CE 340, Fall 2015 Soil Classification 1 / 7

The geotechnical engineer predicts the behavior of soils for his or her clients (structural engineers, architects, contractors, etc). A first step is to classify the soil. Soil is typically classified according to its distribution of grain sizes, its plasticity, and its relative density or stiffness.

Classification by Distribution of Grain Sizes. While an experienced geotechnical engineer can visually examine a soil sample and estimate its grain size distribution, a more accurate determination can be made by performing a sieve analysis.

Sieve Analyis. In a sieve analysis, the dried soil sample is placed in the top of a stacked set of sieves. The sieve with the largest opening is placed on top, and sieves with successively smaller openings are placed below. The sieve number indicates the number of openings per linear inch (e.g. a #4 sieve has 4 openings per linear inch).

The results of a sieve analysis can be used to help classify a soil. Soils can be divided into two broad classes: coarse‐grained soils and fine‐grained soils. Coarse‐grained soils have particles with a diameter larger than 0.075 mm (the mesh size of a #200 sieve). Fine‐grained soils have particles smaller than 0.075 mm. The four basic grain sizes are indicated in Figure 1 below: Gravel, Sand, Silt and Clay.

Sieve Opening, mm Opening, in Soil Type

Cobbles

76.2 mm 3 in

Gravel

#4 4.75 mm ~0.2 in

[# 10 for AASHTO) (2.0 mm) (~0.08 in)

Coarse Sand #10 2.0 mm ~0.08 in Grained Medium Sand

#40 0.425 mm ~0.017 in Coarse Fine Sand #200 0.075 mm ~0.003 in

Silt

0.002 mm to

Grained 0.005 mm Fine Clay

Figure 1. Unified Soil Classification by Grain Size CE 340, Fall 2015 Soil Classification 2 / 7

Hydrometer Analysis. The grain size distribution of fine‐grained soil is determined using a hydrometer analysis. In a hydrometer analysis, fine‐grained soil is dispersed in water and its relative density is measured over time. The relative density of a soil with a high percentage of larger‐grained soils will decrease rapidly, since the larger‐grained soils settle more rapidly than fine‐grained soils. The text provides more details regarding hydrometer analysis. Also, you will perform a hydrometer analysis in the laboratory portion of this course.

Particle‐Size Distribution Curve. The results of sieve and hydrometer analyses can be plotted to show the distribution of grain sizes. Soils with an even‐distribution of grain sizes are called “well‐graded” and soils with predominantly one grain size are called “poorly‐graded” (see Figure 4 below). Well‐graded soils, having a variety of particles sizes, can be packed “tighter” than poorly‐graded soils leading to higher unit weight and therefore higher strength and lower settlement potential.

poorly‐graded Finer

60

30 well‐graded Percent 30 60 well‐graded poorly‐graded D D (variety of particle sizes) (particles all one size) Particle Size (mm) ‐log scale

Figure 4. Particle‐Size Distribution Curve

Soils can be classified as either “well‐graded” or “poorly‐graded” according to the following criteria:

Soil is well‐graded if: 4 <= Cu for gravel or Where: 6 <= Cu for sand D60 – and – Cu = Uniformity Coefficient = D 1 <= Cc <= 3 10 Cc = Coefficient of Gradation = 2 Else soil is poorly‐graded. D 30 D60  D10

Classification of Fine‐grained Soils by Plasticity. The engineering properties of fine‐grained soils are affected more by their plasticity than by their grain size distribution. Clay particles are mostly flake‐shaped particles that exhibit plasticity (become putty‐like) when mixed with a small amount of water. The mineral composition of the clay has a large effect on its plasticity.

CE 340, Fall 2015 Soil Classification 3 / 7

In the presence of very little or no water, the clay soil behaves like a solid. With increasing moisture content, the soil behaves like a semi‐solid, then like a plastic putty, and finally like a liquid. The moisture contents representing the transition points between these four states are the shrinkage limit, the plastic limit and the liquid limit, as indicated in Figure 4 below.

Solid SemiSolid Plastic Liquid

Shrinkage Plastic Liquid Limit Limit Limit Plasticity Index (PI)

Figure 4. Atterberg Limits

These limits are collectively known as the Atterberg Limits after Swedish scientist Albert Atterberg who invented them circa 1900. The Atterberg Limits are performed on soil passing the #40 sieve (fine sand and fines). The liquid limit is the moisture content at which the soil flows (the groove closes along a ~½‐inch length when the brass cup holding the soil is tapped 25 times). The plastic limit is the moisture content at which 1/8‐inch diameter threads crumble. These tests are described in more detail in the text and you will perform them in the laboratory portion of this course. The Plasticity Index (PI) is defined as the difference between the Liquid Limit and the Plastic Limit, and is used to help classify fine‐grained soils in the Unified Classification System.

Plastic Limit: moisture content at which 1/8‐inch threads crumble Liquid Limit: moisture content at which groove closes over ~1/2‐inch length

Arthur Casagrande noticed that soils could be classified according to their liquid limit (LL) and plasticity index (PI). Soils that plot on similar regions of his Plasticity Chart had similar engineering properties. The primary feature of the Plasticity Chart is the “A‐line” which divides inorganic clay from inorganic silts and organic soils (see Figure 5 below).

80 70 Low Plasticity HighHigh PlasticityPlasticity (Lean) (Fat) 60 A‐line PI = 0.73 ( LL –20) 50 Index

40 30

Plasticity 20 10 0 0 20406080100120 Liquid Limit

Figure 5. Plasticity Chart CE 340, Fall 2015 Soil Classification 4 / 7

Classification (Unified System) by Grain Size Only. Two classification systems exist for classifying soils according to their engineering properties: The AASHTO (American Association of State Highway and Transportation Officials) system and the Unified Soil Classification System (USCS). The AASHTO systems is used by departments of transportation for highway construction; the Unified System is favored by geotechnical engineers. In both systems, soils are classified according to their grain size distribution and their plasticity (described in the next section). We will focus on the Unified System in these notes (the text describes the AASHTO system.)

The basic procedure for classifying soils according to grain size distribution is as follows: 1. Perform a sieve analysis 2. Determine the percentages of gravel, sand and fines using the criteria shown in Figure 1 3. Describe the soil using the USCS criteria shown in Figure 2.

1. Identify primary component (largest percentage) 2. Identify second‐largest percentage If primary component is coarse‐grained, then If secondary component is course‐grained, then If 15% < secondary, WITH SAND or WITH GRAVEL

If secondary component is fine‐grained, then If % fines < 5%, CLEAN If 5% < % fines < 12%, WITH FINES If 12% < % fines, SILTY or CLAYEY

If primary component is fine‐grained then If secondary component is course‐grained, then If 30% < sand + gravel, If sand > gravel SANDY If gravel > sand GRAVELLY If 15% < sand + gravel < 30%, If sand > gravel WITH SAND If gravel > sand WITH GRAVEL

If tertiary component is course‐grained, then If 15% < tertiary, WITH SAND or WITH GRAVEL

3. Repeat Step 2 for third‐largest component if necessary. Figure 2. Procedure for Classifying Soils by Grain Size Using USCS CE 340, Fall 2015 Soil Classification 5 / 7

Example #1, Soil Classification Using Results of Sieve Analysis.

Write the soil classification based on the following sieve analysis results: 40 g retained on the #4 sieve, 72 g retained on the #200 sieve, 35 g retained in the pan.

Solution: Sieve Wt. Retained, g % Retained Soil Type

4 40 27% gravel 200 72 49% sand pan 35 24% fines

Total 147

Figure 3. Results of Sieve Analysis

Assume for this example that the fines are classified as “clay” by Atterberg Limits tests (see next section).

Step Soil Description 1. Primary component is sand sand 2. Secondary component is gravel Since primary component is coarse‐grained And 15% < secondary component sand with gravel 3. Tertiary (3rd‐largest) component is clay Since primary component is coarse‐grained And tertiary component is fine‐grained And 12% < % fines clayey sand with gravel

Soil Classification: clayey sand with gravel

Example #2: Write the soil classification based on the following sieve analysis results and Atterberg Limit test results:  110 g retained on the #4 sieve, 200 g retained on the #200 sieve, 50 g retained in the pan.  Cu = 6.5, Cc = 3.5  Liquid Limit (LL) = 52, Plasticity Index (PI) = 21

Solution: Sieve Wt. Retained, g % Retained Soil Type 4 110 31% gravel 200 200 56% sand pan 50 13% fines Total 360 CE 340, Fall 2015 Soil Classification 6 / 7

Primary component = sand, Secondary component = gravel, P2 = 31% Tertiary component = fines, P3 = 13%

Gradation: Cu= 6.5, 6 <= Cu for sand, so WG (well‐graded) Cc = 3.5, 1 <= Cc <= 3 not true, so PG (poorly‐graded) Therefore soil is: poorly graded

Plasticity: PIA_Line = 0.73 ( LL – 20) = 0.73 ( 52 – 20 ) = 23.4 PI = 21 < 23.4 = PIA_Line , therefore fines = silt

LL = 52 > 50, therefore, silt is fat

Primary component = course‐grained (sand), sand Secondary component = course‐grained (gravel, P2 = 31%) Since 15% < 31% = P2, with gravel

Tertiary component = fines (silt, P3 = 13%) Since 12% < 13% = P3, silty

Soil Classification: Silty poorly‐graded sand with gravel

CE 340, Fall 2015 Soil Classification 7 / 7

Example #3: Write the soil classification based on the following sieve analysis results and Atterberg Limit test results:  10 g retained on the #4 sieve, 40 g retained on the #200 sieve, 130 g retained in the pan.  LL = 40, PI = 30

Solution: Sieve Wt. Retained, g % Retained Soil Type 4 10 6% gravel 200 40 22% sand pan 130 72% fines

Total 190

Primary component = fines, therefore: Secondary component = sand, P2 = 22% Tertiary component = gravel, P3 =6%

Plasticity: PIA_Line = 0.73 ( LL – 20) = 0.73 ( 40 – 20 ) = 14.6 PI = 30 > 14.6 = PIA_Line , therefore fines = clay

LL = 40 < 50, therefore, clay is lean

Secondary component = course‐grained (sand, P2 = 22%) Since 15% < 22% + 5% = P2 + P3 < 30% with sand

Tertiary component = course‐grained (gravel, P3 = 6%) Since P3 = 6% < 15%, leave blank

Soil Classification: lean clay with sand