--- title: "ACI mix design for M35 concrete (numerical)" subject: "Concrete Technology and Masonry Structure" chapter: 3 key: "semester_5/concrete_technology_and_masonry_structure:aci-mix-design-m35-numerical" source: https://www.bctengineer.com/q/semester_5/concrete_technology_and_masonry_structure/aci-mix-design-m35-numerical site: BCT Engineering --- # ACI mix design for M35 concrete (numerical) Concrete Technology and Masonry Structure · Chapter 3: Mix design of concrete and property of green concrete - Asked in: 2082 Baisakh, 8 marks - Video: [M35 grade of concrete | Design mix proportion of concrete | Volume batching](https://www.youtube.com/watch?v=5JuhHxRX1M4) — ENGINEERING TACTICS - 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-m35-numerical - Whole chapter: https://www.bctengineer.com/md/semester_5/concrete_technology_and_masonry_structure/chapter-3 ## Question Design a concrete mix for M35 for fresh water exposure condition using the ACI method. The specific gravity of FA and CA is 2.6 and 2.7 respectively. The dry rodded bulk density of CA is 1600 kg/m$^3$ and the fineness modulus of FA is 2.8. The nominal size of aggregate used is 20 mm and the slump value required for concreting is 50 mm. ## Short answer Using ACI 211.1 absolute volume method for M35 (target strength 43.3 MPa): water 190 kg/m³, w/c 0.39, cement 487 kg/m³, coarse aggregate 992 kg/m³, fine aggregate 697 kg/m³. Mix ratio 1:1.43:2.04. ## Answer **Method: ACI 211.1 (absolute volume).** **Data:** $f'_c = 35$ MPa; fresh-water exposure; $G_{FA} = 2.6$, $G_{CA} = 2.7$; dry-rodded bulk density of CA = 1600 kg/m$^3$; FM of sand = 2.8; nominal size 20 mm (taken as 19 mm in the ACI tables); slump 50 mm; non-air-entrained; cement SG assumed 3.15. **Step 1: Target strength.** The standard deviation is not given, so $f'_{cr} = f'_c + 8.3$ (ACI 318, for $f'_c$ from 21 to 35 MPa when $s$ is unknown): $$ f'_{cr} = 35 + 8.3 = 43.3\ \text{MPa} $$ **Step 2: Water and air** (ACI 211.1 Table 6.3.3, 25-50 mm slump, 19 mm size, non-air-entrained): water = **190 kg/m$^3$**, entrapped air = **2%**. **Step 3: w/c ratio** (Table 6.3.4(a), non-air-entrained): 40 MPa gives 0.42, and 45 MPa gives 0.38 (extended beyond the table, as is commonly done); interpolating for 43.3 MPa: $$ w/c = 0.42 - \frac{(43.3 - 40)}{5}(0.04) = 0.394 \approx 0.39 $$ Durability check: for fresh-water exposure the maximum w/c is 0.50 (ACI 318 Table 4.2.2), so 0.39 governs. **Step 4: Cement** = $190/0.39 = 487$ kg/m$^3$. **Step 5: Coarse aggregate.** Bulk volume per unit volume of concrete (Table 6.3.6, 19 mm, FM 2.8) = 0.62: $$ CA = 0.62 \times 1600 = 992\ \text{kg/m}^3 $$ **Step 6: Fine aggregate by absolute volume.** | Ingredient | Mass (kg) | Volume (m$^3$) | |---|---|---| | Water | 190 | 0.190 | | Cement | 487 | 0.1547 | | Coarse aggregate | 992 | 0.3674 | | Air (2%) | - | 0.020 | | **Sum** | | **0.7321** | $V_{FA} = 1 - 0.7321 = 0.2679$ m$^3$, so $FA = 0.2679 \times 2.6 \times 1000 = 697$ kg/m$^3$. **Answer (per m$^3$): water 190 kg, cement 487 kg, fine aggregate 697 kg, coarse aggregate 992 kg; w/c = 0.39; ratio 1 : 1.43 : 2.04** (total mass about 2366 kg/m$^3$). Corrections for surface and absorbed moisture, and trial mixes, follow. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With s unknown and fc' = 35 MPa, the ACI required average strength is A. 35 + 1.34 s B. 35 + 4 = 39 MPa C. 35 x 1.15 D. 35 + 8.3 = 43.3 MPa Answer: D. When s is unknown, ACI adds 8.3 MPa for fc' up to 35 MPa. (Key point: Target strength) 2. For target 43.3 MPa between 40 MPa (0.42) and 45 MPa (0.38), the w/c is about A. 0.39 B. 0.42 C. 0.38 D. 0.50 Answer: A. 0.42 - (3.3/5)(0.04) = 0.394, about 0.39; this is below the 0.50 durability limit. (Key point: w/c) 3. With water 190 kg/m3 and w/c 0.39, cement is found; and CA = 0.62 x 1600. These give A. 190 kg and 1600 kg B. 487 kg and 992 kg C. 487 kg and 1600 kg D. 390 kg and 992 kg Answer: B. Cement = 190/0.39 = 487 kg and CA = 0.62 x 1600 = 992 kg. (Key point: Cement and CA) 4. How is fine aggregate found? A. FA = CA / 2 B. FA = cement x 1.43 C. FA = remaining volume (1 - sum of water, cement, CA and air) x SG x 1000 D. FA from bulk density Answer: C. V_FA = 1 - 0.7321 = 0.2679 m3, so FA = 0.2679 x 2600 = 697 kg. 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