Ch 03 05 Soil Clasification

8/21/2019 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

CE 353 Dr. Talat 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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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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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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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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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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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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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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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


subgrade rating


Stone


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


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


(more than 35% passing No.200)

Granular materials


General

Classification




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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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subgrade rating


Stone


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


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



Granular materials


General

Classification

Red =

Green =% passing Sieve No. 200 = 95%

LL = 60 & PI = 40.



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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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;


subgrade rating


Stone


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


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



Granular materials


GeneralClassification

% passing No.200 = 58LL = 30; PI = 10.



Nonof these

No No No

Nextis A-4.


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Thank youQuestion Time

Dr. Talat BaderCE 353 Dr. Talat Bader

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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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Dr.Talat Bader




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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

Ch 03 05 Soil Clasification

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