---
title: "Biaxial short square column, 800 kN"
subject: "Design of RCC Structure"
chapter: 10
key: "semester_7/design_of_rcc_structure:biaxial-short-square-column-800-kn"
source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/biaxial-short-square-column-800-kn
site: BCT Engineering
---

# Biaxial short square column, 800 kN

Design of RCC Structure · Chapter 10: Design of compression members: Columns

- Asked in: 2078 Bhadra, 10 marks
- Video: [Design of a Square Column subjected to combined axial loading and biaxial bending](https://www.youtube.com/watch?v=ZTv--3b0gVs) — 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-short-square-column-800-kn
- Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-10

## Question

Determine the longitudinal and transverse reinforcements to be provided in a biaxially loaded short square shaped RCC column with following data: Size of column = 400 x 400 mm. Ultimate factored axial load = 800 kN. Inclusive of live load at an eccentricity of 80 mm in both X and Y direction. Use concrete grade of M20 and steel grade of Fe415.

## Short answer

For 400x400 mm column, Pu=800 kN with 80 mm eccentricity in both directions, provide 8-16 mm bars (Asc=1608 mm2, p=1.01%) and 8 mm ties at 250 mm c/c; interaction ratio is 0.821.

## Answer

#### Data and assumptions
- Column 400 mm (width $b$) × 400 mm (depth $D$); $P_u = 800$ kN; $f_{ck} = 20$ 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) $M_{ux}$ = 64 kN·m; about minor axis (y-y) $M_{uy}$ = 64 kN·m.
- Moments from eccentricity: $M_{ux} = M_{uy} = P_u e = 800 \times 0.08 = 64$ kN·m.
- Column is short; unsupported length 3 m assumed for the minimum-eccentricity check.
- 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 | 400 | 7.50 | 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: $e_x = 3000/500 + 400/30 = 19.3$ mm (min. 20 mm); adopt 20.0 mm; $M_{ux,min} = P_u e = 16.0$ kN·m
- About y-y: $e_y = 3000/500 + 400/30 = 19.3$ mm (min. 20 mm); adopt 20.0 mm; $M_{uy,min} = 16.0$ kN·m

#### Design moments
- Design moments: $M_{ux} = 64.0$ kN·m, $M_{uy} = 64.0$ kN·m (not less than the minimum-eccentricity values).

#### Longitudinal reinforcement
$\dfrac{P_u}{f_{ck} b D} = \dfrac{800000}{20 \times 400 \times 400} = 0.250$
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 | 1.01 | 0.140 | 123.2 | 123.2 | 0.415 | 1.36 | 0.821 ✓ |

#### Check
$P_{uz} = 0.45 f_{ck} A_c + 0.75 f_y A_{sc} = 0.45 \times 20 \times 158392 + 0.75 \times 415 \times 1608 = 1926$ kN
$\dfrac{P_u}{P_{uz}} = \dfrac{800}{1926} = 0.415$, so $\alpha_n = 1.36$ (IS 456 cl. 39.6; linear between 1.0 at 0.2 and 2.0 at 0.8).
$$
\left(\frac{M_{ux}}{M_{ux1}}\right)^{\alpha_n} + \left(\frac{M_{uy}}{M_{uy1}}\right)^{\alpha_n} = \left(\frac{64.0}{123.2}\right)^{1.36} + \left(\frac{64.0}{123.2}\right)^{1.36} = 0.821 \le 1.0
$$
The interaction condition is satisfied; the section is **safe**.

Provide **8-16 mm** bars ($A_{sc} = 1608$ mm², $p = 1.01\%$), 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 400
    +-------------------------+
    |                         |
    |   o        o        o   |
    |                         |
    |                         |
    |                         |
    |   o                 o   |
    |                         |
    |                         |
    |                         |
    |   o        o        o   |
    |                         |
    +-------------------------+
   o = 16 mm bars, ties 8 mm @ 250 c/c
```

**Answer:** 400 × 400 mm column with 8-16 mm longitudinal bars ($p = 1.01\%$) 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. For Pu = 800 kN acting at 80 mm eccentricity in both X and Y directions, what design moments are used?
   A. Mux = 80 kN·m and Muy = 80 kN·m
   B. Mux = 64 kN·m and Muy = 64 kN·m
   C. Mux = 64 kN·m and Muy = 0
   D. Mux = 10 kN·m and Muy = 10 kN·m
   Answer: B. Moment = Pu x e = 800 x 0.08 = 64 kN·m about each axis, since the load is eccentric in both directions. (Key point: Moment from eccentricity)

2. With le = 3000 mm and a 400 mm square section, why is the column treated as short?
   A. le/D = 7.5, which is below 12
   B. le/D = 7.5, which is above 3
   C. Because the load is below 1000 kN
   D. Because the minimum eccentricity is below 20 mm
   Answer: A. A column is short when le/D is less than 12 on both axes, so no additional slenderness moment is added. (Key point: Short column check)

3. Which condition must be satisfied for a biaxially loaded short column to be safe?
   A. Mux/Mux1 + Muy/Muy1 <= 1 exactly with power 1
   B. (Mux/Mux1)^αn + (Muy/Muy1)^αn <= 1
   C. Mux1 + Muy1 <= Pu
   D. Pu/Puz must be at least 0.8
   Answer: B. IS 456 cl. 39.6 requires the sum of the moment ratios each raised to αn to be not more than 1.0. (Key point: Biaxial interaction check)

4. Pu/Puz = 0.415 gives αn = 1.36. How is αn obtained?
   A. It is fixed at 2.0 for all columns
   B. It is taken equal to Pu/Puz
   C. Linear interpolation from 1.0 at Pu/Puz = 0.2 to 2.0 at 0.8
   D. Linear interpolation from 0 at Pu/Puz = 0 to 1.0 at 1.0
   Answer: C. IS 456 varies αn linearly between 1.0 (Pu/Puz <= 0.2) and 2.0 (Pu/Puz >= 0.8). (Key point: Exponent αn)

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