--- title: "Biaxial column 400x600, Fe500" subject: "Design of RCC Structure" chapter: 10 key: "semester_7/design_of_rcc_structure:biaxial-column-400x600-fe500" source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/biaxial-column-400x600-fe500 site: BCT Engineering --- # Biaxial column 400x600, Fe500 Design of RCC Structure · Chapter 10: Design of compression members: Columns - Asked in: 2079 Baisakh, 10 marks - Video: [Design of a Rectangular Column subjected to combined axial loading and biaxial bending](https://www.youtube.com/watch?v=VL2MoB0n9Tg) — Stan Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/biaxial-column-400x600-fe500 - Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-10 ## Question Determine the reinforcement in short column having 4 m length fixed in both sides with bi-axially loaded having the following parameters: Size of column = 400 mm x 600 mm; Factored load, Pu = 1000 kN; Factored moment Mux = 125 kNm; Factored moment Muy = 200 kNm; M25 concrete and Fe500 steel ## Short answer Short column with Pu=1000 kN, Mux=125, Muy=200 kNm is safe. Provide 12-20 mm bars (Asc=3770 mm2, p=1.57%) and 8 mm ties at 300 mm c/c; biaxial interaction ratio is 0.95. ## Answer #### Data and assumptions - Column 400 mm (width $b$) × 600 mm (depth $D$); $P_u = 1000$ kN; $f_{ck} = 25$ MPa, $f_y = 500$ MPa. Clear cover 40 mm (IS 456 cl. 26.4.2.1), 8 mm ties. - Moments: about major axis (x-x) $M_{ux}$ = 125 kN·m; about minor axis (y-y) $M_{uy}$ = 200 kN·m. - Both ends fixed: $l_e = 0.65L = 2.6$ m. $M_{ux}=125$ kN·m acts about the major axis (depth 600 mm), $M_{uy}=200$ kN·m about the minor axis (depth 400 mm). - Sizes are given as $b \times D$ with $D \ge b$; moments $M_{ux}$ act about the major axis (depth $D$) and $M_{uy}$ about the minor axis (depth $b$). #### Slenderness | Axis | $l_e$ (mm) | Dimension (mm) | $l_e/$dim | Type | |---|---|---|---|---| | x-x (about major) | 2600 | 600 | 4.33 | short | | y-y (about minor) | 2600 | 400 | 6.50 | short | Both ratios are below 12, so the column is **short**; no additional moment is required. #### Minimum eccentricity (cl. 25.4) $$ e_{min} = \max\left(\frac{L}{500} + \frac{D}{30},\ 20\right) $$ - About x-x: $e_x = 4000/500 + 600/30 = 28.0$ mm (min. 20 mm); adopt 28.0 mm; $M_{ux,min} = P_u e = 28.0$ kN·m - About y-y: $e_y = 4000/500 + 400/30 = 21.3$ mm (min. 20 mm); adopt 21.3 mm; $M_{uy,min} = 21.3$ kN·m #### Design moments - Design moments: $M_{ux} = 125.0$ kN·m, $M_{uy} = 200.0$ kN·m (not less than the minimum-eccentricity values). #### Longitudinal reinforcement $\dfrac{P_u}{f_{ck} b D} = \dfrac{1000000}{25 \times 400 \times 600} = 0.167$ Bars are chosen so that $p$ lies between 0.8% (minimum) and 4% (practical maximum; IS 456 cl. 26.5.3.1 allows 6%). The uniaxial capacities $M_{ux1}$, $M_{uy1}$ at $P_u$ are read from the SP-16 interaction charts for rectangular sections; here they are computed by strain compatibility with the IS 456 stress block and the design stress-strain curve of steel (same basis as the charts), with effective cover $d' = 40 + 8 + \phi/2$. | Trial | $p$ (%) | $d'/D$ | $M_{ux1}$ | $M_{uy1}$ | $P_u/P_{uz}$ | $\alpha_n$ | $(M_{ux}/M_{ux1})^{\alpha_n}+(M_{uy}/M_{uy1})^{\alpha_n}$ | |---|---|---|---|---|---|---|---| | 10-20 φ | 1.31 | 0.097 | 395.3 | 267.7 | 0.260 | 1.10 | 1.007 ✗ | | 16-16 φ | 1.34 | 0.093 | 418.8 | 256.0 | 0.258 | 1.10 | 1.028 ✗ | | 12-20 φ | 1.57 | 0.097 | 457.9 | 277.6 | 0.246 | 1.08 | 0.950 ✓ | #### Check $P_{uz} = 0.45 f_{ck} A_c + 0.75 f_y A_{sc} = 0.45 \times 25 \times 236230 + 0.75 \times 500 \times 3770 = 4071$ kN $\dfrac{P_u}{P_{uz}} = \dfrac{1000}{4071} = 0.246$, so $\alpha_n = 1.08$ (IS 456 cl. 39.6; linear between 1.0 at 0.2 and 2.0 at 0.8). $$ \left(\frac{M_{ux}}{M_{ux1}}\right)^{\alpha_n} + \left(\frac{M_{uy}}{M_{uy1}}\right)^{\alpha_n} = \left(\frac{125.0}{457.9}\right)^{1.08} + \left(\frac{200.0}{277.6}\right)^{1.08} = 0.950 \le 1.0 $$ The interaction condition is satisfied; the section is **safe**. Provide **12-20 mm** bars ($A_{sc} = 3770$ mm², $p = 1.57\%$), within 0.8% to 4%. #### Lateral ties - Diameter $\ge \max(\phi_{max}/4,\ 6) = \max(20/4, 6) = 6$ mm → provide 8 mm. - Pitch $\le \min$(least lateral dimension, $16\phi_{min}$, 300 mm) $= \min(400, 320, 300) = 300$ mm → provide 8 mm ties @ 300 mm c/c. #### Detailing ```text CROSS-SECTION 400 x 600 +-------------------------+ | o o o o | | | | | | o o | | | | | | | | o o | | | | | | o o o o | +-------------------------+ o = 20 mm bars, ties 8 mm @ 300 c/c ``` **Answer:** 400 × 600 mm column with 12-20 mm longitudinal bars ($p = 1.57\%$) and 8 mm ties @ 300 mm c/c. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is no additional moment added for this biaxially loaded column? A. Both ends are fixed, so le = 0.65L = 2.6 m and the ratios 4.33 and 6.50 are below 12 B. Biaxial columns never have slenderness moments C. Because the moments are large D. Because Fe500 steel is used Answer: A. Fixed ends give le = 2.6 m. The column is short about both axes, so only the applied moments are used. (Key point: Short column) 2. For Pu/Puz = 0.246 (trial 12-20φ), what is αn? A. 1.00 B. 2.00 C. 1.08, interpolated between 1.0 at 0.2 and 2.0 at 0.8 D. 0.246 Answer: C. αn = 1 + (0.246 - 0.2)/0.6 = 1.08. (Key point: Interaction exponent) 3. Why was 12-20φ adopted over 10-20φ and 16-16φ? A. It uses the fewest bars B. The first two gave interaction sums just above 1 (1.007 and 1.028), while 12-20φ gives 0.950 C. It exceeds 4% steel D. It has the smallest d'/D Answer: B. The interaction sum must be <= 1. Only 12-20φ (p = 1.57%) satisfies it. 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