--- title: "Short column 400x500, 1600 kN" subject: "Design of RCC Structure" chapter: 10 key: "semester_7/design_of_rcc_structure:short-column-400x500-1600-kn" source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/short-column-400x500-1600-kn site: BCT Engineering --- # Short column 400x500, 1600 kN Design of RCC Structure · Chapter 10: Design of compression members: Columns - Asked in: 2073 Shrawan, 15 marks - Video: [Design of a Axially Loaded Rectangular Short Column](https://www.youtube.com/watch?v=QJQv-WeEBMM) — 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/short-column-400x500-1600-kn - Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-10 ## Question Design the longitudinal reinforcements to be provided for a short column 400 x 500 mm subjected to the following forces: Pu = 1600 kN, $M_{ux}$ = 20.0 kN-m, $M_{uy}$ = 150 kN-m. Use M25 concrete and Fe415 steel. Unsupported length = 3 m. ## Short answer Column is short. Minimum eccentricity raises Mux to 36.3 kNm, Muy is 150 kNm. Provide 12-16 mm bars (Asc 2413 mm2, 1.21%) and 8 mm ties at 250 mm c/c; biaxial check is safe. ## Answer #### Data and assumptions - Column 400 mm (width $b$) × 500 mm (depth $D$); $P_u = 1600$ kN; $f_{ck} = 25$ MPa, $f_y = 415$ MPa. Clear cover 40 mm (IS 456 cl. 26.4.2.1), 8 mm ties. - Moments: about major axis (x-x) $M_{ux}$ = 20 kN·m; about minor axis (y-y) $M_{uy}$ = 150 kN·m. - 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) | 3000 | 500 | 6.00 | short | | y-y (about minor) | 3000 | 400 | 7.50 | 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: $e_x = 3000/500 + 500/30 = 22.7$ mm (min. 20 mm); adopt 22.7 mm; $M_{ux,min} = P_u e = 36.3$ kN·m - About y-y: $e_y = 3000/500 + 400/30 = 19.3$ mm (min. 20 mm); adopt 20.0 mm; $M_{uy,min} = 32.0$ kN·m #### Design moments - Design moments: $M_{ux} = 36.3$ kN·m (the given 20 kN·m is less than $P_u e_{x,min}$, so the minimum-eccentricity moment governs) and $M_{uy} = 150.0$ kN·m. #### Longitudinal reinforcement $\dfrac{P_u}{f_{ck} b D} = \dfrac{1600000}{25 \times 400 \times 500} = 0.320$ 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 | $p$ (%) | $d'/D$ | $M_{ux1}$ | $M_{uy1}$ | $P_u/P_{uz}$ | $\alpha_n$ | $(M_{ux}/M_{ux1})^{\alpha_n}+(M_{uy}/M_{uy1})^{\alpha_n}$ | |---|---|---|---|---|---|---|---| | 8-16 φ | 0.80 | 0.112 | 180.1 | 139.6 | 0.586 | 1.64 | 1.197 ✗ | | 10-16 φ | 1.01 | 0.112 | 191.9 | 155.9 | 0.561 | 1.60 | 1.010 ✗ | | 12-16 φ | 1.21 | 0.112 | 214.7 | 165.0 | 0.538 | 1.56 | 0.924 ✓ | #### Check $P_{uz} = 0.45 f_{ck} A_c + 0.75 f_y A_{sc} = 0.45 \times 25 \times 197587 + 0.75 \times 415 \times 2413 = 2974$ kN $\dfrac{P_u}{P_{uz}} = \dfrac{1600}{2974} = 0.538$, so $\alpha_n = 1.56$ (IS 456 cl. 39.6; linear between 1.0 at 0.2 and 2.0 at 0.8). $$ \left(\frac{M_{ux}}{M_{ux1}}\right)^{\alpha_n} + \left(\frac{M_{uy}}{M_{uy1}}\right)^{\alpha_n} = \left(\frac{36.3}{214.7}\right)^{1.56} + \left(\frac{150.0}{165.0}\right)^{1.56} = 0.924 \le 1.0 $$ The interaction condition is satisfied; the section is **safe**. Provide **12-16 mm** bars ($A_{sc} = 2413$ mm², $p = 1.21\%$), within 0.8% to 4%. #### Lateral ties - Diameter $\ge \max(\phi_{max}/4,\ 6) = \max(16/4, 6) = 6$ mm → provide 8 mm. - Pitch $\le \min$(least lateral dimension, $16\phi_{min}$, 300 mm) $= \min(400, 256, 300) = 256$ mm → provide 8 mm ties @ 250 mm c/c. #### Detailing ```text CROSS-SECTION 400 x 500 +-------------------------+ | o o o o | | | | | | o o | | | | | | | | o o | | | | | | o o o o | +-------------------------+ o = 16 mm bars, ties 8 mm @ 250 c/c ``` **Answer:** 400 × 500 mm column with 12-16 mm longitudinal bars ($p = 1.21\%$) and 8 mm ties @ 250 mm c/c. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Mux given is 20 kN·m but the design Mux is 36.3 kN·m. Why? A. A calculation error B. The minimum-eccentricity moment Pu x emin is larger than the applied moment C. Mux is doubled for safety D. Slenderness adds moment Answer: B. emin = 22.7 mm gives 1600 x 0.0227 = 36.3 kN·m, which exceeds 20 kN·m and so governs. (Key point: Minimum moment governs) 2. What is Pu/(fck b D) for this column? A. 0.320 B. 0.538 C. 1.21 D. 0.150 Answer: A. 1600000/(25 x 400 x 500) = 0.320. (Key point: Axial load ratio) 3. Why was 12-16φ finally chosen? A. It was the first trial tried B. It is the first trial whose interaction sum (0.924) is within 1 C. It has the minimum 0.8% steel D. It has the lowest αn Answer: B. 8-16φ (1.197) and 10-16φ (1.010) exceeded 1.0; 12-16φ gives 0.924. (Key point: Trial progression) 4. With Pu/Puz = 0.538, αn = 1.56. Which interpolation gives this? A. 1 + (0.538 - 0.2)/0.6 B. 2 - 0.538 C. 0.538 x 2 D. 1.0 always Answer: A. αn = 1 + (0.538 - 0.2)/(0.8 - 0.2) = 1.56. 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