---
title: "Biaxial column 400x500, equal steel all sides"
subject: "Design of RCC Structure"
chapter: 10
key: "semester_7/design_of_rcc_structure:biaxial-column-400x500-equal-steel-all"
source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/biaxial-column-400x500-equal-steel-all
site: BCT Engineering
---

# Biaxial column 400x500, equal steel all sides

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

- Asked in: 2066 Bhadra (old course), 15 marks
- 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-400x500-equal-steel-all
- Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-10

## Question

Determine the reinforcement equal in all sides of a biaxially loaded column with the following parameters. Size of column = 400 mm x 500 mm; Factored load, $P_u$ = 1200 kN; Factored moment $M_{ux}$ = 120 kNm; Factored moment $M_{uy}$ = 100 kNm. M20 concrete and Fe 415 steel. d'/D = 0.15 for both axes.

## Short answer

Provide 14-16 mm bars (1.41%) equally on all sides with 8 mm ties at 250 mm c/c for a 400×500 mm biaxially loaded column under 1200 kN and moments 120 and 100 kNm, with d'/D=0.15.

## Answer

#### Data and assumptions
- Column 400 mm (width $b$) × 500 mm (depth $D$); $P_u = 1200$ 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) $M_{ux}$ = 120 kN·m; about minor axis (y-y) $M_{uy}$ = 100 kN·m.
- Unsupported length not given; a short column with $L = 3$ m is assumed.
- $d'/D = 0.15$ in both directions: $d' = 0.15 \times 500 = 75$ mm about the major axis and $0.15 \times 400 = 60$ mm about the minor axis.
- 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; $M_{ux,min} = P_u e = 27.2$ kN·m
- About y-y: $e_y = 3000/500 + 400/30 = 19.3$ mm (min. 20 mm); adopt 20.0 mm; $M_{uy,min} = 24.0$ kN·m

#### Design moments
- Design moments: $M_{ux} = 120.0$ kN·m, $M_{uy} = 100.0$ kN·m (not less than the minimum-eccentricity values).

#### Longitudinal reinforcement
$\dfrac{P_u}{f_{ck} b D} = \dfrac{1200000}{20 \times 400 \times 500} = 0.300$
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), using the given $d'/D = 0.15$ about both axes.

| Trial | $p$ (%) | $d'/D$ | $M_{ux1}$ | $M_{uy1}$ | $P_u/P_{uz}$ | $\alpha_n$ | $(M_{ux}/M_{ux1})^{\alpha_n}+(M_{uy}/M_{uy1})^{\alpha_n}$ |
|---|---|---|---|---|---|---|---|
| 12-16 φ | 1.21 | 0.150 | 189.3 | 151.4 | 0.474 | 1.46 | 1.061 ✗ |
| 8-20 φ | 1.26 | 0.150 | 196.1 | 156.9 | 0.469 | 1.45 | 1.012 ✗ |
| 14-16 φ | 1.41 | 0.150 | 198.9 | 168.6 | 0.453 | 1.42 | 0.963 ✓ |

#### Check
$P_{uz} = 0.45 f_{ck} A_c + 0.75 f_y A_{sc} = 0.45 \times 20 \times 197185 + 0.75 \times 415 \times 2815 = 2651$ kN
$\dfrac{P_u}{P_{uz}} = \dfrac{1200}{2651} = 0.453$, so $\alpha_n = 1.42$ (IS 456 cl. 39.6; linear between 1.0 at 0.2 and 2.0 at 0.8).
$$
\left(\frac{M_{ux}}{M_{ux1}}\right)^{\alpha_n} + \left(\frac{M_{uy}}{M_{uy1}}\right)^{\alpha_n} = \left(\frac{120.0}{198.9}\right)^{1.42} + \left(\frac{100.0}{168.6}\right)^{1.42} = 0.963 \le 1.0
$$
The interaction condition is satisfied; the section is **safe**.

Provide **14-16 mm** bars ($A_{sc} = 2815$ mm², $p = 1.41\%$), 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     o   |
    |                         |
    +-------------------------+
   o = 16 mm bars, ties 8 mm @ 250 c/c
```

**Answer:** 400 × 500 mm column with 14-16 mm longitudinal bars ($p = 1.41\%$) 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. d'/D = 0.15 is given in both directions. What are d' about each axis?
   A. 75 mm about the major axis and 60 mm about the minor axis
   B. 60 mm about both
   C. 75 mm about both
   D. 40 mm about both
   Answer: A. d' = 0.15 x 500 = 75 mm (D = 500) and 0.15 x 400 = 60 mm (b = 400). (Key point: Using d'/D)

2. Equal reinforcement on all sides means
   A. Bars only on two faces
   B. Bars distributed around the whole perimeter, so uniaxial capacities exist about both axes
   C. Bars only at corners
   D. Bars only along the long face
   Answer: B. Biaxial bending needs steel on all faces so that both Mux1 and Muy1 can be provided. (Key point: Steel on all sides)

3. Pu/Puz = 1200/2651 = 0.453. alpha_n is
   A. 1.42
   B. 1.00
   C. 1.85
   D. 0.45
   Answer: A. alpha_n = 1 + (0.453 - 0.2)/0.6 = 1.42. (Key point: Alpha_n value)

4. Which trial satisfies the biaxial check?
   A. 12-16 (1.061)
   B. 8-20 (1.012)
   C. 14-16 (0.963)
   D. None
   Answer: C. Only 14-16 mm gives a sum <= 1.0 (0.963), with p = 1.41%. (Key point: Trial acceptance)

5. The tie pitch limit is min(400, 256, 300) = 256. What is provided?
   A. 300 mm
   B. 250 mm
   C. 400 mm
   D. 256 mm exactly with a 6 mm tie
   Answer: B. Pitch must not exceed 256 mm, so a practical 250 mm is provided with 8 mm ties. (Key point: Tie pitch with 16 mm bars)

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