Particle Size Distribution & Classification of Soil

According to the US classification standards, soil particles are divided into seven grades: clay particles <0.002 mm, silt particles 0.002–0.05 mm, very fine sand 0.05–0.1 mm, fine sand 0.1–0.25 mm, medium sand 0.25–0.5 mm, coarse sand 0.5–1.0 mm, and very coarse sand 1–2 mm.

1. Particle Size Distribution
& Classification of Soil
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2. Particle Size Distribution
This classification test determines the range of sizes of particles in the soil and the percentage of
particles in each of these size ranges. This is also called ‘grain-size distribution’; ‘mechanical
analysis’ means the separation of a soil into its different size fractions:
The particle-size distribution is found in two stages:
(i) Sieve analysis, for the coarse fraction.
(ii) Sedimentation analysis or wet analysis (Hydrometer analysis), for the fine fraction
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3. Nomenclature of Grain Sizes
The Indian standard nomenclature is as follows:
Gravel ... 80 mm to 4.75 mm
Sand ... 4.75 mm to 0.075 mm
Silt ... 0.075 mm to 0.002 mm
Clay ... Less than 0.002 mm
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4. Sedimentation Analysis (Wet Analysis)
•The soil particles less than 75–µ size can be further analysed for the distribution of the various
grain-sizes of the order of silt and clay be ‘sedimentation analysis’ or ‘wet analysis’.
• The soil fraction is kept in suspension in a liquid medium, usually water. The particles descend at
velocities, related to their sizes, among other things.
•The analysis is based on ‘Stokes Law’ for what is known as the ‘terminal velocity’ of a sphere
falling through an infinite liquid medium. If a single sphere is allowed to fall in an infinite liquid
medium without interference, its velocity first increases under the influence of gravity, but soon
attains a constant value.
•This constant velocity, which is maintained indefinitely unless the boundary conditions change, is
known as the ‘terminal velocity’. The principle is obvious; coarser particles tend to settle faster
than finer ones.
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5. Stoke’s Law
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2
180
1
Dv
w
ws










Here,
D in mm
γ in kN/m
3.
µ in N-sec/m
2.
v in cm/sec
2
91Dv
Note: Above formula is only applicable for practical
uses not for numericals
wsG .
We know,
2)1.(
180
1
D
G
v
w
w






 




6. A soil sample consists of particles ranging in size from 0.6 mm to 0.02 mm. The average specific
gravity of the particles is 2.66. Determine the time of settlement of the coarsest and finest of these
particles through a depth of 1 metre. Assume the viscosity of water as 0.001 N-sec/m
2 and the unit
weight as 9.8 kN/m
3. NOTE: Units of Viscosity are important (Remmeber 1 poise = 0.1 N-sec/m
2)
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Given Data:
γ
w = 9.8 kN/m
3
.
D = 0.6 mm to 0.02 mm.
G = 2.66
µ = 0.001 N-sec/m
2
t= ?
d = 1 m = 100 cm

7. Step I: Calculation of velocities
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2)1.(
180
1
D
G
v
w
w






 



Given Data:
γ
w = 9.8 kN/m
3
. u
D = 0.6 mm to 0.02 mm.
G = 2.66
d = 100 cm
µ = 0.001 N-sec/m
2
t = ?
2
001.0
)166.2(81.9
180
1
Dv 





 

2
90Dv
For the coarsest particle, D = 0.6 mm
2
6.090v
sec/4.32cmv
From this velocity we can determine the time
of settlement using formula v = d/t

8. Step I: Calculation of velocities
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Given Data:
γ
w = 9.8 kN/m
3
. u
D = 0.6 mm to 0.02 mm.
G = 2.66
µ = 0.001 N-sec/m
2
d = 100 cm
t = ?
2
90Dv
For the fine particle, D = 0.02 mm
2
02.090v
sec/036.0 cmv

9. Wasim Shaikh
Given Data:
v
1 = 32.4 cm/sec.
v
2 = 0.036 cm/sec.
d = 100 cm
µ = 0.001 N-sec/m
2
t = ?
v
d
t
Step II: Calculation of time
4.32
100
1t
For the coarsest particle, D = 0.6 mm
sec086.3
1
t
sec78.2777
2t

10. Limitations of Stoke’s Law
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11. Classification Criteria for Fine-grained Soils (IPM)
•The plasticity chart (Fig.) forms the basis for the classification of
fine-grained soils, based on the laboratory tests.
•Fine grain soil is divided into two groups, by A-line.
• clays above the line and silts and organic soils below the line.
•The equation of A-Line is PI = 0.73 (LL-20) %
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12. Dec 2019: 10 Marks
A soil sample has a liquid limit 40% and Plastic limit 20%. The following data are also avaible from
sieve analysis
Classify the soil as per IS classififcation system.
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Sieve Size (mm) % Passing
4.75 90
0.425 85
0.075 38

13. Theory of IS Classification (No need to write in exam)
Soils shall be broadly divided into three divisions:
•1) Coarse-grained Soils: Less than 50% passing through 75- µ IS Sieve size.
•2) Fine-grained Soils: More than 50% passing through 75- µ IS Sieve size.
•3). Highly Organic Soils and Other Miscellaneous Soil Materials: These soils contain
large percentages of fibrous organic matter, such as peat, and particles of decomposed vegetation.
Coarse-grained soils shall be divided into two sub-divisions :
(a) Gravels: Less than 50% passing through 4.75 mm IS Sieve size.
(b) Sands: More than 50% passing through 4.75 mm IS Sieve size.
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14. Step I: Determine if soil is fine or course grain soil
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Since less than 50% of the material is larger than 75-µ size, the soil is a coarse-grained one.
Step II: Deciding if given course grain soil is Gravel or sand
Coarse-grained soils shall be divided into two sub-divisions :
(a) Gravels: Less than 50% of coarse fraction is larger than 4.75 mm IS Sieve size.
(b) Sands: More than 50% of Coarse fraction is smaller than 4.75 mm IS Sieve size.
Sieve Size % Passing
4.75 90
0.425 85
0.075 38
Given Soil is Sand as more than 50% passing thourgh 4.75 mm Sieve
Sands are designated with symbol S

15. Wasim Shaikh
Step III: Borderline Classification.
Criteria I: Coarse-grained soils with greater 5% fines (+75µ) are considered as border-line cases
Sieve Size % Passing
4.75 90
0.425 85
0.075 38
As 38% soil is smaller than 75 size, it is a borderline case
The given soil could be SM or SC ( M stand for Silt C stand for Clay)
SM or SC can be decided using Plasticity Chart in next step

16. Wasim Shaikh
Step IV: Classification based upon plasticity chart
The equation of A-Line is I
p= 0.73 (LL-20) %
I
p= 0.73 (40-20) %
I
p= 14.6 %
From Plasticity Chart, it looks like soil is above line A.
Hence given soil is a SC

17. Dec 2018-10 marks
Sieve analysis was performed on 1000 gm dry soil dry soil sample and following observations were
made.
If the liquid limit and plasticity index of the sample is 15% and 20% respectively, classify the soil
as per IS classification.
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Sieve Size
(mm)
20104.7521 0.60.4250.30.2120.150.075
Mass
Retained
(gm)
334985 1401601421188256 3523

18. Wasim Shaikh
Sieve Size
(mm)
20 104.75 2 1 0.60.4250.3 0.2120.150.075
Mass Retained
(gm)
33 4985 140160142118 82 56 35 23
% Retained
%3.3
1000
33100


X
Cross multiplication %
retained:
1000 gm - 100%
33 gm - X%

% Retained 3.34.98.5 14 16 14.211.8 8.2 5.6 3.5 2.3
% Passing 96.791.883.3 69.353.339.127.3 19.113.5 10 7.7
% Passing
= 100%- % cummulative retain
% Cummulative
retained
3.38.216.7 30.746.760.972.7 80.986.5 90 92.3

19. Step I: Classification
Less than 50% of soil passing through 75 micron sieve hence soil is
coarse grain soil
More than 50% of soil passing through 4.75 mm sieve hence the given
soil is Sandy soil (S)
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20. Wasim Shaikh
Step II: Classification based
upon plasticity chart
From Plasticity Chart, it looks like soil is above line A.
Hence given soil is a SC
The liquid limit and plasticity index of the sample is 15%
and 20% respectively
20%, 15%

21. From the results of a sieve analysis, shown below, determine: (a) the percent finer
than each sieve and plot a grain-size distribution curve, (b) D
10, D
30, D
60 from the
grain-size distribution curve, (c) the uniformity coefficient, Cu, and (d) the
coefficient of gradation, Cc.
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Sieve Number Diamter (mm)
Weight retained on each sieve
(g)
4 4.750 28
10 2.000 42
20 0.850 48
40 0.425 128
60 0.250 221
100 0.150 86
200 0.075 40
Pan -- 24
21
Solution is available in the
description

22. Step I: Calculation of Percent retained on each sieve
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Sieve Number Diamter (mm)
Weight retained on
each sieve (g)
4 4.750 28
10 2.000 42
20 0.850 48
40 0.425 128
60 0.250 221
100 0.150 86
200 0.075 40
Pan -- 24
Calculation of Total Weight of soil
22
617244086221128484228 
Total -- 617
Percent retained
on each sieve (%)
Calculation of % retained

23. Step I: Calculation of Percent retained on each sieve
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Sieve Number Diamter (mm)
Weight retained on
each sieve (g)
4 4.750 28
10 2.000 42
20 0.850 48
40 0.425 128
60 0.250 221
100 0.150 86
200 0.075 40
Pan -- 24
Calculation of Total Weight of soil
23
617244086221128484228 
Total -- 617
Percent retained
on each sieve (%)
X= 4.53
X2= 6.81
Calculation of % retained
Cross multiplication for first value
617 g - 100%
28 g - X%
53.4
617
10028


X
81.6
617
10042
2 

X

24. Step I: Calculation of Percent retained on each sieve
Wasim Shaikh
Sieve Number Diamter (mm)
Weight retained on
each sieve (g)
4 4.750 28
10 2.000 42
20 0.850 48
40 0.425 128
60 0.250 221
100 0.150 86
200 0.075 40
Pan -- 24
Calculation of Total Weight of soil
24
617244086221128484228 
Total -- 617
Percent retained
on each sieve (%)
4.53
6.81
7.78
20.75
35.82
19.93
6.48
3.89
Calculation of % retained
Cross multiplication for first value
617 g - 100%
28 g - X%
53.4
617
10028


X
81.6
617
10042
2 

X

25. Step II: Calculation of Cumulative Percent retained on each sieve
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Sieve
Number
Diamter
(mm)
Weight
retained
on each
sieve (g)
Percent
retained
on each sieve
(%)
Cumulative
Percent
retained on
each sieve (%)
4 4.750 28 4.53
10 2.000 42 6.81
20 0.850 48 7.78
40 0.425128 20.75
60 0.250221 35.82
100 0.150 86 19.93
200 0.075 40 6.48
Pan -- 24 3.89
Total -- 617 --
25
Cumulative Percent retained = ∑ Percentage retained on top sives

26. Step II: Calculation of Percent retained on each sieve
Wasim Shaikh
Sieve
Number
Diamter
(mm)
Weight
retained
on each
sieve (g)
Percent
retained
on each sieve
(%)
Cumulative
Percent
retained on
each sieve (%)
4 4.750 28 4.53
10 2.000 42 6.81 11.34
20 0.850 48 7.78
40 0.425128 20.75
60 0.250221 35.82
100 0.150 86 19.93
200 0.075 40 6.48
Pan -- 24 3.89
Total -- 617 --
26
Cumulative Percent retained = ∑ Percentage retained on top sives
Example calculation of sieve no 10:
Cumulative Percent retained on sieve No 10. = 4.53+6.81 = 11.34

27. Step II: Calculation of Percent retained on each sieve
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Sieve
Number
Diamter
(mm)
Weight
retained
on each
sieve (g)
Percent
retained
on each sieve
(%)
Cumulative
Percent
retained on
each sieve (%)
4 4.750 28 4.53
10 2.000 42 6.81 11.35
20 0.850 48 7.78 19.13
40 0.425128 20.75
60 0.250221 35.82
100 0.150 86 19.93
200 0.075 40 6.48
Pan -- 24 3.89
Total -- 617 --
27
Cumulative Percent retained = ∑ Percentage retained on top sives
Example calculation of sieve no 20:
Cumulative Percent retained on sieve No 20. = 4.53+6.81+7.78 = 19.13
OR
Cumulative Percent retained on sieve No 20. = 11.35 + 7.78 = 19.13

28. Step II: Calculation of Percent retained on each sieve
Wasim Shaikh
Sieve
Number
Diamter
(mm)
Weight
retained
on each
sieve (g)
Percent
retained
on each sieve
(%)
Cumulative
Percent
retained on
each sieve (%)
4 4.750 28 4.53
10 2.000 42 6.81 11.35
20 0.850 48 7.78 19.13
40 0.425128 20.75
60 0.250221 35.82
100 0.150 86 19.93
200 0.075 40 6.48
Pan -- 24 3.89
Total -- 617 --
28
Cumulative Percent retained = ∑ Percentage retained on top sives
Example calculation of sieve no 20:
Cumulative Percent retained on sieve No 20. = 4.53+6.81+7.78 = 19.13
OR
Cumulative Percent retained on sieve No 20. = 11.35 + 7.78 = 19.13

29. Step II: Calculation of Percent retained on each sieve
Wasim Shaikh
Sieve
Number
Diamter
(mm)
Weight
retained
on each
sieve (g)
Percent
retained
on each sieve
(%)
Cumulative
Percent
retained on
each sieve (%)
4 4.750 28 4.53 4.53
10 2.000 42 6.81 11.35
20 0.850 48 7.78 19.13
40 0.425128 20.75
60 0.250221 35.82
100 0.150 86 19.93
200 0.075 40 6.48
Pan -- 24 3.89
Total -- 617 --
29
Cumulative Percent retained = ∑ Percentage retained on top sives
Example calculation of sieve no 20:
Cumulative Percent retained on sieve No 4. = 0 + 4.53 = 4.53

30. Step II: Calculation of Percent retained on each sieve
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Sieve
Number
Diamter
(mm)
Weight
retained
on each
sieve (g)
Percent
retained
on each sieve
(%)
Cumulative
Percent
retained on
each sieve (%)
4 4.750 28 4.53 4.53
10 2.000 42 6.81 11.35
20 0.850 48 7.78 19.13
40 0.425128 20.75 39.88
60 0.250221 35.82 75.70
100 0.150 86 19.93 89.63
200 0.075 40 6.48 96.11
Pan -- 24 3.89 100.
Total -- 617 --
30
Cumulative Percent retained = ∑ Percentage retained on top sives

31. Step III: Calculation of Percent passing (% finer than sieve)
Wasim Shaikh
Sieve
Number
Diamter
(mm)
Weight
retained on
each sieve
(g)
Percent
retained
on each
sieve (%)
Cumulative
Percent
retained on
each sieve
(%)
Percentage
passing (%)
4 4.750 28 4.53 4.53 95.47
10 2.000 42 6.81 11.35 88.65
20 0.850 48 7.78 19.13
40 0.425 128 20.75 39.88
60 0.250 221 35.82 75.70
100 0.150 86 19.93 89.63
200 0.075 40 6.48 96.11
Pan -- 24 3.89 100.
Total -- 617 --
31
Percent passing through a sieve = 100% - Cumulative %
Percent passing through sieve 4 = 100- 4.53 =
Percent passing through sieve 20 = 100- 11.35 = 88.65

32. Step III: Calculation of Percent passing (% finer than sieve)
Wasim Shaikh
Sieve
Number
Diamter
(mm)
Weight
retained on
each sieve
(g)
Percent
retained
on each
sieve (%)
Cumulative
Percent
retained on
each sieve
(%)
Percentage
passing (%)
4 4.750 28 4.53 4.53 95.47
10 2.000 42 6.81 11.35 88.65
20 0.850 48 7.78 19.13 80.87
40 0.425 128 20.75 39.88 60.12
60 0.250 221 35.82 75.70 24.30
100 0.150 86 19.93 89.63 10.37
200 0.075 40 6.48 96.11 3.89
Pan -- 24 3.89 100. 0
Total -- 617 --
32
Percent passing through a sieve = 100% - Cumulative %
Percent passing through sieve 10 = 100- 4.53 =
Percent passing through sieve 20 = 100- 11.35 = 88.65

33. Step IV: Curve Plotting
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Sieve
Number
Diamter
(mm)
Weight
retained on
each sieve
(g)
Percent
retained
on each
sieve (%)
Cumulative
Percent
retained on
each sieve
(%)
Percentage
passing (%)
4 4.750 28 4.53 4.53 95.47
10 2.000 42 6.81 11.35 88.65
20 0.850 48 7.78 19.13 80.87
40 0.425 128 20.75 39.88 60.12
60 0.250 221 35.82 75.70 24.30
100 0.150 86 19.93 89.63 10.37
200 0.075 40 6.48 96.11 3.89
Pan -- 24 3.89 100. 0
33
X Axis Y Axis
Values of Pan are not plotted on the graph

34. Step IV: Curve Plotting
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Diamter
(mm)
Percentage
passing (%)
4.750 95.47
2.000 88.65
0.850 80.87
0.425 60.12
0.250 24.30
0.150 10.37
0.075 3.89
34
X Axis Y Axis
log scale
arithmetic scale

35. 35
Wasim ShaikhUnderstanding Semi Log Plotting
Diamter
(mm)
Percentage
passing (%)
4.750 95.47
2.000 88.65
0.850 80.87
0.425 60.12
0.250 24.30
0.150 10.37
0.075 3.89
0
20
40
60
80
100
Scale on Y Axis
1 unit = 4 %
Scale on X Axis
Log Scale
0.001
origin = 0.001, 0
0.010.1110

36. Wasim Shaikh 36
Step IV: Determination of D
10, D
30, D
60, Uniformity coefficient, Cu Coefficient of gradation, Cc.
10%
30%
60%
D
10 = 0.14
D
30 = 0.27
D
60 = 0.42
Step V: Calculation of uniformity coefficient, Cu
10
60
D
D
Cu
0.3
14.0
42.0
Cu
Step VI: Calculation of coefficient of gradation, Cc.
1060
30
2
DD
D
Cc
14.042.0
27.0
2

Cc
2.1Cc
D
10 is also called as effective size