---
title: "Cantilever 3 m: bending stress 25 mm below top"
subject: "Strength of Materials"
chapter: 7
key: "semester_3/strength_of_materials:cantilever-3m-udl-20kn-t-section"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/cantilever-3m-udl-20kn-t-section
site: BCT Engineering
---

# Cantilever 3 m: bending stress 25 mm below top

Strength of Materials · Chapter 7: Theory of Flexure

- Asked in: 2069 Asar, 8 marks
- Video: [Understanding Stresses in Beams](https://www.youtube.com/watch?v=f08Y39UiC-o) — The Efficient Engineer
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/cantilever-3m-udl-20kn-t-section
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-7

## Question

A cantilever beam 3m in length is subjected to load as shown below. Determine maximum bending stress at 25mm below from the top surface of the beam. [Figure: cantilever AB fixed at A, length 3 m, with 5 kN/m UDL over the span and a 20 kN point load at B; T-section, flange 20 cm wide, web 5 cm, stem 12 cm and flange depth 5 cm as marked.]

## Short answer

Maximum hogging moment 82.5 kNm. NA is 11.31 cm from base. At 25 mm below top (31.9 mm above NA) bending stress = 72.3 N/mm² tensile.

## Answer

**Section:** flange 20 cm × 5 cm on top, stem 5 cm × 12 cm below; overall depth 17 cm.

#### Maximum bending moment (at fixed end A, hogging)
$$
M = \frac{5\times3^2}{2} + 20\times3 = 22.5 + 60 = 82.5\ \text{kNm}
$$

Top fibres are in tension.

#### Neutral axis (from base, cm)
$$
\bar y = \frac{20\times5\times14.5 + 5\times12\times6}{100+60} = \frac{1450+360}{160} = 11.31\ \text{cm}
$$

#### Moment of inertia
$$
I = \frac{20\times5^3}{12} + 100(14.5-11.31)^2 + \frac{5\times12^3}{12} + 60(6-11.31)^2 = 3637.7\ \text{cm}^4 = 3.638\times10^7\ \text{mm}^4
$$

#### Stress 25 mm below the top
Top fibre is $17 - 11.31 = 5.69$ cm $= 56.9$ mm above the NA. The point 25 mm below the top is $y = 56.9 - 25 = 31.9$ mm above the NA (tension).

$$
\sigma = \frac{My}{I} = \frac{82.5\times10^6\times 31.9}{3.638\times10^7} = 72.3\ \text{N/mm}^2\ \text{(T)}
$$

For reference, the extreme fibres are 129.0 N/mm² (top, tension) and 256.6 N/mm² (bottom, compression).

**Answer: bending stress at 25 mm below the top = 72.3 N/mm² (tensile).**

## Check yourself

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

1. What is M at the fixed end?
   A. 82.5 kNm
   B. 105 kNm
   C. 22.5 kNm
   D. 60 kNm
   Answer: A. wL^2/2 + PL = 22.5 + 60. (Key point: Max moment)

2. The neutral axis lies
   A. 8.5 cm above base
   B. At the flange bottom
   C. 11.31 cm above the base
   D. 5.69 cm above base
   Answer: C. Sum(Ay)/Sum(A) = 1810/160. (Key point: NA position)

3. Top fibre is 56.9 mm above NA; what is y for a point 25 mm below the top?
   A. 25 mm above NA
   B. 81.9 mm above NA
   C. 31.9 mm below NA
   D. 31.9 mm above NA
   Answer: D. 56.9 - 25 = 31.9 mm. (Key point: y for point)

4. Stress at that point is
   A. 129 N/mm^2 tension
   B. 72.3 N/mm^2 tension
   C. 256.6 N/mm^2 tension
   D. 72.3 N/mm^2 compression
   Answer: B. 82.5e6 x 31.9/3.638e7, above NA in a hogging beam is tension. (Key point: Result)

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