Sieve Opening, mm Opening, in Soil Type. Cobbles mm 3 in. Gravel mm (2.0 mm) #4 [# 10 for AASHTO) ~0.2 in (~0.

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1 CE 340, Sumer 2015 Soil Classification 1 / 6 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 mm (the mesh size of a #200 sieve). Fine grained soils have particles smaller than 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 [# 10 for AASHTO) 4.75 mm (2.0 mm) ~0.2 in (~0.08 in) # mm ~0.08 in # mm ~0.017 in # mm ~0.003 in Coarse Sand Medium Sand Fine Sand Coarse Grained mm to mm Silt Clay Fine Grained Figure 1. Unified Soil Classification by Grain Size

2 CE 340, Sumer 2015 Soil Classification 2 / 6 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 largergrained 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. 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 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

3 CE 340, Sumer 2015 Soil Classification 3 / 6 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 % gravel % 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 (3 rd 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

4 CE 340, Sumer 2015 Soil Classification 4 / 6 Particle Size Distribution Curve. The results of a sieve analysis 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. Percent Finer poorly graded well graded D 60 D 30 Particle Size (mm) log scale well graded (variety of particle sizes) poorly graded (particles all one size) 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: Else soil is poorly graded. 4 < Cu for gravel or 6 < Cu for sand and 1 < Cc < 3 Where: D C u = Uniformity Coefficient = D C c = Coefficient of Gradation = 2 D30 D60 D 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. 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. These limits are collectively known as the Atterberg Limits after Swedish scientist Albert Atterberg who invented them circa The Atterberg Limits are performed on soil passing the #40 sieve (fine sand and fines). Solid SemiSolid Plastic Liquid Shrinkage Plastic Liquid Limit Limit Limit Plasticity Index (PI) Figure 4. Atterberg Limits 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

5 CE 340, Sumer 2015 Soil Classification 5 / 6 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. 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). Plasticity Index Low Plasticity (Lean) Liquid Limit High Plasticity High Plasticity (Fat) A line PI = 0.73 ( LL 20) Figure 5. Plasticity Chart 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. C u = 6.5, C c = 3.5 Liquid Limit (LL) = 52, Plasticity Index (PI) = 21 Solution: Sieve Wt. Retained, g % Retained Soil Type % gravel % sand pan 50 13% fines Total 360 Primary component = sand, Secondary component = gravel, P 2 = 31% Tertiary component = fines, P 3 = 13% Gradation: C u = 6.5, 6 < C u for sand, so WG (well graded) C c = 3.5, 1 < C c < 3 not true, so PG (poorly graded) Therefore soil is: poorly graded

6 CE 340, Sumer 2015 Soil Classification 6 / 6 Plasticity: PI A_Line = 0.73 ( LL 20) = 0.73 ( ) = 23.4 PI = 21 < 23.4 = PI A_Line, therefore LL = 52 > 50, therefore, silt is Primary component = course grained (sand), Secondary component = course grained (gravel, P 2 = 31%) Since 15% < 31% = P 2, Tertiary component = fines (silt, P 3 = 13%) Since 12% < 13% = P 3, fines = silt fat sand with gravel silty Soil Classification: Silty poorly graded sand with gravel Example #2: 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 % gravel % sand pan % fines Total 190 Primary component = fines, therefore: Secondary component = sand, P 2 = 26% Tertiary component = gravel, P 3 =5% Plasticity: PI A_Line = 0.73 ( LL 20) = 0.73 ( ) = 14.6 PI = 30 > 14.6 = PI A_Line, therefore LL = 40 < 50, therefore, clay is Primary component = fine grained, (silt) Secondary component = course grained (sand, P 2 = 26%) Since 30% < 26% + 5% = P 2 + P 3, fines = clay lean silt sandy Tertiary component = course grained (gravel, P 3 = 5%) Since P 3 = 5% < 15%, leave blank Soil Classification: Sandy lean clay