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University of Tennessee at Chattanooga�ENCE 3610 Soil Mechanics

Lecture 2

Size Gradation of Soil Particles�Soil Classification

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Importance of Particle Size and Gradation

  • Of all of the characteristics of soils, particle size is the single most important one
  • Soils respond differently to loads and many other actions (compaction, addition of water, etc.) based on their particle size
  • Unfortunately virtually any soil sample is a composite of different particle sizes, and the proportion of the different particle sizes is also important in determining soil properties
  • We must be able to determine the distribution of different particle sizes in a given soil sample

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Gradation Tests

ASTM #

Description

Sample Size

Comments

D6913

Particle-Size Distribution (Gradation) of Soils Using Sieve Analysis

50 g – 70 kg (depends upon particle size)

For maximum particle size (99% passing):

0.425 mm: 50 g required

2 mm: 50 g

50 g – 70 kg

4.75 mm: 75 g

9.5 mm: 165 g

19 mm: 1.3 kg

25.4 mm: 3 kg

38.1 mm: 10 kg

50.8 mm: 25 kg

76.2 mm: 70 kg

D7928

Particle- Size Distribution (Gradation) of Fine-Grained Soils Using the Sedimentation (Hydrometer) Analysis

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50 g (dry)

Passing No. 10 sieve and retained on No. 200 sieve

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Sieve Analysis

  • Primarily applied to granular (cohesionless) soils
  • Passes soil sample through a series of sieves of varying mesh fineness
  • Different portions of soil with different grain size pass through each mesh
  • Distribution of grain sizes constructed and plotted

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U.S. Standard Sieve Sizes

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Sieve Analysis for Cohesionless Soils

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Results of Sieve Analysis

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Blank Gradation Chart from DD-1207 (the best to use)

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Sample Results of Sieve Analysis

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Gradation

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Soil Gradation

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Angularity

Angular particled soils generally exhibit better engineering properties; also can frequently pass larger particles through a given sieve size

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Example of Gradation

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Hydrometer Test (for fraction smaller than #200 sieve)

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Properties of Clay Soils

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States of Clay Soils

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Atterberg Limits

  • Shrinkage Limit (SL)
  • Plastic Limit (PL)
  • Liquid Limit (LL)
  • Plasticity Index PI = LL – PL
  • Liquidity Index LI = (w-PL)/PI
  • Consistency Index CI or Ic = (LL-w)/PI
  • The more plastic a soil, it will:
    • Be more compressible
    • Have higher shrink-swell potential
    • Be less permeable
  • Help identify and classify the soil.
  • PI (plasticity index) is an indicator of soil compressibility and potential for volume change.
  • PL (plastic limit) can indicate if clay has been preconsolidated. Most soils are deposited at or near their liquid limit. If the in situ natural water content (w) is near the plastic limit (PL), then the soil is probably preconsolidated. Some stress has been applied in the past to squeeze that water out.
  • PL, LL, PI, LI and CI are always expressed as percentages

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Liquid Limit Test

  • Several samples are prepared with varying moisture contents
  • Specimens placed in bottom of cup and split with grooving tool
  • Crank turned and cup is impacted until groove is closed
  • LL is moisture content at which groove closes with 25 blows

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Plastic Limit Test

  • Samples prepared at varying moisture contents
  • Sample rolled out on glass plate to 4.2 mm (1/8”) diameter
  • PL is defined as the moisture content at which rolled sample crumbles

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Liquid and Plastic Limit Test Sample Results

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Atterberg Limit Example

  • Given soil with following Atterberg Limits:
    • LL = 46
    • PL = 18
  • Find
    • Plasticity Index
    • Shrinkage Limit Using Holtz and Kovacs’ “Quick and Dirty Method”
  • Solution
    • PI = LL – PL
    • PI = 46 – 18
    • PI = 28
    • Holtz and Kovacs Method for SL in equation form:

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Consistency

  • Refers to the texture and strength of a cohesive soil
  • Can be measured in the field with pocket penetrometers, vane shear or torvane testers

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Methods of Classifying Soils

  • USDA Method
    • Developed primarily for agricultural and surface soil purposes
    • Not used often in soil mechanics
  • Unified Classification System
    • Developed by Arthur Casagrande during World War II for the U.S. Army Corps of Engineers
    • Most widely used classification system
  • AASHTO System
    • Originally developed in the 1920’s as the Bureau of Public Roads system
    • Primarily aimed at classification for pavement purposes
  • All methods similar, but differences are significant enough that they should be understood

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USDA System

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Unified Classification System

  • Primary Characteristics
    • G: gravels
    • S: sands
    • C: clays
    • M: silts
    • O: organic soils
    • Pt: peat
  • Secondary Characteristics
    • W: well graded
    • P: poorly graded
    • M: silty (as opposed to a predominant silt in the left column)
    • C: clayey (as opposed to a predominant clay in the left column)
    • L: lean (LL < 50)
    • H: fat (LL > 50)

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SM

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First Decision for Unified System:�Coarse or Fine Grained?

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Soil Classification Chart (Laboratory Method)

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Plasticity Chart

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Importance of Plasticity Index and “A” Line

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Location of clay minerals on the Casagrande Plasticity Chart

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Unified Classification Example�(Soil A)

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Unified Classification Example

  • Given
    • Results of Sieve Test
      • Uniformity Coefficient Cu = 8.1
      • Curvature Coefficient Cc = 0.9
      • Percentage Passing #200 Sieve = 10%
      • Percentage Passing #4 sieve = 89%
    • Result of Atterberg Limit Tests (for portion passing #40 sieve)
      • Liquid Limit LL = 63
      • Plastic Limit PL = 42
      • Plasticity Index = LL-PL=21
  • Find
    • Unified Soil Classification

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Unified Classification Example

  • Question 1:
    • What is the percentage of the material passing the #200 (0.074 mm opening) sieve?
  • Answer:
    • 10%. Since this is < 50%, the soil is a cohesionless (coarse grained) soil
  • Remaining: “G” or “S” classification soils
  • Question 2:
    • What is the percentage of the coarse fraction which is gravel?
  • Answer:
    • 11% of this sample is retained on the #4 sieve
    • This represents 11/(100-10) = 12.2% of coarse fraction
    • Since this is < 50%, this eliminates all of the “G” classification soils
  • Remaining: “S” classification soils

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Unified Classification Example

  • Question 3:
    • How “clean” are the sands?
  • Answer:
    • “Clean” sands or gravels have less than 5% of the material passing the #200 sieve
    • Sands (or gravels) “with fines” have more than 12% of material passing the #200 sieve
    • Since 5% < 10% < 12%, no classifications are eliminated
  • Remaining: “S” classification soils
  • Question 4:
    • How is the soil graded?
  • Answer:
    • Uniformity Coefficient Cu = 8.1, Curvature Coefficient Cc = 0.9.
    • For SW, Cu > 6 and 1 < Cc < 3, so this is eliminated
  • Remaining: SP, SM and SC

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Unified Classification Example

  • Question 5:
    • What are the Atterberg Limits?
  • Answer:
    • Liquid Limit LL = 63, Plastic Limit PL = 42, Plasticity Index = LL-PL=21
    • A-Line Analysis: PI = 0.73(LL -20) = 31.39 > 21, so below the “A” Line
    • “A” Line analysis eliminates SC classification
  • Remaining: SP and SM
  • Final Classification
    • Atterberg Limit is below the “A” line, so SM is possible
    • Cu and Cc do not meet the classification for SW, so SP is possible
    • Soil in question is subject to a dual classification, or SM-SP

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Unified Soil Classification Example, Using Original Gradation Information

  • Since less than 5% of the sample passes the #200 sieve, we can say the following:
    • Soil is a sand (S) or gravel (G)
    • We can eliminate any soils but “W” (well graded) or “P” (poorly graded) soils
  • Since all of the sample passes the #4 sieve, there is no gravel, thus the sample is sand, thus we are left with SW or SP
  • D60 = 0.35, D30 = 0.1, D10 = 0.08 from chart, thus Cu = D60/D10 = 0.35/0.08 = 4.375 and Cc = D302/(D10D60) = (0.1)2/((0.08)(0.35) = 0.357
  • We can see that neither Cu is nor Cc is in the range for well-graded status, thus it is a poorly graded sand (SP)

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AASHTO System

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Notes on AASHTO System

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AASHTO System (ODOT)

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AASHTO Classification Example

  • Given (same soil as before)
    • Results of Sieve Test
      • Uniformity Coefficient Cu = 8.1
      • Curvature Coefficient Cc = 0.9
      • Percentage Passing #10 Sieve: 82%
      • Percentage Passing #40 Sieve: 51%
      • Percentage Passing #200 Sieve = 10%
    • Result of Atterberg Limit Tests (for portion passing #40 sieve)
      • Liquid Limit LL = 63
      • Plastic Limit PL = 42
      • Plasticity Index = LL-PL=21
  • Find
    • AASHTO Soil Classification

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AASHTO Classification Example

  • Examine Sieve Passing Points
    • #10: is greater than 50%, so move past A-1-a
    • #40: is greater than 50%, so eliminate A-1, but A-2 possible
    • #200: is equal to 10%, so A-3 is possible, but unlikely since LL > PL , thus soil has plasticity
    • #200: is less than 35%, so A-2 is possible
  • Examine Liquid Limit and Plasticity Index
    • LL > 40, so eliminate A-2-4 and A-2-6
    • PI > 10, so eliminate A-2-5
    • This leaves A-2-7, which meets all three criteria (#200 sieve, LL and PI)
  • Because this is an A-2-7, we must use partial group index
    • PGI = 0.01(F200 – 15)(PI – 10) = (0.01(10-15)(21-10) < 0, so PGI = 0
  • Classification is A-2-7(0)

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AASHTO Soil Classification Example, Using Original Gradation Information

  • Sieve Analysis Cumulative Passing Results:
    • #10, 98.58%
    • #40, 72.72%
    • #200, 1.7%
  • These number exceed all of the maximum values for A-1-a and A-1-b
  • For A-3, #40 = 72.72% > 51% and #200 = 1.7% < 10%. Additionally no Atterberg Limits are required, so soil can be classified as A-3.

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Differences between Unified and AASHTO Classification Systems

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Questions?