---
title: "DOE mix design for M25 concrete (numerical)"
subject: "Concrete Technology and Masonry Structure"
chapter: 3
key: "semester_5/concrete_technology_and_masonry_structure:doe-mix-design-m25-numerical"
source: https://www.bctengineer.com/q/semester_5/concrete_technology_and_masonry_structure/doe-mix-design-m25-numerical
site: BCT Engineering
---

# DOE mix design for M25 concrete (numerical)

Concrete Technology and Masonry Structure · Chapter 3: Mix design of concrete and property of green concrete

- Asked in: 2072 Chaitra, 10 marks
- Video: [DOE Method  of concrete Mix Design](https://www.youtube.com/watch?v=3eWQlhxhlWY) — Anand kudoli
- 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/doe-mix-design-m25-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_5/concrete_technology_and_masonry_structure/chapter-3

## Question

Design the mix proportion for concrete with the help of the following given data: concrete grade M 25; maximum size of aggregate 25 mm; specific gravity of C.A. 2.7; specific gravity of F.A. 2.6; degree of exposure moderate; fineness modulus of F.A. 3.00; method of design: DOE method. Based on the mix ratio obtained, calculate the quantity of ingredients of concrete for 2 m$^3$ concrete production. (Assume all necessary relevant data)

## Short answer

DoE method, w/c 0.50, water 205 kg/m³. Mix per m³: cement 410 kg, water 205 kg, FA 706 kg, CA 1059 kg. For 2 m³: cement 820 kg, water 410 L, FA 1412 kg, CA 2118 kg.

## Answer

**Method: DoE (BRE 1988).**

**Data:** M25; maximum size 25 mm; $G_{CA} = 2.7$, $G_{FA} = 2.6$; moderate exposure; FM of FA = 3.0 (zone 2/1 sand).

**Assumptions:** OPC 43, crushed aggregate, medium workability (slump 30-60 mm), $s = 4$ MPa (IS 456 Table 8 for M25), cement SG 3.15, no admixture.

**Step 1: Target mean strength**

$$
f_m = f_{ck} + 1.64 s = 25 + 1.64 \times 4 = 31.6\ \text{MPa}
$$

**Step 2: w/c ratio.** From Table 2 (49 MPa at w/c 0.5 for crushed aggregate with OPC at 28 days) and Fig 1, 31.6 MPa corresponds to w/c $\approx$ 0.7. Durability (IS 456 Table 5, moderate exposure, RCC): maximum w/c = **0.50**, minimum cement 300 kg/m$^3$. So adopt **w/c = 0.50**.

**Step 3: Free water** (DoE Table 3, crushed aggregate, 30-60 mm slump): 20 mm gives 210 and 40 mm gives 190 kg/m$^3$; interpolating for 25 mm gives **205 kg/m$^3$**.

**Step 4: Cement** = $205/0.50 = 410$ kg/m$^3$ (above 300 and below 550 kg/m$^3$).

**Step 5: Wet density** (Fig 3, water 205, combined aggregate SG about 2.65) $\approx$ 2380 kg/m$^3$; total aggregate = $2380 - 205 - 410 = 1765$ kg/m$^3$.

**Step 6: Fine aggregate proportion** (Fig 4, size 25 mm, 30-60 mm slump, w/c 0.5, coarse sand) $\approx$ 40%:

$$
FA = 0.40 \times 1765 = 706\ \text{kg},\quad CA = 1765 - 706 = 1059\ \text{kg}
$$

Volume check: $0.205 + 410/3150 + 706/2600 + 1059/2700 = 0.999$ m$^3$ - satisfactory.

**Mix per m$^3$: cement 410 kg : water 205 kg : FA 706 kg : CA 1059 kg**, ratio 1 : 1.72 : 2.58, w/c = 0.50.

#### Quantities for 2 m$^3$
| Material | Quantity (kg) |
|---|---|
| Cement | 820 |
| Water | 410 (litres) |
| Fine aggregate | 1412 |
| Coarse aggregate | 2118 |

**Answer: for 2 m$^3$, cement 820 kg (about 16.4, i.e. 17 bags of 50 kg), water 410 L, FA 1412 kg, CA 2118 kg.** Correct for moisture and absorption at site and verify by a trial mix.

## Check yourself

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

1. The strength-based w/c is about 0.7 but durability allows max 0.50. Which is adopted?
   A. 0.70
   B. 0.60
   C. Average 0.60
   D. 0.50
   Answer: D. The lower value, 0.50, is adopted to satisfy the moderate exposure limit. (Key point: w/c adopted)

2. Water for 25 mm size, interpolated between 230 (20 mm) and 250 (40 mm), is
   A. 235 kg/m3
   B. 240 kg/m3
   C. 230 kg/m3
   D. 250 kg/m3
   Answer: A. 25 mm is a quarter of the way: 230 + 5 = 235 kg/m3. (Key point: Free water)

3. Cement per m3 is
   A. 235 kg
   B. 470 kg
   C. 350 kg
   D. 500 kg
   Answer: B. 235/0.50 = 470 kg/m3. (Key point: Cement)

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