Ch 03 05 Soil Clasification
Transcript of Ch 03 05 Soil Clasification
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CE353 Soil Mechanics
Dr.Talat Bader
Civil Engineering Department
King Fahad University of Petroleum & Minerals
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CE353Geotechnical
Engineering Laboratory
SOILCLASSIFICATION
Dr. Talat Bader
Lecture
5
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Lecture 5SOIL
CLASSIFICATION
Dr. Talat A Bader
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Why do we need to
classify soils ?
To describe various soil types
encountered in the nature in a
systematic way and gathering
soils that have distinct physicalproperties in groups and units.
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General Requirements of a
soil Classification Systems
Based on a scientific method
Simple Permit classification by visual and
manual tests
Describe certain engineeringproperties
Should be accepted to all engineers
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Classification Systemso Geologic Soil Classification System
o Agronomic Soil Classification System
Textural Soil Classification System (USDA)
Unified Soil Classification System (USCS)
American Association of State HighwayTransportation Officials System (AASHTO)
American Society for Testing and MaterialsSystem (ASTM)
Federal Aviation Agency System (FAA)
Others
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The Simplest
Classification
0.075-4.754.75-75Unified
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CE353 Soil Mechanics
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Civil Engineering Department
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ChartCassagrande's
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Textural Soil Classification
System (USDA)
USDA considers only:
No. Gravel in the System
Sand Silt Clay
If you encounter gravel in the soil ------- Subtract the % of gravel from the100%.
12 Subgroups in the system
Also Called Agriculture Classification System
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Basses for the
Classification
d 0.002mmlays0.002mm d < 0.05mmilts
0.05mm d < 2mmands2mm d 75mmravels
d > 75mmobblesBouldersDiameter RangeSoil Type
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mm2d%with100%
2mmd0.05mmwith%Sand%Relative
=
mm2d%with100%
mm05.d0.002mmwith%Silt%Relative
=
mm2d%with100%
0.002mmdwith%Clay%Relative
=
The USDA classification system
is based on
the relative % sand, % silt, and % clay.
These
%age
Are
Based
On
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Textural Soil ClassificationSystem (USDA)
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Example 1
Classify the following soil by the
USDA Textural Classification
System. Given Data
% gravel = 18; % sand = 51;% silt = 22; %clay = 9
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CE353 Soil Mechanics
Dr.Talat Bader
Civil Engineering Department
King Fahad University of Petroleum & Minerals
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Solution:Example 1Since we have 18% Gravel
the soil to be classifiedis = 100 18 = 82%
Calculating the % sand = 51/ 82 = 62%
Calculating the % silt = 22/ 82 = 27%
Calculating the % clay = 09/ 82 = 11%
Using these values in the USAC
Given Data
% gravel = 18;
% sand = 51;
% silt = 22;
% clay = 9
sandy loam
Then the Soil is Classified as
However, due to the presence of 18% gravel inthe soil, it is
called " gravell y sandy loam."CE 353 Dr. Talat Bader
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Textural Soil Classification
System (USDA)
Sand=62%Silt =27%Clay=11%
Back
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Example 2
Classify the following soil by the
USDA Textural Classification
System.
Given:
% gravel = 0 ; % sand = 30;% silt = 30; % clay = 40;
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Solution: Example 2
% sand Present = 30/ 100 = 30%
% silt Present = 30/ 100 = 30%
% clay Present = 40/ 100 = 40%
Using these values in the USAC
clay
Then the Soil is Classified as
Given Data
% gravel = 0;
% sand = 30;
% silt = 30;
% clay = 40
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Textural Soil Classification
System (USDA)
sand=30%
silt =30%
clay =40%
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Unified Classification System
(UCS)
Divides soils into two categories Coarse &Fine
Atterberg Limitschecks for Law and HighCompressibility
Used by Geotechnical engineers for selectingappropriate soils in non-highway projects
Last modified in 1991
First Developed by the Army Corps ofEngineers Based in 1942.
G = Gravel
S = Sand
M = Silt
C = ClayO = Organic
USC Main Groups
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CE353 Soil Mechanics
Dr.Talat Bader
Civil Engineering Department
King Fahad University of Petroleum & Minerals
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Unified Soil Classification
(USC) System
Fine Grained
More than 50% Passing
Sieve 200
Coarse Grained
Less than 50% Passing
Sieve 200
More than 50%
Passing Sieve 4
Less than 50%
Passing Sieve 4
Clay Silt Organic
Gravel Sand
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Cassagrande's Chart
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Unified Soil Classification (USC) System
Course Grained
>50% passing Sieve 200
>5% Passing
Sieve 200
= 4
Cz=1-3
Cu< 6
Cz=1-3
GM GC
PI < 4
Below
A-Line
PI > 7
Above
A-Line
>5% Passing
Sieve 200
=6
Cz=1-3
Cu>=6
Cz=1-3
SM SC
PI 7
Above
A-Line
Gravelly Soils
Less than 50%
passing
Sieve 4
Sandy Soils
More than 50%
passing
Sieve 4
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Cassagrande's Chart
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Unified Soil Classification (USC) System
Fine Grained>50% passing Sieve 200
75.0LL
LLRatioLimitLiquid
Undried
DriedOven= Then the Soil is Organic
MLCL MH CH OL OH Pt
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ChartCassagrande's
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Cu>= 4
Cz=1-3
Cu< 6
Cz=1-3
GPGW
PI < 4
Below
A-Line
PI > 7
Above
A-Line
GM GC
Cu>=6
Cz=1-3
Cu>=6
Cz=1-3
SPSW
PI 7
Above
A-Line
SM SC
ML
MH
OH
CL
CH
OL
Pt
>5% Passing
Sieve 200
5% Passing
Sieve 200
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King Fahad University of Petroleum & Minerals
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Coarse = 100 - 76 = 24%, soFINE-GRAINED SOIL
LL = 60, and PI = 32
Classification CH, inorganic claywith high plasticity
Using Gasagrandi Chart
32607699B
(cumulative % passing)
PILLNo. 200
Sieve
No.4
Sieve
Soil
Solution
Soil B
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ChartCassagrande's
3260Soil B
PILL
Soil
B
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Coarse = 100 - 35 = 65%, so
COARSE-GRAINED SOIL
20% retained on No. 4, vs. 65%coarse, 20/65 = 31% (
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Civil Engineering Department
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AASHTO System
Soils are classified into 7 major groups
A-1 to A-7 Granular
A-1 {A-1-a - A-1-b}
Gravel & Sand
A-2 {A-2-4 - A-2-5 - A-2-6 - A-2-6}
A-3
More than 35% pass # 200
A-4
A-5 Fine
A-6 (Silt & Clay)
A-7CE 353 Dr. Talat Bader
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AASHTO
Soil Classification System
Fair to poorExcellent to goodGeneral
subgrade rating
Clayey SoilsSilty SoilsSilty or clayey gravel and sandFinesand
Stone
fragments,gravel, and
sand
Usual types of
significant
constituent
materials
41 min
11 min
40 max
11 min
41 min
10 max
40 max
10 max
41 min
11 min
40 max
11 min
41 min
10 max
40 max
10 max
-
NP
-
6 max.
Characteristic of
Fraction passingNo. 40
Liquid Limit
Plasticity index
36 min.36 min.36 min.36 min.35 max.35 max.35 max.35 max.
51 min.
10 max
50 max.
25 max.
50 max.
30 max.
15 max.
Sieve Analysis
% Passing
No. 10
No. 40
No.200
A-7-5
A-7-6A-2-7A-2-6A-2-5A-2-4
A-1-bA-1-a
A-7
A-6A-5A-4
A-2
A-3
A-1Group
Classification
Silty-Clay materials
(more than 35% passing No.
200)
Granular materials
(35% or less of Total Sample passing No. 200)
General
Classification
Plasticity index of A-7-5 subgrade is equal to or less than LL minus 30
Plasticity index of A-7-6 subgrade is grater than LL minus 30
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Basses of
AASHTO Classification
I. Grain size
a) Boulders (larger than 75mm), notclassified, but recorded
b) Gravel: material passing 75mm sieve,retained on 2mm sieve
c) Sand: material passing 2mm sieve,retained on 0.075mm sieve
d) Silt and Clay: pass 0.075mm sieve
II. Plasticitya) Plasticity index =10, soils are
silty
b) Plasticity index =11, soils areclayey
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ChartCassagrande's
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Basses of AASHTO
Classification
III. Categories range A-1 to A-7
a) A-1 is most suitable for building
material
b) A-1 through A-3 are sands and
gravels
c) A-4 and A-5 are silty soils
d) A-6 and A-7 are clayey soils
e) A-7 is least suitable
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AASHTO Soil Classification System
Fair to poorExcellent to goodGeneral
subgrade rating
Clayey SoilsSilty SoilsSilty or clayey gravel and sandFinesand
Stone
fragments,gravel, and
sand
Usual types of
significantconstituent
materials
41 min
11 min
40 max
11 min
41 min
10 max
40 max
10 max
41 min
11 min
40 max
11 min
41 min
10 max
40 max
10 max
-
NP
-
6 max.
Characteristic of
Fraction passing
No. 40
Liquid Limit
Plasticity index
36 min.36 min.36 min.36 min.35 max.35 max.35 max.35 max.
51 min.
10 max
50 max.
25 max.
50 max.
30 max.
15 max.
Sieve Analysis
% Passing
No. 10
No. 40
No.200
A-7-5
A-7-6A-2-7A-2-6A-2-5A-2-4
A-1-bA-1-a
A-7
A-6A-5A-4
A-2
A-3
A-1Group
Classification
Silty-Clay materials
(more than 35% passing No.200)
Granular materials
(35% or less of Total Sample passing No. 200)
General
Classification
Plasticity index of A-7-5 subgrade is equal to or less than LL minus 30
Plasticity index of A-7-6 subgrade is grater than LL minus 30
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CE353 Soil Mechanics
Dr.Talat Bader
Civil Engineering Department
King Fahad University of Petroleum & Minerals
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43
AASHTO
Group Index (GI)
GI = (F-35) [0.2 + 0.005( LL-40)] + 0.01( F-15)( PI-10)
Where:
GI can be computed to furtherclassify soils within a given group.
F= % Passing Sieve 200
LL= Liquid Limits
PI= Plasticity Index
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Basses of AASHTO
Classification
IV. Subcategories , or Group
Index Numbers
a) Range from 0 (best engineering
value)
b) to 20 (poorest engineering
value)
c) In parentheses following
category
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Example 4
Classify the following soil Using
AASHTO System.
Given:
% passing Sieve No. 200 = 95%
LL = 60
PI = 40.
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Solution: Example 4
Is A-7-6.
From Table, the group classification
Now we Calculate the Group IndexGI= (F-35) [0.2 + 0.005(LL-40)] + 0.01(F-15)(PI-10)
Using the given data, F= 95, LL=60 & PI=40GI= (95-35)[ 0.2 + 0.005( 60-40)] + 0.01( 95-15)(40-10)GI= (35)[ 0.2 + 0.005(60)] + 0.01( 60-15)(40-10)GI= (35)[ 0.5] + 13.5
GI= 31.0Then GI= 31.0 --> 31
Thus, the AASHTO Classification is
A - 7 - 6 (31); Clayey Soil;Fair to poor As subgrade material
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Fair to poorExcellent to goodGeneral
subgrade rating
Clayey SoilsSilty SoilsSilty or clayey gravel and sandFinesand
Stone
fragments,gravel, and
sand
Usual types of
significant
constituentmaterials
41 min
11 min
40 max
11 min
41 min
10 max
40 max
10 max
41 min
11 min
40 max
11 min
41 min
10 max
40 max
10 max
-
NP
-
6 max.
Characteristic of
Fraction passing
No. 40
Liquid Limit
Plasticity index
36 min.36 min.36 min.36 min.35 max.35 max.35 max.35 max.
51 min.
10 max
50 max.
25 max.
50 max.
30 max.
15 max.
Sieve Analysis
% Passing
No. 10
No. 40
No.200
A-7-5
A-7-6A-2-7A-2-6A-2-5A-2-4
A-1-bA-1-a
A-7
A-6A-5A-4
A-2
A-3
A-1Group
Classification
Silty-Clay materials
(more than 35% passing No.200)
Granular materials
(35% or less of Total Sample passing No. 200)
General
Classification
Red =
Green =% passing Sieve No. 200 = 95%
LL = 60 & PI = 40.
Plasticity index of A-7-5 subgrade is equal to or less than LL minus 30
Plasticity index of A-7-6 subgrade is grater than LL minus 30
Nonof these
No No No
PI(40) is grater than LL-30=(30)
Is A-7-6.Next
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Example # 5Classify the following soil Using
AASHTO System.
% passing No. 10 = 100;
% passing No. 40 = 80;
% passing No.200 = 58
LL = 30; PI = 10.
Given:
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CE353 Soil Mechanics
Dr.Talat Bader
Civil Engineering Department
King Fahad University of Petroleum & Minerals
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49
is A-4.
Solution: Example 5From Table, the group classification
Using the given data, F= 58, LL=30 & PI=10
GI= (58-23)[ 0.2 + 0.005( 30-40)] + 0.01( 58-15)(10-10)GI= (35)[ 0.2 + 0.005(-10)] + 0.01( 58-15)(0)GI= 5.25
Then GI= 5.25 --> 5
Thus, the AASHTO Classification is
A- 4 (5); Silty Soil; Fair to poor As subgrade material
Now we Calculate the Group IndexGI= (F-35) [0.2 + 0.005(LL-40)] + 0.01(F-15)(PI-10)
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% passing No. 10 = 100;
% passing No. 40 = 80;
Fair to poorExcellent to goodGeneral
subgrade rating
Clayey SoilsSilty SoilsSilty or clayey gravel and sandFinesand
Stone
fragments,gravel, and
sand
Usual types of
significantconstituent
materials
41 min
11 min
40 max
11 min
41 min
10 max
40 max
10 max
41 min
11 min
40 max
11 min
41 min
10 max
40 max
10 max
-
NP
-
6 max.
Characteristic of
Fraction passing
No. 40
Liquid Limit
Plasticity index
36 min.36 min.36 min.36 min.35 max.35 max.35 max.35 max.
51 min.
10 max
50 max.
25 max.
50 max.
30 max.
15 max.
Sieve Analysis
% Passing
No. 10
No. 40
No.200
A-7-5
A-7-6A-2-7A-2-6A-2-5A-2-4
A-1-bA-1-a
A-7A-6A-5A-4
A-2A-3
A-1GroupClassification
Silty-Clay materials
(more than 35% passing No.200)
Granular materials
(35% or less of Total Sample passing No. 200)
GeneralClassification
% passing No.200 = 58LL = 30; PI = 10.
Plasticity index of A-7-5 subgrade is equal to or less than LL minus 30
Plasticity index of A-7-6 subgrade is grater than LL minus 30
Nonof these
No No No
Nextis A-4.
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Thank youQuestion Time
Dr. Talat BaderCE 353 Dr. Talat Bader
52
Talat
This is a reminder that thefollowing slides are just for maybe future use as a secondexample or a quiz
http://www.engr.uaa.alaska.edu/Sc
hool_of_Engineering/Courses/eqe694f/694F_F97_Lecture2.html Ask Sam
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Example Classify this
Soil
(sum = 450)
0.06.931.2panpan
6.913.460.40.075200
20.421.295.60.150100
41.619.889.10.25060
61.422.8102.60.42540
84.211.049.50.85020
95.24.821.62.00010
100.00.004.7504
Cumulative
passing
%
retained
Mass
retainedDiameter
US
Sieve
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Calculations
D10 = 0.11
D30 = 0.19
D60 = 0.38
Cu = D60/D10 = 3.45
Cc = (D30)2/(D60
. D10) = 0.86
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CE353 Soil Mechanics
Dr.Talat Bader
Civil Engineering Department
King Fahad University of Petroleum & Minerals
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55
Example #4
Classify the following soil usingUCS
Given: gravel fraction (% retainedon #4) = 30%
Sand fraction (passing #4, retainedon #200) = 40%
Silt and clay fraction (passing#200) = 30%
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Solution example 4
Gravel = 30%,
Sand = 40% and (100-Silts & Clay)/2 = 35%
Since Sand > (100-F)/2, coarse fraction ismore sandy than gravelly
Then
Group symbol is SC
Then, group name is "Clayey sand withgravel"
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Example 5
Classify the following soil using UCS
Given: gravel fraction (% retained on#4) = 0%
sand fraction (passing #4, retainedon #200) = 14%
silt and clay fraction (passing #200)= 86%
LL = 55; PI = 28
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Solution Example 6
Silts & Clay = 86%,
From Chart
Group symbol is CH Inorganic
Clay
Group name is "Fat Clay"
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Unified soil
ClassificationGROUP SYMBOL GROUP NAME
GW WELL-GRADED GRAVEL, FINE TO COARSE GRAVEL
GP POORLY-GRADED GRAVEL
GM SILTY GRAVEL
GC CLAYEY GRAVEL
SW WELL-GRADED SAND, FINE TO COARSE SAND
SP POORLY-GRADED SAND
SM SILTY SAND
SC CLAYEY SAND
ML SILT
CL CLAY
OL ORGANIC SILT, ORGANIC CLAY
MH SILT OF HIGH PLASTICITY, ELASTIC SILT
CH CLAY OF HIGH PLASTICITY, FAT CLAY
OH ORGANIC CLAY, ORGANIC SILT
PT PEAT
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Thank youQuestion Time
Dr. Talat Bader