p. 1

(1)

<-- 200 Lebar balok (mm), b : <-- 400 Tinggi balok (mm), h : <-- 25 Selimut beton (mm), c_v : <-- 20 Mutu beton (MPa), f_c :

<-- 400 Kuat leleh baja tul. pokok (MPa), f_y :

<-- 240 tul. begel (MPa),f_ys :

<-- 16.0 Diameter tul. pokok (mm), D_tul :

<-- 8 Diameter tul. begel(mm), d_tul :

Gaya dalam terfaktor

<-- 100 Momen (kN.m), M_u : <-- 50 Geser (kN), V_u : <-- 25 Aksial (kN), P_u :

(2)

Faktor reduksi kekuatan

<-- Yes

digunakan default SKSNI-T15-1991-03(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 1 5 - 1 9 9 1 - 0 3 Oleh : Suyono Nt., 2001 ======================================================================= Data-data Dimensi balok : Lebar, b = 200.00 mm Tinggi, h = 400.00 mm Selimut beton, c_v = 25.00 mm Material :

Mutu beton silinder, f_c = 20.00 MPa

kubus, K = (f_c / 0.83) * 9.81 = 236.39 Kgf/cm^2 Kuat leleh baja tul. pokok, f_y = 400.00 MPa

tul. begel, f_ys = 240.00 MPa Diameter tul. pokok, D_tul = 16.00 mm

begel, d_tul = 8.00 mm Gaya dalam terfaktor :

Momen, M_u = 100.00 kN.m Geser, V_u = 50.00 kN Aksial, P_u = 25.00 kN

Faktor reduksi kekuatan, tinjauan :

- lentur, phi_b = 0.80 (sesuai SKSNI-T15-1991-03) - geser, phi_v = 0.60 (sesuai SKSNI-T15-1991-03) - aksial /+ lentur, phi_a = 0.65 (sesuai SKSNI-T15-1991-03) Faktor beta_1 yang tergantung mutu beton :

f_c = 20.00 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 = 201.06 mm^2

Perkiraan kebutuhan tulangan

Luas tul., A_sp = (M_u * 10^6 / (240 * h))) * (400/f_y)

= (100.00 * 10^6 / (240 * 400.00))) * (400/400.00) = 1041.67 mm^2

Jumlah, N_tulp = A_sp / A_sd = 1041.67 / 201.06 = 5.18 ~ 6 btg.

(3)

Jumlah tulangan maksimum dalam satu baris (s >= 25 mm)

N_maks1 = ((b - 2 * (c_v + d_tul + (D_tul/2))) / (25.0 + D_tul)) + 1 = ((200.00 - 2 * (25.00 + 8.00 + (16.00/2))) / (25.0 + 16.00) + 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*(30+8.00+(16.00/2)))+((6-3) *(25.00+8.00+16.00+30+(16.00/2))))/6 = 66.50 mm

Tingi effektif balok, d = h - d_s

= 400.00 - 66.50 = 333.50 mm Titik berat tulangan tekan,

d_s1 = c_v + d_tul + (D_tul/2) = 25.00 + 8.00 + (16.00/2) = 41.00 mm

Faktor momen pikul:

K_t = M_u * 10^6 / (phi_b * b * d^2) = 100.00 * 10^6 / (0.80 * 200.00 * 333.50^2) = 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*(600+400.00-225*0.85))/(600+400.00)^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))) * 333.50 = 96.49 mm

A_1 = (0.85 * f_c * a_1 * b) / f_y

= (0.85 * 20.00 * 96.49 * 200.00) / 400.00 = 820.19 mm^2

A_2 = ((K_t - K_1)*b*d^2)/((d-d_s1)*f_y)

= ((5.62 - 4.21)*200.00*333.50^2)/((333.50-41.00)*400.00) = 268.50 mm^2

Tulangan tarik, A_su = A_1 + A_2

(4)

digunakan, A_st = 6 D 16 mm

= 6 * (1/4) * pi * D_tul^2

= 1206.37 mm^2 > A_su (1088.70 mm^2) ... Okey Tulangan tekan, A_su1 = A_2

= 268.50 mm^2 digunakan, A_st = 2 D 16 mm

= 2 * (1/4) * pi * D_tul^2

= 402.12 mm^2 > A_su (268.50 mm^2) ... Okey Menghitung momen tersedia, Mt

A_st = 6 D 16 = 1206.37 mm^2 A_st1 = 2 D 16 = 402.12 mm^2 Rasio tulangan

rho = (A_st - A_st1) / (b * d)

= (1206.37 - 402.12) / (200.00 * 333.50) = 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) / ((600 + 400.00) * 400.00) = 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)/(600 - 400.00) = 2.55

Tinggi blok tekan beton, a = ((A_st - A_st1)*f_y)/(0.85 * f_c * b)

= ((1206.37 - 402.12)*400.00)/(0.85 * 20.00 * 200.00) = 94.62 mm

a_leleh = nu_l * d_s1 = 2.55 * 41.00 = 104.55 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 * 402.12 - 1206.37 * 400.00) / (1.7 * 20.00 * 200.00) = -35.48

q = (600 * beta_1 *d_s1 * A_st1 ) / (0.85 * f_c * b)

= (600 * 0.85 *41.00 * 402.12 ) / (0.85 * 20.00 * 200.00) = 2473.06

Tinggi blok tekan beton, a = sqrt (p^2 + q) - ( p )

= sqrt (-35.48 + 2473.06) - ( -35.48 ) = 96.57 mm

Tegangan baja tul., f_s1 = ((a - beta_1 * d_s1) / a) * 600

= ((96.57 - 0.85 * 41.00) / 96.57) * 600 = 383.48 MPa

Momen nominal dari beton, M_nc = 0.85 * f_c * a * b * (d - 0.5*a)

= 0.85 * 20.00 * 96.57 * 200.00 * (333.50 - 0.5*96.57) = 93648221.79 N.mm = 93.65 kN.m

(5)

dari baja, M_ns = A_st1 * f_s1 * (d - d_s1) = 402.12 * 383.48 * (333.50 - 41.00) = 45105005.92 N.mm = 45.11 kN.m dijumlahkan, M_n = M_nc + M_ns = 93.65 + 45.11 = 138.75 kN.m

Momen tersedia positif, M_t(+) = phi_b * M_n = 0.80 * 138.75

= 111.00 kN.m > M_u (100.00 kN.m) ... Okey Momen tersedia negatif :

Tulangan tarik (atas), A_st = 2 D 16.00 mm = 402.12 mm^2 Tulangan tekan (bawah), A_st1 = 6 D 16.00 mm = 1206.37 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 * 1206.37 - 402.12 * 400.00) / (1.7 * 20.00 * 200.00) = 82.79

q = (600 * beta_1 *d_s1 * A_st1 ) / (0.85 * f_c * b)

= (600 * 0.85 *41.00 * 1206.37 ) / (0.85 * 20.00 * 200.00) = 7419.19

Tinggi blok tekan beton, a = sqrt (p^2 + q) - ( p )

= sqrt (82.79 + 7419.19) - ( 82.79 ) = 36.68 mm

Tegangan baja tul., f_s1 = ((a - beta_1 * d_s1) / a) * 600

= ((36.68 - 0.85 * 41.00) / 36.68) * 600 = 29.95 MPa

Momen nominal dari beton, M_nc = 0.85 * f_c * a * b * (d - 0.5*a)

= 0.85 * 20.00 * 36.68 * 200.00 * (333.50 - 0.5*36.68) = 39305388.09 N.mm = 39.31 kN.m

dari baja, M_ns = A_st1 * f_s1 * (d - d_s1)

= 1206.37 * 29.95 * (333.50 - 41.00) = 10569094.20 N.mm = 10.57 kN.m dijumlahkan, M_n = M_nc + M_ns

= 39.31 + 10.57 = 49.87 kN.m

Momen tersedia negatif, M_t(-) = phi_b * M_n = 0.80 * 49.87 = 39.90 kN.m

Panjang penyaluran tul. deform tarik D_tul <= 36 mm

16.00 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 * 400.00 / sqrt(20.00) = 359.67 mm l_db > 0.06 * D_tul * f_y > 0.06 * 16.00 * 400.00 > 384.00 mm dipakai, l_db = 384.00 mm

(6)

Faktor pengali :

Jarak tul. ke tepi beton s = c_v = 25.00 mm s < 300 mm K_1 = 1.00 K_2 = 2 - 400/f_y = 2 - 400/400.00 = 1.00

s_tul < 150 and s_tepi < 70 K_3 = 1.00

K_4 = A_su / A_st = 1088.70 / 1206.37 = 0.90

Digunakan panjang penyaluran tulangan : l_d = l_db * K_1 * K_2 * K_3 * K_4 = 384.00 * 1.00 * 1.00 * 1.00 * 0.90 = 346.54 mm

Perhitungan tul. begel untuk Gaya geser

Gaya geser yg ditahan beton, V_c = sqrt(f_c)/6 * b * d

= sqrt(20.00)/6 * 200.00 * 333.50 = 49715.24 N

phi_v * V_c = 0.60 * 49715.24 = 29829.15 N nilai V_u >= phi_v * V_c 50000.00 > 29829.15

Gaya geser yang ditahan begel : V_s = (V_u - phi_v * V_c) / phi_v

= (50000.00 - 0.60 * 49715.24) / 0.60 = 33618.09 N

V_smaks = (2/3) * sqrt(f_c) * b * d

= (2/3) * sqrt(20.00) * 200.00 * 333.50 = 198860.98 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) = (33618.09 * 1000.00) / (240.00 * 333.50) = 420.02 mm^2 A_vmin = (b * S)/(3 * f_ys) = (200.00 * 1000.00)/(3 * 240.00) = 277.78 mm^2

dipilih A_vu = max(A_v,A_vmin) = max(420.02,277.78) = 420.02 mm^2

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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 = 100.53 mm^2

Jarak begel, s = (A_vs * 1000) / A_vu = (100.53 * 1000) / 420.02 = 239.35 mm

Persyaratan jarak begel V_s < sqrt(f_c) / 3 * b * d V_s < sqrt(20.00) / 3 * 200.00 * 333.50 33618.09 N < 99430.49 N, maka s = d / 2 = 333.50 / 2 = 166.75 mm s harus <= 600 mm

dipilih jarak s = min (239.35,166.75,600.00)

= 166.75 mm ~ 150.00 mm (dibulatkan)

Jadi digunakan begel 2 kaki d_tul = 8.00 mm jarak 150.00 mm

Tinjau gaya aksial terfaktor maksimum yg bekerja pada balok Batasan gaya aksial balok, P_maks = 0.1 * f_c * A_g

= 0.1 * 20.00 * 200.00 * 400.00 = 160000.00 N = 160.00 kN Kontrol :

P_u < P_maks

25.00 kN < 160.00 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(+) = 111.00 kN.m M_t(-) = 39.90 kN.m

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+---+ | O O | ---- 2 D 16 | | | || h (400 mm) | || | +--- Begel 2 d8 - 150 mm | || | | | O O O | | O O O | ---- 6 D 16 +---+ b (200 mm) c_v = 25.00 mm rho_tot = 2.01 % Daftar Pustaka :

* Ali Asroni, 1997 "Hitungan Beton Bertulang Berdasarkan SK SNI T-15-1991-03 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-15-1991-03", Seri Beton I, Penerbit Erlangga, Jakarta.