---
title: "ACI mix design for fck 30 MPa (numerical)"
subject: "Concrete Technology and Masonry Structure"
chapter: 3
key: "semester_5/concrete_technology_and_masonry_structure:aci-mix-design-fck30-numerical"
source: https://www.bctengineer.com/q/semester_5/concrete_technology_and_masonry_structure/aci-mix-design-fck30-numerical
site: BCT Engineering
---

# ACI mix design for fck 30 MPa (numerical)

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

- Asked in: 2070 Chaitra, 8 marks
- Video: [Mix Design M30 Grade | Concrete mix design M30 grade | M30 concrete mix design | Concrte mix design](https://www.youtube.com/watch?v=TZPoRXEM4YQ) — Civil Engineering Mastery
- 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-fck30-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 particulars using the American Concrete Institute (ACI) method: characteristic compressive strength, $f_{ck}$ = 30 MPa; water cement ratio based on the compressive strength = 0.48. Assume all necessary data.

## Short answer

ACI method with given w/c 0.48: water 205 kg/m³, cement 427 kg/m³, coarse aggregate 992 kg/m³, fine aggregate 721 kg/m³. Mix ratio1:1.69:2.32.

## Answer

**Method: ACI 211.1**, with the w/c given as 0.48.

**Assumed data:** maximum size 20 mm (19 mm in the ACI tables); slump 75-100 mm; non-air-entrained, entrapped air 2%; $G_{FA} = 2.65$, $G_{CA} = 2.70$, cement 3.15; dry-rodded density of CA = 1600 kg/m$^3$; FM of sand = 2.8.

**Step 1: Water** (ACI Table 6.3.3, 75-100 mm, 19 mm) = **205 kg/m$^3$**, air = 2%.

**Step 2: w/c** = 0.48 (given). From the ACI w/c table (0.47 at 35 MPa, 0.54 at 30 MPa), this corresponds to an average strength of about 34 MPa, a margin of about 4 MPa over $f_{ck}$ = 30 MPa.

**Step 3: Cement** = $205/0.48 = 427$ kg/m$^3$.

**Step 4: Coarse aggregate** = $0.62 \times 1600 = 992$ kg/m$^3$ (Table 6.3.6, 19 mm, FM 2.8).

**Step 5: Fine aggregate by absolute volume**

| Ingredient | Mass (kg) | Volume (m$^3$) |
|---|---|---|
| Water | 205 | 0.2050 |
| Cement | 427 | 0.1356 |
| CA | 992 | 0.3674 |
| Air | - | 0.0200 |
| **Sum** | | **0.7280** |

$V_{FA} = 1 - 0.7280 = 0.2720$ m$^3$, $FA = 0.2720 \times 2650 = 721$ kg/m$^3$.

**Answer (per m$^3$): cement 427 kg, water 205 kg, FA 721 kg, CA 992 kg; ratio 1 : 1.69 : 2.32 with w/c = 0.48** (total about 2345 kg/m$^3$). Correct for the moisture and absorption of the aggregates at site, then confirm by trial batches.

## Check yourself

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

1. Since w/c = 0.48 is given, cement content with water 205 kg/m3 is
   A. 205 x 0.48 = 98 kg
   B. 205 + 0.48
   C. 300 kg
   D. 205/0.48 = 427 kg
   Answer: D. Cement = water / (w/c) = 427 kg/m3. (Key point: Given w/c)

2. Coarse aggregate with 0.62 and 1600 kg/m3 bulk density is
   A. 1600 kg/m3
   B. 992 kg/m3
   C. 620 kg/m3
   D. 1024 kg/m3
   Answer: B. 0.62 x 1600 = 992 kg/m3. (Key point: CA)

3. FA is found from
   A. half of CA
   B. cement x 1.69
   C. V_FA = 1 - (water + cement + CA + air volumes) multiplied by SG x 1000
   D. CA / SG
   Answer: C. 0.272 m3 x 2650 = 721 kg. (Key point: FA by volume)

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