Determination of grain size distribution by sieve analysis

Theory

The main purpose of this experiment is to determine grain size distribution of coarse grained soil by sieving. The test for both coarse as well as fine soil sieve analysis, hence it can be defined as the process of analysis coarse of soil is known as sieve analysis. A soil may obtain particle ranging in size from 0.01 mircon to 200 mm depending on size of particle. The soil can be classified into following group.

Boulder more than 300 mm

Gravel in between = 4.75 m – 80 m

Coarse gravel = 20 mm - 80 mm

Fine gravel = 4.75 mm  - 20 mm

Sand = 4.5 micron to 4.75 mm

Coarse sand = 2 mm to 4.75 mm

Medium sand = 425 micron to 2 mm

Fine sand = 95 micron to 425 micron

Silt = 2 micron to 75 micron

Clay less than 2 micron

The particle size distribution of soil is important in compaction, consolidation and seepage of soil.

Items required

Balance set of I.S. sieve from 100 micron to 75 micron sieve, sieve shaker, oven, sieve brush and soil sample

Diagram

Application

Sieve analysis is found to know the grading of soil. If the soil is well graded or not is found by finding sieve analysis.

Procedure

1.    Weigh the dry soil taken for the analysis in kg. sieve the sample through 2mm I.S. sieve.

2. Collect the soil passing through the sieve and keep it separate in the container.

3.    Now fill the above soil in the sieve arranging the I.S. sieves. The largest sieve at bottom.

4.   Pour the soil sample on the top most sieve put the liq. On it keep the below the sieve.

5.  Switch on the bottom so that the motor of the sieve shekar starts working and shake the whole assembly 10 – 20 min.

6.    Finally weight the soil mass with retained on each sieve, so the calculation on in triangular form.

7.     Draw the graph sieve size against communicative percentage weight passing (percentage finer) and to find out the value of coefficient of uniformity and coefficient of curvature and its grade.

Precaution

1.  Given sieve’s should be cleaned carefully no particle should remain in the sieve.

2.    The sample should be carefully sieved thoroughly the various sizes.

3.    The weight retaining on the sieve should carefully weighted.

4.    Calculation of the problem should be done correctly.

Observation table

I.S. sieve in mm	Wt. of soil retained	% wt. retained	Comm. % wt. retained	Comm % wt. passing
100	0	0	0	0
80	0	0	0	0
40	600	8.57	600	91.43
31.5	950	22.14	1550	77.86
12.5	2823	62.47	4373	37.53
10	85	63.68	4458	36.37
8	312	68.14	4770	31.86
6.3	299	72.41	5069	27.59
4.75	0	0	5069	100
3.5	13.7	74.37	5206	25.63
2.36	97	75.67	5297	24.33
2	472	82.41	5769	17.59
1	12	82.58	5781	17.42
850 µ	86	83.81	5867	16.19
600 µ	75	84.88	5942	15.12
450 µ	510	92.17	6452	7.83
425 µ	204	95.08	6056	4.92
250 µ	9	95.21	6665	4.79
212 µ	93	96.41	6758	3.46
180 µ	144	98.60	6902	1.4
150 µ	5	98.67	6907	1.33
100 µ	41	99.25	6948	0.75
75 µ	25	99.61	6973	0.39
pan	27	100	7000	0

Calculation

F.M. of soil = Total cumulative wt. retained / 100

                    = 5.43

D₁₀ = 0.60 mm

D₃₀ = 8.2 mm

D₆₀ = 10 mm

C.C = ( D₃₀ )² / D₁₀  X D₆₀

       = 11.20

Cu = D₆₀ / D₁₀ = 16.66

F.M. of soil is 5.43

CC = 11.20 and Cu = 16.66

Result

Cu is not nearly equal to unity therefore the soil is not uniformly graded and c.c. is not lies between 1 and 3, therefore the soil is not well graded but the value of Cu is greater than 6. The soil is sandy soil.