---
title: "Design of a simply supported timber beam"
subject: "Design of Timber and Masonry Structures"
chapter: 3
key: "semester_5/design_of_timber_and_masonry_structures:design-timber-floor-beam"
source: https://www.bctengineer.com/q/semester_5/design_of_timber_and_masonry_structures/design-timber-floor-beam
site: BCT Engineering
---

# Design of a simply supported timber beam

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

- Asked in: year not given, 8 marks
- Video: [Timber Beam Design | According to Eurocode | Tutorial](https://www.youtube.com/watch?v=GwNKz6XvDZg) — Structural Basics
- 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/design-timber-floor-beam
- Whole chapter: https://www.bctengineer.com/md/semester_5/design_of_timber_and_masonry_structures/chapter-3

## Question

Design a rectangular timber beam of width 100 mm to carry a floor over a clear span of 3.6 m. The total uniformly distributed load including self-weight is 7.5 kN/m. Take permissible bending stress 11 N/mm², permissible shear stress 1.0 N/mm², $E = 9800$ N/mm², and limit deflection to span/360. (Assume the span between the centres of bearings equals 3.6 m.)

## Short answer

M=12.15 kN·m, V=13.5 kN. Depth from deflection controls: 272 mm, provide 100×275 mm. Checks: bending 9.64 N/mm², shear 0.74 N/mm², deflection 9.66 mm ≤10 mm; all within limits.

## Answer

#### Design loads
$w = 7.5$ kN/m, $L = 3.6$ m.

$$
\begin{aligned}
M &= \frac{wL^2}{8} = \frac{7.5 \times 3.6^2}{8} = 12.15\ \text{kN·m} \\
V &= \frac{wL}{2} = \frac{7.5 \times 3.6}{2} = 13.5\ \text{kN}
\end{aligned}
$$

#### Step 1: Depth for bending
$$
Z_{req} = \frac{M}{\sigma_{bc}} = \frac{12.15\times10^6}{11} = 1.105\times10^6\ \text{mm}^3
$$
$$
d = \sqrt{\frac{6 Z_{req}}{b}} = \sqrt{\frac{6 \times 1.105\times10^6}{100}} = 257\ \text{mm}
$$

#### Step 2: Depth for deflection
$$
\delta_{allow} = \frac{3600}{360} = 10\ \text{mm}
$$
$$
I_{req} = \frac{5wL^4}{384 E \delta} = \frac{5 \times 7.5 \times 3600^4}{384 \times 9800 \times 10} = 1.674\times10^8\ \text{mm}^4
$$
$$
d = \left(\frac{12 I_{req}}{b}\right)^{1/3} = \left(\frac{12 \times 1.674\times10^8}{100}\right)^{1/3} = 272\ \text{mm}
$$

Deflection controls. Provide $100 \times 275$ mm.

#### Step 3: Checks for 100 mm × 275 mm
$$
\begin{aligned}
Z &= \frac{100 \times 275^2}{6} = 1.260\times10^6\ \text{mm}^3 \\
\sigma_b &= \frac{12.15\times10^6}{1.260\times10^6} = 9.64\ \text{N/mm}^2 < 11 \quad \text{OK} \\
\tau &= \frac{3V}{2bd} = \frac{3 \times 13\,500}{2 \times 100 \times 275} = 0.74\ \text{N/mm}^2 < 1.0 \quad \text{OK} \\
I &= \frac{100 \times 275^3}{12} = 1.733\times10^8\ \text{mm}^4 \\
\delta &= \frac{5 \times 7.5 \times 3600^4}{384 \times 9800 \times 1.733\times10^8} = 9.66\ \text{mm} < 10\ \text{mm} \quad \text{OK}
\end{aligned}
$$

#### Step 4: Bearing and lateral stability
Bearing length is chosen so that bearing stress $V/(b \times l_b)$ does not exceed the permissible compression perpendicular to grain; a bearing length of 100 mm gives about 1.35 N/mm². Since $d/b = 2.75$, the ends must be held in position and the compression edge held by floor boards.

**Answer: Provide a 100 mm × 275 mm timber beam (bending stress 9.64 N/mm², shear 0.74 N/mm², deflection 9.66 mm).**

## Check yourself

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

1. Required depth is 257 mm from bending and 272 mm from deflection. What is provided?
   A. 257 mm, since bending is the main check
   B. 272 mm, exactly as the minimum
   C. Either, as the smaller satisfies both
   D. 100 x 275 mm, since deflection controls
   Answer: D. The larger depth governs, and deflection (272 mm) controls, so 275 mm is provided. (Key point: Controlling criterion)

2. Which expression gives the required I for deflection of a simply supported beam under UDL?
   A. I = wL^2/(8 sigma)
   B. I = 5wL^4/(384 E delta)
   C. I = 3V/(2bd)
   D. I = M/sigma
   Answer: B. From delta = 5wL^4/(384EI), the required I is 5wL^4/(384 E delta_allow), using delta_allow = L/360 = 10 mm. (Key point: Deflection design)

3. How is the maximum shear stress of the rectangular beam found?
   A. V/(bd)
   B. tau = 3V/(2bd) = 0.74 N/mm^2, below 1.0
   C. tau = 2V/(3bd)
   D. tau = M/Z
   Answer: B. For a rectangle, the peak shear stress is 1.5 V/(bd) = 3 x 13,500/(2 x 100 x 275) = 0.74 N/mm^2. (Key point: Shear check)

4. Since d/b = 2.75, what extra provision is needed?
   A. No provision, as d/b is small
   B. Increase the width to 275 mm
   C. Reduce the span to 3 m
   D. The ends must be held in position and the compression edge held by floor boards
   Answer: D. A deep narrow beam can buckle sideways, so the ends are held and the compression edge is restrained by flooring. (Key point: Bearing and lateral stability)

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