---
title: "Design of a timber beam"
subject: "Strength of Materials"
chapter: 9
key: "semester_4/strength_of_materials_mechanical:timber-beam-design-udl-point-load"
source: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/timber-beam-design-udl-point-load
site: BCT Engineering
---

# Design of a timber beam

Strength of Materials · Chapter 9: Design of Beams and shafts

- 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 Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/timber-beam-design-udl-point-load
- Whole chapter: https://www.bctengineer.com/md/semester_4/strength_of_materials_mechanical/chapter-9

## Question

A simply supported timber beam of span 5 m carries a uniformly distributed load of 6 kN/m, which includes its self weight, over the whole span and a concentrated load of 12 kN at mid-span. The allowable bending stress is 10 MPa and the allowable shear stress is 1.0 MPa. Design a rectangular section with depth equal to twice the width, and check it for shear.

## Short answer

Required Z = 3.375×10⁶ mm³. With depth = twice width, adopt 175 mm × 350 mm section. Bending stress = 9.45 MPa (< 10 MPa); maximum shear stress = 0.51 MPa (< 1.0 MPa). Bending governs.

## Answer

#### Reactions and maximum bending moment
```text
          12 kN
  6 kN/m   |
 _vvvvvvvvvvvvvv_
 A                B     L = 5 m
```

Total load $= 6\times5 + 12 = 42$ kN, so $R_A = R_B = 21$ kN. Maximum shear force $V_{max} = 21$ kN at the supports. Maximum bending moment at mid-span:

$$
M_{max} = \frac{wL^2}{8} + \frac{PL}{4} = \frac{6\times25}{8} + \frac{12\times5}{4} = 18.75 + 15 = 33.75\ \text{kN m}
$$

#### Design for bending
$$
Z_{req} = \frac{M_{max}}{\sigma_{all}} = \frac{33.75\times10^6}{10} = 3.375\times10^6\ \text{mm}^3
$$

With $h = 2b$:

$$
Z = \frac{bh^2}{6} = \frac{b(2b)^2}{6} = \frac{2b^3}{3}
$$

$$
\frac{2b^3}{3} = 3.375\times10^6 \;\Rightarrow\; b^3 = 5.0625\times10^6 \;\Rightarrow\; b = 171.7\ \text{mm}
$$

Adopt $b = 175$ mm and $h = 350$ mm.

Check bending stress with the adopted section:

$$
Z = \frac{175\times350^2}{6} = 3.573\times10^6\ \text{mm}^3, \qquad \sigma = \frac{33.75\times10^6}{3.573\times10^6} = 9.45\ \text{MPa} < 10\ \text{MPa}\ \checkmark
$$

#### Check for shear
$$
\tau_{max} = \frac32\,\frac{V}{bh} = \frac32\times\frac{21\times10^3}{175\times350} = 0.514\ \text{MPa} < 1.0\ \text{MPa}\ \checkmark
$$

The shear stress is only about half the allowable value, so bending governs the design (the beam is long, $L/h = 14$).

**Answer: provide a 175 mm $\times$ 350 mm section ($\sigma_b = 9.45$ MPa, $\tau_{max} = 0.51$ MPa). Deflection should also be checked.**

## Check yourself

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

1. What is Mmax for the 5 m beam with 6 kN/m and 12 kN at midspan?
   A. 18.75 kN m
   B. 33.75 kN m
   C. 15 kN m
   D. 67.5 kN m
   Answer: B. wL²/8 + PL/4 = 18.75 + 15. (Key point: Maximum moment)

2. For a rectangle with h = 2b, what is Z?
   A. b³/3
   B. 4b³/3
   C. 2b³/3
   D. b³/6
   Answer: C. Z = bh²/6 = b(4b²)/6 = 2b³/3. (Key point: Section modulus h = 2b)

3. What section is adopted?
   A. b = 150 mm, h = 300 mm
   B. b = 171.7 mm, h = 171.7 mm
   C. b = 350 mm, h = 175 mm
   D. b = 175 mm, h = 350 mm, from b = 171.7 mm rounded up
   Answer: D. 2b³/3 = 3.375×10⁶ gives b = 171.7 mm; rounding up keeps σ = 9.45 MPa. (Key point: Sizing)

4. What is the maximum shear stress formula for a rectangle and the result?
   A. τmax = 3V/2bh = 0.514 MPa
   B. τmax = V/bh = 0.34 MPa
   C. τmax = 2V/bh = 0.69 MPa
   D. τmax = 3V/2bh = 1.03 MPa
   Answer: A. Rectangular maximum shear is 1.5 times average: 1.5×21000/(175×350) = 0.514 MPa. (Key point: Shear check)

## Related in Strength of Materials

- [Bending stresses in a T-section beam](https://www.bctengineer.com/question.md?key=semester_4%2Fstrength_of_materials_mechanical%3Acast-iron-t-beam-bending-stresses) — Strength of Materials, chapter 3
- [Elongation of tapered bar and self-weight](https://www.bctengineer.com/question.md?key=semester_4%2Fstrength_of_materials_mechanical%3Atapered-bar-elongation-derivation) — Strength of Materials, chapter 2

## Related in other subjects

- [Design of a simply supported timber beam](https://www.bctengineer.com/question.md?key=semester_5%2Fdesign_of_timber_and_masonry_structures%3Adesign-timber-floor-beam) — Design of Timber and Masonry Structures, chapter 3
- [Simply supported beam 6 m, 60 kN/m](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Abeam-6m-60knm-udl) — Design of Steel and Timber Structure, chapter 7
- [Simply supported I-beam: stress at flange-web junction](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Ai-beam-flange-web-junction-stress) — Strength of Materials, chapter 7
- [Design simply supported beam, 6 m span](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Adesign-simply-supported-beam-6-m) — Design of RCC Structure, chapter 4