---
title: "ACI mix design for 25 MPa concrete (numerical)"
subject: "Concrete Technology and Masonry Structure"
chapter: 3
key: "semester_5/concrete_technology_and_masonry_structure:aci-mix-design-25mpa-numerical"
source: https://www.bctengineer.com/q/semester_5/concrete_technology_and_masonry_structure/aci-mix-design-25mpa-numerical
site: BCT Engineering
---

# ACI mix design for 25 MPa concrete (numerical)

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

- Asked in: 2076 Ashwin, 8 marks
- Video: [Calculation of Quantities for M25 concrete mix](https://www.youtube.com/watch?v=L5hhOcykwu0) — howYwat
- 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/aci-mix-design-25mpa-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_5/concrete_technology_and_masonry_structure/chapter-3

## Question

Design a concrete mix using the ACI method. The specified strength of concrete is 25 MPa at 28 days. The specific gravity of FA and CA is 2.6 and 2.7 respectively. Use standard deviation as 4 MPa. The dry rodded bulk density of CA is 1650 kg/m$^3$ and the fineness modulus of FA is 2.8. Approximate air entrapped is 2%. Assume suitable data if required.

## Short answer

ACI method with standard deviation4 MPa, required average strength 30.87 MPa: water 205 kg/m³, w/c 0.53, cement 387 kg/m³, coarse aggregate 1023 kg/m³, fine aggregate 710 kg/m³. Mix ratio1:1.83:2.64.

## Answer

**Method: ACI 211.1.**

**Data:** $f'_c = 25$ MPa (28 days); $s = 4$ MPa; $G_{FA} = 2.6$, $G_{CA} = 2.7$; dry-rodded density of CA = 1650 kg/m$^3$; FM = 2.8; air = 2%.

**Assumptions:** maximum size 20 mm (19 mm in ACI table), slump 75-100 mm (beams and columns), non-air-entrained, cement SG 3.15.

**Step 1: Required average strength** (ACI 318, $f'_c \leq 35$ MPa), larger of:

$$
\begin{aligned}
f'_{cr} &= f'_c + 1.34 s = 25 + 1.34 \times 4 = 30.36\ \text{MPa} \\
f'_{cr} &= f'_c + 2.33 s - 3.45 = 25 + 9.32 - 3.45 = 30.87\ \text{MPa}
\end{aligned}
$$

So $f'_{cr} = 30.87$ MPa.

**Step 2: Water** (Table 6.3.3, 75-100 mm, 19 mm) = **205 kg/m$^3$**; air = 2% as given.

**Step 3: w/c** (Table 6.3.4(a)): 30 MPa to 0.54, 35 MPa to 0.47; for 30.87 MPa:

$$
w/c = 0.54 - \frac{0.87}{5}(0.07) = 0.528 \approx 0.53
$$

**Step 4: Cement** = $205/0.53 = 387$ kg/m$^3$.

**Step 5: CA** = $0.62 \times 1650 = 1023$ kg/m$^3$ (Table 6.3.6, 19 mm, FM 2.8).

**Step 6: FA (absolute volume)**

| Ingredient | Mass (kg) | Volume (m$^3$) |
|---|---|---|
| Water | 205 | 0.2050 |
| Cement | 387 | 0.1228 |
| CA | 1023 | 0.3789 |
| Air | - | 0.0200 |
| **Sum** | | **0.7268** |

$V_{FA} = 0.2732$ m$^3$, $FA = 0.2732 \times 2600 = 710$ kg/m$^3$.

**Answer (per m$^3$): cement 387 kg, water 205 kg, FA 710 kg, CA 1023 kg; w/c = 0.53; ratio 1 : 1.83 : 2.64.** Moisture corrections and trial batches are required.

## Check yourself

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

1. For fc' <= 35 MPa with s = 4, f'cr is the
   A. smaller of two equations
   B. fc' + s
   C. fc' + 8.3
   D. larger of fc' + 1.34 s and fc' + 2.33 s - 3.45
   Answer: D. The larger value, 30.87 MPa, is adopted. (Key point: Required strength rule)

2. The larger required average strength here is
   A. 30.87 MPa
   B. 30.36 MPa
   C. 33.3 MPa
   D. 29 MPa
   Answer: A. 25 + 2.33(4) - 3.45 = 30.87 MPa exceeds 30.36 MPa. (Key point: Result)

3. For 75-100 mm slump and 19 mm aggregate, the water content used is
   A. 190 kg/m3
   B. 205 kg/m3
   C. 215 kg/m3
   D. 175 kg/m3
   Answer: B. Table 6.3.3 gives 205 kg/m3 for this slump. (Key point: Water for slump)

4. With w/c 0.53, cement and CA (0.62 x 1650) are
   A. 205 kg and 1650 kg
   B. 387 kg and 1650 kg
   C. 387 kg and 1023 kg
   D. 430 kg and 1023 kg
   Answer: C. Cement = 205/0.53 = 387; CA = 0.62 x 1650 = 1023 kg. (Key point: CA and cement)

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