Calculate the percentage voids in the aggregate when the dry unit weight is 112.3 lb/ft and
the specific gravity of the aggregate is 2.63?
Answers
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%.
From the given question the correct answer is:
Percentage air void is 31.78453%.
Given:
(γ↓d) = dry density or dry unit weight = 112 .3 pounds / cubic ft
G = specific gravity of the aggregate = 2.63
To find :
(n↓a) percentage air void
Solution:
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%.