DOE mix design for M25 concrete (numerical)

Method: DoE (BRE 1988).

Data: M25; maximum size 25 mm; $G_{CA} = 2.7$, $G_{FA} = 2.6$; moderate exposure; FM of FA = 3.0 (zone 2/1 sand).

Assumptions: OPC 43, crushed aggregate, medium workability (slump 30-60 mm), $s = 4$ MPa (IS 456 Table 8 for M25), cement SG 3.15, no admixture.

Step 1: Target mean strength

$$f_m = f_{ck} + 1.64 s = 25 + 1.64 \times 4 = 31.6\ \text{MPa}$$

Step 2: w/c ratio. From Table 2 (49 MPa at w/c 0.5 for crushed aggregate with OPC at 28 days) and Fig 1, 31.6 MPa corresponds to w/c $\approx$ 0.7. Durability (IS 456 Table 5, moderate exposure, RCC): maximum w/c = 0.50, minimum cement 300 kg/m$^3$. So adopt w/c = 0.50.

Step 3: Free water (DoE Table 3, crushed aggregate, 30-60 mm slump): 20 mm gives 210 and 40 mm gives 190 kg/m$^3$; interpolating for 25 mm gives 205 kg/m$^3$.

Step 4: Cement = $205/0.50 = 410$ kg/m$^3$ (above 300 and below 550 kg/m$^3$).

Step 5: Wet density (Fig 3, water 205, combined aggregate SG about 2.65) $\approx$ 2380 kg/m$^3$; total aggregate = $2380 - 205 - 410 = 1765$ kg/m$^3$.

Step 6: Fine aggregate proportion (Fig 4, size 25 mm, 30-60 mm slump, w/c 0.5, coarse sand) $\approx$ 40%:

$$FA = 0.40 \times 1765 = 706\ \text{kg},\quad CA = 1765 - 706 = 1059\ \text{kg}$$

Volume check: $0.205 + 410/3150 + 706/2600 + 1059/2700 = 0.999$ m$^3$ - satisfactory.

Mix per m$^3$: cement 410 kg : water 205 kg : FA 706 kg : CA 1059 kg, ratio 1 : 1.72 : 2.58, w/c = 0.50.

Quantities for 2 m$^3$

Answer: for 2 m$^3$, cement 820 kg (about 16.4, i.e. 17 bags of 50 kg), water 410 L, FA 1412 kg, CA 2118 kg. Correct for moisture and absorption at site and verify by a trial mix.