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.