Perhitungan Bagian Tapak Dinding

4.6.4.2. Perhitungan Bagian Tapak Dinding

1. Bagian Tapak Depan Toe q = q toe - X x B min maks    q = σ min + Uplift - γ 3 x H 4 - 4 b B min maks    Uplift = P = H uplift . γ w = 0 x 1 = 0 tm 2 Sehingga : q = [6,29 + 0 - 1,688 x 1,6] - X 7 , 8 29 , 6 744 , 40  = 3,588 - 3,96 X = 3,588 - 3,96 3,95 = -12,055 tm v =  9 , 1 qdx = 3,588 X - 1,98 X 2 = -16,721 t M =  9 , 1 vdx = 1,794 X 2 - 0,66 X 3 = -12,685 tm a. Perhitungan Tulangan Mu = 12,685 tm = 126852034,028 Nmm Mn =  Mu = 8 , 028 126852034, = 158565038,785 Nmm H 4 = 1600 mm f’c = 225 kgcm 2 = 22,5 Nmm 2 f y = 2400 kgcm 2 = 240 Nmm 2 Dicoba tulangan pokok Ø 28 mm d’ = tebal selimut beton = 50 mm Tabel 3. Dasar-Dasar Perencanaan Beton Bertulang, Seri Beton -1, W. C. Vis, Gideon Kusuma d = H 4 - d’ - ½ Ø = 1600 - 50 - 28 2 = 1536 mm K = R . d . b Mn 1 2 → R 1 = β 1 x f’c = 0,85 x 22,5 = 19,125 N mm 2 = 125 , 19 x 1536 x 1000 785 , 158565038 2 = 3,51 x 10 -3 F = 1 - √1 - 2k = 1 - √ 1 - 2 x 3,51 x 10 -3 = 3,52 x 10 -3 F maks = f 600 450 . y 1   = 240 600 450 x 85 ,  = 0,455 F F maks → tulangan tunggal under reinforced As = y 1 f R . d . b . F = 240 125 , 19 x 1536 x 1000 x 10 x 3,51 -3 = 430,894 mm 2 ρ = d . b As = 1536 x 1000 894 , 430 = 2,8 x 10 -4 ρ min = y f 4 , 1 = 240 4 , 1 = 5,833 x 10 -3 ρ maks = 0,036295 → Tabel-8 Dasar-Dasar Perencanaan Beton Bertulang, Seri Beton- 1, W.C. Vis, Gideon Kusuma atau ρ maks = β 1 . f R . f 600 450 y 1 y  = 0,85 x 240 600 450  x 19,125 240 = 0,03629 ρ ρ min → dipakai ρ min As min = ρ min .b.d = 5,833 x 10 -3 x 1000 x 1536 = 8960 mm 2 Tulangan pokok terpasang = Ø 28 - 70 As = 9594 mm 2 Cek : ρ = d . b As terpasang = 1536 x 1000 9594 = 6,25 x 10 -3 ρ min ρ ρ maks Dicoba tulangan bagi Ø 12 Tulangan bagi = 20 x tulangan pokok = 20 x 9594 = 1918,8 mm 2 Tulangan bagi terpasang = Ø 12 - 50 As = 2262 mm 2 2. Bagian Tapak Belakang Heel Konstruksi pelat yang ditumpu pada ketiga sisinya, dimana beban yang bekerja di atas heel mencakup beberapa beban sebagai berikut : a. Beban Merata q = 1,000 tm 2 b. Tanah 1 = γ 1 .h 1 = 1,664 x 4,5 = 7,489 tm 2 c. Tanah 2 = γ 2 .h 2 = 1,676 x 5 = 8,382 tm 2 d. Tanah 3 = γ 3 .h 3 = 1,688 x 0,85 = 1,435 tm 2 e. Berat Sendiri = γ 3 .h 3 - H 4 = 1,688 x 2,45 - 1,6 = 15,168 tm 2 q heel = 1 + 7,489 + 8,382 + 15,168 = 33,474 tm 2 q u = φ x q heel = 1,2 x 33,474 = 40,168 tm 2 Jarak antar counterfort = 0,3 - 0,6 H → H = 11,85 m → diambil = 7,11 m l y = panjang heel = jarak antar counterfort = 7,11 m l x = lebar heel = 3,95 m l y l x = 7,11 3,95 = 1,8 Maka momen untuk pelat yang terjepit pada tiga sisi adalah : M lx = 0,001.q u .l x 2 .x 1 M ly = 0,001.q u .l x 2 .x 2 M tx = -0,001.q u .l x 2 .x 3 M ty = -0,001.q u .l x 2 .x 4 M tiy = ½.M lx Dimana nilai x berturut-turut berdasarkan perbandingan l y l x = 1,8 adalah 54, 17, 82, dan 53 → Tabel Hal 26 Grafik dan Tabel Perhitungan Beton Bertulang, Seri Beton- 4, W.C. Vis, Gideon Kusuma M Ix = 0,001 x 40,168 x 3,95 2 x 54 = 33,843 tm M Iy = 0,001 x 40,168 x 3,95 2 x 17 = 10,654 tm M tx = -0,001 x 40,168 x 3,95 2 x 82 = -51,391 tm M ty = -0,001 x 40,168 x 3,95 2 x 53 = -33,216 tm M tiy = ½ x 33,843 = 16,922 tm Diambil momen yang maksimum a. Tulangan Arah X Tumpuan dan Lapangan Mu = M maks = M tx = 51,391 tm = 513914573,324 Nmm Mn =  Mu = 8 , 324 513914573, = 642393216,656 Nmm H 4 = 1600 mm f’c = 225 kgcm 2 = 22,5 Nmm 2 Iy Ix f y = 2400 kgcm 2 = 240 Nmm 2 Dicoba tulangan pokok Ø 28 mm d’ = tebal selimut beton = 50 mm Tabel 3. Dasar-Dasar Perencanaan Beton Bertulang, Seri Beton -1, W. C. Vis, Gideon Kusuma d = H 4 - d’ - ½ Ø = 1600 - 50 - 282 = 1536 mm K = R . d . b Mn 1 2 → R 1 = β 1 x f’c = 0,85 x 22,5 = 19,125 N mm 2 = 125 , 19 x 1536 x 1000 642393216 2 = 0,01424 F = 1 - √1 - 2k = 1 - √1 - 2 x 0,01424 = 0,01434 F maks = f 600 450 . y 1   = 240 600 450 x 85 ,  = 0,455 F F maks → tulangan tunggal under reinforced As = y 1 f R . d . b . F = 240 125 , 19 x 1536 x 1000 x 0,01434 = 1755,188 mm 2 ρ = d . b As = 1536 x 1000 188 , 1755 = 0,00114 ρ min = y f 4 , 1 = 240 4 , 1 = 5,833 x 10 -3 ρ maks = 0,036295 → Tabel-8 Dasar-Dasar Perencanaan Beton Bertulang, Seri Beton- 1, W.C. Vis, Gideon Kusuma atau ρ maks = β 1 . f R . f 600 450 y 1 y  = 0,85 x 240 600 450  x 19,125 240 = 0,03629 ρ ρ min → dipakai ρ min As min = ρ min .b.d = 5,833 x 10 -3 x 1000 x 1536 = 8960 mm 2 Tulangan pokok terpasang = Ø 28 - 70 As = 9594 mm 2 Cek : ρ = d . b As terpasang = 1536 x 1000 9594 = 6,25 x 10 -3 ρ min ρ ρ maks → OK b. Tulangan Arah Y Tumpuan dan Lapangan Mu = M ty = 33,216 tm = 332164297,393 Nmm Mn =  Mu = 8 , 393 332164297, = 415205371,741 Nmm H 4 = 1600 mm f’c = 225 kgcm 2 = 22,5 Nmm 2 f y = 2400 kgcm 2 = 240 Nmm 2 Dicoba tulangan pokok Ø 20 mm d’ = tebal selimut beton = 50 mm Tabel 3. Dasar-Dasar Perencanaan Beton Bertulang, Seri Beton -1, W. C. Vis, Gideon Kusuma d = H 4 - d’ - Ø tulangan arah X - ½ Ø = 1600 - 50 - 28 - 282 = 1508 mm K = R . d . b Mn 1 2 → R 1 = β 1 x f’c = 0,85 x 22,5 = 19,125 N mm 2 = 125 , 19 x 1508 x 1000 741 , 415205371 2 = 9,55 x 10 -3 F = 1 - √1 - 2k = 1 - √1 - 2 x 9,55 x 10 -3 = 9,59 x 10 -3 F maks = f 600 450 . y 1   = 240 600 450 x 85 ,  = 0,455 F F max → tulangan tunggal under reinforced As = y 1 f R . d . b . F = 240 125 , 19 x 1508 x 1000 x 10 x 9,59 -3 = 1152,729 mm 2 ρ = d . b As = 1508 x 1000 729 , 1152 = 7,6 x 10 -4 ρ min = y f 4 , 1 = 240 4 , 1 = 5,833 x 10 -3 ρ maks = 0,036295 → Tabel-8 Dasar-Dasar Perencanaan Beton Bertulang, Seri Beton- 1, W.C. Vis, Gideon Kusuma atau ρ maks = β 1 . f R . f 600 450 y 1 y  = 0,85 x 240 600 450  x 19,125 240 = 0,03629 ρ ρ min → dipakai ρ min As min = ρ min .b.d = 5,83 x 10 -3 x 1000 x 1508 = 8796,667 mm 2 Tulangan pokok terpasang = Ø 28 - 70 As = 9594 mm 2 Cek : ρ = d . b As terpasang = 1508 x 1000 9594 = 6,36 x 10 -3 ρ min ρ ρ maks → OK

4.6.4.3. Perhitungan Konstruksi Dinding Tegak