---
title: "Column 650x450 in double curvature"
subject: "Design of RCC Structure"
chapter: 10
key: "semester_7/design_of_rcc_structure:column-650x450-in-double-curvature"
source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/column-650x450-in-double-curvature
site: BCT Engineering
---

# Column 650x450 in double curvature

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

- Asked in: 2082 Bhadra, 12+2 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/column-650x450-in-double-curvature
- Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-10

## Question

Design a column of size 650 mm x 450 mm, subjected to a factored axial load of 1600 kN and factored moments of 90 kNm and 55 kNm at the top end and 35 kNm and 30 kNm at the bottom end with respect to the major axis and minor axis respectively. The effective length in respect of major and minor axis is 7 m and 6.5 m respectively. The column bends in double curvature. Use M30 concrete and Fe 500 TMT steel. Make construction drawing.

## Short answer

Column is slender about minor axis only; design moments are 90 kNm about major and 87.1 kNm about minor after adding reduced additional moment. Use 12-16 mm bars (A_sc=2413 mm²); biaxial interaction ratio 0.326 ≤ 1, so safe.

## Answer

#### Data and assumptions
- Column 450 mm (width $b$) × 650 mm (depth $D$); $P_u = 1600$ kN; $f_{ck} = 30$ MPa, $f_y = 500$ MPa. Clear cover 40 mm (IS 456 cl. 26.4.2.1), 8 mm ties.
- Moments: major-axis (x-x) moments 90 kN·m (top) and 35 kN·m (bottom); minor-axis (y-y) moments 55 kN·m (top) and 30 kN·m (bottom). The column bends in double curvature, so the smaller end moment is taken as negative.
- The unsupported length for the minimum eccentricity is taken equal to the larger effective length, 7 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) | 7000 | 650 | 10.77 | short |
| y-y (about minor) | 6500 | 450 | 14.44 | slender |
The limit for a short column is 12 (IS 456 cl. 25.1.2). The column is slender only about the minor axis, so an additional moment is added about that axis only (cl. 39.7.1).

#### Minimum eccentricity (cl. 25.4)
$$
e_{min} = \max\left(\frac{L}{500} + \frac{D}{30},\ 20\right)
$$
- About x-x: $e_x = 7000/500 + 650/30 = 35.7$ mm (min. 20 mm); adopt 35.7 mm; $M_{ux,min} = P_u e = 57.1$ kN·m
- About y-y: $e_y = 7000/500 + 450/30 = 29.0$ mm (min. 20 mm); adopt 29.0 mm; $M_{uy,min} = 46.4$ kN·m

#### Design moments
- Major axis (short about x-x): no additional moment; the design moment is the larger end moment, $M_{ux} = 90.0$ kN·m ($> P_u e_{x,min} = 57.1$ kN·m).
- Minor axis (slender): $M_i = 0.6M_2 + 0.4M_1 = 0.6(55) + 0.4(-30) = 21.0$ kN·m, not less than $0.4M_2 = 22.0$; adopt 22.0 kN·m (cl. 39.7.1).
Additional moment due to slenderness: $M_a = \dfrac{P_u b}{2000}\left(\dfrac{l_{ey}}{b}\right)^2 = \dfrac{1600 \times 0.45}{2000} \times 14.44^2 = 75.1$ kN·m, reduced by $k = \dfrac{P_{uz}-P_u}{P_{uz}-P_b} \le 1$.
- For the selected steel: $P_{uz} = 4821$ kN; $P_{by} = 1107$ kN, $k_y = 0.87$, $M_{ay} = 65.1$ kN·m.
- Design moments: $M_{ux} = 90.0$ kN·m, $M_{uy} = 22.0 + 65.1 = 87.1$ kN·m (not less than $M_2 = 55$ kN·m or the minimum-eccentricity values).

#### Longitudinal reinforcement
$\dfrac{P_u}{f_{ck} b D} = \dfrac{1600000}{30 \times 450 \times 650} = 0.182$
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 | $p$ (%) | $d'/D$ | $M_{ux1}$ | $M_{uy1}$ | $P_u/P_{uz}$ | $\alpha_n$ | $(M_{ux}/M_{ux1})^{\alpha_n}+(M_{uy}/M_{uy1})^{\alpha_n}$ |
|---|---|---|---|---|---|---|---|
| 12-16 φ | 0.82 | 0.086 | 492.3 | 325.0 | 0.332 | 1.22 | 0.326 ✓ |

#### Check
$P_{uz} = 0.45 f_{ck} A_c + 0.75 f_y A_{sc} = 0.45 \times 30 \times 290087 + 0.75 \times 500 \times 2413 = 4821$ kN
$\dfrac{P_u}{P_{uz}} = \dfrac{1600}{4821} = 0.332$, so $\alpha_n = 1.22$ (IS 456 cl. 39.6; linear between 1.0 at 0.2 and 2.0 at 0.8).
$$
\left(\frac{M_{ux}}{M_{ux1}}\right)^{\alpha_n} + \left(\frac{M_{uy}}{M_{uy1}}\right)^{\alpha_n} = \left(\frac{90.0}{492.3}\right)^{1.22} + \left(\frac{87.1}{325.0}\right)^{1.22} = 0.126 + 0.201 = 0.326 \le 1.0
$$
The interaction condition is satisfied; the section is **safe**.

Provide **12-16 mm** bars ($A_{sc} = 2413$ mm², $p = 0.82\%$), 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) $= \min(450, 256, 300) = 256$ mm → provide 8 mm ties @ 250 mm c/c.

#### Detailing
```text
CROSS-SECTION 450 x 650
    +-------------------------+
    |  o      o     o      o  |
    |                         |
    |                         |
    |  o                   o  |
    |                         |
    |                         |
    |                         |
    |  o                   o  |
    |                         |
    |                         |
    |  o      o     o      o  |
    +-------------------------+
   o = 16 mm bars, ties 8 mm @ 250 c/c
```

**Answer:** 450 × 650 mm column with 12-16 mm longitudinal bars ($p = 0.82\%$) 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 the 650 x 450 column with le = 7 m (major) and 6.5 m (minor), about which axis is the column slender?
   A. Major axis x-x only (le/D = 10.77)
   B. Minor axis y-y only (le/b = 14.44)
   C. Both axes
   D. Neither axis
   Answer: B. 7000/650 = 10.77 < 12 (short about x-x), while 6500/450 = 14.44 > 12 so it is slender about y-y. (Key point: Slender axis)

2. For the minor axis (M2 = 55, M1 = 30 kN·m, double curvature), what initial moment is adopted?
   A. 0.6(55) + 0.4(30) = 51 kN·m
   B. 21 kN·m
   C. 22 kN·m
   D. 55 kN·m
   Answer: C. In double curvature M1 is negative: 0.6(55) + 0.4(-30) = 21 kN·m. This is less than 0.4M2 = 22 kN·m, so 22 kN·m is adopted. (Key point: Initial moment in double curvature)

3. Why is the design moment Mux about the major axis taken as 57.1 kN·m?
   A. The axis is short, so no Ma is added, and the initial moment of 40 kN·m is below the minimum eccentricity moment Pu*emin = 57.1 kN·m
   B. It is the sum of 40 kN·m and the additional moment
   C. It is the larger end moment 90 kN·m reduced by 35
   D. It is 0.4 times 90 plus 35
   Answer: A. x-x is short so no additional moment applies. The design moment cannot be less than Pu*emin = 1600 x 0.0357 = 57.1 kN·m, which exceeds 40 kN·m. (Key point: Design moment about short axis)

4. How does the design moment Muy = 87.1 kN·m about the minor axis arise?
   A. Pu * emin = 46.4 kN·m
   B. 55 + 65.1 kN·m
   C. 21.0 + 65.1 kN·m
   D. 22.0 + 65.1 kN·m
   Answer: D. Muy = adopted initial moment (22.0) + reduced additional moment May (65.1) = 87.1 kN·m, which exceeds the minimum eccentricity moment. (Key point: Total moment for slender axis)

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