---
title: "Rectangular beam dimensions, depth twice breadth"
subject: "Strength of Materials"
chapter: 7
key: "semester_3/strength_of_materials:rect-beam-udl-point-load-depth"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/rect-beam-udl-point-load-depth
site: BCT Engineering
---

# Rectangular beam dimensions, depth twice breadth

Strength of Materials · Chapter 7: Theory of Flexure

- Asked in: 2076 Chaitra, 8 marks
- Video: [Rectangular Beam loading | Anna university April May 2024 question](https://www.youtube.com/watch?v=IEDyMTodQtE) — Kings Mech
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/rect-beam-udl-point-load-depth
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-7

## Question

A simply supported beam of span 10m is to carry uniformly distributed load 20 KN/m over the entire span and a point load 50 KN at its center. Determine the dimension of beam, if the beam is rectangular in cross section and the maximum permissible stress in bending tension and compression are $120\ \text{N/mm}^2$ and $100\ \text{N/mm}^2$ respectively. Take depth of beam two times its breadth.

## Short answer

Maximum bending moment is 375 kNm; permissible stress is 100 N/mm². Required section modulus 3.75×10⁶ mm³, with d = 2b gives breadth 180 mm and depth 360 mm.

## Answer

**Given:** $L = 10$ m, $w = 20$ kN/m, $P = 50$ kN at centre, $d = 2b$, permissible $\sigma_t = 120$, $\sigma_c = 100\ \text{N/mm}^2$. Self-weight is ignored.

#### Maximum bending moment

$$
M = \frac{wL^2}{8} + \frac{PL}{4} = \frac{20\times100}{8} + \frac{50\times10}{4} = 250 + 125 = 375\ \text{kNm}
$$

#### Governing stress
A rectangle is symmetrical, so the tension and compression fibres have equal stress. The lower permissible value, $100\ \text{N/mm}^2$ (compression), governs.

$$
Z = \frac{M}{\sigma} = \frac{375\times10^6}{100} = 3.75\times10^6\ \text{mm}^3
$$

#### Dimensions
$$
Z = \frac{bd^2}{6} = \frac{b(2b)^2}{6} = \frac{2b^3}{3} \Rightarrow b^3 = 5.625\times10^6 \Rightarrow b = 177.8\ \text{mm}
$$

$d = 2b = 355.6$ mm.

**Check** with $180\times360$ mm: $Z = 3.888\times10^6\ \text{mm}^3$, $\sigma = 96.5\ \text{N/mm}^2 < 100$. Safe.

**Answer: $b \approx 180$ mm, $d \approx 360$ mm (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 M_max for UDL plus central point load?
   A. wL^2/2 + PL
   B. wL^2/8 only
   C. 250 kNm
   D. wL^2/8 + PL/4 = 375 kNm
   Answer: D. 250 + 125 = 375 kNm. (Key point: Max moment)

2. Which permissible stress is used for a rectangle?
   A. 220 N/mm^2
   B. 120 N/mm^2
   C. 110 N/mm^2
   D. 100 N/mm^2, the lower
   Answer: D. A symmetrical section has equal stress at both faces, so the lower limit governs. (Key point: Governing stress)

3. With d = 2b, Z equals
   A. b^3/3
   B. 2b^3/3
   C. b^3/6
   D. 4b^3
   Answer: B. b(2b)^2/6 = 2b^3/3. (Key point: Z in terms of b)

4. What breadth results from Z = 3.75e6 mm^3?
   A. b about 177.8 mm, d about 355.6 mm
   B. b about 355.6 mm, d about 177.8 mm
   C. b about 250 mm, d about 500 mm
   D. b about 88.9 mm, d about 177.8 mm
   Answer: A. b^3 = 5.625e6 mm^3, b = 177.8 mm. (Key point: Dimensions)

## Related in Strength of Materials

- [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
- [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
- [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
- [Beam AFD, SFD, BMD with hinge (80 kN)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Abeam-afd-sfd-bmd-20knm-80kn) — Strength of Materials, chapter 1

## Related in other subjects

- [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 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
- [Castigliano: central deflection of simply supported beam](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Acastigliano-central-deflection-ss-beam-udl) — Theory of Structures II, chapter 2