---
title: "I-section beam: moment of resistance and UDL"
subject: "Strength of Materials"
chapter: 7
key: "semester_3/strength_of_materials:i-section-moment-resistance-udl-10m"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/i-section-moment-resistance-udl-10m
site: BCT Engineering
---

# I-section beam: moment of resistance and UDL

Strength of Materials · Chapter 7: Theory of Flexure

- Asked in: 2076 Asoj, 8 marks
- Video: [SECTION MODULUS, MOMENT OF RESISTANCE& BENDING STRESS NUMERICAL PROBLEMS I-MODULE-IV-LECTURE-06](https://www.youtube.com/watch?v=OTeYTjd7NZk) — CIVILTECH STUDIES
- 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-moment-resistance-udl-10m
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-7

## Question

A simply supported beam of span 10 m, subjected to UDL w throughout the length. If permissible bending stress in tension and compression are 150 MPa and 180 MPa respectively. Calculate Moment of resistance and value of UDL by assuming the I-section as shown in figure. [Figure: I-section: top and bottom flanges 100 mm wide and 25 mm thick each, web 50 mm thick, web height 50 mm.]

## Short answer

Moment of resistance: 23.44 kNm (governed by 150 MPa stress). Safe uniformly distributed load for the 10 m span: w = 1.875 kN/m.

## Answer

**Section:** flanges 100 mm × 25 mm, web 50 mm thick, web height 50 mm. Overall depth $= 25 + 50 + 25 = 100$ mm. The section is symmetrical, so the neutral axis is at mid-depth, $y = 50$ mm from top and bottom.

```text
 |<--100-->|
 +---------+  25
    | | |
    | | |     50
 +---------+  25
```

#### Moment of inertia

$$
I = \frac{100\times100^3}{12} - \frac{(100-50)\times 50^3}{12} = 8.333\times10^6 - 0.521\times10^6 = 7.8125\times10^6\ \text{mm}^4
$$

#### Moment of resistance
Both extreme fibres are 50 mm from the neutral axis. The permissible stress in the weaker fibre governs (the smaller value, 150 MPa):

$$
M_r = \frac{\sigma I}{y} = \frac{150\times 7.8125\times10^6}{50} = 23.44\times10^6\ \text{Nmm} = 23.44\ \text{kNm}
$$

(At 180 MPa the other face would allow 28.1 kNm, which is more, so 150 MPa governs.)

#### Safe UDL
$$
M_{max} = \frac{wL^2}{8} = M_r \Rightarrow w = \frac{8\times 23.44}{10^2} = 1.875\ \text{kN/m}
$$

**Answer: moment of resistance = 23.44 kNm; safe UDL $w = 1.875$ kN/m.**

## Check yourself

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

1. Where is the neutral axis of the symmetrical I-section?
   A. At mid-depth, 50 mm from top and bottom
   B. At 25 mm
   C. At the top flange
   D. At the web bottom
   Answer: A. Symmetry puts the NA at mid-depth of the 100 mm section. (Key point: Neutral axis)

2. How is I computed?
   A. bd^3/3
   B. Outer rectangle minus the two side voids
   C. Sum of flange areas
   D. Web only
   Answer: B. I = 100 x 100^3/12 - 50 x 50^3/12 = 7.8125e6 mm^4. (Key point: Moment of inertia method)

3. Why does 150 MPa govern?
   A. Because compression is weaker
   B. Average is used
   C. It is the smaller permissible stress and the section has equal distances
   D. It is larger
   Answer: C. Both fibres are at 50 mm, so the lower limit is reached first. (Key point: Governing stress)

4. How is w found from Mr?
   A. Mr = wL/4
   B. Mr = wL^2/8
   C. Mr = wL/2
   D. Mr = wL^2/2
   Answer: B. w = 8 x 23.44/100 = 1.875 kN/m. (Key point: Safe UDL)

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