DoE mix design for fck 35 MPa (numerical)

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.