[ K_n = \frac{P_u}{\phi f'_c A_g} = \frac{1800 \times 10^3}{0.65 \times 28 \times 160000} = \frac{1.8 \times 10^6}{2.912 \times 10^6} \approx 0.62 ] [ R_n = \frac{M_u}{\phi f'_c A_g h} = \frac{120 \times 10^6}{0.65 \times 28 \times 160000 \times 400} = \frac{1.2 \times 10^8}{1.1648 \times 10^9} \approx 0.103 ]
[ A_g - 0.015 A_g = 0.985 A_g ] [ 0.85 \times 21 \times 0.985 A_g = 17.57 A_g ] [ 420 \times 0.015 A_g = 6.3 A_g ] Sum = (17.57 A_g + 6.3 A_g = 23.87 A_g) diseno de columnas de concreto armado ejercicios resueltos
[ P_u = 0.80 \phi [0.85 f' c (A_g - A {st}) + f_y A_{st}] ] [ 850 \times 10^3 = 0.80 \times 0.65 \left[0.85 \times 21 (A_g - 0.015A_g) + 420 \times 0.015 A_g \right] ] [ K_n = \frac{P_u}{\phi f'_c A_g} = \frac{1800
[ A_g = 300 \times 300 = 90,000 , \text{mm}^2 ] [ A_{st} = 0.015 \times 90,000 = 1350 , \text{mm}^2 ] Use 4 #19 bars (4 × 284 mm² = 1136 mm²) – slightly less, adjust to 4 #22 (4 × 387 = 1548 mm²). Check adequacy
[ h = \sqrt{A_g} = \sqrt{68492} \approx 262 , \text{mm} ] Use 300 mm × 300 mm (common practical size).
Section adequate. 4. Solved Exercise 3: Biaxial Bending (Approximate Method – Bresler’s Formula) Problem: A 500×500 mm column with (P_u = 1500 , \text{kN}), (M_{ux} = 100 , \text{kN·m}), (M_{uy} = 80 , \text{kN·m}). (f' c = 35 , \text{MPa}), (f_y = 420 , \text{MPa}), (A {st} = 3000 , \text{mm}^2) (symmetrical). Check adequacy.
[ 850 \times 10^3 = 0.80 \times 0.65 \times 23.87 A_g ] [ 850 \times 10^3 = 12.41 A_g ] [ A_g = 68,492 , \text{mm}^2 ]