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
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