Question

Calculate the percentage voids in the aggregate when the dry unit weight is 112.3 lb/ft an
the specific gravity of the aggregate is 2.63?​

Answer 1

Answer:

31.7845%

Explanation:

The relationship goes like:

(γ↓d) = [{1 - (n↓a)} * G * (γ↓w) / (1 + G * w)]

where (γ↓d) = dry density or dry unit weight = 112 pounds / cubic ft

(n↓a) = proportion of air void = ?

G = specific gravity of the aggregate = 2.63

(γ↓w) = density or unit weight of water = 62.428 pounds / cubic ft

w = water content = (weight of water / weight of solid) = not provided in this question.

As water content of the aggregate mixture is not provided; so, we can assume that no water was needed to prepare the aggregate mixture.

If we assume, there is no trace of water in the aggregate mixture; w = 0.

(γ↓d) = [{1 - (n↓a)} * G * (γ↓w) / (1 + G * 0)]

Or, 112 = {1 - (n↓a)} * 2.63 * 62.428 / 1

Or, {1 - (n↓a)} = 0.6821547

Or, (n↓a) = 0.3178453

So, percentage air void = (0.3178453 * 100)% = 31.78453%.