---
title: "Fineness modulus of coarse aggregate (numerical)"
subject: "Concrete Technology and Masonry Structure"
chapter: 1
key: "semester_5/concrete_technology_and_masonry_structure:fineness-modulus-numerical"
source: https://www.bctengineer.com/q/semester_5/concrete_technology_and_masonry_structure/fineness-modulus-numerical
site: BCT Engineering
---

# Fineness modulus of coarse aggregate (numerical)

Concrete Technology and Masonry Structure · Chapter 1: Introduction to concrete and concrete materials

- Asked in: 2081 Bhadra, 5 marks
- Video: [Concrete Technology  Fineness Modulus of Coarse Aggregates](https://www.youtube.com/watch?v=I8EBIo3NZjo) — Nitin Bhopale
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/concrete_technology_and_masonry_structure/fineness-modulus-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_5/concrete_technology_and_masonry_structure/chapter-1

## Question

A sieve analysis conducted on a sample of coarse aggregate reveals the following data. Determine the fineness modulus of the coarse aggregate.

| IS sieve size | 80 mm | 40 mm | 20 mm | 10 mm | 4.75 mm | 2.36 mm | 1.18 mm | 600 micron | 300 micron | 150 micron | Lower than 150 micron |
|---|---|---|---|---|---|---|---|---|---|---|---|
| Weight retained (kg) | 0 | 0 | 6 | 5 | 4 | 0 | 0 | 0 | 0 | 0 | 0 |

## Short answer

Sum of cumulative percentages retained on sieves 80mm to 150μm is 713.33; dividing by 100 gives a fineness modulus of 7.13.

## Answer

**Fineness modulus** = (sum of the cumulative percentages retained on the standard sieves 80, 40, 20, 10, 4.75, 2.36 and 1.18 mm, and 600, 300 and 150 micron) / 100 (IS 2386 Part 1).

Total weight of sample = 0 + 0 + 6 + 5 + 4 = **15 kg**.

| IS sieve | Weight retained (kg) | % retained | Cumulative % retained |
|---|---|---|---|
| 80 mm | 0 | 0 | 0 |
| 40 mm | 0 | 0 | 0 |
| 20 mm | 6 | 40.00 | 40.00 |
| 10 mm | 5 | 33.33 | 73.33 |
| 4.75 mm | 4 | 26.67 | 100.00 |
| 2.36 mm | 0 | 0 | 100.00 |
| 1.18 mm | 0 | 0 | 100.00 |
| 600 micron | 0 | 0 | 100.00 |
| 300 micron | 0 | 0 | 100.00 |
| 150 micron | 0 | 0 | 100.00 |
| Total | 15 | 100 | 713.33 |

$$
FM = \frac{0 + 0 + 40 + 73.33 + 100 \times 6}{100} = \frac{713.33}{100} = 7.13
$$

**Answer: Fineness modulus of the coarse aggregate = 7.13** (all particles lie between 40 mm and 4.75 mm, and the value is within the usual range of about 5.5-8.0 for coarse aggregate).

## Check yourself

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

1. Fineness modulus is:
   A. Sum of cumulative % retained on standard sieves / 100
   B. Sum of % passing / 100
   C. Sum of weights retained
   D. Total weight / 100
   Answer: A. FM = sum of cumulative percent retained on standard sieves / 100. (Key point: FM definition)

2. Total sample is 15 kg and 6 kg is retained on 20 mm. The % retained is:
   A. 60
   B. 33.33
   C. 40
   D. 26.67
   Answer: C. 6/15 x 100 = 40%. (Key point: Percent retained)

3. Cumulative % retained on 10 mm sieve is:
   A. 33.33
   B. 73.33
   C. 100
   D. 40
   Answer: B. 40 + 33.33 = 73.33. (Key point: Cumulative at 10 mm)

4. The fineness modulus of the sample is:
   A. 7.13
   B. 6.13
   C. 5.13
   D. 8.13
   Answer: A. Sum = 0+0+40+73.33+100x6 = 713.33, divided by 100 = 7.13. (Key point: Final FM)

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