---
title: "Column design, 1350 kN, 3.4 m"
subject: "Design of RCC Structure"
chapter: 10
key: "semester_7/design_of_rcc_structure:column-design-1350-kn-3-4"
source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/column-design-1350-kn-3-4
site: BCT Engineering
---

# Column design, 1350 kN, 3.4 m

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

- Asked in: 2082 Baisakh, 12 marks
- Video: [Complete manual calculation of RCC column design by using IS 456 2000 code | structural design |](https://www.youtube.com/watch?v=ozWFZs3Cato) — Civil Engineering by Shravan
- 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/column-design-1350-kn-3-4
- Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-10

## Question

Design a column member to transfer the factored load of $P_u$ = 1350 kN, $M_{ux}$ = 150 kN m and $M_{uy}$ = 100 kNm having length 3.4 m unsupported. Use M25 concrete and Fe 500 steel. Use reinforcement distributed equally on all four sides and assume that the column is fully restrained at both the supports for displacement and rotation.

## Short answer

Short column with l_e=2.21 m; design moments 150 and 100 kNm. Use 8-25 mm bars (A_sc=3927 mm², p=2.45%) with 8 mm ties at 300 mm c/c; biaxial interaction 0.977 ≤ 1, safe.

## Answer

#### Data and assumptions
- Column 400 mm (width $b$) × 400 mm (depth $D$); $P_u = 1350$ 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) $M_{ux}$ = 150 kN·m; about minor axis (y-y) $M_{uy}$ = 100 kN·m.
- Both ends fully restrained for displacement and rotation: $l_e = 0.65L = 0.65 \times 3.4 = 2.21$ m (IS 456 Table 28).
- Section size is not given and is selected by trial.
- 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) | 2210 | 400 | 5.53 | short |
| y-y (about minor) | 2210 | 400 | 5.53 | 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: $e_x = 3400/500 + 400/30 = 20.1$ mm (min. 20 mm); adopt 20.1 mm; $M_{ux,min} = P_u e = 27.2$ kN·m
- About y-y: $e_y = 3400/500 + 400/30 = 20.1$ mm (min. 20 mm); adopt 20.1 mm; $M_{uy,min} = 27.2$ kN·m

#### Design moments
- Design moments: $M_{ux} = 150.0$ kN·m, $M_{uy} = 100.0$ kN·m (not less than the minimum-eccentricity values).

#### Longitudinal reinforcement
$\dfrac{P_u}{f_{ck} b D} = \dfrac{1350000}{25 \times 400 \times 400} = 0.338$
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 | $p$ (%) | $d'/D$ | $M_{ux1}$ | $M_{uy1}$ | $P_u/P_{uz}$ | $\alpha_n$ | $(M_{ux}/M_{ux1})^{\alpha_n}+(M_{uy}/M_{uy1})^{\alpha_n}$ |
|---|---|---|---|---|---|---|---|
| 16-16 φ | 2.01 | 0.140 | 184.6 | 184.6 | 0.455 | 1.42 | 1.162 ✗ |
| 12-20 φ | 2.36 | 0.145 | 204.9 | 204.9 | 0.426 | 1.38 | 1.024 ✗ |
| 8-25 φ | 2.45 | 0.151 | 212.9 | 212.9 | 0.418 | 1.36 | 0.977 ✓ |

#### Check
$P_{uz} = 0.45 f_{ck} A_c + 0.75 f_y A_{sc} = 0.45 \times 25 \times 156073 + 0.75 \times 500 \times 3927 = 3228$ kN
$\dfrac{P_u}{P_{uz}} = \dfrac{1350}{3228} = 0.418$, so $\alpha_n = 1.36$ (IS 456 cl. 39.6; linear between 1.0 at 0.2 and 2.0 at 0.8).
$$
\left(\frac{M_{ux}}{M_{ux1}}\right)^{\alpha_n} + \left(\frac{M_{uy}}{M_{uy1}}\right)^{\alpha_n} = \left(\frac{150.0}{212.9}\right)^{1.36} + \left(\frac{100.0}{212.9}\right)^{1.36} = 0.977 \le 1.0
$$
The interaction condition is satisfied; the section is **safe**.

Provide **8-25 mm** bars ($A_{sc} = 3927$ mm², $p = 2.45\%$), within 0.8% to 4%.

#### Lateral ties
- Diameter $\ge \max(\phi_{max}/4,\ 6) = \max(25/4, 6) = 6$ mm → provide 8 mm.
- Pitch $\le \min$(least lateral dimension, $16\phi_{min}$, 300 mm) $= \min(400, 400, 300) = 300$ mm → provide 8 mm ties @ 300 mm c/c.

#### Detailing
```text
CROSS-SECTION 400 x 400
    +-------------------------+
    |                         |
    |   o        o        o   |
    |                         |
    |                         |
    |                         |
    |   o                 o   |
    |                         |
    |                         |
    |                         |
    |   o        o        o   |
    |                         |
    +-------------------------+
   o = 25 mm bars, ties 8 mm @ 300 c/c
```

**Answer:** 400 × 400 mm column with 8-25 mm longitudinal bars ($p = 2.45\%$) 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. For a column fully restrained against displacement and rotation at both ends with L = 3.4 m, what is le and the column type?
   A. le = 0.8L = 2.72 m, slender
   B. le = 1.0L = 3.4 m, slender
   C. le = 0.65L = 2.21 m, short (ratio 5.53 < 12)
   D. le = 1.2L = 4.08 m, short
   Answer: C. Fixed-fixed ends give le = 0.65L = 2.21 m. le/D = 5.53 < 12, so the column is short and no additional moment is required. (Key point: Effective length and short column)

2. Why were the 16-16φ and 12-20φ trials rejected?
   A. Their steel percentage exceeded 4%
   B. The interaction sum was greater than 1 (1.162 and 1.024)
   C. The ties could not be provided
   D. Pu/Puz was above 0.8
   Answer: B. The biaxial check requires the sum of the ratios to be <= 1. Only 8-25φ gives 0.977, so it is adopted. (Key point: Trial rejection)

3. For Pu/Puz = 0.418, what value of αn is used?
   A. 1.00
   B. 2.00
   C. 0.418
   D. About 1.36 (linear between 1.0 at 0.2 and 2.0 at 0.8)
   Answer: D. αn = 1 + (0.418 - 0.2)/0.6 = 1.36. (Key point: Interaction exponent)

4. Why are the applied moments 150 and 100 kN·m used rather than the minimum eccentricity moments?
   A. Minimum eccentricity moments (27.2 kN·m) are smaller than the applied moments
   B. Minimum eccentricity is ignored for short columns
   C. Minimum eccentricity moments are larger and are added to the applied moments
   D. emin applies only to circular columns
   Answer: A. Design moments must be not less than Pu*emin = 27.2 kN·m. The given moments are larger, so they govern. (Key point: Minimum eccentricity vs applied moments)

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