ACI mix design for M35 concrete (numerical)

Method: ACI 211.1 (absolute volume).

Data: $f'_c = 35$ MPa; fresh-water exposure; $G_{FA} = 2.6$, $G_{CA} = 2.7$; dry-rodded bulk density of CA = 1600 kg/m$^3$; FM of sand = 2.8; nominal size 20 mm (taken as 19 mm in the ACI tables); slump 50 mm; non-air-entrained; cement SG assumed 3.15.

Step 1: Target strength. The standard deviation is not given, so $f'_{cr} = f'_c + 8.3$ (ACI 318, for $f'_c$ from 21 to 35 MPa when $s$ is unknown):

$$f'_{cr} = 35 + 8.3 = 43.3\ \text{MPa}$$

Step 2: Water and air (ACI 211.1 Table 6.3.3, 25-50 mm slump, 19 mm size, non-air-entrained): water = 190 kg/m$^3$, entrapped air = 2%.

Step 3: w/c ratio (Table 6.3.4(a), non-air-entrained): 40 MPa gives 0.42, and 45 MPa gives 0.38 (extended beyond the table, as is commonly done); interpolating for 43.3 MPa:

$$w/c = 0.42 - \frac{(43.3 - 40)}{5}(0.04) = 0.394 \approx 0.39$$

Durability check: for fresh-water exposure the maximum w/c is 0.50 (ACI 318 Table 4.2.2), so 0.39 governs.

Step 4: Cement = $190/0.39 = 487$ kg/m$^3$.

Step 5: Coarse aggregate. Bulk volume per unit volume of concrete (Table 6.3.6, 19 mm, FM 2.8) = 0.62:

$$CA = 0.62 \times 1600 = 992\ \text{kg/m}^3$$

Step 6: Fine aggregate by absolute volume.

$V_{FA} = 1 - 0.7321 = 0.2679$ m$^3$, so $FA = 0.2679 \times 2.6 \times 1000 = 697$ kg/m$^3$.

Answer (per m$^3$): water 190 kg, cement 487 kg, fine aggregate 697 kg, coarse aggregate 992 kg; w/c = 0.39; ratio 1 : 1.43 : 2.04 (total mass about 2366 kg/m$^3$). Corrections for surface and absorbed moisture, and trial mixes, follow.