[ 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)
[ h = \sqrt{A_g} = \sqrt{68492} \approx 262 , \text{mm} ] Use 300 mm × 300 mm (common practical size). diseno de columnas de concreto armado ejercicios resueltos
300×300 mm column, 4#22 longitudinal bars, #10 ties at 300 mm spacing. 3. Solved Exercise 2: Column Under Combined Axial Load and Uniaxial Bending Problem: Check if a 400×400 mm tied column with 8#25 bars (total (A_{st} = 8 \times 491 = 3928 , \text{mm}^2)) can resist: [ P_u = 1800 , \text{kN}, \quad M_u = 120 , \text{kN·m} ] Given: (f'_c = 28 , \text{MPa}), (f_y = 420 , \text{MPa}), cover = 40 mm. [ A_g - 0
[ 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 ] Solved Exercise 2: Column Under Combined Axial Load
Let (\rho_g = 0.015) (1.5% of (A_g)). [ A_{st} = 0.015 A_g ]