Balok Anak Balok Anak Daerah Tumpuan Balok Anak

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BAB 6 Balok Anak

3. Beban berfaktor q U qU 1 = 1,2. qD + 1,6. qL = 1,2 . 1009,32 + 1,6.530 = 2059,18 kgm qU 2 = 1,2. qD + 1,6. qL = 1,2 . 1033,98 + 1,6.545 = 2112,77 kgm

6.2.2. Perhitungan Tulangan

Tulangan Lentur Balok Anak Data Perencanaan : h = 350 mm Ø t = 16 mm b = 250 mm Ø s = 8 mm p = 40 mm d = h - p - 12 Ø t - Ø s fy = 360 Mpa = 350 – 40 - 12.16 - 8 f’c = 25 Mpa = 294 Tulangan Lentur Daerah Lapangan ρb = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + fy 600 600 fy c. β 0,85.f = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + 360 600 600 85 , 360 25 . 85 , = 0,031 ρ max = 0,75 . ρb = 0,75 . 0,031 = 0,0232 ρ min = 0038 , 360 4 , 1 4 , 1 = = fy commit to user Tugas Akhir Perencanaan Struktur Factory Outlet dan Resto 2 Lantai

BAB 6 Balok Anak Daerah Tumpuan

Dari perhitungan SAP 2000 diperoleh : Mu = 2800,35 kgm = 2,8003 . 10 7 Nmm Mn = φ Mu = 7 7 10 . 5 , 3 8 , 10 . 8003 , 2 = Nmm Rn = = 2 .d b Mn = × 2 7 294 250 10 . 5 , 3 1,62 Nmm 2 m = = = 0,85.25 360 c 0,85.f fy 17 ρ perlu = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − − fy Rn . m 2 1 1 . m 1 = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ × × − − 360 62 , 1 17 2 1 1 . 17 1 = 0,0046 ρ ρ max ρ ρ min , di pakai ρ perlu = 0,0046 As = ρ perlu . b . d = 0,0046 . 250 . 294 = 338,1 mm 2 Digunakan tulangan D 16 = ¼ . π . 16 2 = 200,96 mm 2 Jumlah tulangan = 68 , 1 96 , 200 1 , 338 = ~ 2 buah. Dipakai 2 D 16 mm As ada = 2 . ¼ . π . 16 2 = 401,92 mm 2 As ……… aman a = 250 25 85 , 360 92 , 401 b c f 0,85 fy ada As × × × = × × × = 27,23 commit to user Tugas Akhir Perencanaan Struktur Factory Outlet dan Resto 2 Lantai

BAB 6 Balok Anak

Mn ada = As ada × fy ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − 2 a d = 401,92 × 360 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − 2 23 , 27 294 = 4,0569 . 10 7 Nmm Mn ada Mn ......... aman Kontrol Spasi : S = 1 - n sengkang 2 - tulangan n - 2p - b φ φ = 1 2 8 . 2 - 16 . 2 - 40 . 2 - 50 2 − = 122 25 mm......oke Jadi dipakai tulangan D 16 mm Daerah Lapangan Dari perhitungan SAP 2000 diperoleh : Mu = 2789.97 kgm = 2,789 . 10 7 Nmm Mn = φ Mu = 7 7 10 . 48 , 3 8 , 10 . 789 , 2 = Nmm Rn = = 2 .d b Mn = × 2 7 294 250 10 . 48 , 3 1,61 Nmm 2 m = = = 0,85.25 360 c 0,85.f fy 17 ρ perlu = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − − fy Rn . m 2 1 1 . m 1 = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ × × − − 360 61 , 1 17 2 1 1 . 17 1 = 0,004 ρ ρ max ρ ρ min , di pakai ρ perlu = 0,004 commit to user Tugas Akhir Perencanaan Struktur Factory Outlet dan Resto 2 Lantai

BAB 6 Balok Anak