Experiment 1: Sleve Analysis Report Examples

Letter of transmittal

Lab Performed: Sleeve Analysis
Description of tests or measurement performed:
The sleeve analysis experiment was conducted in order to determine the distribution of grain size in a mass of soil. For the experiment, a representative dry soil sample was collected and broken down using a pestle and mortar. The sample’s weight was measured and recorded and the soil sample was poured through a stack of sleeves. The stack was shaken and the amount of soul retained in each sleeve was measured as well as the soil at the bottom of the pan. The original weight of the stack was also measured and recorded.

Description of report contents:

This report details a description of the soil sample used, the test methodology, test results, discussion and conclusion of the experiment conducted.

Description of soil sample

The soil sample used was pinkish white.
Description of test methodology
For the experiment, a representative dry soil sample was collected
Using a motor and pestle the soil sample was broken down.
The sample’s weight was measured and recorded
A stack of sleeves was created by placing sleeves with larger openings above the sleeve with smaller openings
The prepared soil sample was poured through a stack of sleeves
The stack was shaken for about 10 to 15 minutes
The amount of soul retained in each sleeve was measured as well as the soil at the bottom of the pan. This was recorded n a table for further analysis.

Test results

The table below shows the results of the tests carried out in the laboratory.
∑ =498.7 = W1
Loss during sleeve analysis = (W – W1) / W X 100 = 0.76%

Using the data collected, the chart shown below was developed.

Chart 1: Chart showing the % finer vs. sieve opening
Discussion
Using the chart above, it is possible to determine D10, D30 and D60 as shown below.
D60 = 0.5 mm
D30= 0.6 mm
D10 = 0.9 mm

The uniformity coefficient CU and coefficient of gradation Cc has been determined as shown below:

Cu = D60 / D10
= 0.5 / 0.9
= 0.556
CC = (D230 / D60) x D10
= (0.62/0.5) x 0.9
= 0.69192
Based on the uniformity coefficient CU and coefficient of gradation Cc it is evident that the soil can be considered to be not well graded.

Conclusion

In conclusion, the experiment conducted was successful since the uniformity coefficient CU and coefficient of gradation Cc have been
Experiment 2: Specific Gravity
Lab Performed: Specific Gravity
Description of tests or measurement performed:
For this experiment, the specific gravity of soil was determined. Specific gravity refers to the specific gravity of mineral grains. It is determined as:
Gs= (unit weight of soil solids only / unit weight of water)
Different soils have different ranges of specific gravity with sand having a range of between 2.63 and 2.67. Silt on the other hand has a range of between 2.65 and 2.7. Clay has a specific gravity range of between 2.67 and 2.9. Lastly, organic soil has a specific gravity of less than two. For this experiment, the specific gravity of a soil sample will be determined experimentally as described on description of test methodology section.
Description of report contents:
This report details a description of the soil sample used, the test methodology, test results, discussion and conclusion of the specific gravity experiment carried out.
Description of soil sample

The soil sample used for the experiment was pinkish white.

Description of test methodology

A volumetric flask was cleaned and dried

The flask was filled carefully with de-aired, distilled water up to the 500ml mark
The weight of the flask filled with water up to the 500mm mark was measured and recorded
A thermometer was inserted into the flask with the water in order to determine the temperature of the water in the flask
100 grams of air dry soil was placed into an evaporating dish
Water was added to the soil in order to make it less cohesive and make it into a smooth paste. The soil paste was soaked in the evaporating dish for an hour

The mixture was transferred into the volumetric flask

Distilled water was added into the volumetric flask containing the soil to make it two-thirds full
Air was removed from the soil-water mixture by boiling the flask for 15 minutes4
Test results
The table below shows the results of the tests carried out in the laboratory.
Discussion
Using the data collected, it is possible to determine the specific gravity of the soil sample. As stated earlier, specific gravity can be determined as:
Gs= (unit weight of soil solids only / unit weight of water)

Where,

Weight of soil = W3
Weigh of equal volume of water = (W1 + W3) – W2
G s(at 24)= W3 / (W1 + W3) – W2
G s(at 24)= 852.8 / (563.3 + 852.8) – 625.1
= 1.078
G s(at 23)= W3 / (W1 + W3) – W2
G s(at 24)= 852.8 / (563.3 + 852.8) – 625.1
= 1.078
Average = (1.078 + 1.078) / 2 = 1.078
The calculated value of specific gravity is equal to 1.078. From the table of general ranges for various soils it is evident that the soil sample is an organic soil. This is mainly because its specific gravity is less than two.
The data collected also indicates that the specific gravity at 240C and 200C is 1.078 and 1.077 respectively. This implies that the specific gravity of the sample varies with changes in temperature. However, it is not possible to determine whether these changes are significant over a range of temperatures.

Conclusion:

The experiment conducted was successful as the specific gravity of the soil sample used for the experiment was determined to be 1.078. As stated earlier, it can be concluded that the soil sample is an organic soil due to this value.