Note on Particle Size Analysis

Title of Experiment

Particle Size Analysis or Sieve Analysis

Introduction

A given soil sample has many particles of different sizes. For the purpose of construction, a particular size of the soil sample will be preferable to other sizes to enhance the durability and stability of the structure whose construction the soil sample will constitute. Therefore, it is necessary to classify the soil sample into different sizes so as to get the size needed for construction purposes.

Particles in the soil can vary from 100 mm to less than 0.001 mm. The British Standard specifies the following size ranges-add the post here. Like concrete aggregate, the particle size analysis of soil samples involves determining the percentage by weight of particles within the different size ranges in the given soil sample. The method of sieving is applied to coarse-grained soils (sand + gravel ranges) while the sedimentation method is used for fine-grained soils (clay + silt ranges). The soil sample is passed through a series of standard sieve sizes having successively smaller mesh sizes. The weight of the soil retained on each sieve is determined and the cumulative percentage passing each sieve is calculated. The particle distribution curve of the soil sample is plotted on a semi-logarithmic graph, the ordinates (vertical axis) being the percentage by weight of particles smaller than the size given on the abscissa (horizontal axis).

Objective of Test

To determine the different sizes of soil particles in a given soil sample

Theoretical Background

Particle size distribution or sieve analysis is one of the methods employed in engineering to classify soil. Soil is defined as the end-product of the weathering of rocks. There are different types of soils such as sandy soils, clay soils etc depending on the mode of formation and environment of deposition. The soil is one of the most important components of concrete which is the main construction material used by the Civil Engineers. To produce good quality and durable concrete, it is expedient to use a good quality aggregate (fine and coarse) and suitable sizes. To determine whether a given soil sample has the required sizes of particles needed for construction purposes, it is necessary to conduct a sieve analysis on the given soil sample so as to classify it.

The object of soil classification is to divide the soil into groups such that all the soils in a particular group have similar characteristics by which they may be identified and exhibit similar behaviours in given engineering situations. The classification system also provides a common language for the exchange of information and experience regarding soils. it is regarded as the first step in the evaluation of soils because classification tests (particle size analysis, liquid limit and plastic limit tests etc) use samples of soil in disturbed form. In particle size analysis, the shape of the particle size distribution curve (grading curve) gives a sound knowledge of the different sizes and levels of grading of particles in a given soil sample. The flatter the distribution curve, the larger the range of particle sizes while the steeper the curve, the smaller the range of particle sizes. The particle distribution curve, together with the use of certain grading parameters, such as coefficient of uniformity (C_u = D₆₀/D₁₀), and coefficient of curvature (C_c = (D₃₀)²/ (D₆₀ x D₁₀) can be used to know whether a given soil is well-graded, poorly graded, gap-graded or uniformly graded. For instance, soil with C_u of 5 or more is well-graded while uniformly-graded soils have C_u of 2 or less. These terms when employed by the engineer in the choice of aggregate for a given construction work (in addition to other classification tests) enable him to select the best sizes and grades/grading of aggregates to produce the most stable and durable structure that is always anticipated.

 Apparatus

1. Mechanical sieve shaker with different sieve sizes (4.75 mm, 2 mm, 1.18 mm, 0.85 mm, 0.6 mm, 0.425 mm, 0.3 mm, 0.15 mm, 0.075 mm)
2. A bowel
3. A weighing balance readable and accurate to 0.1g
4. A wire brush and a fine brush
5. An Oven

Procedure

1. Measure the mass of the empty bowel and record
2. Pulverize the soil sample in a mortar with a pestle to reduce lumps in the soil
3. Measure out the mass of 200 – 500 g in the bowel and wash the sample using a 0.075 mm sieve to remove the fines
4. Dry the washed sample in the oven and record the new mass.
5. Arrange the sieves in ascending order of sizes in the sieve shaker with the lowest aperture sizes coming at the bottom above the tray e.g. tray …. 0.075 mm …. 0.15 mm …etc
6. Pour the soil sample into the topmost sieve with the largest aperture size and close the sieves tight with the lid so as to prevent the soil from spilling during shaking
7. Connect the mechanical sieve shaker to a power source and put it on to start the shaking. Shake the soil for 7 – 10 minutes.
8. When the time elapses, stop the power source, and remove the lid of the sieves.
9. Clean the bowel very well, weigh out the mass retained on each sieve accordingly and record the values of each sieve size. When this is done, put together the sieves again and keep them in a safe place.
10. Work on the recorded masses to get the percentage retained on each sieve, the cumulative percentage retained on each sieve and the percentage passing each sieve using appropriate formulas.
11. Plot a graph of the percentage passing each sieve size against the corresponding sieve sizes on a semi-log graph with the percentage passing on the y-axis (natural axis) and the sieve sizes on the x-axis (logarithmic axis).

Example

Sieve analysis was carried out for a given soil mass.

Mass of soil before washing = 500 g

Mass of soil after washing in clean water and dried in an oven = 249.97g

Mass of fines = 500 – 249.97 = 250.03g

Calculations

Percentage Retained

Percentage (%) of mass retained on each sieve = (mass retained on each sieve/total mass of soil before washing) x 100

For 4.75 mm sieve, % retained = (0.27/500) x 100 = 0.054%

For 2.00 mm sieve, % retained = (0.57/500) x 100 = 0.114%

For 1.18 mm sieve, % retained = (3.14/500) x 100 = 0.628%

For 0.85 mm sieve, % retained = (11.05/500) x 100 = 2.21%

For 0.6 mm sieve, % retained = (32.25/500) x 100 = 6.45%

For 0.425 mm sieve, % retained = (53.85/500) x 100 = 10.77%

For 0.30 mm sieve, % retained = (62.29/500) x 100 = 12.46%

For 0.15 mm sieve, % retained = (62.59/500) x 100 = 12.52%

For 0.075 mm sieve, % retained = (23.96/500) x 100 = 4.79%

Cumulative Percentage Retained

For 4.75 mm sieve = 0.054%

For 2.00 mm sieve = 0.054 + 0.114 = 0.168%

For 1.18 mm sieve = 0.168 + 0.628 = 0.796%

For 0.85 mm sieve = 0.796 + 2.21 = 3.006%

For 0.6 mm sieve = 3.006 + 6.45 = 9.456%

For 0.425 mm sieve = 9.456 + 10.77 = 20.226%

For 0.30 mm sieve = 20.226 + 12.46 = 32.686%

For 0.15 mm sieve = 32.686 + 12.52 = 45.206%

For 0.075 mm sieve = 45.206 + 4.79 = 49.996%

Percentage Passing

For 4.75 mm sieve = 100 – 0.054 = 99.946%

For 2.00 mm sieve = 100 – 0.168 = 99.832%

For 1.18 mm sieve = 100 – 0.796 = 99.204%

For 0.85 mm sieve = 100 – 3.006 = 96.994%

For 0.6 mm sieve = 100 – 9.456 = 90.544%

For 0.425 mm sieve = 100 – 20.226 = 79.774%

For 0.30 mm sieve = 100 – 32.686 = 67.314%

For 0.15 mm sieve = 100 – 45.206 = 54.794%

For 0.075 mm sieve = 100 – 49.996 = 50.004%

Note: 50.004% which is the percentage passing sieve No. 200 (0.075 mm) represents the fine content of the soil which usually comprises clay and silt.

The results are summarized in the Table below while the Figure below shows the grading curve. From the grading curve, we can neither obtain D₃₀ nor D₁₀. Therefore, such soil can only be classified based on Atterberg limit properties because it is obvious it is a high fines content soil and its behaviour would depend largely on the behaviour of the fines present.

Thanks