Penulangan Plat Lantai Penulangan tumpuan arah x

commit to user Tugas Akhir Perencanaan St rukt ur Gedung Sekolah 2 L ant ai BAB 5 Plat Lant ai 128

5.5. Penulangan Plat Lantai

Tabel 5.1. Perhitungan Plat Lantai Tipe Plat LyLx m Mlx kgm Mly kgm Mtx kgm Mty kgm 4,003,00 = 1,33 361,746 209,009 787,803 618,988 3,002,50 = 1,2 212,135 156,310 474,513 413,105 3,002,50 = 1,2 173,058 156,310 413,105 385,193 4,002,50 = 1,6 256,795 139,563 552,668 429,853 4,002,50 = 1,6 206,553 89,320 441,018 318,203 3,253,00 = 1,08 209,009 217,048 522,522 522,522 3,253,00 = 1,08 200,970 168,815 474,289 434,095 4,003,25 = 1,23 292,485 179,265 651,015 537,795 4,003,25 = 1,23 330,225 169,830 698,190 537,795 2,002,00 = 1,0 75,029 75,029 185,786 185,786 4,001,25 = 3,2 58,603 11,162 118,602 79,534 3,003,00 = 1,0 168,815 209,009 442,134 482,328 3,003,00 = 1,0 168,534 168,534 418,018 418,018 A1 A A2 A3 B1 B2 B3 B4 B5 B6 C1 C2 A4 commit to user Tugas Akhir Perencanaan St rukt ur Gedung Sekolah 2 L ant ai BAB 5 Plat Lant ai 129 4,003,00 = 1,33 273,319 144,698 586,832 458,212 3,253,00 = 1,08 265,280 225,086 618,988 578,794 3,253,00 = 1,08 233,125 160,775 530,561 458,212 4,003,25 = 1,23 339,660 188,700 773,670 679,32 2,002,00 = 1,0 75,029 92,893 196,504 214,368 4,002,50 = 1,6 173,138 157,748 - 380,903 Dari perhitungan momen diambil momen terbesar yaitu: Mlx = 361,746 kgm Mly = 225,086 kgm Mtx = 787,803 kgm Mty = 679,320 kgm Data – data plat : Tebal plat h = 12 cm = 120 mm Diameter tulangan ∅ = 10 mm fy = 240 MPa f’c = 25 MPa b = 1000 mm p = 20 mm Tebal penutup d’ = p + ½ ∅ tul = 20 + 5 = 25 mm C3 D1 D2 D3 D4 P.atp commit to user Tugas Akhir Perencanaan St rukt ur Gedung Sekolah 2 L ant ai BAB 5 Plat Lant ai 130 Tinggi Efektif d = h - d’ = 120 – 25 = 95 mm Tingi efektif Gambar 5.4 Perencanaan Tinggi Efektif dx = h – p - ½Ø = 120 – 20 – 5 = 95 mm dy = h – d’ – Ø - ½ Ø = 120 – 20 - 10 - ½ . 10 = 85 mM ρb = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + fy fy fc 600 600 . . . 85 , β = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + 240 600 600 . 85 , . 240 25 . 85 , = 0,0538 ρ max = 0,75 . ρb = 0,75 . 0,0538 = 0,04035 ρ min = 0,0025 h dy dx d commit to user Tugas Akhir Perencanaan St rukt ur Gedung Sekolah 2 L ant ai BAB 5 Plat Lant ai 131

5.6. Penulangan tumpuan arah x

Mu = 787,803 kgm = 7,878x10 6 Nmm Mn = φ Mu = = 8 , 7,878x106 9,8475x10 6 Nmm Rn = = 2 .dx b Mn = 2 6 95 . 1000 10 8475 , 9 x 1,091 Nmm 2 m = 294 , 11 25 . 85 , 240 . 85 , = = c f fy ρ perlu = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − − fy Rn . m 2 1 1 . m 1 = . 294 , 11 1 ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − − 240 091 , 1 . 294 , 11 . 2 1 1 = 0,00467 ρ ρ max ρ ρ min , di pakai ρ perlu = 0,00467 As perlu = ρ perlu . b . dx = 0,00467. 1000 . 95 = 443,65mm 2 Digunakan tulangan ∅ 10 As = ¼ . π . 10 2 = 78,5 mm 2 Jumlah tulangan dalam 1m 1 = 5 . 78 443.65 = 5,65 ~ 6 buah Jarak tulangan dalam 1m 1 = 6 1000 = 166,67mm ~ 160mm Jarak maksimum tulangan = 2 x h = 2 x 120 = 240mm As yang timbul = 6. ¼ . π.10 2 = 471 mm 2 As perlu …ok Dipakai tulangan ∅ 10 – 160 mm commit to user Tugas Akhir Perencanaan St rukt ur Gedung Sekolah 2 L ant ai BAB 5 Plat Lant ai 132

5.7. Penulangan tumpuan arah y