---
title: "I-section: bending stress 80 mm above base"
subject: "Strength of Materials"
chapter: 7
key: "semester_3/strength_of_materials:i-section-bending-stress-80mm-above"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/i-section-bending-stress-80mm-above
site: BCT Engineering
---

# I-section: bending stress 80 mm above base

Strength of Materials · Chapter 7: Theory of Flexure

- Asked in: 2074 Chaitra, 8 marks
- Video: [Symmetrical section (I) - Bending stress](https://www.youtube.com/watch?v=jeuX8FzkIZg) — Mounika Ganta
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/i-section-bending-stress-80mm-above
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-7

## Question

For the simply supported beam of 4m span loaded with UDL of 3 kN/m, determine the value of bending stress 80 mm above the base of the cross section. The cross section of the beam is I section and the dimensions are shown below. [Figure: I-section with top and bottom flanges each 30 cm wide and 3 cm thick, web 2 cm thick and 14 cm high.]

## Short answer

For the symmetrical I-section NA is at 100 mm. Maximum bending moment 6 kNm; at 80 mm above base (20 mm below NA) tensile stress = 0.88 N/mm².

## Answer

**Section:** flanges 300 mm × 30 mm, web 20 mm × 140 mm, overall depth 200 mm. It is symmetrical, so the neutral axis is at mid-depth, 100 mm above the base.

#### Bending moment
$$
M = \frac{wL^2}{8} = \frac{3\times 4^2}{8} = 6\ \text{kNm}
$$

#### Moment of inertia (about neutral axis)
$$
I = 2\left[\frac{300\times30^3}{12} + 300\times30\times(85)^2\right] + \frac{20\times140^3}{12}
$$

$$
I = 2\,(0.675\times10^6 + 65.025\times10^6) + 4.573\times10^6 = 135.97\times10^6\ \text{mm}^4
$$

#### Stress 80 mm above the base
Distance from neutral axis $y = 100 - 80 = 20$ mm, below the NA, so the fibre is in tension (sagging beam).

$$
\sigma = \frac{My}{I} = \frac{6\times10^6\times 20}{135.97\times10^6} = 0.88\ \text{N/mm}^2
$$

**Answer: bending stress at 80 mm above base = 0.88 N/mm² (tensile).**

## Check yourself

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

1. Where is the NA of the symmetric I-section?
   A. 80 mm
   B. 140 mm
   C. 100 mm above base (mid-depth)
   D. 70 mm
   Answer: C. Overall depth 200 mm and symmetry give 100 mm. (Key point: NA for symmetric section)

2. In the parallel axis term, why 85 mm?
   A. Flange width/3
   B. Half the web
   C. Depth minus 115
   D. Flange centroid is 70 + 15 mm from NA
   Answer: D. Web half height 70 plus half flange thickness 15 = 85 mm. (Key point: Flange distance)

3. What is y for the point 80 mm above the base?
   A. 180 mm
   B. 80 mm
   C. 100 mm
   D. 20 mm below NA
   Answer: D. 100 - 80 = 20 mm below the NA. (Key point: y at 80 mm)

4. Which is the stress there?
   A. 0.88 N/mm^2 compression
   B. 0.88 N/mm^2 tension
   C. 4.4 N/mm^2 tension
   D. 0 N/mm^2
   Answer: B. 6e6 x 20/135.97e6 = 0.88 N/mm^2, below NA in a sagging beam so tension. (Key point: Result)

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