The fineness modulus of fine aggregate is 2.78 and of coarse aggregate is 7.82 and the desired fineness modulus of mixed aggregate is 6.14. What is the amount of fine aggregate to be mixed with one part of coarse aggregate?
50%
The fineness modulus is an empirical figure obtained by adding the cumulative percentages of aggregate retained on each of a specified series of sieves and dividing the sum by 100. It provides an index of the fineness of the aggregate. A lower fineness modulus indicates a finer aggregate, while a higher fineness modulus indicates a coarser aggregate.
When mixing fine aggregate and coarse aggregate, the fineness modulus of the resulting mixture depends on the proportions of each aggregate used. We can determine the required proportions to achieve a desired fineness modulus for the mixed aggregate.
We are given the following values:
We need to find the amount of fine aggregate to be mixed with one part of coarse aggregate. Let \(W_f\) be the weight (or volume) of fine aggregate and \(W_c\) be the weight (or volume) of coarse aggregate. The fineness modulus of the mixture (\(F_m\)) is the weighted average of the fineness moduli of the individual aggregates:
\(F_m = \frac{W_f \cdot F_f + W_c \cdot F_c}{W_f + W_c}\)
We are asked to find the amount of fine aggregate per one part of coarse aggregate, which is the ratio \(W_f / W_c\). Let's denote this ratio by \(r\), so \(r = W_f / W_c\). We can rewrite the formula by dividing the numerator and denominator by \(W_c\):
\(F_m = \frac{(W_f / W_c) \cdot F_f + (W_c / W_c) \cdot F_c}{(W_f / W_c) + (W_c / W_c)}\)
\(F_m = \frac{r \cdot F_f + F_c}{r + 1}\)
Now, substitute the given values into this equation:
\(6.14 = \frac{r \cdot 2.78 + 7.82}{r + 1}\)
Multiply both sides by \((r + 1)\) to remove the denominator:
\(6.14 \cdot (r + 1) = r \cdot 2.78 + 7.82\)
Distribute 6.14 on the left side:
\(6.14r + 6.14 = 2.78r + 7.82\)
Rearrange the terms to group \(r\) terms on one side and constant terms on the other:
\(6.14r - 2.78r = 7.82 - 6.14\)
Perform the subtractions:
\(3.36r = 1.68\)
Solve for \(r\):
\(r = \frac{1.68}{3.36}\)
\(r = 0.5\)
The ratio \(r = W_f / W_c = 0.5\). This means the weight of fine aggregate is 0.5 times the weight of coarse aggregate. If we consider one part of coarse aggregate (\(W_c = 1\)), the amount of fine aggregate (\(W_f\)) required is 0.5 parts.
To express this as a percentage of the coarse aggregate, we multiply the ratio by 100%:
Amount of fine aggregate = \(0.5 \times 100\% = 50\%\)
Therefore, the amount of fine aggregate to be mixed with one part of coarse aggregate is 50% of the weight of the coarse aggregate.
| Term | Definition/Meaning | Typical Range (Approx.) |
|---|---|---|
| Fineness Modulus (FM) | An index representing the average size of aggregate particles; higher FM means coarser aggregate. | Fine Aggregate: 2.0 to 3.5 Coarse Aggregate: 5.5 to 8.0 |
| Fine Aggregate | Aggregate mostly passing a 4.75 mm sieve. | FM typically 2.0 - 3.5 |
| Coarse Aggregate | Aggregate mostly retained on a 4.75 mm sieve. | FM typically 5.5 - 8.0 |
| Mixed Aggregate FM | Weighted average FM when fine and coarse aggregates are combined. | Depends on desired mix properties. |
The fineness modulus of aggregates is a crucial parameter in concrete mix design. It influences the workability, proportioning of cement paste, and overall economy of the mix.
Achieving a specific target fineness modulus for the total aggregate blend is a common step in concrete mix design procedures, ensuring the combined aggregate has suitable grading characteristics.
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