Pembebanan Perhitungan Tulangan Perhitungan Balok Anak as

commit to user 194 Tugas A khir Perencanaan St rukt ur dan Rencana A nggaran Biaya Gedung Puskesmas 2 L ant ai Bab 6 Perencanaan Balok A nak F G H 5 0 0 5 0 0 2 2 2 2

6.9. Perhitungan Balok Anak as

2 F-H

6.9.1. Pembebanan

Gambar 6.23. Lebar Equivalen Balok Anak as 2 F-H a. Beban Mati qD Pembebanan balok as 2 F-G = 2 G-H • Beban Plat = 2 x 1,32 x 404 kgm 2 = 1066,56 kgm qD 1 = 1066,56 kgm b. Beban hidup qL Beban hidup digunakan 250 kgm 2 qL 1 = 2 x 1,32 x 250 kgm = 660 kgm Bidang momen: Gambar 6.24. Bidang momen Balok Anak as 2 F-H Bidang geser: Gambar 6.25. Bidang geser Balok Anak as 2 F-H commit to user 195 Tugas A khir Perencanaan St rukt ur dan Rencana A nggaran Biaya Gedung Puskesmas 2 L ant ai Bab 6 Perencanaan Balok A nak

6.9.2. Perhitungan Tulangan

a Tulangan Lentur Balok Anak Data Perencanaan : h = 400 mm D t = 16 mm b = 200 mm` Ø s = 8 mm p = 40 mm d = h - p -12 Ø t - Ø s f’c = 20 MPa = 400 – 40 – ½ . 16– 8 fy = 400 Mpa = 344 mm ρb = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + fy 600 600 fy c. β 0,85.f = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + 400 600 600 85 , 400 20 . 85 , = 0,022 ρ max = 0,75 . ρb = 0,75 . 0,022 = 0,0165 ρ min = 0035 , 400 4 , 1 4 , 1 = = fy ¾ Daerah Tumpuan : Dari Perhitungan SAP 2000 diperoleh : Mu = 7267,76 kgm = 7,267×10 7 Nmm Mn = φ Mu = 8 , 10 267 , 7 7 × = 9,08×10 7 Nmm Rn = 8 , 3 344 00 2 10 08 , 9 d . b Mn 2 7 2 = × × = m = 53 , 23 20 . 85 , 400 . 85 , = = c f fy ρ = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − − fy 2.m.Rn 1 1 m 1 commit to user 196 Tugas A khir Perencanaan St rukt ur dan Rencana A nggaran Biaya Gedung Puskesmas 2 L ant ai Bab 6 Perencanaan Balok A nak = 0108 , 400 8 , 3 53 , 23 2 1 1 53 , 23 1 = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ × × − − ρ ρ min ρ ρ max Pakai tulangan tunggal Digunakan ρ = 0,0108 As perlu = ρ . b . d = 0,0108 × 200 × 344 = 743,04 mm 2 n = 2 16 π . 4 1 perlu As = tulangan 4 6 , 3 96 , 200 04 , 743 ≈ = As ada = n . ¼ . π . d 2 = 4 . ¼ . π . 16 2 = 803,84 As perlu → Aman.. a = = b c f fy Asada . . 85 , . 57 , 94 200 20 85 , 400 84 , 803 = × × × Mn ada = As ada . fy d – a2 = 803,84 . 400 344 – 94,572 = 9,54×10 7 Nmm Mn ada Mn → Aman.. Jadi dipakai tulangan 4 D 16 mm Kontrol Spasi : S = 1 - n sengkang 2 - tulangan n - 2p - b φ φ = 1 4 8 . 2 - 16 4. - 40 . 2 - 200 − = 13,3 mm 25 mm dipakai tulangan 2 lapis Karena jarak antar tulangan 25 mm maka kita rancang dengan tulangan berlapis dengan cara mencari d yang baru. commit to user 197 Tugas A khir Perencanaan St rukt ur dan Rencana A nggaran Biaya Gedung Puskesmas 2 L ant ai Bab 6 Perencanaan Balok A nak 4D 16 ∅8−100 40 20 2D 16 d 1 = h - p - 12 D t - Ø s = 400 – 40 – ½ . 16– 8 = 344 mm d 2 = h - p - Ø t - 12 D t – s - Ø s = 400 – 40 – 16 – ½ 16 – 30 - 8 = 298 mm d = n . d 2 1 n d n + = 4 298.2 344.2 + = 321 mm T = As ada . fy = 803,84. 400 = 321536 Mpa C = 0,85 . f’c . a . b T = C As . fy = 0,85 . f’c . a . b a = b c f fy As . . 85 , . = 200 . 20 . 85 , 321536 = 94,57 ØMn = Ø . T d – a2 = 0,85 . 321536 321 – 94,572 = 7,48 × 10 7 Nmm ØMn Mu → Aman.. 7,48 × 10 7 Nmm 7,267×10 7 Nmm Jadi dipakai tulangan 4 D 16 mm commit to user 198 Tugas A khir Perencanaan St rukt ur dan Rencana A nggaran Biaya Gedung Puskesmas 2 L ant ai Bab 6 Perencanaan Balok A nak ¾ Daerah Lapangan : Dari Perhitungan SAP 2000 diperoleh : Mu = 4102,10 kgm = 4,102×10 7 Nmm Mn = φ Mu = 8 , 10 102 , 4 7 × = 5,13×10 7 Nmm Rn = 17 , 2 44 3 00 2 10 13 , 5 d . b Mn 2 7 2 = × × = m = 53 , 23 20 . 85 , 400 . 85 , = = c f fy ρ = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − − fy 2.m.Rn 1 1 m 1 = 0058 , 400 17 , 2 53 , 23 2 1 1 53 , 23 1 = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ × × − − ρ ρ min ρ ρ max Pakai tulangan tunggal Digunakan ρ ada = 0,0058 As perlu = ρ . b . d = 0,0058 × 200 × 344 = 399,04 mm 2 n = 2 .16 π . 4 1 perlu As = tulangan 2 98 , 1 96 , 200 04 , 399 ≈ = As ada = n . ¼ . π . d 2 = 2 . ¼ . π . 16 2 = 401,92 As perlu → Aman.. a = = b c f fy Asada . . 85 , . 28 , 47 200 20 85 , 400 92 , 401 = × × × Mn ada = As ada . fy d – a2 = 401,92. 400 344 – 47,282 = 5,15×10 7 Nmm commit to user 199 Tugas A khir Perencanaan St rukt ur dan Rencana A nggaran Biaya Gedung Puskesmas 2 L ant ai Bab 6 Perencanaan Balok A nak 2D 16 ∅8−100 40 20 2D 16 Mn ada Mn → Aman.. 5,15×10 7 Nmm 5,13×10 7 Nmm Jadi dipakai tulangan 2 D 16 mm Kontrol Spasi : S = 1 - n sengkang 2 - tulangan n - 2p - b φ φ = 1 2 8 . 2 - 16 2. - 40 . 2 - 00 2 − = 72 mm 25 mmdipakai tulangan 1 lapis b Tulangan Geser Balok anak Dari perhitungan SAP 2000 Diperoleh : Vu = 7294,55 kgm = 72945,5 N f’c = 20 Mpa fy = 240 Mpa d = 344 mm Vc = 16 . c f .b .d = 16 . 20 .200 .344 = 51280,49 N Ø Vc = 0,6 . 51280,49 = 30768,29 N 3 Ø Vc = 3 . 30768,29 = 92304,88 N 5 Ø Vc = 5 . 30768,29 = 153841,45 N Syarat tulangan geser : ∅Vc Vu 3Ø Vc : 30768,29 N 80138,9 N 153841,45 N Jadi diperlukan tulangan geser: ∅ Vs = Vu - ∅ Vc commit to user 200 Tugas A khir Perencanaan St rukt ur dan Rencana A nggaran Biaya Gedung Puskesmas 2 L ant ai Bab 6 Perencanaan Balok A nak 500 500 500 400 500 500 500 A B C D E F G H 2 1 2 1 7 3 3 5 9 8 3 5 7 3 2 1 2 1 = 80138,9 – 30768,29 = 49370,61 N Vs perlu = 6 , Vsp φ = 6 , 49370,61 = 82284,35 N Av = 2 . ¼ π 8 2 = 2 . ¼ . 3,14 . 64 = 100,48 mm 2 S = 82 , 100 35 , 82284 344 240 48 , 100 perlu Vs d . fy . Av = × × = mm S max = d2 = 2 344 = 172 mm Dipakai Ø 8 – 100 mm : Vs ada = 29 , 82956 100 344 240 48 , 100 S d . fy . Av = × × = N Vs ada Vs perlu 82956,29 N 82284,35 N...... aman Jadi dipakai sengkang dengan tulangan Ø 8 – 100 mm

6.10. Perhitungan Balok Anak as 4’ A-H