ACI mix design for M25 concrete (numerical)

Method: ACI 211.1 (absolute volume).

Data: $f'_c = 25$ MPa; fresh-water exposure; $G_{FA} = 2.6$, $G_{CA} = 2.9$; dry-rodded bulk density of CA = 1550 kg/m$^3$; FM = 2.9; maximum size 20 mm (19 mm in the tables); slump 50 mm; non-air-entrained; cement SG 3.15 (assumed).

Step 1: Target strength ($s$ not given): $f'_{cr} = 25 + 8.3 = 33.3$ MPa.

Step 2: Water and air (Table 6.3.3: 25-50 mm slump, 19 mm): water = 190 kg/m$^3$, air = 2%.

Step 3: w/c (Table 6.3.4(a)): 30 MPa gives 0.54 and 35 MPa gives 0.47; interpolating for 33.3 MPa:

$$w/c = 0.54 - \frac{3.3}{5}(0.07) = 0.494 \approx 0.49$$

Fresh-water exposure limit: w/c $\leq 0.50$; satisfied, so adopt 0.49.

Step 4: Cement = $190/0.49 = 388$ kg/m$^3$.

Step 5: Coarse aggregate. From Table 6.3.6 (19 mm): FM 2.8 gives 0.62 and FM 3.0 gives 0.60, so for FM 2.9 the bulk volume = 0.61:

$$CA = 0.61 \times 1550 = 945.5 \approx 946\ \text{kg/m}^3$$

Step 6: Fine aggregate (absolute volume)

$V_{FA} = 1 - 0.6594 = 0.3406$ m$^3$, so $FA = 0.3406 \times 2600 = 886$ kg/m$^3$.

Answer (per m$^3$): water 190 kg, cement 388 kg, fine aggregate 886 kg, coarse aggregate 946 kg; w/c = 0.49; ratio 1 : 2.28 : 2.44 (total about 2410 kg/m$^3$). Adjust for moisture and verify by a trial mix.