--- title: "DoE mix design for fck 35 MPa (numerical)" subject: "Concrete Technology and Masonry Structure" chapter: 3 key: "semester_5/concrete_technology_and_masonry_structure:doe-mix-design-fck35-numerical" source: https://www.bctengineer.com/q/semester_5/concrete_technology_and_masonry_structure/doe-mix-design-fck35-numerical site: BCT Engineering --- # DoE mix design for fck 35 MPa (numerical) Concrete Technology and Masonry Structure · Chapter 3: Mix design of concrete and property of green concrete - Asked in: 2068 Baisakh (old course), 7 marks - Video: [Concrete Mix Design M40 Grade | Mix design for M40 grade concrete as per IS-456:2000&IS-10262:2019](https://www.youtube.com/watch?v=O5dBR5dii2c) — 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/doe-mix-design-fck35-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 Department of Environment (DoE) method: characteristic compressive strength, $f_{ck}$ = 35 MPa; water cement ratio based on the compressive strength = 0.46. [Tables supplied with the paper: (1) water content in kg/m$^3$ of concrete for maximum aggregate size 10, 20 and 40 mm, uncrushed/crushed, at workability extremely low/very low/low/medium/high; (2) standard deviation (MPa) for M25 to M50 for very good, good and fair degree of control.] ## Short answer DoE method with given w/c 0.46: free water 210 kg/m³, cement 457 kg/m³, fine aggregate 659 kg/m³, coarse aggregate 1074 kg/m³. Mix ratio1:1.44:2.35. ## Answer **Method: DoE (BRE 1988)**; the w/c is given as 0.46. **Assumptions** (stated clearly, since the paper's tables are only partly reproduced): OPC; crushed aggregate of maximum size 20 mm; medium workability (slump 30-60 mm); combined aggregate SG = 2.7; sand in grading zone 2 (about 60% passing 600 micron). **Step 1: Target mean strength** (check): the standard deviation for M35 with good control is $s \approx 5$ MPa, so $f_m = 35 + 1.64 \times 5 = 43.2$ MPa; the w/c of 0.46 given from the strength curve is accepted. **Step 2: Free-water content** (DoE Table 3, 20 mm crushed aggregate, slump 30-60 mm) = **210 kg/m$^3$**. **Step 3: Cement** = $210/0.46 = 457$ kg/m$^3$ (within the 300-550 kg/m$^3$ limits). **Step 4: Wet density** (Fig 3, free water 210 kg/m$^3$ and SG 2.7) $\approx$ 2400 kg/m$^3$. **Step 5: Total aggregate** = $2400 - 210 - 457 = 1733$ kg/m$^3$. **Step 6: Fine aggregate proportion** (Fig 4: 20 mm, slump 30-60 mm, w/c 0.46, zone 2) $\approx$ 38%: $$ \begin{aligned} FA &= 0.38 \times 1733 = 659\ \text{kg/m}^3 \\ CA &= 1733 - 659 = 1074\ \text{kg/m}^3 \end{aligned} $$ Check by absolute volume: $210/1000 + 457/3150 + 659/2700 + 1074/2700 = 0.997$ m$^3$, which is close to 1 m$^3$ - acceptable. **Answer (per m$^3$): cement 457 kg, water 210 kg, FA 659 kg, CA 1074 kg; w/c 0.46; ratio 1 : 1.44 : 2.35.** If the table values supplied with the paper differ, replace the water content in Step 2 with the value read from them and repeat Steps 3-6. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For M35 with s about 5 MPa, DoE target mean is A. 35 + 5 B. 35 - 8 C. 35 + 1.65 x 8 D. 35 + 1.64 x 5 = 43.2 MPa Answer: D. fm = fck + 1.64 s = 43.2 MPa. (Key point: Target mean) 2. With free water 230 kg/m3 and w/c 0.46, cement is A. 500 kg/m3 B. 230 kg/m3 C. 106 kg/m3 D. 420 kg/m3 Answer: A. 230/0.46 = 500 kg/m3, within the 300-550 kg/m3 limits. (Key point: Cement) 3. With wet density 2350, water 230, cement 500, total aggregate is A. 2350 kg/m3 B. 1620 kg/m3 C. 1850 kg/m3 D. 1120 kg/m3 Answer: B. 2350 - 230 - 500 = 1620 kg/m3. (Key point: Total aggregate) 4. With 38% fine proportion, FA and CA are A. 1004 kg and 616 kg B. 700 kg and 920 kg C. 616 kg and 1004 kg D. 38 kg and 1582 kg Answer: C. FA = 0.38 x 1620 = 616; CA = 1620 - 616 = 1004 kg. 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