Penulangan Plat Lantai Penulangan tumpuan arah y

Perencanaan Struktur Hotel 2 lantai BAB I Pendahuluan Gambar 5.5 Plat tipe B2 2,5 2 5 Lx Ly   Mlx = 0,001.qu .Lx 2 .x = 0.001 x 893,2 x 2 2 x 60 = 214,37 kg m Mly = 0,001.qu .Lx 2 .x = 0.001 x 893,2 x 2 2 x 18 = 64,31 kg m Mtx = 0,001.qu .Lx 2 .x = 0.001 x 893,2 x 2 2 x 120 = 428,74 kg m Mty = 0,001.qu .Lx 2 .x = 0.001 x 893,2 x 2 2 x 79 = 282,25 kg m

5.4. Penulangan Plat Lantai

Tabel 5.1. Perhitungan Plat Lantai Tipe Plat LyLx m Mlx kgm Mly kgm Mtx kgm Mty kgm A 52,5 = 2 36,62 10,72 296,54 203,65 B 2,51,5 = 1,7 76,37 28,14 162,79 114,55 C 3,52,5 = 1,4 189,81 100,49 407,52 318,2 D 52 = 2,5 214,37 64,31 428,74 282,25 Dari perhitungan momen diambil momen terbesar yaitu: Mlx = 214,37 kgm Mly = 100,49 kgm Mtx = 428,74 kgm Mty = 318,2 kgm Data – data plat : Perencanaan Struktur Hotel 2 lantai BAB I Pendahuluan Tebal plat h = 12 cm = 120 mm Diameter tulangan  = 12 mm fy = 240 MPa f’c = 25 MPa b = 1000 mm tebal penutup d’ = 20 mm Gambar 5.8 Perencanaan Tinggi Efektif Tinggi efektif : dx = h – d’ - ½Ø = 120 – 20 – 6 = 94 mm dy = h – d’ – Ø - ½ Ø = 120 – 20 - 12 - ½ . 12 = 82 mm b =      fy fy fc 600 600 . . . 85 ,  f’c = 25 Mpa 30 Mpa →  = 0,85 =        240 600 600 . 85 , . 240 25 . 85 , = 0,054  max = 0,75 . b = 0,75 . 0,054 = 0,0405  min = 0,0025 untuk plat 5.5. Penulangan tumpuan arah x Mu = 428,74 kgm = 4,287.10 6 Nmm h d dy dx Perencanaan Struktur Hotel 2 lantai BAB I Pendahuluan Mn =  M u =  8 , 10 . 287 , 4 6 5,359.10 6 Nmm Rn =  2 . dx b Mn    2 6 94 . 1000 10 . 359 , 5 0,606 Nmm 2 m = 3 , 11 25 . 85 , 240 . 85 ,   c f fy  perlu =       fy Rn . m 2 1 1 . m 1 = . 3 , 11 1       240 606 , . 3 , 11 . 2 1 1 = 0,00256  perlu  max  perlu  min , di pakai  perlu = 0,00256 As perlu =  perlu . b . dx = 0,00256 . 1000 . 94 = 240,64 mm 2 Digunakan tulangan  12 As = ¼ .  . 12 2 = 113,04 mm 2 S = per lu As b As . = 64 , 240 1000 . 04 , 113 = 469,75 mm S max = 2 x h = 240 mm n = s b = 240 1000 = 5 buah As yang timbul = 5. ¼ .  . 12 2 = 339,12 mm 2 As perlu …..…ok Dipakai tulangan  12 – 240 mm Perencanaan Struktur Hotel 2 lantai BAB I Pendahuluan

5.6. Penulangan tumpuan arah y

Mu = 318,2 kgm = 3,182.10 6 Nmm Mn =  M u =  8 , 10 . 182 , 3 6 3,978.10 6 Nmm Rn =  2 .dy b Mn    2 6 82 . 1000 10 . 798 , 3 0,592 Nmm 2 m = 3 , 11 25 . 85 , 240 . 85 ,   c f fy  perlu =       fy Rn . m 2 1 1 . m 1 = . 3 , 11 1       240 592 , . 3 , 11 . 2 1 1 = 0,002502  perlu  max  perlu  min , di pakai  perlu = 0,002502 As perlu =  perlu . b . dy = 0,002502. 1000 . 82 = 205,164 mm 2 Digunakan tulangan  12 As = ¼ .  . 12 2 = 113,04 mm 2 S = per lu As b As . = 164 , 205 1000 . 04 , 113 = 550,97 mm S max = 2 x h = 240 mm n = s b = 240 1000 Perencanaan Struktur Hotel 2 lantai BAB I Pendahuluan = 5 buah As yang timbul = 5. ¼ .  . 12 2 = 565,2 mm 2 As perlu …..…ok Dipakai tulangan  12 – 240 mm

5.7. Penulangan lapangan arah x