Portal Portal Daerah Lapangan Portal

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BAB 7 Portal

m = 17 25 0,85 360 c 0,85.f fy = × = ρ = ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ − − fy 2.m.Rn 1 1 m 1 = ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ × × − − 360 85 , 2 17 2 1 1 17 1 = 0,0085 ρ ρ min ρ ρ max → dipakai tulangan tunggal Digunakan ρ = 0,0085 As perlu = ρ. b . d = 0,0085 × 300 × 440,5 = 1123,27 mm 2 Digunakan tulangan D 19 n = 385 , 283 27 , 1123 19 . 4 1 perlu As 2 = π = 3,964 ≈ 4 tulangan As’ = 2 19 . 4 1 π = 2 19 . 14 , 3 4 1 = 283,385 mm As = 4 × 283,385 = 1133,54 mm 2 As’ As………………….aman Ok Dipakai tulangan 4 D 19 Kontrol Spasi : S = 1 - n sengkang 2 - tulangan n - 2p - b φ φ = 1 4 10 . 2 - 19 . 4 - 40 . 2 - 00 3 − = 120 25 mm......oke Jadi dipakai tulangan D 19 mm commit to user Tugas Akhir Perencanaan Struktur Kafe dan Resto 2 Lantai

BAB 7 Portal Daerah Lapangan

Dari Perhitungan SAP 2000 diperoleh momen terbesar pada batang nomor 278. Mu = 11375,55 = 11,375 × 10 7 Nmm Mn = φ Mu = 8 , 10 375 , 11 7 × = 14,22 × 10 7 Nmm Rn = 44 , 2 440,5 300 10 14,22 d . b Mn 2 7 2 = × × = m = 17 25 0,85 360 c 0,85.f fy = × = ρ = ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ − − fy 2.m.Rn 1 1 m 1 = ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ × × − − 360 44 , 2 17 2 1 1 17 1 = 0,0072 ρ ρ min ρ ρ max → dipakai tulangan tunggal Digunakan ρ perlu = 0,0072 As perlu = ρ. b . d = 0,0072 × 300 × 440,5 = 951,48 mm 2 Digunakan tulangan D 19 n = 385 , 283 48 , 951 19 . 4 1 perlu As 2 = π = 3,357 ≈ 4 tulangan As’ = 2 19 . 4 1 π = 2 19 . 14 , 3 4 1 = 283,385 mm As ada = 4 × 283,385 = 1133,54 mm 2 As’ As………………….aman Ok Dipakai tulangan 4 D 19 commit to user Tugas Akhir Perencanaan Struktur Kafe dan Resto 2 Lantai

BAB 7 Portal

Kontrol Spasi : S = 1 - n sengkang 2 - tulangan n - 2p - b φ φ = 1 4 10 . 2 - 19 . 4 - 40 . 2 - 00 3 − = 120 25 mm......oke Jadi dipakai tulangan D 19 mm

7.6.4. Perhitungan Tulangan Geser Balok Portal Melintang

Dari Perhitungan SAP 2000 diperoleh momen terbesar pada batang nomor 278: Vu = 13607,10 kg = 136071,0 N f’c = 25 Mpa fy = 360 Mpa d = 440,5 Vc = 1 6 . c f .b .d = 1 6 . 25 .300.440,5 = 110125 N φ Vc = 0,75 . 110125 = 82593,75 N ½ Ø Vc = 0,5 . 82593,75 = 41296,87 N 3 φ Vc = 3 . 82593,75 = 247781,25 N Syarat tulangan geser : Ø Vc Vu 3Ø Vc : 82593,75 N 136071,0 N 247781,25 N Jadi diperlukan tulangan geser Ø Vs = Vu – Ø Vc = 136071,0 - 82593,75 = 53477,25 N Vs perlu = 75 , 53477,25 75 , = Vs φ = 71303 N Av = 2 . ¼ π 10 2 = 2 . ¼ . 3,14 . 100 = 157 mm 2 S = 17 , 349 71303 5 , 440 . 360 . 157 perlu Vs d . fy . Av = = mm commit to user Tugas Akhir Perencanaan Struktur Kafe dan Resto 2 Lantai

BAB 7 Portal