The document provides examples of classifying soils using the AASHTO and USCS soil classification systems. Key steps include determining the particle size distribution, plasticity characteristics (liquid limit, plastic limit, plasticity index), and using this data on classification charts to identify the appropriate soil type symbols. Soils are classified as sand, silt, clay or combinations based on their grain size and plasticity properties.

1. Example 1: Classify the soil samples shown below using AASHTO and
USCS Systems:
Soil, % passing
Sieve No. A B C
4 - - 69.3
10 68.5 79.5 59.1
20 - - 48.3
40 36.1 69.0 38.5
60 - - 28.4
100 - - 19.8
200 21.9 54.3 4.5
LL 34.1 53.5 Non-plastic
(NP)PL 16.5 31.6
Solution: 1) AASHTO System:
Classifying Soil A
1. compute PI, PI=LL-PL=34.1-16.5=17.6
2. Since 21.9 % passes the No. 200 sieve, the soil is an A-2 with
subgroup to be determined from PI and LL.
3. proceeding across AASHTO table from left to right with
LL=34.1<40 and PI=17.6>11
the first soil which satisfies these criteria is an A-2-6 soil.
4. next compute the GI, because the soil is A-2-6, then
GI=0.01(F-15)(PI-10)=0.01(21.6-15)(17.6-10)=0.52
rounding to the nearest whole number, obtain GI=1
Therefore the final classification of soil is A-2-6(1)
Classifying Soil B
1. The PI is, PI= 53.5- 31.6 = 21.9, therefore, 11 minimum controls.
2. The percent passing sieve No. 200 is 54.3 > 35, therefore, the soil is
an
A-4, A-5, A-6 or A-7.
3. With LL = 53.5 and PI = 21.9 the soil is an A-7, but we must still find
if it is an A-7-5 or A-7-6.
4. For LL = 53.5
PI = 53.5 – 30 = 23.5 > 21.9, therefore the soil is A-7-5.
5. Compute the G.I.

2. GI = (F200-35)[0.2+0.005(LL-40)]+0.01(F200-15)(PI-10)
= (54.3-35)[0.2+0.005(53.5-40]+0.01(54.3-15)(21.9-10)
=9.84, use 10
6. The final classification of the soil B is: A-7-5(10)
Classifying Soil C
The soil is either A-1 or A-3.with the percent passing the sieve No. 40 of
38.5 < 51, but 38.5 > 30, the soil must be A-1-b.
There is no group index for this soil.
The final classification of the soil C is A-1-b
Solution: 2) USCS System:
Classifying Soil A
1. we have 21.9<50 % passing the No. 200 sieve and more than 50%
passing the No. 4 sieve (since 68.5 % passed the No. 10) therefore, the
soil is either SM or SC
2. based on LL of 34.1 and PI=34.1-16.5=17.6, we obtained the
coordinates on the plasticity chart that the soil is a CL. Taking C for
"clay" the soil is classified as , SC
Classifying Soil B
a) Since 54.3% passes the No. 200 sieve, the soil is immediately fine-
grained and is MH, OH, or CH because the LL = 53.5 > 50 percent.
b) From the plasticity chart at LL = 53.5 and PI = 21.9, the soil
coordinates are difficult to read so we will compute PI = 0.73(LL-20)
= 0.73(53.5-20) = 24.4 > 21.9 therefore, the soil is an MH from the
coordinate location.
Classifying Soil C
a) with 4.5 % passing the No. 200 sieve, soil is either GW, GP, SW, SP
b) it will be necessary to plot a grain size distribution curve and obtained
D60=2.00 mm
D30=0.29 mm
D10=0.086 mm
To compute
Cu=2/0.086=23.3 > 6
Cc=0.292
/(2*0.086)=0.5 < 1 (not between 1 and 3)
Since the gradation criteria are not met for well-graded, the soil is
poorly graded (P suffix)

3. c) with 95.5 % retained on the No. 200 sieve and with 69.3 % passing
the No. 4 sieve, the percent between the No. 4 and No. 200 is
69.3.4.5 = 64.8
% retained on No. 4 = 100-69.3 = 30.7
%passing No. 200 = 4.5
100 %
With 64.8 % as sand the soil is classified as, SP
Example: 2) Explain the engineering properties of SP-SM soil group
Solution:
1. The soil is poorly graded sand with slight silt.
2. The soil is coarse grained (i.e. less than 50% of the soil sample
was passed sieve No. 200).
3. The soil is sand that means more than 50% of the coarse
fraction was passed sieve No. 4.
4. Poorly graded means that cu or cc or both of them not meet the
specifications.
5. The soil is carry double symbol SP-SM means that the percent
passing No. 200 sieve is between 5-12%.
6. The soil is located below the A-line of plasticity chart because
the soil is content silt.
Example: 3) Classify the soil according to the USCS and determine
percentage of gravel and sand fraction.
% passing sieve No. 4 = 86 %, D10 (mm) = 0.1, D60 (mm) = 0.9
% passing sieve No. 200 = 12 %, D30 (mm) = 0.32, PL = 26 %, PI = 10 %
Solution:
Less than 50 % of the soil sample is passed sieve No. 200; therefore the soil
is coarse-grained (G or S). And more than 50 % of the coarse fraction passed
sieve No. 4 therefore, the soil is sand (S).
69
0.1
0.9
D
D
c
10
60
u >=== ,
( ) 14.1
0.1*0.9
32.0
DD
D
c
2
1060
30
2
c === between 1 and 3
Therefore the soil is W
% passing sieve No. 4 = 86 %, therefore, the % gravel = 14 %
% passing sieve No. 200 = 12 %, the soil requires dual symbols SW-SM,
SW-SC, SP-SM, or SP-SC
LL = PL + PI = 26 + 10 = 36%, from plasticity chart the soil is located
below A-line, therefore the final soil classification is SW-SM

4. Example: 4) Ninety-five percent of a soil passes through the No. 200 sieve
and has a LL = 60 % and PL = 20 %. Classify the soil by AASHTO System.
Solution:
PI = 60-20 = 40 %
Passing sieve No. 200 = 95 %
Therefore, the soil is A-7
For A-7-5, PI ≤ LL-30 = 40-30 = 10 not correct
For A-7-6, PI > LL-30 = 40-30 = 10 correct
Therefore the soil classified as A-7-6, hence
GI = (F200-35) [0.2+0.005(LL-40)] +0.01(F200-15) (PI-10)
= (95-35) [0.2+0.005(60-40)] +0.01(95-15) (40-10)
= 42
So, the final soil classification is A-7-6(42)
Example: 5) Classify the following soil using AASHTO system.
% passing sieve # 10 = 100% LL = 30 %
sieve # 40 = 80% PL = 20 %
sieve # 200 = 58%
Solution:
1. PI = 30 – 20 = 10 %
2. % passing sieve # 200 > 35, therefore the soil is A-4, A-5, A-6 or A-7.
3. LL = 30 % and PI = 10%, therefore the soil is A-4.
4. GI = (F200-35) [0.2+0.005(LL-40)] +0.01(F200-15) (PI-10)
= (58-35) [0.2+0.005(30-40)] +0.01(58-15) (10-10)
= 3.45 round to (3.0)
Therefore the final soil classification is A-4 (3).
Example: 6) For, a soil specimen, given the following
% passing sieve # 4 = 92% sieve # 10 = 81%
sieve # 40 = 78% sieve # 200 = 65%
LL = 48 % PI = 32 %
Classify the soil by AASHTO and USCS .
Solution: A. AASHTO system
1. PI = 32 %
2. % passing sieve # 200 > 35, therefore the soil is A-4, A-5, A-6 or A-7.
3. LL = 48 % and PI = 32%, therefore the soil is A-7.
4. LL – 30 = 48 – 30 = 18 % < PI, therefore the soil is A-7-5.
5. GI = (F200-35) [0.2+0.005(LL-40)] +0.01(F200-15) (PI-10)
= (65-35) [0.2+0.005(48-40)] +0.01(65-15) (32-10)

5. = 18.2 = 18
The final soil classification is A-7-6 (18).
B. USCS
Since more than 50% of the soil sample passing through a sieve #
200, it is a fine-grained soil, i.e., it could be ML, CL, MH, or CH. Now, if
we plot LL=48 and PI=32 on the plasticity chart, it falls in the zone CL. So
the soil is classified as CL.
Example: 7) Using AASHTO system classify the soil with properties of
% passing sieve # 4 = 79% LL = 35 %
sieve # 200 = 24% PL = 27 %
Solution:
Because the % passing sieve # 200< 35, therefore the soil either
A-1, A-2, or A-3
From sieve analysis the A-1a and A-3 are canceled.
From atterberg's limits PI = 35 – 27 = 8 > 6, therefore A-1b also is canceled
LL = 35%, therefore A-2-5 and A-2-7 are canceled
PI = 8 < 11, therefore A-2-6 canceled also. Then the soil is A-2-4
For A-2 the GI = 0, therefore
The final soil classification is A-2-4(0)
Example: 8) Explain the engineering properties of soil CL.
Solution:
1. Clay soil with low plasticity
2. More than 50% of soil sample passes sieve #200
3. it is located above the A-line from the plasticity chart
4. Low plasticity means that the LL is less than 50%.
Example: 9) classify the soil using USCS
Sieve # 1 in 3/4 1/2 4 10 40 200
Size, mm 25.4 19.05 12.7 4.75 2 0.42 0.075
% passing 100 90 70 60 30 10 7
LL = 45% and PL = 33%
Solution:
Less than 50% of the soil passes sieve #200, then it is coarse
grained
Gravel = 100 – 60 = 40%
Sand = 60 – 7 = 53%
Sand > gravel, therefore the soil is sand (S)

6. 7% is between 5 – 12%, therefore use double symbols
SW-SM, SW-SC, or SP-SM, SP-SC
From soil distribution curve
611.31
0.42
4.75
D
D
c
10
60
u >===
( ) 005.2
0.42*4.75
2
DD
D
c
2
1060
30
2
c === between 1 and 3
Therefore, SW-SM or SW-SC
From plasticity chart with LL = 45 and PI = 45-33 = 12, the point is located
below A-line, therefore is M
The final classification of the soil is SW-SM
Example: 10)
Classify the following soil using USCS
% passing sieve # 4 = 70% LL = 30 %
sieve # 200 = 30% PL = 18 %
Solution:
1. F200 = 30 % < 50% , therefore the soil is coarse grained soil (gravel or
sand)
2. Sand fraction = 70 – 30 = 40% > gravel fraction 100 – 70 =30%,
therefore the soil is sand (S).
3. More than 12% of the soil passes sieve # 200, the soil is either SM or
SC
4. From plasticity chart with LL=30% and PI=12%, the soil is located
above the A-line, therefore the symbol is C
5. The final classification of the soil is SC (Clayey sand with gravel)
Example: 11)
Classify the following soil using USCS
% passing sieve # 4 = 100% LL = 55 %
sieve # 200 = 86% PL = 27 %
Solution:
1. F200 = 86 % > 50% , therefore the soil is fine grained soil (clay or silt)
2. From plasticity chart with LL=55% and PI=28%, the soil is
located above the A-line, therefore the group symbol is CH (clay with
high plasticity)
(inorganic clay) or (fat clay).

7. Example: 12)
Classify the following soil using USCS and AASHTO.
% passing sieve # 4 = 71% LL = 41 %
sieve # 200 = 43% PL = 27 %
Solution: A. AASHTO system
1. PI = 41 – 27 =14 %
2. % passing sieve # 200 > 35, therefore the soil is A-4, A-5, A-6
or A-7.
3. LL = 41 % and PI = 14%, therefore the soil is A-7.
4. LL – 30 = 41 – 30 = 11 % < PI, therefore the soil is A-7-5.
5. GI = (F200-35) [0.2+0.005(LL-40)] +0.01(F200-15) (PI-10)
= (43-35) [0.2+0.005(41-40)] +0.01(43-15) (14-10)
= 2.76 = 3
The final soil classification is A-7-6 (3).
B. USCS
1. F200 = 43 % < 50% , therefore the soil is coarse grained soil
(gravel or sand)
2. Gravel fraction = 100 – 71 = 29% > Sand fraction 71 – 43
=28%, therefore the soil is sand (G).
3. More than 12% of the soil passes sieve # 200, the soil is either
GM or GC
4. From plasticity chart with LL=41% and PI=14%, the soil is
located below the A-line, therefore the symbol is M.
5. Therefore the final soil classification is GM.