Absolute Volume Mix Design

V_{i} = W_{i} / (r h o_{i}) = W_{i} / (ma t h r m S G_{i}, r h o_{w}) q q u a d s u m V_{i} = 1 ma t h r m m^{3}

Each constituent occupies volume = mass / (SG × ρw). The volumes of cement, water, SCM, aggregate (SSD), and air sum to 1 m³ (or 27 ft³/yd³). SSD aggregate does not add or steal mix water. Air is a volume fraction (e.g. 6%).

$V_{i}$ (m³): Absolute volume of constituent i
$W_{i}$ (kg): Mass
$SG_{i}$: Specific gravity

Step-by-step solved example

Cement 400 kg, SGc = 3.15; water 180 kg. Absolute volumes in 1 m³ mix? ρw = 1000 kg/m³.

Absolute volumes of cement, water, aggregate, and air fill 1 m³.

1. Cement
Vc = 400/(3.15×1000) = 0.127 m³.
2. Water
Vw = 180/1000 = 0.180 m³.

Answer: Vc = 0.127 m³, Vw = 0.180 m³

10 practice questions

0/10 correct

1.Absolute volume of a material is

2.Volumes in 1 m³ of concrete must

3.SSD aggregate moisture means

4.If aggregate is wetter than SSD you must

5.6% air in 1 m³ is

6.SGagg = 2.65, mass 1100 kg. Vagg =

7.Yield of a batch is

8.Cement SG is typically about

9.Adding 50 kg extra cement without removing volume from others would

10.Fine and coarse aggregate volumes are computed from