Limits of Consistency

Soil was taken and water was eventually added to every sample in order to ascertain the Plastic Limit (PL) and the Liquid Limit (LL) and which moistures contents depicted the soil’s consistency boundaries. The table below displays the first four samples which were tested for the Liquid limit while the last three samples were tested for the PL. The samples were then placed in an oven and subjected to temperatures of 105 degrees Celsius. After being removed from the oven after 24 hours, the values indicated ion red was obtained:

Introduction

In order to perform an Atterberg Limit Test, a sample of soil was obtained. This comprised the testing of the PL where two of the tests from it had a penetration which was less than 20mm while the other two had a penetration which was more than 20mm. Three more tests were performed for the LL state of the soil sample. The sample of soil obtained in each of the test was weighed and then placed in an oven with 105 degrees Celsius. The results obtained from the experiment were later discussed and a conclusion made.

Procedure

A soil sample was obtained and put in a tray and gradually, water was mixed into it. A portion of the sample was then placed in a tin, assigned with a number and weighed. Eventually, additional water was added to the sample and then mixed. The process was then repeated. The tins having the sample soils was then taken and placed in a oven, temperatures set to 105 degrees Celsius and left for 24 hours. Each tin was then removed from the oven and weighted.

Results and Analysis of the tests

The equation m=Mw/Ms (mass of wet sample/mass of dry sample) is used in calculating moisture content. For each sample, the moisture content was calculated and the results obtained depicted below:

Tin 110m=(33.78-28.57/28.57-16.06)*100=41.6%

Tin 72m=(34.74-29.49/29.49-16.24)*100=39.6%

Tin 26m=(29.95-26.20/26.20-15.88)*100=36.3%

Tin 11 m=(28.03-25.08/25.08-15.91)*100=32.2%

Tin 129 m=(22.65-21.8/21.8-17.25)*100=18.7%

Tin 181 m=(21.47-20.57/20.57-15.64)*100=18.3%

Tin 148 m=(21.14-20.33/20.33-15.89)*100=18.2%

In order to obtain a value that is directly down from 20mm penetration the 20mm LL test displays the graph showing it as 0.37 or 37%. The PL is calculated through the remaining three samples in Tins 181, 129, 148 as displayed below

PL= (18.7+18.2+18.3)/3=18.4

It is now easier to ascertain the Plasticity Index after calculating the Plasticity Limit. The Plasticity Index is required to ascertain the two letters to categorize the soil.

PI=LL-PL=37-18.4=18.1

The plasticity chart is used in comparing soils at equal liquid limit and the dryness and toughness strength is known to increase with the increasing plasticity index. It can be ascertained from the chart that the soil sample is CL and thus it can be correctly referred to be as low Plasticity Clay.

LAB REPORT ON SOIL COMPACTION

A proctor test was conducted and it comprised compaction through the use of a metal hammer which was dropped at a height of 300mm. The empty proctor’s weight was then determined and the base and collar plate was then fixed. The soil is then compacted into three layers receiving 25 blows each at specific moisture content. The process was then repeated with contents of different moisture.

Table showing results from the compaction test

Used Formulas

Moisture Content (m) %= x100

Bulk Unit Weight ( bulk)

Dry Unit Weight (dry) = 

A=0:  Dry =

A=10%:  Dry =     

Specific Volume v=                           

Figure showing the compaction curve

The air Voids Curve

Compaction curve displaying 0 and 10% Voids Line

Conclusion

The soil under test has a high compaction that makes it suitable for motorways and embankments. The PL and LL tests imply that the soil is low plasticity or CL which can be used in construction in putting on foundations because it is capable of resisting compression. This idea, however is not used in all nations because it might result in cracking issues potential shrink problems and greater pressure on the walls of basements.