Perhitungan Tulangan Balok Anak as C-

14 Perencanaan Struktur Hotel 2 lantai BAB I Pendahuluan d = h – p – ½ Ø = 250 – 40 – ½ 8 = 206 mm Vc = 1 6 . c f .b .d = 1 6 . 25 .170.206 = 29183,333 N Ø Vc = 0,6 . 29183,33 N = 17510 N 3 Ø Vc = 3 . 17510 = 52530 N Ø Vc Vu 3 Ø Vc, diperlukan tulangan geser Ø Vs = Vu - Ø Vc = 25850 – 17510 = 8340 Vs Perlu = 13900 6 , 8340 6 ,    s V Av = 2. ¼  8 2 = 2. ¼ . 3,14. 64 = 100,48 mm 2 S = 12 , 791 13900 456 . 240 . 48 , 100 . .   Vsper lu d fy Av mm S max = h2 = 125 2 250  mm Jadi dipakai sengkang dengan tulangan Ø 8 – 125

6.3.3 Perhitungan Tulangan Balok Anak as C-

C’ Data Perencanaan : h = 500 mm Ø t = 12 mm b = 340 mm` Ø s = 8 mm p = 40 mm d = h - p - 12 Ø t - Ø s fy = 380 Mpa = 500 – 40 – ½ .12 – 8 = 446 mm f’c = 25 MPa  Daerah Lapangan 15 Perencanaan Struktur Hotel 2 lantai BAB I Pendahuluan b =      fy 600 600 fy c. β 0,85.f =        380 600 600 85 , 380 25 . 85 , = 0,0291  max = 0,75 . b = 0,75 . 0,0291 = 0,0219  min = 00368 , 380 4 , 1 4 , 1   fy Dari Perhitungan SAP 2000 diperoleh : Mu = 4225,6 kgm= 4,226.10 7 Nmm Mn = φ M u = 8 , 10 . 226 , 4 7 = 5,283.10 7 Nmm Rn = 781 , 446 . 340 10 5,283. d . b Mn 2 7 2   m = 8824 , 17 25 . 85 , 380 . 85 ,   c f fy  =       fy 2.m.Rn 1 1 m 1 = 00174 , 380 781 , . 8824 , 17 . 2 1 1 8824 , 17 1          min   max  dipakai tulangan tunggal Digunakan  min = 0,00368 As perlu =  . b . d = 0,00368. 340 .446 = 558,035 mm 2 n = 2 12 4 1 perlu As  16 Perencanaan Struktur Hotel 2 lantai BAB I Pendahuluan = tulangan 5 937 , 4 04 , 113 035 , 558   Dipakai tulangan 5 D 10 mm As ada = 5 . ¼ .  . 12 2 = 5 . ¼ . 3,14 . 12 2 = 565,2mm 2 As perlu  Aman.. a =  b c f fy Asada . , 85 , . 727 , 29 340 . 25 . 85 , 380 . 2 , 565  Mn ada = As ada . fy d – a2 = 565,2 .380 446 – 29,7272 = 9,26.10 7 Nmm Mn ada Mn  Aman.. Jadi dipakai tulangan 5 D 12 mm  Daerah Tumpuan Dari Perhitungan SAP 2000 diperoleh : Mu = 8451,2 kgm= 8,451.10 7 Nmm Mn = φ M u = 8 , 10 . 451 , 8 7 = 10,564.10 7 Nmm Rn = 562 , 1 446 . 340 10 10,564. d . b Mn 2 7 2   m = 8824 , 17 25 . 85 , 380 . 85 ,   c f fy  =       fy 2.m.Rn 1 1 m 1 = 00428 , 380 562 , 1 . 8824 , 17 . 2 1 1 8824 , 17 1          min   max  dipakai tulangan tunggal Digunakan  = 0,00428 17 Perencanaan Struktur Hotel 2 lantai BAB I Pendahuluan As perlu =  . b . d = 0,00428. 340 . 446 = 649,019 mm 2 n = 2 10 . 4 1 perlu As  = tulangan 6 74 , 5 04 , 113 019 , 649   Dipakai tulangan 6 D 12 mm As ada = 6 . ¼ .  . 12 2 = 6 . ¼ . 3,14 . 12 2 = 678,24 mm 2 As perlu  Aman.. a =  b c f fy Asada . , 85 , . 135 , 34 340 . 25 . 85 , 380 . 24 , 678  Mn ada = As ada . fy d – a2 = 678,24 . 380 446 – 34,1352 = 11,055.10 7 Nmm Mn ada Mn  Aman.. Jadi dipakai tulangan 6 D 12 mm

4. Tulangan Geser Balok anak