---
title: "Solid Sal column, 500 kN + 50 kNm, 2 m"
subject: "Design of Steel and Timber Structure"
chapter: 11
key: "semester_6/design_of_steel_and_timber_structure:timber-column-sal-select-500kn-50knm"
source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/timber-column-sal-select-500kn-50knm
site: BCT Engineering
---

# Solid Sal column, 500 kN + 50 kNm, 2 m

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

- Asked in: 2073 Magh, 7 marks
- 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-column-sal-select-500kn-50knm
- Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-11

## Question

Design a solid sal (Select grade) wood column to resist an axial load of 500 KN and moment of 50 KN-m. The length of column is 2 m.

## Short answer

Provide a solid Sal column of 225 mm × 375 mm (Select grade). S/d=8.9 (short), fc=12.30 N/mm²; combined stress ratio 0.99 < 1.

## Answer

**Approach.** Working stress design to IS 883:1994 (cl. 7.6 for columns, cl. 7.7 for combined axial compression and bending). Ends are taken as pin-ended, so the effective length equals the unsupported length $S$ (cl. 7.6.1.5).

#### Data and permissible stresses

Length $S = 2.00$ m = 2000 mm. Axial load $N = 500$ kN, bending moment $M = 50$ kN·m.

**Species:** Sal (Shorea robusta), Group A. **Location:** inside. **Grade:** Select grade (Grade I values multiplied by 1.16, IS 883:1994 cl. 6.3).

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 | 19.60 |
| Compression parallel to grain $f_{cp}$ | 10.6 | 12.30 |
| Compression perpendicular to grain $f_{cn}$ | 4.6 | 5.34 |
| Horizontal shear $H$ | 0.94 | 1.09 |
| Modulus of elasticity $E$ | 12670 | 12670 |

Location not stated: **inside**. Bending about the major axis; buckling about the minor axis.

#### Trial section

Try **225 mm × 375 mm**. Area $A = 84,375$ mm²; least dimension $d = 225$ mm.

#### Slenderness and permissible axial stress (cl. 7.6.1)

$$
\begin{aligned} S/d &= \frac{2000}{225} = 8.9 \\ K_8 &= 0.584\sqrt{\frac{E}{f_{cp}}} = 0.584\sqrt{\frac{12670}{12.30}} = 18.75 \end{aligned}
$$

$S/d \le 11$, so it is a **short column** and $f_c = f_{cp} = 12.30$ N/mm².

$S/d = 8.9 < 50$ (cl. 7.6.1.4). OK.

#### Combined axial compression and bending (cl. 7.7.1)

$$
\begin{aligned} f_{ac} &= \frac{N}{A} = \frac{500000}{84,375} = 5.926\ \text{N/mm}^2 \\ f_{ab} &= \frac{6M}{bd^2}\ \text{(bending about the strong axis)} = \frac{6\times50000000}{225\times375^2} = 9.481\ \text{N/mm}^2 \end{aligned}
$$

Permissible bending stress $f_b = 19.60$ N/mm²; depth 375 mm $> 300$ mm so the form factor $K = 0.81\,(d^2 + 89400)/(d^2 + 55000) = 0.952$ is applied: $f_b = 18.67$ N/mm².

$$
\frac{f_{ac}}{f_c} + \frac{f_{ab}}{f_b} = \frac{5.926}{12.30} + \frac{9.481}{18.67} = 0.990 \le 1.0
$$

Safe.

#### Answer

Provide a solid Sal column of **225 mm × 375 mm** cross-section (Select grade, inside location). Combined stress ratio $= 0.99 < 1$.

## Check yourself

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

1. S/d = 8.9. What is fc?
   A. fc = fcp = 12.30 N/mm^2, since S/d <= 11
   B. fc = 0.329E/(S/d)^2
   C. fc from the intermediate formula
   D. fc = 0.5 fcp
   Answer: A. For S/d <= 11 the column crushes and fc = fcp, 12.30 N/mm^2 for Select grade. (Key point: Short column)

2. Why is a form factor applied to fb?
   A. Depth 375 mm exceeds 300 mm, so K = 0.81(d^2 + 89400)/(d^2 + 55000) = 0.952 reduces fb
   B. The column is short
   C. The grade is Select
   D. The moment exceeds 40 kN-m
   Answer: A. Deeper members get a form factor; fb = 19.6 x 0.952 = 18.67 N/mm^2. (Key point: Form factor)

3. How is fab = 9.481 N/mm^2 obtained?
   A. 6M/(bd^2) with b = 225, d = 375
   B. N/A
   C. 6M/(db^2)
   D. M/(bd)
   Answer: A. 6 x 50 x 10^6/(225 x 375^2) = 9.481 N/mm^2. (Key point: Bending stress)

4. What interaction value is found?
   A. 0.99 <= 1, safe but close
   B. 1.35, unsafe
   C. 0.50, over-designed
   D. 0.99 > 1, unsafe
   Answer: A. 5.926/12.30 + 9.481/18.67 = 0.482 + 0.508 = 0.990. (Key point: Interaction)

## Related in Design of Steel and Timber Structure

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