---
title: "Simply supported I-beam: stress at flange-web junction"
subject: "Strength of Materials"
chapter: 7
key: "semester_3/strength_of_materials:i-beam-flange-web-junction-stress"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/i-beam-flange-web-junction-stress
site: BCT Engineering
---

# Simply supported I-beam: stress at flange-web junction

Strength of Materials · Chapter 7: Theory of Flexure

- Asked in: 2080 Baisakh, 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/i-beam-flange-web-junction-stress
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-7

## Question

A simply supported I-beam of flange 40 cm $\times$ 6 cm each and web 60 cm $\times$ 6 cm, is used over a span of 6 m. It carries a uniformly distributed load of intensity 25 kN/m over its entire length. Find the maximum stress induced in the beam at the flange-web junction due to bending.

## Short answer

Bending stress at the flange-web junction is 5.34 N/mm² (about 5.34 MPa), compressive at the top junction and tensile at the bottom.

## Answer

**Data:** flanges 40 cm $\times$ 6 cm each (width $\times$ thickness), web 60 cm deep $\times$ 6 cm thick, so total depth $= 6 + 60 + 6 = 72$ cm. Span $L = 6$ m, UDL $w = 25$ kN/m.

#### Maximum bending moment
$$
M_{max} = \frac{wL^2}{8} = \frac{25(6)^2}{8} = 112.5\ \text{kN·m} = 1.1250\times10^{8}\ \text{N·mm}
$$

#### Moment of inertia about the neutral axis (symmetrical section, in mm)
$$
I = 2\left[\frac{400(60)^3}{12} + (400\times60)(330)^2\right] + \frac{60(600)^3}{12}
$$

where the flange centroid is $300 + 30 = 330$ mm from the neutral axis.

$$
I = 2\left[7.2000\times10^{6} + 2.6136\times10^{9}\right] + 1.0800\times10^{9} = 6.3216\times10^{9}\ \text{mm}^4
$$

#### Stress at the flange-web junction
The junction is at $y = 300$ mm from the neutral axis:

$$
\sigma = \frac{My}{I} = \frac{1.1250\times10^{8}\times300}{6.3216\times10^{9}} = 5.339\ \text{N/mm}^2
$$

The stress is compressive at the top junction and tensile at the bottom junction. For comparison, the extreme fibre stress is $\dfrac{M(360)}{I} = 6.407\ \text{N/mm}^2$.

**Answer: bending stress at the flange-web junction $= 5.34\ \text{N/mm}^2$ (about 5.34 MPa).**

## Check yourself

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

1. For a UDL on a simply supported beam, M_max is
   A. wL^2/8 = 112.5 kN·m
   B. wL/4
   C. wL^2/12
   D. wL^2/2
   Answer: A. 25 x 36/8 = 112.5 kN·m. (Key point: Maximum moment)

2. What y is used at the flange-web junction?
   A. 300 mm (half the web depth)
   B. 330 mm
   C. 600 mm
   D. 360 mm
   Answer: A. The web is 600 mm deep, so the junction is 300 mm from the NA. (Key point: Distance at junction)

3. In the parallel axis term, why is 330 mm used?
   A. Flange centroid is 300 + 30 mm from NA
   B. Half the overall depth
   C. Flange thickness x 11
   D. Web depth plus flange width
   Answer: A. The flange centroid is 30 mm beyond the junction at 300 mm. (Key point: Flange centroid)

4. How does the junction stress compare with the extreme fibre stress?
   A. Zero
   B. Greater
   C. Smaller (5.34 vs 6.41 N/mm^2)
   D. Equal
   Answer: C. Stress is proportional to y, 300 vs 360 mm. (Key point: Stress comparison)

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