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.

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

[ 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

[ 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).

From standard interaction curves, for (K_n = 0.62), (R_n \approx 0.12) is allowable. Our (R_n = 0.103 < 0.12) → OK . 300×300 mm column, 4#22 longitudinal bars, #10 ties

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

[ \gamma = \frac{\text{distance between bar centers}}{\text{total depth}} \approx \frac{400 - 2 \times 40}{400} = 0.80 ] [ \rho_g = \frac{A_{st}}{A_g} = \frac{3928}{160000} = 0.0245 = 2.45% ] Let (\rho_g = 0

[ 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 ]