---
title: "Timber beam: value of W for stress limit"
subject: "Strength of Materials"
chapter: 7
key: "semester_3/strength_of_materials:timber-beam-w-point-loads-udl"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/timber-beam-w-point-loads-udl
site: BCT Engineering
---

# Timber beam: value of W for stress limit

Strength of Materials · Chapter 7: Theory of Flexure

- Asked in: 2066 Chaitra (old course), 16 marks
- Video: [Theory of Structures: Flexure Formula (Problem 2)](https://www.youtube.com/watch?v=vGSN79nt42g) — Lumagbas Production
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/timber-beam-w-point-loads-udl
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-7

## Question

A timber beam 160mm wide and 300mm deep is simply supported on a span of 5m. It carries an uniformly distributed load of 1 KN/m run over the whole span and three equal concentrated loads W each placed at mid span and quarter span points. If the stress in timber is not to exceed $10\ \text{N/mm}^2$, find value of W.

## Short answer

Moment of resistance is 24 kNm. The maximum bending moment at mid-span is 3.125 + 2.5W kNm. Equating gives W = 8.35 kN.

## Answer

**Given:** 160 mm × 300 mm timber beam, $L = 5$ m, UDL $w = 1$ kN/m, three equal point loads $W$ at 1.25 m, 2.5 m and 3.75 m (quarter points and mid-span). Permissible $\sigma = 10\ \text{N/mm}^2$. Self-weight is ignored.

```text
         W     W     W
 1 kN/m  v     v     v
 vvvvvvvvvvvvvvvvvvvvvvvv
 A ^-----|-----|-----|-----^ B
     1.25  1.25  1.25  1.25
```

#### Moment of resistance
$$
Z = \frac{bd^2}{6} = \frac{160\times300^2}{6} = 2.4\times10^6\ \text{mm}^3
$$

$$
M_r = \sigma Z = 10\times2.4\times10^6 = 24\times10^6\ \text{Nmm} = 24\ \text{kNm}
$$

#### Maximum bending moment (at mid-span, by symmetry)
Reaction:

$$
R_A = R_B = \frac{1\times5 + 3W}{2} = 2.5 + 1.5W
$$

$$
M_{max} = R_A\times2.5 - W\times1.25 - \frac{1\times2.5^2}{2} = 6.25 + 3.75W - 1.25W - 3.125
$$

$$
M_{max} = 3.125 + 2.5W
$$

#### Value of W
$$
3.125 + 2.5W = 24 \Rightarrow W = \frac{24 - 3.125}{2.5} = 8.35\ \text{kN}
$$

**Answer: $W = 8.35$ kN.**

## Check yourself

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

1. What is the moment of resistance of the timber section?
   A. 10 x 160x300/6
   B. sigma Z = 10 x 2.4x10^6 = 24 kNm
   C. 24 N m
   D. 10 x 300x160^2/6 = 12.3 kNm
   Answer: B. Z = bd^2/6 = 2.4x10^6 mm^3, so M_r = 24 kNm. (Key point: Moment of resistance)

2. What is each reaction?
   A. 5 + 3W
   B. 2.5 + W
   C. 2.5 + 1.5W
   D. 2.5 + 0.5W
   Answer: C. Total load is 5 + 3W, shared equally by symmetry. (Key point: Reactions)

3. Where is the maximum bending moment?
   A. At the quarter points
   B. At the supports
   C. At 1.25 m from A only
   D. At mid span by symmetry
   Answer: D. Symmetry loading gives zero shear at mid span, so M is maximum there. (Key point: Critical section)

4. What is the resulting W?
   A. 8.35 kN, from 3.125 + 2.5W = 24
   B. 9.6 kN
   C. 6.25 kN
   D. 24 kN
   Answer: A. M_max = 6.25 + 3.75W - 1.25W - 3.125 = 3.125 + 2.5W; equate to 24. (Key point: Value of W)

## Related in Strength of Materials

- [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
- [Simply supported T-beam: max bending stresses](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3At-beam-6m-100kn-10kn) — Strength of Materials, chapter 7
- [Mid-span deflection of beam, 10 kN load](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Ass-beam-deflection-mid-span-4m) — Strength of Materials, chapter 7
- [T-beam: maximum tensile and compressive stress](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3At-beam-udl-40kn-30cm-tension) — Strength of Materials, chapter 7
- [Timber joist dimensions under UDL and point load](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Atimber-joist-5kn-m-15kn) — Strength of Materials, chapter 7
- [T-beam: bending stress 15 cm above base](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3At-beam-udl-5m-bending-stress) — Strength of Materials, chapter 7

## 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
- [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
- [Design of a timber beam](https://www.bctengineer.com/question.md?key=semester_4%2Fstrength_of_materials_mechanical%3Atimber-beam-design-udl-point-load) — Strength of Materials, chapter 9