SOIL CLASSIFICATION
Engr. Ma. Gracia Antoinette T. Ley-Lorenzo, RMP, MEng CE
USDA Textural Classification Chart
To use the USDA soil textural triangle, plot the percentages of sand, silt, and clay for a
soil sample on the triangle and determine the soil type where the lines representing
those percentages intersect.
Steps to Read the USDA Soil Textural Triangle:
1. Identify the triangle's axes:
The triangle has three sides, each representing a different soil particle size: sand, silt,
and clay.
2. Locate the percentages:
Find the percentages of sand, silt, and clay for your soil sample on the corresponding
sides of the triangle.
3. Draw lines:
From each percentage, draw a line parallel to the opposite side of the triangle.
4. Find the intersection:
The point where all three lines intersect within the triangle indicates the soil's textural
class.
5. Determine the soil type:
The labeled area where the lines intersect represents the soil's texture, such as "Loam,"
"Sandy Loam," etc.
6. Verify the result:
Ensure that the intersection point corresponds to the percentages of sand, silt, and clay
you entered on the triangle.
Key Points about the USDA Soil Textural Triangle:
•12 Textural Classes:
The USDA recognizes 12 soil textural classes, each with a specific combination of sand,
silt, and clay percentages.
Soil Properties:
•Soils within the same textural class tend to have similar physical and chemical
properties.
Practical Use:
•The triangle is a helpful tool for soil scientists, agronomists, and anyone working with
soils.
SAMPLE PROBLEMS:
1. A soil sample has the following composition: 60% sand, 20% silt, and 20% clay.
2. A soil sample has the following composition: 20% sand, 30% silt, and 50% clay.
3. A soil sample has the following composition: 70% sand, 10% silt, and 20% clay.
Activity: Answer 2 & 3
( 1 & 2)
Sieve Analysis
Atterberg limits are critical moisture content values that define the boundaries
between different states of fine-grained soil, such as silt and clay. These limits,
specifically the liquid limit (LL), plastic limit (PL), and shrinkage limit (SL), help classify
and characterize the behavior of fine-grained soils.
1. Liquid Limit (LL):
The LL is the moisture content at which a fine-grained soil can no longer flow like a
liquid.
It's the upper boundary between the liquid and plastic states.
At moisture contents above the LL, the soil is considered liquid and will flow freely.
2. Plastic Limit (PL):
The PL is the moisture content at which a fine-grained soil can no longer be remolded
without cracking.
It's the lower boundary between the plastic and semi-solid states.
At moisture contents below the PL, the soil is considered semi-solid and can't be
easily remolded.
3. Shrinkage Limit (SL):
The SL is the moisture content below which the soil will no longer shrink in
volume upon drying.
It's the lower boundary between the semi-solid and solid states.
At moisture contents below the SL, the soil is considered solid.
Importance of Atterberg Limits:
Soil Classification:
Atterberg limits, along with grain size distribution, are used to classify soils according
to their consistency and plasticity.
Engineering Properties:
They provide valuable insights into the engineering properties of soils, such as their
compressibility, shear strength, and potential for settlement or liquefaction.
Distinguishing Between Silts and Clays:
The LL and PL are particularly useful in distinguishing between silts and clays.
Predicting Soil Behavior:
The Atterberg limits help predict how a soil will behave under different moisture
conditions and loading scenarios.
Determining Plasticity Index (PI):
The difference between the LL and PL, known as the plasticity index (PI), is a measure
of the soil's plasticity.
Sieve No.
4
10
20
40
60
100
200
LL
PL
A
68.5
36.1
21.9
34.1
16.5
Soil, % passing
B
C
69.3
79.5
59.1
48.3
69.0
38.5
28.4
19.8
54.3
4.5
53.5
Non-plastic
31.6
(NP)
Example 1: Classify the soil samples shown below using AASHTO and USCS Systems:
Solution: 1) AASHTO System:
Classifying Soil A
- compute PI, PI=LL-PL=34.1-16.5=17.6
Since 21.9 % passes the No. 200 sieve, the soil is an A-2 with subgroup to be determined from PI and LL.
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.
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)
Sieve No.
4
10
20
40
60
100
200
LL
PL
A
68.5
36.1
21.9
34.1
16.5
Soil, % passing
B
C
69.3
79.5
59.1
48.3
69.0
38.5
28.4
19.8
54.3
4.5
53.5
Non-plastic
31.6
(NP)
Activity: Classify Soil B & C using AASHTO and compute for GI
(3, 4, 5 & 6)
Sieve No.
4
10
20
40
60
100
200
LL
PL
A
68.5
36.1
21.9
34.1
16.5
Soil, % passing
B
79.5
69.0
54.3
53.5
31.6
C
69.3
59.1
48.3
38.5
28.4
19.8
4.5
Non-plastic
(NP)
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
Sieve No.
4
10
20
40
60
100
200
LL
PL
A
68.5
36.1
21.9
34.1
16.5
Soil, % passing
B
79.5
69.0
54.3
53.5
31.6
Classify Soil B & C using USCS System.
( 7 & 8)
Hand written on A4 paper.
Submit your Activity on Monday, until 5pm at TC 102.
C
69.3
59.1
48.3
38.5
28.4
19.8
4.5
Non-plastic
(NP)