---
title: "Slender column 400x500, single curvature"
subject: "Design of RCC Structure"
chapter: 10
key: "semester_7/design_of_rcc_structure:slender-column-400x500-single-curvature"
source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/slender-column-400x500-single-curvature
site: BCT Engineering
---

# Slender column 400x500, single curvature

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

- Asked in: 2081 Bhadra, 12 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/slender-column-400x500-single-curvature
- Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-10

## Question

Design a RC slender column of size 400 mm x 500 mm with the following data: Effective lengths, $l_{ex}$ = 7.0 m, $l_{ey}$ = 6.0 m; Factored axial load = 1600 kN; Factored moments at top, $M_{ux}'$ = 180 kNm, $M_{uy}'$ = 110 kNm; Factored moments at bottom, $M_{ux}''$ = 100 kNm, $M_{uy}''$ = 50 kNm. The column bends in single curvature. Use M20 concrete and Fe 415 steel.

## Short answer

Slender about both axes; design moments after additional moments are 195.8 kNm and 128.8 kNm. Use 12-25 mm bars (A_sc=5890 mm², p=2.95%); biaxial interaction 0.900 ≤ 1, safe.

## Answer

#### Data and assumptions
- Column 400 mm (width $b$) × 500 mm (depth $D$); $P_u = 1600$ 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) $M_{ux}$ = 180 kN·m; about minor axis (y-y) $M_{uy}$ = 110 kN·m.
- The unsupported length for the minimum eccentricity is taken equal to the larger effective length, 7 m.
- Single curvature: both end moments have the same sign.
- 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) | 7000 | 500 | 14.00 | slender |
| y-y (about minor) | 6000 | 400 | 15.00 | slender |
The limit for a short column is 12 (IS 456 cl. 25.1.2). The column is slender about both axes, so additional moments are added about both (cl. 39.7.1).

#### Minimum eccentricity (cl. 25.4)
$$
e_{min} = \max\left(\frac{L}{500} + \frac{D}{30},\ 20\right)
$$
- About x-x: $e_x = 7000/500 + 500/30 = 30.7$ mm (min. 20 mm); adopt 30.7 mm; $M_{ux,min} = P_u e = 49.1$ kN·m
- About y-y: $e_y = 7000/500 + 400/30 = 27.3$ mm (min. 20 mm); adopt 27.3 mm; $M_{uy,min} = 43.7$ kN·m

#### Design moments
- Major axis: $M_i = 0.6M_2 + 0.4M_1 = 0.6(180) + 0.4(100) = 148.0$ kN·m, not less than $0.4M_2 = 72.0$; adopt 148.0 kN·m (cl. 39.7.1).
- Minor axis: $M_i = 0.6(110) + 0.4(50) = 86.0$ kN·m, not less than $0.4M_2 = 44.0$; adopt 86.0 kN·m.
Additional moment due to slenderness: $M_a = \dfrac{P_u D}{2000}\left(\dfrac{l_e}{D}\right)^2$ (78.4 kN·m about x-x, 72.0 kN·m about y-y), reduced by $k = \dfrac{P_{uz}-P_u}{P_{uz}-P_b} \le 1$.
- For the selected steel: $P_{uz} = 3580$ kN, $P_{bx} = 331$ kN, $k_x = 0.61$, $M_{ax} = 47.8$ kN·m; $P_{by} = 247$ kN, $k_y = 0.59$, $M_{ay} = 42.8$ kN·m.
- $P_b$ is much lower than the concrete force alone ($0.36 f_{ck} b x_{u,max} \approx 600$ kN) because, at the balanced strain, the bars in the tension half (the far face and the inner rows) carry a large tensile force that outweighs the compression steel.
- Design moments: $M_{ux} = 148.0 + 47.8 = 195.8$ kN·m, $M_{uy} = 86.0 + 42.8 = 128.8$ kN·m (not less than the larger end moments or the minimum-eccentricity values).

#### Longitudinal reinforcement
$\dfrac{P_u}{f_{ck} b D} = \dfrac{1600000}{20 \times 400 \times 500} = 0.400$
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 | $p$ (%) | $d'/D$ | $M_{ux1}$ | $M_{uy1}$ | $P_u/P_{uz}$ | $\alpha_n$ | $(M_{ux}/M_{ux1})^{\alpha_n}+(M_{uy}/M_{uy1})^{\alpha_n}$ |
|---|---|---|---|---|---|---|---|
| 10-25 φ | 2.45 | 0.121 | 267.4 | 217.5 | 0.487 | 1.48 | 1.071 ✗ |
| 16-20 φ | 2.51 | 0.116 | 282.4 | 211.7 | 0.482 | 1.47 | 1.050 ✗ |
| 12-25 φ | 2.95 | 0.121 | 325.3 | 241.4 | 0.447 | 1.41 | 0.900 ✓ |

#### Check
$P_{uz} = 0.45 f_{ck} A_c + 0.75 f_y A_{sc} = 0.45 \times 20 \times 194110 + 0.75 \times 415 \times 5890 = 3580$ kN
$\dfrac{P_u}{P_{uz}} = \dfrac{1600}{3580} = 0.447$, so $\alpha_n = 1.41$ (IS 456 cl. 39.6; linear between 1.0 at 0.2 and 2.0 at 0.8).
$$
\left(\frac{M_{ux}}{M_{ux1}}\right)^{\alpha_n} + \left(\frac{M_{uy}}{M_{uy1}}\right)^{\alpha_n} = \left(\frac{195.8}{325.3}\right)^{1.41} + \left(\frac{128.8}{241.4}\right)^{1.41} = 0.900 \le 1.0
$$
The interaction condition is satisfied; the section is **safe**.

Provide **12-25 mm** bars ($A_{sc} = 5890$ mm², $p = 2.95\%$), 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 500
    +-------------------------+
    |   o     o     o     o   |
    |                         |
    |                         |
    |   o                 o   |
    |                         |
    |                         |
    |                         |
    |   o                 o   |
    |                         |
    |                         |
    |   o     o     o     o   |
    +-------------------------+
   o = 25 mm bars, ties 8 mm @ 300 c/c
```

**Answer:** 400 × 500 mm column with 12-25 mm longitudinal bars ($p = 2.95\%$) 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 the 400 x 500 column with lex = 7.0 m and ley = 6.0 m, which statement is correct?
   A. Short about both axes
   B. Slender about x-x only
   C. Slender about y-y only
   D. Slender about both axes (14.0 and 15.0)
   Answer: D. 7000/500 = 14 and 6000/400 = 15, both above 12, so additional moments are required about both axes. (Key point: Slenderness check)

2. How is the initial moment Mi found about the major axis in single curvature (M2 = 180, M1 = 100)?
   A. Mi = 0.6(180) + 0.4(100) = 148 kN·m
   B. Mi = 0.6(180) - 0.4(100) = 68 kN·m
   C. Mi = 180 kN·m
   D. Mi = 100 kN·m
   Answer: A. Mi = 0.6M2 + 0.4M1 with both moments the same sign gives 148 kN·m, which is above 0.4M2 = 72. (Key point: Initial moment in single curvature)

3. How is Mux = 195.8 kN·m obtained?
   A. 180 + 47.8
   B. 148 + 47.8
   C. 148 + 49.1
   D. 100 + 47.8
   Answer: B. Mux = Mi + reduced Max = 148.0 + 47.8 = 195.8 kN·m. (Key point: Total design moment)

4. In the additional moment Ma = (Pu D/2000)(le/D)^2 * k, what does k = (Puz - Pu)/(Puz - Pb) do?
   A. Increases Ma when Pu exceeds Pb
   B. Equals 1 only for short columns
   C. Reduces Ma (k <= 1) because Pu is above the balanced load Pb
   D. Replaces the minimum eccentricity
   Answer: C. k is at most 1 and scales down Ma. Here kx = 0.61 and ky = 0.59, giving Max = 47.8 and May = 42.8 kN·m. (Key point: Reduction factor k)

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