Diseno De Columnas De Concreto Armado Ejercicios Resueltos < WORKING — Hacks >

From standard interaction curves, for (K_n = 0.62), (R_n \approx 0.12) is allowable. Our (R_n = 0.103 < 0.12) → OK .

[ 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 ] diseno de columnas de concreto armado ejercicios resueltos

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. From standard interaction curves, for (K_n = 0

Let (\rho_g = 0.015) (1.5% of (A_g)). [ A_{st} = 0.015 A_g ] Check adequacy

[ h = \sqrt{A_g} = \sqrt{68492} \approx 262 , \text{mm} ] Use 300 mm × 300 mm (common practical size).

[ 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)