---
title: "Box Sal column 5 m, 400 kN"
subject: "Design of Steel and Timber Structure"
chapter: 11
key: "semester_6/design_of_steel_and_timber_structure:timber-box-column-sal-5m"
source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/timber-box-column-sal-5m
site: BCT Engineering
---

# Box Sal column 5 m, 400 kN

Design of Steel and Timber Structure · Chapter 11: Design of Compression Members

- Asked in: 2077 Chaitra, 10 marks
- Video: [Design of a box column (sal) || Timber structure || design of steel & timber structure || most VVI..](https://www.youtube.com/watch?v=Tl1k2yuH7YA) — Rabi 100
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/timber-box-column-sal-5m
- Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-11

## Question

Design a 5m long rectangular box columns built up by 5cm thick Sal wood planks to carry on axial load 400KN.

## Short answer

Provide box column 275 mm × 275 mm overall built from 50 mm thick Sal planks (inside 175 mm × 175 mm). Permissible compressive stress 9.04 N/mm², area 45000 mm², capacity 407 kN > 400 kN.

## Answer

**Approach.** Box (built-up) column designed to IS 883:1994 cl. 7.6.2: the planks are nailed or bolted to form a hollow square section (a square is a special case of the rectangular box); the strength depends on $S/\sqrt{d_1^2 + d_2^2}$, where $d_1$ is the overall side and $d_2$ the inside side of the box.

#### Data and permissible stresses

Length $S = 5.00$ m, axial load $N = 400$ kN, plank thickness $t = 50$ mm. Pin ends, effective length $= S$.

**Species:** Sal (Shorea robusta), Group A. **Location:** inside. **Grade:** Grade I (standard) (Table 1 values are for Grade I).

Permissible stresses (IS 883:1994 Table 1, inside location):

| Property | Table 1 value (N/mm²) | Used (N/mm²) |
|---|---|---|
| Bending / tension along grain $f_b$ | 16.9 | 16.90 |
| Compression parallel to grain $f_{cp}$ | 10.6 | 10.60 |
| Compression perpendicular to grain $f_{cn}$ | 4.6 | 4.60 |
| Horizontal shear $H$ | 0.94 | 0.94 |
| Modulus of elasticity $E$ | 12670 | 12670 |

Location is not stated, so **inside** and Grade I are assumed.

For plank thickness 50 mm (cl. 7.6.2.5): $U = 0.6$, $q = 1.0$ (values given for 25 mm and 50 mm planks).

#### Trial section

Square box, overall side $d_1 = 275$ mm, inside side $d_2 = 275 - 2\times50 = 175$ mm.
$$
A = d_1^2 - d_2^2 = 275^2 - 175^2 = 45,000\ \text{mm}^2
$$

$$
\begin{aligned} \frac{S}{\sqrt{d_1^2 + d_2^2}} &= \frac{5000}{\sqrt{275^2 + 175^2}} = \frac{5000}{326.0} = 15.34 \\ K_9 &= \frac{\pi}{2}\sqrt{\frac{UE}{5qf_{cp}}} = \frac{\pi}{2}\sqrt{\frac{0.6\times12670}{5\times1.0\times10.60}} = 18.81 \end{aligned}
$$

$8 < 15.34 < K_9 = 18.81$, so it is an **intermediate column**:

$$
f_c = q f_{cp}\left[1 - \frac{1}{3}\left(\frac{S}{K_9\sqrt{d_1^2 + d_2^2}}\right)^4\right] = 1.0\times10.60\left[1 - \frac{1}{3}\left(\frac{15.34}{18.81}\right)^4\right] = 9.04\ \text{N/mm}^2
$$

#### Check

$$
P = f_cA = 9.04\times45,000/10^3 = 407\ \text{kN} > N = 400\ \text{kN}
$$

Safe.

#### Arrangement

```text
   +------------------------------+
   |  plank 50 mm thick           |
   |   +----------------------+   |
   |   |        hollow        |   |   outside 275 x 275 mm
   |   |      175 x 175       |   |
   |   +----------------------+   |
   +------------------------------+
```
The 50 mm thick planks are fixed with staggered nails or bolts at a pitch not exceeding about 150 mm to 200 mm so that they act as one unit; the ends are fitted with solid end blocks.

#### Answer

Provide a box column of overall size **275 mm × 275 mm** formed with 50 mm thick Sal planks (inside 175 mm). Capacity 407 kN $>$ 400 kN.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For a 275 mm box with 50 mm planks, what is the net area?
   A. 275^2 - 225^2 = 25,000 mm^2
   B. 275^2 - 175^2 = 45,000 mm^2
   C. 275^2 - 175^2 = 90,000 mm^2
   D. 4 x 275 x 50 = 55,000 mm^2
   Answer: B. d2 = 275 - 100 = 175 mm, so A = 75,625 - 30,625 = 45,000 mm^2. (Key point: Inside side and area)

2. S/sqrt(d1^2 + d2^2) = 15.34, K9 = 18.81. What type is the column?
   A. Short
   B. Intermediate
   C. Long
   D. Spaced
   Answer: B. The ratio is between 8 and K9, so it is an intermediate column. (Key point: Classification)

3. Which formula gives fc for this column?
   A. q fcp[1 - (1/3)(S/(K9 sqrt(d1^2+d2^2)))^4]
   B. 0.329E/(S/d)^2
   C. fcp/(1 + S/d)
   D. q fcp only
   Answer: A. Intermediate box columns use the fourth-power reduction with K9, giving 9.04 N/mm^2. (Key point: Permissible stress formula)

4. What conclusion is reached?
   A. 407 kN > 400 kN, safe
   B. 407 kN > 400 kN, but only if U = 0.8
   C. 380 kN < 400 kN, unsafe
   D. 450 kN, so planks are unnecessary
   Answer: A. P = 9.04 x 45,000 = 407 kN, just above the 400 kN load. (Key point: Final check)

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