ACI mix design for 25 MPa concrete (numerical)

Method: ACI 211.1.

Data: $f'_c = 25$ MPa (28 days); $s = 4$ MPa; $G_{FA} = 2.6$, $G_{CA} = 2.7$; dry-rodded density of CA = 1650 kg/m$^3$; FM = 2.8; air = 2%.

Assumptions: maximum size 20 mm (19 mm in ACI table), slump 75-100 mm (beams and columns), non-air-entrained, cement SG 3.15.

Step 1: Required average strength (ACI 318, $f'_c \leq 35$ MPa), larger of:

$$\begin{aligned} f'_{cr} &= f'_c + 1.34 s = 25 + 1.34 \times 4 = 30.36\ \text{MPa} \\ f'_{cr} &= f'_c + 2.33 s - 3.45 = 25 + 9.32 - 3.45 = 30.87\ \text{MPa} \end{aligned}$$

So $f'_{cr} = 30.87$ MPa.

Step 2: Water (Table 6.3.3, 75-100 mm, 19 mm) = 205 kg/m$^3$; air = 2% as given.

Step 3: w/c (Table 6.3.4(a)): 30 MPa to 0.54, 35 MPa to 0.47; for 30.87 MPa:

$$w/c = 0.54 - \frac{0.87}{5}(0.07) = 0.528 \approx 0.53$$

Step 4: Cement = $205/0.53 = 387$ kg/m$^3$.

Step 5: CA = $0.62 \times 1650 = 1023$ kg/m$^3$ (Table 6.3.6, 19 mm, FM 2.8).

Step 6: FA (absolute volume)

$V_{FA} = 0.2732$ m$^3$, $FA = 0.2732 \times 2600 = 710$ kg/m$^3$.

Answer (per m$^3$): cement 387 kg, water 205 kg, FA 710 kg, CA 1023 kg; w/c = 0.53; ratio 1 : 1.83 : 2.64. Moisture corrections and trial batches are required.