---
title: "Modulus of rupture of beam (numerical)"
subject: "Concrete Technology and Masonry Structure"
chapter: 5
key: "semester_5/concrete_technology_and_masonry_structure:modulus-of-rupture-numerical"
source: https://www.bctengineer.com/q/semester_5/concrete_technology_and_masonry_structure/modulus-of-rupture-numerical
site: BCT Engineering
---

# Modulus of rupture of beam (numerical)

Concrete Technology and Masonry Structure · Chapter 5: Testing of concrete and quality control

- Asked in: 2070 Chaitra, 4 marks
- Video: [DETERMINATION OF FLEXURAL STRENGTH OF CONCRETE](https://www.youtube.com/watch?v=l6tnBsdTf34) — HIGHWAY GUIDE
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/concrete_technology_and_masonry_structure/modulus-of-rupture-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_5/concrete_technology_and_masonry_structure/chapter-5

## Question

Calculate the modulus of rupture of the concrete beam under single and two point loading for the following data: size of beam = 150 mm $\times$ 150 mm, length of beam = 750 mm. Failure load for single point loading is 100 kN and for two point loadings each of 50 kN.

## Short answer

For single-point loading on a 150×150 mm beam with 600 mm effective span, modulus of rupture = 3Pl/(2bd²) = 26.67 N/mm². For two-point loading, modulus of rupture = Pl/(bd²) = 17.78 N/mm².

## Answer

**Assumption:** the beam of length 750 mm rests on supports 75 mm in from each end, so the effective span is $l = 750 - 2\times75 = 600$ mm (IS 516 standard span for a 150 mm beam). $b = d = 150$ mm.

#### Single-point (centre) loading
$P = 100$ kN $= 100000$ N. Maximum moment $M = Pl/4$, section modulus $bd^2/6$:

$$
f_{cr} = \frac{3Pl}{2bd^2} = \frac{3\times100000\times600}{2\times150\times150^2} = 26.67\ \text{N/mm}^2
$$

#### Two-point (third-point) loading
Total load $P = 50 + 50 = 100$ kN, with loads at $l/3$ from each support. The moment is constant, $M = (P/2)(l/3) = Pl/6$, in the middle third (the crack is assumed to occur within it):

$$
f_{cr} = \frac{Pl}{bd^2} = \frac{100000\times600}{150\times150^2} = 17.78\ \text{N/mm}^2
$$

**Answer: modulus of rupture = 26.67 N/mm$^2$ (single point) and 17.78 N/mm$^2$ (two point).** If the full 750 mm is taken as the span, the values would be 33.3 and 22.2 N/mm$^2$. The centre-point test gives the higher value because the maximum stress acts at only one section. These values are far above the usual $0.7\sqrt{f_{ck}}$ (about 3 to 4 N/mm$^2$ for ordinary concrete), so the data are unrealistic, but they are used as given.

## Check yourself

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

1. What effective span is assumed for the 750 mm beam?
   A. 750 mm
   B. 700 mm
   C. 600 mm (75 mm in from each end)
   D. 450 mm
   Answer: C. Supports 75 mm from each end give l = 600 mm. (Key point: Effective span)

2. Which formula applies to single-point central loading?
   A. fcr = Pl/(bd^2)
   B. fcr = 3Pa/(bd^2)
   C. fcr = 2P/(pi L d)
   D. fcr = 3Pl/(2bd^2)
   Answer: D. Centre loading M = Pl/4 gives 3Pl/(2bd^2). (Key point: Single point formula)

3. What is the modulus of rupture under single-point loading of 100 kN?
   A. 26.67 N/mm2
   B. 17.78 N/mm2
   C. 33.3 N/mm2
   D. 13.3 N/mm2
   Answer: A. 3 x 100000 x 600/(2 x 150 x 150^2) = 26.67. (Key point: Single point result)

4. What is the modulus of rupture under two-point loading (total 100 kN)?
   A. 26.67 N/mm2
   B. 17.78 N/mm2
   C. 22.2 N/mm2
   D. 8.89 N/mm2
   Answer: B. 100000 x 600/(150 x 150^2) = 17.78. (Key point: Two point result)

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