---
title: "Timber joist dimensions under UDL and point load"
subject: "Strength of Materials"
chapter: 7
key: "semester_3/strength_of_materials:timber-joist-5kn-m-15kn"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/timber-joist-5kn-m-15kn
site: BCT Engineering
---

# Timber joist dimensions under UDL and point load

Strength of Materials · Chapter 7: Theory of Flexure

- Asked in: 2078 Kartik, 8 marks
- Video: [Design of Timber Flexure Member | Joist | IOE TU](https://www.youtube.com/watch?v=Z3nPFb-OoiU) — Parash Joshi - Civil Construction and Tutor
- 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-joist-5kn-m-15kn
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-7

## Question

A simply supported timber joist of 6 m span has to carry uniformly distributed load 5 kN/m over its entire length and a point load of 15 kN at its center. Determine the dimensions of the rectangular joist if the maximum permissible stress in bending is $12\ \text{N/mm}^2$.

## Short answer

Required section modulus from maximum bending moment 45 kNm and allowable stress 12 N/mm² is 3.75×10⁶ mm³. Assuming depth = twice breadth gives b = 180 mm, d = 360 mm.

## Answer

**Given:** $L = 6$ m, $w = 5$ kN/m, $P = 15$ kN at centre, $\sigma_b = 12\ \text{N/mm}^2$. Self-weight is ignored.

#### Maximum bending moment (at mid-span)

$$
M_{max} = \frac{wL^2}{8} + \frac{PL}{4} = \frac{5\times 6^2}{8} + \frac{15\times 6}{4} = 22.5 + 22.5 = 45\ \text{kNm}
$$

#### Required section modulus

$$
Z = \frac{M}{\sigma_b} = \frac{45\times10^6}{12} = 3.75\times10^6\ \text{mm}^3
$$

#### Dimensions
A joist is deeper than it is wide; assume $d = 2b$ (common practice).

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

$$
b^3 = \frac{3\times 3.75\times10^6}{2} = 5.625\times10^6 \Rightarrow b = 177.8\ \text{mm}
$$

$d = 2b = 355.6$ mm.

**Check** with $180\times360$ mm: $Z = 180\times360^2/6 = 3.888\times10^6\ \text{mm}^3$, $\sigma = 45\times10^6/3.888\times10^6 = 11.6\ \text{N/mm}^2 < 12$, safe.

**Answer: provide a joist 180 mm wide × 360 mm deep (calculated 177.8 mm × 355.6 mm).**

## Check yourself

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

1. What is the total maximum moment at mid-span?
   A. 67.5 kNm
   B. 45 kNm
   C. 90 kNm
   D. 22.5 kNm
   Answer: B. wL^2/8 = 22.5 plus PL/4 = 22.5 gives 45 kNm. (Key point: Max moment)

2. How is the required Z found?
   A. M x sigma
   B. M/(sigma^2)
   C. sigma/M
   D. M/sigma
   Answer: D. Z = M/sigma_b = 45e6/12 = 3.75e6 mm^3. (Key point: Section modulus)

3. For depth d = 2b, Z = bd^2/6 becomes
   A. b^3/3
   B. 2b^3/3
   C. b^3/6
   D. 4b^3/6 x 2
   Answer: B. b(2b)^2/6 = 2b^3/3. (Key point: Rectangle Z)

4. What practical size is adopted?
   A. 180 mm x 360 mm
   B. 360 x 180 mm (width x depth)
   C. 90 x 180 mm
   D. 355 x 178 mm
   Answer: A. Calculated 177.8 x 355.6 mm is rounded up to 180 x 360 mm giving sigma = 11.6 N/mm^2. (Key point: Final size)

## Related in Strength of Materials

- [Rectangular beam dimensions, depth twice breadth](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Arect-beam-udl-point-load-depth) — Strength of Materials, chapter 7
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- [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
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- [SS beam 6 m: maximum and point deflection](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Ass-beam-6m-max-deflection-20kn) — Strength of Materials, chapter 7

## Related in other subjects

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- [Moment area: mid-span deflection under UDL](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Amoment-area-ss-udl-midspan) — Theory of Structures I, chapter 4
- [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
- [Intermediate Sal timber joist, 2.5 m span](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Atimber-joist-sal-600mm) — Design of Steel and Timber Structure, chapter 12