati tambahan q sdl Adukan 3 cm = 3 x 21 kgm 2 = 63 kgm 2 Keramik 2cm = 2 x 24 kgm 2 = 48 kgm 2 Berat plafond + penggantung = 14 kgm 2 Mekanikal elektrikal = 20 kgm 2 + q sdl = 145 kgm 2  Beban hidup LL Beban hidup = 250 kgm 2  q ult = 1,2 q dl + q sdl + 1,6 LL q ult = 1,2 360 + 145 + 1,6 250 = 1006 kgm 2  q ek = q ult x l q ek = 1006 x 6 = 6036 kgm  Mult = x q ek x l 2 = x 6036 x 6 2 = 18108 kgm = 181080000 Nmm  M ult =  x Zx x f y = 0,9 x Zx x 410 Zx = 490731,71 mm 3 162 Universitas Kristen Maranatha Diambil profil : IWF 300.300.10.15  Zx = x Luas = 150 – 24,7 x 11980 = 1501094 mm 3 490731,71 mm 3 ………OK 3. Balok lantai 1 sumbu C antara as 4 dan 3  Beban mati q dl Tebal pelat = 0,15 m berat jenis beton = 2400 kgm 3 berat sendiri pelat = 0,15 x 2400 = 360 kgm 2  Beban mati tambahan q sdl Adukan 3 cm = 3 x 21 kgm 2 = 63 kgm 2 Keramik 2cm = 2 x 24 kgm 2 = 48 kgm 2 Berat plafond + penggantung = 14 kgm 2 Mekanikal elektrikal = 20 kgm 2 + q sdl = 145 kgm 2  Beban hidup LL Beban hidup = 250 kgm 2  q ult = 1,2 q dl + q sdl + 1,6 LL q ult = 1,2 360 + 145 + 1,6 250 = 1006 kgm 2  q ek = q ult x l q ek = 1006 x 6 = 6036 kgm  Mult = x q ek x l 2 = x 6036 x 6 2 = 18108 kgm = 181080000 Nmm  M ult =  x Zx x f y = 0,9 x Zx x 410 Zx = 490731,71 mm 3 163 Universitas Kristen Maranatha Diambil profil : IWF 350.350.10.15  Zx = x Luas = 175 – 24,7 x 11980 = 1501094 mm 3 490731,71 mm 3 ………OK

4. Preliminary Kolom