---
title: "Column 35x40 cm, not restrained against rotation"
subject: "Design of RCC Structure"
chapter: 10
key: "semester_7/design_of_rcc_structure:column-35x40-cm-not-restrained-against"
source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/column-35x40-cm-not-restrained-against
site: BCT Engineering
---

# Column 35x40 cm, not restrained against rotation

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

- Asked in: 2074 Asoj, 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/column-35x40-cm-not-restrained-against
- Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-10

## Question

A RC column of size 35 cm x 40 cm with unsupported length of 3.10 m is subjected to a factored axial load of 1500 kN and biaxial moments, $M_{ux}$ = 125 kNm and $M_{uy}$ = 88 kNm. The ends of the column are effectively held in position but not restrained against rotation. Design the column for longitudinal and transverse reinforcements and sketch the details. Use M25 Concrete and Fe500 grade steel.

## Short answer

Column is short since slenderness ratios are 7.75 and 8.86. Biaxial interaction check is safe with 20-16 mm longitudinal bars (Asc 4021 mm2, 2.87%) and 8 mm ties at 250 mm c/c.

## Answer

#### Data and assumptions
- Column 350 mm (width $b$) × 400 mm (depth $D$); $P_u = 1500$ 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) $M_{ux}$ = 125 kN·m; about minor axis (y-y) $M_{uy}$ = 88 kN·m.
- Held in position but not restrained against rotation at both ends: $l_e = L = 3.1$ m (IS 456 Table 28).
- 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) | 3100 | 400 | 7.75 | short |
| y-y (about minor) | 3100 | 350 | 8.86 | 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: $e_x = 3100/500 + 400/30 = 19.5$ mm (min. 20 mm); adopt 20.0 mm; $M_{ux,min} = P_u e = 30.0$ kN·m
- About y-y: $e_y = 3100/500 + 350/30 = 17.9$ mm (min. 20 mm); adopt 20.0 mm; $M_{uy,min} = 30.0$ kN·m

#### Design moments
- Design moments: $M_{ux} = 125.0$ kN·m, $M_{uy} = 88.0$ kN·m (not less than the minimum-eccentricity values).

#### Longitudinal reinforcement
$\dfrac{P_u}{f_{ck} b D} = \dfrac{1500000}{25 \times 350 \times 400} = 0.429$
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 | $p$ (%) | $d'/D$ | $M_{ux1}$ | $M_{uy1}$ | $P_u/P_{uz}$ | $\alpha_n$ | $(M_{ux}/M_{ux1})^{\alpha_n}+(M_{uy}/M_{uy1})^{\alpha_n}$ |
|---|---|---|---|---|---|---|---|
| 12-20 φ | 2.69 | 0.145 | 175.7 | 145.8 | 0.509 | 1.52 | 1.063 ✗ |
| 8-25 φ | 2.80 | 0.151 | 182.7 | 151.1 | 0.499 | 1.50 | 1.011 ✗ |
| 20-16 φ | 2.87 | 0.140 | 184.7 | 153.6 | 0.494 | 1.49 | 0.995 ✓ |

#### Check
$P_{uz} = 0.45 f_{ck} A_c + 0.75 f_y A_{sc} = 0.45 \times 25 \times 135979 + 0.75 \times 500 \times 4021 = 3038$ kN
$\dfrac{P_u}{P_{uz}} = \dfrac{1500}{3038} = 0.494$, so $\alpha_n = 1.49$ (IS 456 cl. 39.6; linear between 1.0 at 0.2 and 2.0 at 0.8).
$$
\left(\frac{M_{ux}}{M_{ux1}}\right)^{\alpha_n} + \left(\frac{M_{uy}}{M_{uy1}}\right)^{\alpha_n} = \left(\frac{125.0}{184.7}\right)^{1.49} + \left(\frac{88.0}{153.6}\right)^{1.49} = 0.995 \le 1.0
$$
The interaction condition is satisfied; the section is **safe**.

Provide **20-16 mm** bars ($A_{sc} = 4021$ mm², $p = 2.87\%$), 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) $= \min(350, 256, 300) = 256$ mm → provide 8 mm ties @ 250 mm c/c.

#### Detailing
```text
CROSS-SECTION 350 x 400
    +-------------------------+
    |                         |
    |   o   o  o   o  o   o   |
    |   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:** 350 × 400 mm column with 20-16 mm longitudinal bars ($p = 2.87\%$) 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. Ends held in position but not restrained against rotation: what effective length is taken?
   A. le = 0.65 L
   B. le = 0.5 L
   C. le = L
   D. le = 2 L
   Answer: C. IS 456 Table 28 gives le = L for both ends held in position but not restrained against rotation, so le = 3.1 m. (Key point: Effective length)

2. The ratios le/D are 7.75 and 8.86. What follows?
   A. The column is slender and needs additional moments
   B. The column is short since both are below 12
   C. The column must be redesigned
   D. Only the major axis is checked
   Answer: B. Below 12 on both axes, so no additional moment due to slenderness is required. (Key point: Short column conclusion)

3. Why was 20-16φ accepted but 8-25φ not?
   A. 20-16φ gave 0.995 <= 1, while 8-25φ gave 1.011 > 1
   B. 20-16φ has fewer bars
   C. 8-25φ exceeded 4% steel
   D. 20-16φ has larger area
   Answer: A. Only the 20-16φ arrangement brings the interaction sum to 1.0 or less. (Key point: Trial selection)

4. Why are 8 mm ties used with 16 mm bars?
   A. Tie diameter must be at least bar diameter
   B. Tie diameter must be at least the larger of φmax/4 and 6 mm
   C. Ties must be 12 mm
   D. Ties must be 16 mm
   Answer: B. Minimum is max(16/4, 6) = 6 mm, so an 8 mm tie is acceptable. (Key point: Tie diameter)

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