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Created: 2024/07/13 15:46
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# Experiment no. 04

# Experiment name: Determination of unit weight and voids in fine and coarse aggregate

## Objective

1. To determine the unit weight and % voids of fine and coarse aggregates using loose and compact methods.

## Significance

1. Provides necessary data for designing concrete mixes including selection of proportions.
2. Helps in determining mass/volume conversion.
3. Aids in optimizing material usage in construction.

## Apparatus

1. Tamping rod
2. Balance
3. Shovel
4. Measure
5. Sample aggregates: fine and coarse

## Data sheet

![](Paper%20copy/IMG_20240714_011259.jpg)

## Calculation

Here,

- Density of water, $W = 1000 kg/m^3$

### For fine aggregate

Here,

- Weight of measure, $T = 4.219\space kg$
- Volume of measure, $V = \pi\times (\frac{160}{2})^2\times 150\times 10^{-9} = 0.003\space m^3$
- Specific gravity, $S = 2.65$
- Absorption capacity, $A = 0.40\%$
- Weight of measure + materials, $G_{loose} = 8.638\space kg$
- Weight of measure + materials, $G_{compact} = 8.924\space kg$

Therefore,
In loose method:

- Weight of materials = $G_{loose} - T = 8.638 - 4.219 = 4.42\space kg$
- Unit weight:
  $$M = \frac{G_{loose} - T}{V} = \frac{4.42}{0.003} = 1473 \equiv 1470\space kg/m^3 = 14.41\space kN/m^3$$
- SSD unit weight:
  $$(1 + \frac{A}{100})\times M = (1 + \frac{0.40}{100})\times 1473 = 1479 \equiv 1480\space kg/m^3 = 14.50\space kN/m^3$$
- % void:
  $$(\frac{S\times W - M}{S\times W})\times 100 = (\frac{2.65\times 1000 - 1473}{2.65\times 1000})\times 100 = 44.42\% \equiv 44\%$$

In compact method:

- Weight of materials = $G_{compact} - T = 8.924 - 4.219 = 4.71\space kg$
- Unit weight:
  $$M = \frac{G_{compact} - T}{V} = \frac{4.71}{0.003} = 1570\space kg/m^3 = 15.39\space kN/m^3$$
- SSD unit weight:
  $$(1 + \frac{A}{100})\times M = (1 + \frac{0.40}{100})\times 1570 = 1576 \equiv 1580\space kg/m^3 = 15.48\space kN/m^3$$
- % void:
  $$(\frac{S\times W - M}{S\times W})\times 100 = (\frac{2.65\times 1000 - 1570}{2.65\times 1000})\times 100 = 40.75\% \equiv 41\%$$

### For coarse aggregate

Here,

- Weight of measure, $T = 12.500\space kg$
- Volume of measure, $V = \pi\times (\frac{230}{2})^2\times 300\times 10^{-9} = 0.012\space m^3$
- Specific gravity, $S = 2.79$
- Absorption capacity, $A = 0.88\%$
- Weight of measure + materials, $G_{loose} = 31.450\space kg$
- Weight of measure + materials, $G_{compact} = 32.550\space kg$

Therefore,
In loose method:

- Weight of materials = $G_{loose} - T = 31.450 - 12.500 = 18.95\space kg$
- Unit weight:
  $$M = \frac{G_{loose} - T}{V} = \frac{18.95}{0.012} = 1579 \equiv 1580\space kg/m^3 = 15.48\space kN/m^3$$
- SSD unit weight:
  $$(1 + \frac{A}{100})\times M = (1 + \frac{0.88}{100})\times 1579 = 1593 \equiv 1590\space kg/m^3 = 15.58\space kN/m^3$$
- % void:
  $$(\frac{S\times W - M}{S\times W})\times 100 = (\frac{2.79\times 1000 - 1579}{2.79\times 1000})\times 100 = 43.41\% \equiv 43\%$$

In compact method:

- Weight of materials = $G_{compact} - T = 32.550 - 12.500 = 20.05\space kg$
- Unit weight:
  $$M = \frac{G_{compact} - T}{V} = \frac{20.05}{0.012} = 1671 \equiv 1670\space kg/m^3 = 16.37\space kN/m^3$$
- SSD unit weight:
  $$(1 + \frac{A}{100})\times M = (1 + \frac{0.88}{100})\times 1671 = 1686 \equiv 1690\space kg/m^3 = 16.56\space kN/m^3$$
- % void:
  $$(\frac{S\times W - M}{S\times W})\times 100 = (\frac{2.79\times 1000 - 1671}{2.79\times 1000})\times 100 = 40.11\% \equiv 40\%$$

## Result

For fine aggregate, in both loose and compact method respectively, the unit weight is $14.41\space kN/m^3$ and $15.39\space kN/m^3$, SSD unit weight is $14.50\space kN/m^3$ and $15.48\space kN/m^3$, and % void is $44\%$ and $41\%$.

For coarse aggregate, in both loose and compact method respectively, the unit weight is $15.48\space kN/m^3$ and $16.37\space kN/m^3$, SSD unit weight is $15.58\space kN/m^3$ and $16.56\space kN/m^3$, and % void is $43\%$ and $40\%$.

## Discussion

The experiment successfully determined the unit weight and % voids in fine and coarse aggregate using both loose and compact methods.

For both fine and coarse aggregates, the compact method results in a higher unit weight and lower percentage of voids compared to the loose method. These differences highlight the effect of compaction on aggregate density.

The SSD unit weight is slightly higher for both fine and coarse aggregates as it account for the absorption capacity, providing a more realistic measure for practical applications.

Understanding these properties helps in selecting appropriate aggregates for specific construction needs. Ensuring proper compaction during construction can improve the strength and durability of the resulting concrete mix.