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Transkripsi

1 ++ Desain Balok Beton Bertulang Pesegi S K S N I - T suyono.nt@gmail.com (2001) ++ *tekan <Enter> untuk nilai default (last sessions) Dimensi dan Mutu bahan < Lebar balok (mm), b : Tinggi balok (mm), h : Selimut beton (mm), c_v : Mutu beton (MPa), f_c : Kuat leleh baja tul. pokok (MPa), f_y : tul. begel (MPa),f_ys : Diameter tul. pokok (mm), D_tul : Diameter tul. begel(mm), d_tul : < <-- 25 <-- 20 < < < <-- 8 Gaya dalam terfaktor < Momen (kn.m), M_u : Geser (kn), V_u : Aksial (kn), P_u : <-- 50 < p. 1

2 Faktor reduksi kekuatan <-- Yes digunakan default SKSNI-T (Y/N) : ======================================================================= D e s a i n B a l o k B e t o n B e r t u l a n g S K S N I - T Oleh : Suyono Nt., 2001 ======================================================================= Data-data Dimensi balok : Lebar, b = mm Tinggi, h = mm Selimut beton, c_v = mm Material : Mutu beton silinder, f_c = MPa kubus, K = (f_c / 0.83) * 9.81 = Kgf/cm^2 Kuat leleh baja tul. pokok, f_y = MPa tul. begel, f_ys = MPa Diameter tul. pokok, D_tul = mm begel, d_tul = 8.00 mm Gaya dalam terfaktor : Momen, M_u = kn.m Geser, V_u = kn Aksial, P_u = kn Faktor reduksi kekuatan, tinjauan : - lentur, phi_b = 0.80 (sesuai SKSNI-T ) - geser, phi_v = 0.60 (sesuai SKSNI-T ) - aksial /+ lentur, phi_a = 0.65 (sesuai SKSNI-T ) Faktor beta_1 yang tergantung mutu beton : f_c = MPa. Untuk f_c <= 30 MPa, nilai beta_1 = 0.85 Luas satu tul. D16.00 mm, A_sd = (1/4)*pi*D_tul^2 = (1/4)*3.14*16.00^2 = mm^2 Perkiraan kebutuhan tulangan Luas tul., A_sp = (M_u * 10^6 / (240 * h))) * (400/f_y) = ( * 10^6 / (240 * ))) * (400/400.00) = mm^2 Jumlah, N_tulp = A_sp / A_sd = / = 5.18 ~ 6 btg. p. 2

3 Jumlah tulangan maksimum dalam satu baris (s >= 25 mm) N_maks1 = ((b - 2 * (c_v + d_tul + (D_tul/2))) / ( D_tul)) + 1 = (( * ( (16.00/2))) / ( ) + 1 = 3.88 ~ 3 btg. Jumlah lapis tulangan : N_lapis = N_tulp / N_maks = 6.00 / 3.00 = 2.00 ~ 2 lapis N_tulp > N_maks, diperkirakan 2 (dua) baris/lapis tulangan (s >= 25 mm) Titik berat tulangan tarik (asumsi 2 lapis tul.) d_s = ((N_maks*(30+d_tul+(D_tul/2)))+((N_tulp-N_maks) *(c_v+d_tul+d_tul+30+(d_tul/2))))/n_tulp = ((3*( (16.00/2)))+((6-3) *( (16.00/2))))/6 = mm Tingi effektif balok, d = h - d_s = = mm Titik berat tulangan tekan, d_s1 = c_v + d_tul + (D_tul/2) = (16.00/2) = mm Faktor momen pikul: K_t = M_u * 10^6 / (phi_b * b * d^2) = * 10^6 / (0.80 * * ^2) = 5.62 MPa K_maks = (382.5*beta_1*f_c*(600+f_y-225*beta_1))/(600+f_y)^2 = (382.5*0.85*20.00*( *0.85))/( )^2 = 5.26 MPa Kontrol : K_t > K_maks 5.62 MPa > 5.26 MPa, Analisa balok dgn tul. GANDA Diambil K_1 = 0.8 * K_maks = 0.8 * 5.26 = 4.21 MPa Tinggi blok tegangan tekan beton, a_t = (1 - sqrt(1-(2*k_1/0.85*f_c))) * d = (1 - sqrt(1-(2*4.21/0.85*20.00))) * = mm A_1 = (0.85 * f_c * a_1 * b) / f_y = (0.85 * * * ) / = mm^2 A_2 = ((K_t - K_1)*b*d^2)/((d-d_s1)*f_y) = (( )*200.00*333.50^2)/(( )*400.00) = mm^2 Tulangan tarik, A_su = A_1 + A_2 = = mm^2 p. 3

4 digunakan, A_st = 6 D 16 mm = 6 * (1/4) * pi * D_tul^2 = mm^2 > A_su ( mm^2)... Okey Tulangan tekan, A_su1 = A_2 = mm^2 digunakan, A_st = 2 D 16 mm = 2 * (1/4) * pi * D_tul^2 = mm^2 > A_su ( mm^2)... Okey Menghitung momen tersedia, Mt A_st = 6 D 16 = mm^2 A_st1 = 2 D 16 = mm^2 Rasio tulangan rho = (A_st - A_st1) / (b * d) = ( ) / ( * ) = 1.21 % rho_maks = 0.75 * rho_b = (382.5 * beta_1 * f_c) / ((600 + f_y) * f_y) = (382.5 * 0.85 * 20.00) / (( ) * ) = 1.63 % Kontrol : rho < rho_maks 1.21 % < 1.63 %... Okey Rasio batas tul. leleh, nu_l = (600 * beta_1)/(600 - f_y) = (600 * 0.85)/( ) = 2.55 Tinggi blok tekan beton, a = ((A_st - A_st1)*f_y)/(0.85 * f_c * b) = (( )*400.00)/(0.85 * * ) = mm a_leleh = nu_l * d_s1 = 2.55 * = mm a < a_leleh, tulangan tekan belum leleh Dihitung nilai a dan f_s1 sebagai berikut : p = (600 * A_st1 - A_st * f_y) / (1.7 * f_c * b) = (600 * * ) / (1.7 * * ) = q = (600 * beta_1 *d_s1 * A_st1 ) / (0.85 * f_c * b) = (600 * 0.85 *41.00 * ) / (0.85 * * ) = Tinggi blok tekan beton, a = sqrt (p^2 + q) - ( p ) = sqrt ( ) - ( ) = mm Tegangan baja tul., f_s1 = ((a - beta_1 * d_s1) / a) * 600 = (( * 41.00) / 96.57) * 600 = MPa Momen nominal dari beton, M_nc = 0.85 * f_c * a * b * (d - 0.5*a) = 0.85 * * * * ( *96.57) = N.mm = kn.m p. 4

5 dari baja, M_ns = A_st1 * f_s1 * (d - d_s1) = * * ( ) = N.mm = kn.m dijumlahkan, M_n = M_nc + M_ns = = kn.m Momen tersedia positif, M_t(+) = phi_b * M_n = 0.80 * = kn.m > M_u ( kn.m)... Okey Momen tersedia negatif : Tulangan tarik (atas), A_st = 2 D mm = mm^2 Tulangan tekan (bawah), A_st1 = 6 D mm = mm^2 A_st1 > A_st, maka tul. tekan belum leleh Dihitung nilai a dan f_s1 sebagai berikut : p = (600 * A_st1 - A_st * f_y) / (1.7 * f_c * b) = (600 * * ) / (1.7 * * ) = q = (600 * beta_1 *d_s1 * A_st1 ) / (0.85 * f_c * b) = (600 * 0.85 *41.00 * ) / (0.85 * * ) = Tinggi blok tekan beton, a = sqrt (p^2 + q) - ( p ) = sqrt ( ) - ( ) = mm Tegangan baja tul., f_s1 = ((a - beta_1 * d_s1) / a) * 600 = (( * 41.00) / 36.68) * 600 = MPa Momen nominal dari beton, M_nc = 0.85 * f_c * a * b * (d - 0.5*a) = 0.85 * * * * ( *36.68) = N.mm = kn.m dari baja, M_ns = A_st1 * f_s1 * (d - d_s1) = * * ( ) = N.mm = kn.m dijumlahkan, M_n = M_nc + M_ns = = kn.m Momen tersedia negatif, M_t(-) = phi_b * M_n = 0.80 * = kn.m Panjang penyaluran tul. deform tarik D_tul <= 36 mm mm < 36 mm, perhitungan sebagai berikut : l_db = 0.02 * A_b * f_y / sqrt(f_c) = 0.02 * (1/4) * pi * D_tul^2 * f_y / sqrt(f_c) = 0.02 * (1/4) * 3.14 * 16.00^2 * / sqrt(20.00) = mm l_db > 0.06 * D_tul * f_y > 0.06 * * > mm dipakai, l_db = mm p. 5

6 Faktor pengali : Jarak tul. ke tepi beton s = c_v = mm s < 300 mm K_1 = 1.00 K_2 = 2-400/f_y = 2-400/ = 1.00 s_tul < 150 and s_tepi < 70 K_3 = 1.00 K_4 = A_su / A_st = / = 0.90 Digunakan panjang penyaluran tulangan : l_d = l_db * K_1 * K_2 * K_3 * K_4 = * 1.00 * 1.00 * 1.00 * 0.90 = mm Perhitungan tul. begel untuk Gaya geser Gaya geser yg ditahan beton, V_c = sqrt(f_c)/6 * b * d = sqrt(20.00)/6 * * = N phi_v * V_c = 0.60 * = N nilai V_u >= phi_v * V_c > Gaya geser yang ditahan begel : V_s = (V_u - phi_v * V_c) / phi_v = ( * ) / 0.60 = N V_smaks = (2/3) * sqrt(f_c) * b * d = (2/3) * sqrt(20.00) * * = N V_s <= V_smaks, ukuran balok telah cukup (dapat dipakai) Luas begel perlu A_vu untuk 1 m. (S = 1000 mm) A_v = (V_s * S) / (f_ys * d) = ( * ) / ( * ) = mm^2 A_vmin = (b * S)/(3 * f_ys) = ( * )/(3 * ) = mm^2 dipilih A_vu = max(a_v,a_vmin) = max(420.02,277.78) = mm^2 p. 6

7 Dipilih begel 2 kaki, d_tul = 8.00 Perhitungan jarak begel Luas begel 2 kaki, A_vs = n * (1/4) * pi * d_tul^2 = 2 * (1/4) * 3.14 * 8.00^2 = mm^2 Jarak begel, s = (A_vs * 1000) / A_vu = ( * 1000) / = mm Persyaratan jarak begel V_s < sqrt(f_c) / 3 * b * d V_s < sqrt(20.00) / 3 * * N < N, maka s = d / 2 = / 2 = mm s harus <= 600 mm dipilih jarak s = min (239.35,166.75,600.00) = mm ~ mm (dibulatkan) Jadi digunakan begel 2 kaki d_tul = 8.00 mm jarak mm Tinjau gaya aksial terfaktor maksimum yg bekerja pada balok - Batasan gaya aksial balok, P_maks = 0.1 * f_c * A_g = 0.1 * * * = N = kn Kontrol : P_u < P_maks kn < kn, pengaruh gaya aksial dapat diabaikan Tinjau batasan perhitungan lanjut lendutan balok rho_def = 0.5 * rho_maks = 0.5 * 1.63 % = 0.81 % Tinjauan (1) rho > rho_def 1.21 % > 0.81 %, tinjauan lanjut perhitungan lendutan balok (mungkin) diperlukan Tinjauan (2) K_t > K_maks Kategori Balok Tul. GANDA, tinjauan lanjut perhitungan lendutan balok (mungkin) diperlukan Sketsa balok beton bertulang : Catatan : * Kategori Balok Tul. GANDA, M_t(+) = kn.m M_t(-) = kn.m p. 7

8 O O D 16 h (400 mm) Begel 2 d8-150 mm O O O O O O D b (200 mm) c_v = mm rho_tot = 2.01 % Daftar Pustaka : * Ali Asroni, 1997 "Hitungan Beton Bertulang Berdasarkan SK SNI T dengan Prinsip Daktilitas Tingkat I", Penerbit Keluarga Mahasiswa Teknik Sipil Universitas Muhammadiyah Surakarta, Surakarta. * Park, R. and Paulay, T., 1974 "Reinforced Concrete Structures", Department of Civil Engineering University of Canterbury New Zealand, John Wiley & Sons, New York. * Vis, W.C dan Kusuma, G.H., 1993 "Dasar-dasar Perencanaan Beton Bertulang Berdasarkan SK SNI T ", Seri Beton I, Penerbit Erlangga, Jakarta. tekan <Enter> untuk keluar. p. 8