---
title: "Safe axial load on a timber column"
subject: "Design of Timber and Masonry Structures"
chapter: 3
key: "semester_5/design_of_timber_and_masonry_structures:axial-capacity-timber-column"
source: https://www.bctengineer.com/q/semester_5/design_of_timber_and_masonry_structures/axial-capacity-timber-column
site: BCT Engineering
---

# Safe axial load on a timber column

Design of Timber and Masonry Structures · Chapter 3: Structural Elements of Timber Structures

- Asked in: year not given, 6 marks
- Video: [Timber Column Design](https://www.youtube.com/watch?v=3EzMBYJbG2E) — Padmalochan Behera
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/design_of_timber_and_masonry_structures/axial-capacity-timber-column
- Whole chapter: https://www.bctengineer.com/md/semester_5/design_of_timber_and_masonry_structures/chapter-3

## Question

A timber column of rectangular section 100 mm × 150 mm and length 3.0 m has both ends hinged. Take the permissible compressive stress parallel to grain $\sigma_{cc} = 7.5$ N/mm² and modulus of elasticity $E = 9500$ N/mm². Using IS 883 type expressions, (a) classify the column and find the safe axial load, and (b) find the safe load if one end is fixed and the other is hinged (effective length factor 0.8).

## Short answer

K8 =24.98. (a) Hinged-hinged: l_e =3000 mm, l_e/d =30 > K8 − long column; σ_c' =0.329×9500/30² =3.47 N/mm²; load =52.1 kN. (b) Fixed-hinged: l_e =2400 mm, l_e/d =24 − intermediate; σ_c' =7.5[1-1/3(24/24.98)⁴] =5.37 N/mm²; load =80.6 kN.

## Answer

#### Formulas (IS 883)
Slenderness ratio $\lambda = l_e/d$, with $d$ the least dimension. Define

$$
K_8 = 0.702\sqrt{\frac{E}{\sigma_{cc}}}
$$

- Short column: $l_e/d \le 11$: $\sigma_c' = \sigma_{cc}$
- Intermediate: $11 < l_e/d < K_8$: $\sigma_c' = \sigma_{cc}\left[1 - \frac{1}{3}\left(\frac{l_e/d}{K_8}\right)^4\right]$
- Long: $l_e/d \ge K_8$: $\sigma_c' = \dfrac{0.329\,E}{(l_e/d)^2}$

#### Value of $K_8$
$$
K_8 = 0.702\sqrt{\frac{9500}{7.5}} = 0.702 \times 35.59 = 24.98
$$

Area $A = 100 \times 150 = 15000$ mm²; least dimension $d = 100$ mm.

#### (a) Both ends hinged
$l_e = L = 3000$ mm, so $l_e/d = 30 > K_8 = 24.98$: **long column**.

$$
\begin{aligned}
\sigma_c' &= \frac{0.329 \times 9500}{30^2} = 3.473\ \text{N/mm}^2 \\
P &= \sigma_c' A = 3.473 \times 15000 \approx 52\,100\ \text{N} \approx 52.1\ \text{kN}
\end{aligned}
$$

#### (b) One end fixed, other hinged
$l_e = 0.8 \times 3000 = 2400$ mm, so $l_e/d = 24 < 24.98$: **intermediate column**.

$$
\begin{aligned}
\sigma_c' &= 7.5\left[1 - \frac{1}{3}\left(\frac{24}{24.98}\right)^4\right] = 7.5\,[1 - 0.2838] = 5.37\ \text{N/mm}^2 \\
P &= 5.3715 \times 15000 = 80\,570\ \text{N} \approx 80.6\ \text{kN}
\end{aligned}
$$

| Case | $l_e$ (mm) | $l_e/d$ | Type | $\sigma_c'$ (N/mm²) | $P$ (kN) |
|---|---|---|---|---|---|
| Hinged–hinged | 3000 | 30 | Long | 3.47 | 52.1 |
| Fixed–hinged | 2400 | 24 | Intermediate | 5.37 | 80.6 |

**Answer: (a) 52.1 kN (long column); (b) 80.6 kN (intermediate column).**

## Check yourself

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

1. For sigma_cc = 7.5 and E = 9500 N/mm^2, K8 = 0.702 sqrt(E/sigma_cc) is about:
   A. 35.6
   B. 11
   C. 25.0
   D. 30.0
   Answer: C. sqrt(9500/7.5) = 35.59, so K8 = 0.702 x 35.59 = 24.98, about 25. (Key point: Slenderness and K8)

2. For the hinged-hinged column (3000 mm long, least dimension 100 mm), the column is:
   A. Short, since 30 > 11
   B. Intermediate, since 11 < 30
   C. Long, since l_e/d = 30 exceeds K8 = 24.98
   D. Long, since the load is large
   Answer: C. l_e/d = 3000/100 = 30 > K8, so it is a long column and the stress is 0.329E/(l_e/d)^2 = 3.47 N/mm^2. (Key point: Classifying the hinged column)

3. The safe axial load for the hinged-hinged case is about:
   A. 112.5 kN
   B. 80.6 kN
   C. 52.1 kN
   D. 26.0 kN
   Answer: C. P = 3.47 x 15000 = 52,090 N, about 52.1 kN. (Key point: Safe load, hinged case)

4. With one end fixed and one hinged (factor 0.8), what changes and what is the result?
   A. l_e = 2400 mm, l_e/d = 24 < K8, so intermediate; P is about 80.6 kN
   B. l_e = 3000 mm, it stays long; P is 52.1 kN
   C. l_e = 2400 mm, l_e/d = 24, so long; P is 52.1 kN
   D. l_e = 3750 mm, so short; P is 112.5 kN
   Answer: A. Reduced effective length gives l_e/d = 24 < 24.98, an intermediate column, with stress 5.37 N/mm^2 and P = 80.6 kN. (Key point: Effective length and intermediate column)

## Related in Design of Timber and Masonry Structures

- [Timber column under axial load and bending](https://www.bctengineer.com/question.md?key=semester_5%2Fdesign_of_timber_and_masonry_structures%3Atimber-column-combined-bending-axial) — Design of Timber and Masonry Structures, chapter 3
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- [Permissible compressive stress of masonry as per IS 1905](https://www.bctengineer.com/question.md?key=semester_5%2Fdesign_of_timber_and_masonry_structures%3Amasonry-permissible-stress-factors-code) — Design of Timber and Masonry Structures, chapter 5
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## Related in other subjects

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- [Rectangular Sal column, 150 kN + 15 kNm](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Atimber-column-sal-4m-150kn-15knm) — Design of Steel and Timber Structure, chapter 11
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