[ \frac{1}{P_{n,bi}} = \frac{1}{P_{nx}} + \frac{1}{P_{ny}} - \frac{1}{P_{n0}} ] [ \frac{1}{P_{n,bi}} = \frac{1}{2200} + \frac{1}{2300} - \frac{1}{6886} ] [ = 0.0004545 + 0.0004348 - 0.0001452 = 0.0007441 ] [ P_{n,bi} = 1344 , \text{kN} ]
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. 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). [ A_g = 300 \times 300 = 90,000 , \text{mm}^2 ] [ A_{st} = 0
From standard interaction curves, for (K_n = 0.62), (R_n \approx 0.12) is allowable. Our (R_n = 0.103 < 0.12) → OK . [ h = \sqrt{A_g} = \sqrt{68492} \approx 262
[ h = \sqrt{A_g} = \sqrt{68492} \approx 262 , \text{mm} ] Use 300 mm × 300 mm (common practical size).
[ 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 ]
(using interaction diagrams or simplified)