Prediksi kekuatan Balok beton bertulang

Menghitung momen tersedia :

Titik berat tulangan tarik,
CG_tul = c_v + d_tul + (D_tul/2)
= 30.00 + 8.00 + (13.00/2)
= 44.00 mm
Tingi effektif balok,

d = h - CG_tul
= 400.00 - 44.00
= 356.00 mm

Rasio tulangan

rho = A_st / (b * d)

= 398.20 / (200.00 * 356.00) = 0.56 %

rho_maks = 0.75 * rho_b = (382.5 * beta_1 * f_c) / ((600 + f_y) * f_y)

= (382.5 * 0.85 * 21.15) / ((600 + 400.00) * 400.00) = 1.72 %

Kontrol : rho < rho_maks 0.56 % < 1.72 % ... Okey

Gaya tarik baja tul.,

T_s = A_st * f_y = 398.20 * 400.00 = 159278.75 N

Gaya tekan beton,

C_c = 0.85 * f_c * a_t * b

= 0.85 * 21.15 * a_t * 200.00 = 3595.50*a_t N

Persamaan kondisi kesetimbangan, T_s = C_c

Tinggi blok tekan beton,

a = 159278.75 / 3595.50 = 44.30 mm

Momen nominal,

M_n = T_s * (d - a/2)
= 159278.75 * (356.00 - 44.30/2)
= 53175252.00 N.mm = 53.18 kN.m



Perbandingan :

M & V dibuat incremental, Gaya N tidak ada

@ maksimum M = 45.8 kN.m, V = 45.8 kN, P = 0.0 kN

Vmaks < (1/6)*Sqrt(21.15)*200*356 = 54.57 kN

Gaya tarik baja atau tekan beton, R1 = 159/151 = ~5%

Lengan momen, R2 = 333.85 / 303.00 = ~10%

Momen nominal, R3 = 53.18/45.7 = ~16%



M dibuat incremental, Gaya V & N tidak ada

@ maksimum M = 71.6 kN.m, V = 0.0 kN, P = 0.0 kN

Gaya tarik baja atau tekan beton, R1 = 159/233 = ~32%

Lengan momen, R2 = 333.85 / 308.00 = ~8%

Momen nominal, R3 = 53.18/71.6 = ~26%



M, V & N dibuat incremental

@ maksimum M = 40.3 kN.m, V = 40.2 kN, P = 40.3 kN

Pmaks < 0.1*21.15*200*400 = 169.20 kN

Momen nominal, R3 = 53.18/40.39 = ~32%



M & N dibuat incremental, Gaya V tidak ada

@ maksimum M = 61.7 kN.m, V = 0.0 kN, P = 61.7 kN

Pmaks < 169.20 kN

Momen nominal, R3 = 53.18/61.7 = ~14%



Steel rebars and concrete properties



Concrete Materials



Rebar materials





... in draft