--- title: "ACI method of mix design steps" subject: "Concrete Technology and Masonry Structure" chapter: 3 key: "semester_5/concrete_technology_and_masonry_structure:aci-mix-design-steps" source: https://www.bctengineer.com/q/semester_5/concrete_technology_and_masonry_structure/aci-mix-design-steps site: BCT Engineering --- # ACI method of mix design steps Concrete Technology and Masonry Structure · Chapter 3: Mix design of concrete and property of green concrete - Asked in: 2080 Bhadra, 8 marks; 2078 Kartik, 8 marks; 2076 Chaitra, 8 marks; 2075 Ashwin, 8 marks; 2074 Ashwin, 8 marks; 2073 Shrawan, 8 marks; 2068 Chaitra, 8 marks; 2067 Ashadh (old course), 5 marks; 2066 Jestha (old course), 6 marks (asked 9 times) - Video: [ACI (American Concrete Institute) Method of Mix Design](https://www.youtube.com/watch?v=lIYXwsscKFc) — Ceylara Academy - 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-steps - Whole chapter: https://www.bctengineer.com/md/semester_5/concrete_technology_and_masonry_structure/chapter-3 ## Question Describe the stepwise process of concrete mix design by the ACI method (key concepts and principle, with suitable example). ## Short answer ACI method selects water, cement, and coarse aggregate from tables based on target strength, slump, and max aggregate size; fine aggregate by absolute volume. w/c from strength/durability; water from slump/size; coarse aggregate from dry-rodded bulk volume. Example: 25 MPa, slump 75-100 mm, 20 mm aggregate yields w/c 0.49, water 205 kg/m³, cement 418 kg/m³, coarse aggregate 992 kg/m³, fine aggregate 715 kg/m³. ## Answer The **ACI method (ACI 211.1)** is an *absolute volume* method: the quantities of water, cement, coarse aggregate and air are chosen from tables, and the fine aggregate makes up the rest of the 1 m$^3$ volume. #### Key concepts - Strength depends on the w/c ratio (Abrams' law); the water content depends on the slump, the maximum aggregate size and the air content. - The coarse aggregate quantity depends on its maximum size and the fineness modulus of the sand (bulk volume of dry-rodded aggregate per unit volume of concrete). - The fine aggregate is found by absolute volume (1 m$^3$ minus the volumes of all other ingredients). #### Steps 1. **Target strength**: $f'_{cr} = f'_c + 1.34 s$ or $f'_c + 2.33 s - 3.45$ (use the larger). If $s$ is unknown, $f'_{cr} = f'_c + 8.3$ MPa (for 21-35 MPa). 2. **Maximum aggregate size** from the section and bar spacing (not more than 1/5 of the narrowest dimension between forms, 1/3 of the slab depth, or 3/4 of the clear spacing of bars). 3. **Slump** from the type of work (ACI Table 6.3.1: 25-75 mm for footings, slabs and pavements; 25-100 mm for beams, walls and columns). 4. **Water content and air** from ACI 211.1 Table 6.3.3 for the slump and size; entrapped air is 3% for 9.5 mm down to 1% for 37.5 mm in non-air-entrained concrete (2% for 20 mm). 5. **w/c ratio** from strength (Table 6.3.4) and from the exposure/durability limit; take the lower. 6. **Cement content** = water / (w/c); check the minimum. 7. **Coarse aggregate**: bulk volume per unit volume of concrete from Table 6.3.6 (by size and sand FM) $\times$ dry-rodded bulk density. 8. **Fine aggregate** by absolute volumes: $$ V_{FA} = 1 - \left(\frac{W}{1000} + \frac{C}{\rho_c \cdot 1000} + \frac{CA}{\rho_{ca} \cdot 1000} + \text{air}\right) $$ where $\rho$ is the specific gravity. Mass of FA = $V_{FA} \times$ SG $\times 1000$. 9. **Adjust for moisture** (absorption, surface moisture) and make a **trial batch**; check the slump and strength, and modify. #### Example (f'c = 25 MPa, slump 75-100 mm, 20 mm aggregate, FM = 2.8, no air entrainment; SG of cement 3.15, CA 2.7, FA 2.6) - $f'_{cr} = 25 + 8.3 = 33.3$ MPa; w/c = 0.49 (interpolated between 0.54 at 30 MPa and 0.47 at 35 MPa). - Water = 205 kg/m$^3$, air = 2%; cement = 205/0.49 = 418 kg. - CA = 0.62 $\times$ 1600 = 992 kg (dry-rodded bulk density 1600 kg/m$^3$). - Volumes: water 0.205, cement 418/3150 = 0.133, CA 992/2700 = 0.367, air 0.02; FA = 1 - 0.725 = 0.275 m$^3$, so FA = 0.275 $\times$ 2600 = 715 kg. - Mix per m$^3$: C 418, W 205, FA 715, CA 992 kg; ratio 1 : 1.71 : 2.37. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In the ACI method, how is fine aggregate found? A. From a table B. As remainder of 1 m3 after water, cement, CA and air volumes C. From w/c D. From slump Answer: B. Absolute volumes of other ingredients are subtracted from 1 m3. (Key point: Absolute volume) 2. If s is unknown for 21-35 MPa, f'cr is... A. f'c + 1.34s B. f'c + 8.3 MPa C. f'c D. f'c + 2.33s Answer: B. The code gives f'c + 8.3 MPa when no data exist. (Key point: Target strength) 3. Coarse aggregate quantity depends on... A. Slump only B. Max size and fineness modulus of sand via bulk volume table C. Cement only D. w/c only Answer: B. Bulk volume per unit volume is read by size and FM, times dry-rodded density. (Key point: Coarse aggregate) 4. How is cement content obtained? A. Water / (w/c) B. Water x (w/c) C. Strength x 10 D. From sand Answer: A. Water content from table divided by the chosen w/c. 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