Soil Classification 1
Transcript of Soil Classification 1
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Standard Practice for Classification of
Soils for Engineering Purposes.
(Unified Soil Classification System)[ASTM:D 248700]
Prepared by: Sitti Aminah
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CONTENT
Scope
Terminology
Significance
Apparatus
Sampling
Examples
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SCOPE
describes a system for classifying mineral and organo-mineral soils
for engineering purposes based on laboratory determination of
particle-size characteristics, liquid limit, and plasticity index and shall
be used when precise classification is required.
The group symbol portion of this system is based on laboratory tests
performed on the portion of a soil sample passing the 3-in. (75-mm)
sieve
This standard is the ASTM version of the Unified Soil Classification
System.
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TERMINOLOGY
Clay
soil passing a No. 200 (75-m) U.S. standard sieve that can be made to exhibit
plasticity (putty-like properties) within a range of water contents and that exhibits
considerable strength when air dry.
Gravel
Particles of rock that will pass a 3-in. (75-mm) sieve and be retained on a No.4
(4.75mm) U.S. standard sieve.
Two types:
> Coarse: passes 3-in. (75-mm) sieve and retained on 34-in. (19-mm)
sieve;
> Fine: passes 34-in. (19-mm) sieve and retained on No. 4 (4.75-mm)
sieve.
Organic
clay
A soil that would be classified as a clay except that its liquid limit value after oven
drying is less than 75 % of its liquid limit value before oven drying.
Organic siltan organic silt is a soil that would be classified as a silt except that its liquid limit value
after oven drying is less than 75 % of its liquid limit value before oven drying.
Peat
a soil composed of vegetable tissue in various stages of decomposition usually with an
organic odor, a dark-brown to black color, a spongy consistency, and a texture ranging
from fibrous to amorphous.
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Sand
Particles of rock that will pass a No.4 (75-mm) sieve and be retained on a
No.200 (75-m) U.S. standard sieve.
Three types:
Coarsepasses No. 4 (4.75-mm) sieve and retained on No.10 (2.00-mm)
sieve,
Mediumpasses No. 10 (2.00-mm) sieve and retained on No. 40 (425-m)
sieve, and
Finepasses No. 40 (425-m) sieve and retained on No. 200 (75-m) sieve.
Silt
a silt is a fine-grained soil, or the fine-grained portion of a soil, with a
plasticity index less than 4 or if the plot of plasticity index versus liquid limit falls
below the A line.
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Soil classification chart5
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SIGNIFICANCE AND USE
classifies soils from any geographic location into categories representing the
results of prescribed laboratory tests to determine the particle-size
characteristics, the liquid limit, and the plasticity index.
The assigning of a group name and symbol(s) can be used to describe a soil
to aid in the evaluation of its significant properties for engineering use. provides a useful first step in any field or laboratory investigation for
geotechnical engineering purposes.
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APPARATUSPlasticity Chart cumulative particle-size distribution
curve
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SAMPLING AND SPECIMEN
Where only the particle-size analysis of the sample is required, specimens
having the following minimum dry weights are required:
*Whenever possible, the field samples should
have weights two to four times larger than
shown.
When the liquid and plastic limit tests must also be performed, additional material
will be required sufficient to provide 150 g to 200 g of soil finer than the No. 40 (425-
m) sieve.8
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Examples
Preliminary classification
Procedure
1)FINE GRAINED:
>= 50% passNo.200 sieve
2) COARSE GRAINED:
> 50% retainedNo. 200 sieve
* In this case, (100-15.2=84.8%),hence >50%retained at No. 200 sieve, COARSE GRAINED
Example 1
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If Coarse- grainedsoil
Retained No 4 (Gravel) = 100 - 76.5= 23.5% of the total
soilor (23.5/84.8)(100) = 27.7%of the coarse-grained soil
Therefore, Sand = 100 - 27.7= 72.3%of the coarse-
grained soil.
Since sand is more than 50% > go to Sands (Primary
letter S)
Sand or gravel?
Check % of fines
(passing No 200)
If % fines < 5% Clean sands
If % fines between 5-12% Dually-
named
If % fines > 12% Sands with fines
*in this case, % fines (passing No 200) = 15.2%, Sands with fines,
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Check Plasticity chart
(type of fines)
* soil is classified as SC(Clayey Sand)
PI = LL- PL = 18, LL = 30
Check plasticity chart CL (clay with low
LL)
Thus, fines are predominantly CLAY
Hence the Secondary letter is C
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Example 2
Given: Sieve Opening (mm) Mass of soil retained on each sieve (kg)4.75 3
2 1
0.425 0.5
0.075 0.3
pan 0.2
Sieve No
Sieve Opening
(mm)
Mass of soil retained on each
sieve (g)
Percent mass
retained
(%)
Cumulative percent
retained % passing
4 4.75 3 60 60 40
9 2 1 20 80 20
35 0.425 0.5 10 90 10
200 0.075 0.3 6 96 4
Pan 0.2 4 100 0
Solution:
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Preliminary classification
Procedure
1)FINE GRAINED:
>= 50% passNo.200 sieve
2) COARSE GRAINED:
> 50% retainedNo. 200 sieve
Fineor coarse?
* In this case, (100-4=96%),hence >50% retained at No. 200 sieve, COARSE GRAINED
Sand or gravel?
Retained No 4 (Gravel) = 60% of the total soilor
(60/96)(100) = 63%of the coarse-grained soil
Therefore, Sand = 100 - 63= 37%of the coarse-grained
soil.
Since gravelis more than 50% > go to Gravel (Primary
letter G)
Check % of fines
(passing No 200)
If % fines < 5% Clean gravel
If % fines between 5-12% Dually-
named
If % fines > 12% Gravel with fines
*in this case, % fines (passing No 200) = 4%, Clean g ravel
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Well Graded (W) or
Poorly Graded (P)?Plot the distribution curve, determine
Cuand Cc
Grained-size distribution curve
0
5
10
15
20
25
30
35
40
45
0.010.1110
Grain Diameter (mm)
(Log Scale)
%P
assing
Find D60, D30, D10,then determine the Cuand Cc
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*D10, D30, and D60= the particle-size diameters corresponding
to 10, 30, and 60 %, respectively, passing on the cumulative
particle-size distribution curve.
Well-graded gravelmust conforming to following conditions:
*For this case, Cu = 2.3, Cc =1.3, Poorly graded gravel GP
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Example 3
Sieve No Mass of soil retained on each sieve
4 2
9 2
35 3
200 2
Pan 1
Given:
Solution:
Sieve No Sieve Opening
Mass of soil retained on
each sieve
Percent mass
retained
Cumulative percent
retained
%
passing
4 4.75 2 20 20 80
9 2 2 20 40 60
35 0.425 3 30 70 30
200 0.075 2 20 90 10
Pan 1 10 100 0
PL =21, LL = 35
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Preliminary classification
Procedure
1)FINE GRAINED:
>= 50% passNo.200 sieve
2) COARSE GRAINED:
> 50% retainedNo. 200 sieve
Fineor coarse?
* In this case, (100-10=90%),hence >50% retained at No. 200 sieve, COARSE GRAINED
Sand or gravel?
Retained No 4 (Gravel) = 20% of the total soilor
(20/90)(100) = 22%of the coarse-grained soil
Therefore, Sand = 100 - 22= 78%of the coarse-grained
soil.
Since gravelis more than 50% > go to Sand (Primary
letter S)
Check % of fines
(passing No 200)
If % fines < 5% Clean sand
If % fines between 5-12% Dually-
named
If % fines > 12% sand with fines
*in this case, % fines (passing No 200) = 10%, Dual ly n amed
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Well Graded (W) or
Poorly Graded (P)?Plot the distribution curve, determine
Cuand Cc
Find D60, D30, D10,then determine the Cuand Cc
Grain size distribution curve
0
10
20
30
40
50
60
70
80
90
0.010.1110
Grain diameter (mm)
(Log scale)
%p
assing
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*D10, D30, and D60= the particle-size diameters corresponding
to 10, 30, and 60 %, respectively, passing on the cumulative
particle-size distribution curve.
Well-graded gravelmust conforming to following conditions:
*For this case, Cu = 2.8, Cc =1.3, Poorly- graded sand SP
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Therefore, the first halfof the dual name is SP-SX
Check plasticity chart for the x
LL =35, PI = 35-21 = 14
From the plasticity chart,
xlocated at CL. Hence,
the soil is classified asSP-SC(Poorly Graded
sand with clay).
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Example 4
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Preliminary classification
Procedure
1)FINE GRAINED:
>= 50% passNo.200 sieve
2) COARSE GRAINED:
> 50% retainedNo. 200 sieve
Fineor coarse?
* In this case, (100-2=98%),hence >50% retained at No. 200 sieve, COARSE GRAINED
Sand or gravel?
Retained No 4 (Gravel) = (100-48)=52% of the total soilor(52/98)(100) = 53%of the coarse-grained soil
Therefore, Sand = 100 - 53= 47%of the coarse-grained
soil.
Since gravelis > 50% go to Gravel (Primary letter
G)
Check % of fines
(passing No 200)
If % fines < 5% Clean gravel
If % fines between 5-12% Dually-
named
If % fines > 12% gravel with fines
*in this case, % fines (passing No 200) = 2%, Clean gravel
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Well Graded (W) or
Poorly Graded (P)?Plot the distribution curve, determine
Cuand Cc
Find D60, D30, D10,then determine the Cuand Cc
Grain size distribution curve
0
10
20
30
40
50
60
70
80
90
0.010.1110
Grain diameter (mm)
(Log scale)
%p
assing
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*D10, D30, and D60= the particle-size diameters corresponding
to 10, 30, and 60 %, respectively, passing on the cumulative
particle-size distribution curve.
Well-graded gravelmust conforming to following conditions:
*For this case, Cu = 4.8, Cc =2.2, Well- graded gravel GP
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Example 5Sieve Opening Mass of soil retained on each sieve
4.75 3
2 1
0.425 0.1
0.075 0.2
0.7
LL = 62, PL= 44
Sieve No Sieve Opening
Mass of soil retained on each
sieve
Percent mass
retained
Cumulative percent
retained % passing
4 4.75 3 60 60 40
10 2 1 20 80 20
40 0.425 0.1 2 82 18
200 0.075 0.2 4 86 14
Pan 0.7 14 100 0
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% passing No. 200 = 14%
LL = 62, PI = 62 -44 = 18, MH.
Hence, GM Silty Gravel
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Example 6
PI= 4216 = 36;
PI plotted above A
line with LL
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Example 7
PI= 6338 = 25;
PI plotted above A
line with LL >50,
MH (Elastic silt)
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Example 8
PI= 7025 = 45;
PI plotted above A
line with LL >50,
CH (Fat Clay)
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Example 9
PI= 2822 = 6;
PI plotted below A
line with LL
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Example 10
PI= 4643 = 3
(less than 4)
PI plotted below A line
with LL
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Thank you