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esab buS edarg buS 5-A

6-A

5-7-A

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5

GI

( Sub grade )

GI = 0 good subgrade soil

GI = 20 poor subgrade soil

EX :- classify soil sample using AASHTO method , the following date were obtained for a soil

sample .

Mechanical Analysis

Plasticity tests Percent finer Sieve No

LL = 48% Pl = 26%

97 4

93 10

88 40

78 100

70 200

Using the AASHTO method for classifying soils , determine the classification of the soil and

state whether this material is suitable in its natural state for use as a sub base material .

SOL /

* since more than 35% of the material passes NO 200 sieve , the soil is

either

A- 4

A -5

A 6

OR

A 7

*LL > 40% and the soil cannot be in group A-4 or A-6 thus , it is either

A-5 or A-7

*The PI is 22% (48 26 ) , which is greater than 10% Thus eliminating

A 5 . the soil is A -7 -5 or A- 7 - 6

*( LL 30 ) = 18 < PI ( 22% ) . Therefore the soil is ( A 7 - 6 ) since the plasticity index

of ( A 7 5 ) soil sub group is less than ( LL 30 ) , the GI is given as :

(70 30 ) [ 0.2 + 0.005 ( 48 40 )] + 0.01 ( 70 15 ) ( 22 10 ) = 8.4 + 6.6 = 15

This soil is ( A 6 5 ( 15 ) ) and is there for unsuitable as a sub base material in its

natural state .

6

2- USCS

1-Coarse grained soils Gravel

Coarse ( 19-75)mm

Fine (4.75-19)mm

Sand

Coarse ( 2-4.75)mm

Medium (0.425-2)mm

Fine (0.075-0.425)mm

2- Fine grained soil

Fine less than NO 200 (0.075mm)

Silt M use Atterberg limits

Clay C use Atterberg limits

3- Organic soils O

4- Peat Pt

7

Well graded W high LL H

Poorly graded p low LL L

Cu=

Cs=

D 60 grain diameter at 60% passing

D 30 grain diameter at 30% passing

D 10 grain diameter at 10% passing

For Gravel Cu > 4

Well graded

1 < Cc < 3

For sand Cu > 6

Well graded

1 < Cc < 3

8

9

EX :- the results obtained from a mechanical analysis and plasticity test an a soil sample are

shown below . classify soil using USCS and state whether or not it can be used in a natural

state as a sub base material

Mechanical Analysis

Plasticity tests Percent passing (by weight)

Sieve No

LL = 40% Pl = 30%

98 4

93 10

85 40

73 100

62 200

SOL /

The grain size distribution is plotted as shown in figure below since more than 50% ( 62%)

of the soil passing the NO 200 sieve , the soil is fine graded plot of the limits on the plasticity

chart is below the "A" line (PI =40 30 = 10 ) Therefor it is either silt or organic clay .

however ,the LL is less than 50% ( 40% ) : therefore it is low LL . this soil can be classified as

ML or OL which is probably equivalent to an A- 4 orA-5 soil in the AASHTO classification

system .

11

EX :- Repeat last example for the data below

Plasticity tests Passing % Sieve No

LL = non plastic Pl = non plastic

95 4

30 10

15 40

8 100

3 200

SOL/

The grain size distribution is plotted as shown in figure below since only 3% passing sieve

NO 200 < 50% , the soil is coarse grained since more than 50% passing sieve NO 4 , the soil is

classified as SAND .because the soil is non-plastic it is necessary to determine its coefficient

of uniformly Cu and coefficient of curvature Cc .

Cu = (D60/D10 ) ( 3.8/0.25) = 15.2 > 6

Cc = ( D30^2/D10*D60 ) (2^2/0.25*3.8) = 4.2 * out of rang (1 3 )

This soil is not well graded is classified as SP and therefore can be used as a sub

base material if properly drained compacted .

11

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21

( )

31

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