Solution:Example 1. Example 2. Solution: Example 2. clay. Textural Soil Classification System (USDA) CE353 Soil Mechanics Dr.

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1 CE353 Soil Mechanics CE353 Lecture 5 Geotechnical Engineering Laboratory SOIL CLASSIFICATION Lecture 5 SOIL CLASSIFICATION Dr. Talat A Bader Dr. Talat Bader 2 Requirements of a soil Systems Why do we need to classify soils? Based on a scientific method To describe various soil types Simple encountered in the nature in a systematic way and gathering soils that have distinct physical properties in groups and units. Permit classification by visual and manual tests. Should be accepted to all engineers Describe certain engineering properties 4 3 Systems The Simplest Unified o Geologic Soil System o Agronomic Soil System Textural Soil System (USDA) Unified Soil System (USCS) Grain Sie (mm) system Gravel Sand Silt Clay AASHTO <0.002 MIT <0.002 ASTM <0.005 USDA < American Association of State Highway Transportation Officials System (AASHTO) American Society for Testing and Materials System (ASTM) Federal Aviation Agency System (FAA) Others

12 Subgroups in the system 8 7 The USDA classification system is based on the relative % sand, % silt, and % clay. % with 0.05mm d 2mm Relative % Sand = 100% %with d 2 mm % with 0.002mm d.

2 Textural Soil System (USDA) Also Called Agriculture System USDA considers only: Sand Silt Clay No. Gravel in the System If you encounter gravel in the soil Subtract the % of gravel from the 100%. 12 Subgroups in the system 8 7 The USDA classification system is based on the relative % sand, % silt, and % clay. % with 0.05mm d 2mm Relative % Sand = 100% %with d 2 mm % with 0.002mm d.05mm Relative % Silt = 100% %with d 2mm These %age Are Basses for the Soil Type Cobbles & Boulders Gravels Sands Diameter Range d > 75mm 2mm d 75mm 0.05mm d < 2mm % with d 0.002mm Relative % Clay = 100% %with d 2mm Based On Silts Clays 0.002mm d < 0.05mm d 0.002mm 10 9 Example 1 Textural Soil System (USDA) Classify the following soil by the USDA Textural System. Given Data % gravel = 18; % sand = 51; % silt = 22; %clay =

3 Textural Soil System (USDA) Back Sand=62% Silt =27% Clay=11% 14 Given Data % gravel = 18; % sand = 51; % silt = 22; % clay = 9 Solution:Example 1 Since we have 18% Gravel the soil to be classified is = = 82% Calculating the % sand = 51/ 82 = 62% Calculating the % silt = 22/ 82 = 27% Calculating the % clay = 09/ 82 = 11% Using these values in the USAC Then the Soil is Classified as sandy loam However, due to the presence of 18% gravel in the soil, it is called "gravelly sandy loam." 13 Given Data % gravel = 0; % sand = 30; % silt = 30; % clay = 40 Solution: Example 2 % sand Present = 30/ 100 = 30% % silt Present = 30/ 100 = 30% % clay Present = 40/ 100 = 40% Using these values in the USAC Example 2 Classify the following soil by the USDA Textural System. Given: % gravel = 0 ; % sand = 30; % silt = 30; % clay = 40; Then the Soil is Classified as clay Unified System (UCS) Textural Soil System (USDA) First Developed by the Army Corps of Engineers Based in Last modified in 1991 Used by Geotechnical engineers for selecting appropriate soils in non-highway projects Divides soils into two categories Coarse & Fine Atterberg Limits checks for Law and High Compressibility USC Main Groups G = Gravel M = Silt O = Organic S = Sand C = Clay 18 sand=30% silt =30% clay =40% 17 3

4 Unified Soil (USC) System Coarse Grained Less than 50% Passing Fine Grained More than 50% Passing Less than 50% Passing Sieve 4 Gravel More than 50% Passing Sieve 4 Sand Clay Silt Organic Unified Soil (USC) System Course Grained >50% passing Gravelly Soils Less than 50% passing Sieve 4 Sandy Soils More than 50% passing Sieve 4 >5% Passing <12% Passing >5% Passing <12% Passing C u >= 4 C =1-3 GW C u < 6 C =1-3 GP PI < 4 Below A-Line PI > 7 Above A-Line C u >=6 C =1-3 SW C u >=6 C =1-3 SP PI <4 Below A-Line PI >7 Above A-Line 22 GM GC SM SC 21 Unified Soil (USC) System Fine Grained >50% passing 24 CL ML MH CH OL OH Pt LL Oven Dried LiquidLimit Ratio= 0.75 Then the Soil is Organic LL Undried 23 4

Unified soil GROUP SYMBOL GROUP NAME GW WELL-GRADED GRAVEL, FINE TO COARSE GRAVEL GP POORLY-GRADED GRAVEL GM SILTY GRAVEL GC CLAYEY GRAVEL SW WELL-GRADED SAND, FINE TO COARSE SAND SP POORLY-GRADED

5 Unified soil GROUP SYMBOL GROUP NAME GW WELL-GRADED GRAVEL, FINE TO COARSE GRAVEL GP POORLY-GRADED GRAVEL GM SILTY GRAVEL GC CLAYEY GRAVEL SW WELL-GRADED SAND, FINE TO COARSE SAND SP POORLY-GRADED SAND SM SILTY SAND SC CLAYEY SAND ML SILT CL CLAY OL ORGANIC SILT, ORGANIC CLAY MH SILT OF HIGH PLASTICITY, ELASTIC SILT CH CLAY OF HIGH PLASTICITY, FAT CLAY OH ORGANIC CLAY, ORGANIC SILT PT PEAT 26 >5% Passing C u >= 4 C u < 6 C =1-3 C =1-3 GW GP Summery Unified Soil System Coarse Grained Fine Grained Less than 50% Passing More than 50% Passing Gravelly Soils Sandy Soils Less than 50% More than 50% passing Sieve 4 passing Sieve 4 <12% Passing PI < 4 PI > 7 Below Above A-Line A-Line GM GC >5% Passing C u >=6 C u >=6 C =1-3 C =1-3 SW SP <12% Passing PI <4 PI >7 Below Above A-Line A-Line SM SC CL ML CH MH OL OH Pt 25 Example 1 Classify the following soils Using Unified System. Summary Unified Soil Soil No. 4 Sieve No. 200 Sieve LL PI (cumulative % passing) A B C LL PI Soil A Soil No.4 No. 200 LL PI Sieve Sieve (cumulative % passing) A Solution Soil A Coarse = = 52% (retained on No. 200), so COARSE-GRAINED SOIL 8% retained on No. 4, vs. 52% coarse, 8/52 = 15% (<50%), so SAND Soil A Using the LL & PL values in the USAC Atterberg limit above line A, so CLAY SC, clayey sand

6 Soil B LL 60 Soil B PI 32 Soil No.4 No. 200 LL PI Sieve Sieve (cumulative % passing) B Solution Soil B Coarse = = 24%, so FINE-GRAINED SOIL LL = 60, and PI = 32 Using Gasagrandi Chart CH, inorganic clay with high plasticity Soil No.4 No. 200 LL PI Sieve Sieve (cumulative % passing) C Solution Soil C LL PI Soil C 24 2 Coarse = = 65%, so COARSE-GRAINED SOIL Soil C 20% retained on No. 4, vs. 65% coarse, 20/65 = 31% (<50%), so SAND Using Gasagrandi Chart SM, Silty sand AASHTO AASHTO SOIL CLASSIFICATION Originally proposed in 1919; the system was last modified in Considers both texture (GSD) and Atterberg Limits.. This system is widely used by highway and transportation engineers. Who is AASHTO? American Association of State Highway and Transportation Officials System What is it used for? Performed on that part of a soil sample that falls in the gravel <--> clay sie range. for design and construction of highways

7 AASHTO Soil System Group Sieve Analysis % Passing No max. 30 max. 50 max. No max. 25 max. Characteristic of Fraction passing Liquid Limit - Plasticity index 6 max. Usual types of Stone significant fragments, constituent gravel, and materials sand subgrade rating Granular materials (35% or less of Total Sample passing No. 200) A-1 A-1-b A-1-a A-3 51 min. - NP Fine sand A max. 35 max. 35 max. 35 max. 40 max A min A-2 Silty or clayey gravel and sand Excellent to good A max A min 36 min. 40 max Silty Soils Plasticity index of A-7-5 subgrade is equal to or less than LL minus 30 Plasticity index of A-7-6 subgrade is grater than LL minus 30 Silty-Clay materials (more than 35% passing No. 200) A-7 A-4 A-5 A-6 A-7-5 A min. 36 min. 36 min. 41 min 40 max Clayey Soils Fair to poor 41 min 38 AASHTO System Soils are classified into 7 major groups A-1 to A-7 Granular A-1 {A-1-a - A-1-b} Gravel & Sand A-2 {A A A A-2-6} A-3 More than 35% pass # 200 A-4 A-5 Fine A-6 (Silt & Clay) A-7 37 Basses of AASHTO I. Grain sie a) Boulders (larger than 75mm), not classified, but recorded b) Gravel: material passing 75mm sieve, retained on 2mm sieve c) Sand: material passing 2mm sieve, retained on 0.075mm sieve d) Silt and Clay: pass 0.075mm sieve 40 II. Plasticity a) Plasticity index =10, soils are silty b) Plasticity index =11, soils are clayey 39 AASHTO Soil System Group Sieve Analysis % Passing No. 10 No.200 Characteristic of Fraction passing Liquid Limit Plasticity index Usual types of significant constituent materials subgrade rating Silty-Clay materials Granular materials (more than 35% passing No. (35% or less of Total Sample passing No. 200) 200) A-1 A-2 A-7 A-1-b A-3 A-4 A-5 A-6 A-7-5 A-1-a A-2-4 A-2-5 A-2-6 A-2-7 A max. 30 max. 50 max. 51 min. 15 max. 25 max. 35 max. 35 max. 35 max. 35 max. 36 min. 36 min. 36 min. 36 min max 41 min 40 max 41 min 40 max 41 min 40 max 41 min 6 max. NP Stone fragments, Fine Silty or clayey gravel and sand Silty Soils Clayey Soils gravel, and sand sand Excellent to good Fair to poor Basses of AASHTO III. Categories range A-1 to A-7 a) A-1 is most suitable for building material b) A-1 through A-3 are sands and gravels c) A-4 and A-5 are silty soils d) A-6 and A-7 are clayey soils e) A-7 is least suitable Plasticity index of A-7-5 subgrade is equal to or less than LL minus 30 Plasticity index of A-7-6 subgrade is grater than LL minus

8 Basses of AASHTO IV. Subcategories, or Group Index Numbers a) Range from 0 (best engineering value) b) to 20 (poorest engineering value) c) In parentheses following category 44 AASHTO Group Index (GI) GI = (F-35) [ ( LL-40)] ( F-15)( PI-10) Where: F= % Passing LL= Liquid Limits PI= Plasticity Index GI can be computed to further classify soils within a given group. 43 Solution: Example 4 From Table, the group classification Is A-7-6. Now we Calculate the Group Index GI= (F-35) [ (LL-40)] (F-15)(PI-10) Using the given data, F= 95, LL=60 & PI=40 GI= (95-35)[ ( 60-40)] ( 95-15)(40-10) GI= (35)[ (60)] ( 60-15)(40-10) GI= (35)[ 0.5] GI= 31.0 Then GI= > 31 Thus, the AASHTO is A (31); Clayey Soil; Fair to poor As subgrade material 46 Example 4 Classify the following soil Using AASHTO System. Given: % passing Sieve No. 200 = 95% LL = 60 PI = Example # 5 Classify the following soil Using AASHTO System. Given: % passing No. 10 = 100; % passing = 80; % passing No.200 = 58 LL = 30; PI = Group Sieve Analysis % Passing No max. 30 max. 50 max. No max. 25 max. Characteristic of Fraction passing Liquid Limit - Plasticity index 6 max. Usual types of Stone significant fragments, constituent gravel, and materials sand subgrade rating % passing Sieve No. 200 = 95% LL = 60 & PI = 40. Granular materials (35% or less of Total Sample passing No. 200) A-1 A-2 A-1-b A-3 A-1-a A-2-4 A-2-5 A-2-6 A min. Non - 40 max 41 min NP of these Fine sand 35 max. 35 max. 35 max. 35 max. Silty or clayey gravel and sand Excellent to good 40 max 41 min Silty-Clay materials (more than 35% passing No. 200) A-7 A-4 A-5 A-6 A-7-5 A min. 40 max Silty Soils Red = Green = 36 min. 36 min. 36 min. No No No 41 min 40 max Clayey Soils Fair to poor 41 min Plasticity index of A-7-5 subgrade is equal to or less than LL minus 30 Plasticity index of A-7-6 subgrade is grater than LL minus 30 PI(40) is grater Next than LL-30=(30) Is A

9 CE353 Soil Mechanics % passing No. 10 = 100; % passing = 80; Group % passing No.200 = 58 LL = 30; PI = 10. Granular materials (35% or less of Total Sample passing No. 200) A-1 A-1-a A-2 A-1-b A-3 A-2-4 A-2-5 A-2-6 A-2-7 Silty-Clay materials (more than 35% passing No. 200) A-7 A-4 A-5 Sieve Analysis % Passing 50 max. No max. 50 max. 51 min. 15 max. 25 max. 35 max. 35 max. 35 max. 35 max. 36 min. No.200 Characteristic of Fraction passing 40 max 41 min 40 max 41 min 40 max Liquid Limit 6 max. NP Plasticity index Usual types of significant constituent materials Stone fragments, gravel, and sand subgrade rating Non of these Fine Silty or clayey gravel and sand sand Excellent to good A-6 Solution: Example 5 From Table, the group classification is A-4. A-7-5 A-7-6 Now we Calculate the Group Index Using the given data, F= 58, LL=30 & PI=10 GI= (F-35) [ (LL-40)] (F-15)(PI-10) 36 min. 36 min. 36 min. No No No GI= GI= GI= 41 min 40 max 41 min Silty Soils Clayey Soils Then Thus, Fair to poor Plasticity index of A-7-5 subgrade is equal to or less than LL minus 30 Plasticity index of A-7-6 subgrade is grater than LL minus 30 is A-4. Â Next Â 50 (58-23)[ ( 30-40)] ( 58-15)(10-10) (35)[ (-10)] ( 58-15)(0) 5.25 GI= > 5 the AASHTO is A- 4 (5); Silty Soil; Fair to poor As subgrade material49 Talat Question Time This is a reminder that the following slides are just for may be future use as a second example or a qui hool_of_engineering/courses/eqe694 f/694f_f97_lecture2.html Ask Sam Thank you Dr. Talat Bader Example Classify this Soil Calculations D10 = 0.11 D30 = 0.19 D60 = 0.38 Cu = D60/D10 = 3.45 Cc = (D30)2/(D60. D10) = 0.86 US Sieve Diameter Mass retained % Cumulative retained passing pan pan (sum = 450)

10 CE353 Soil Mechanics Example #4 Solution example 4 Gravel = 30%, Sand = 40% and (100-Silts & Clay)/2 = 35% Since Sand > (100-F)/2, coarse fraction is more sandy than gravelly Then Group symbol is SC Then, group name is "Clayey sand with Classify the following soil using UCS Given: gravel fraction (% retained on #4) = 30% gravel" 56 Sand fraction (passing #4, retained on #200) = 40% Silt and clay fraction (passing #200) = 30% Example 5 Solution Example 6 Classify the following soil using UCS Given: gravel fraction (% retained on #4) = 0% sand fraction (passing #4, retained on #200) = 14% silt and clay fraction (passing #200) = 86% LL = 55; PI = 28 Silts & Clay = 86%, From Chart Group symbol is CH Inorganic Clay Group name is "Fat Clay" Unified soil Question Time GROUP NAME GROUP SYMBOL Thank you Dr. Talat Bader GW WELL-GRADED GRAVEL, FINE TO COARSE GRAVEL GP POORLY-GRADED GRAVEL GM SILTY GRAVEL GC CLAYEY GRAVEL SW WELL-GRADED SAND, FINE TO COARSE SAND SP POORLY-GRADED SAND SM SILTY SAND SC CLAYEY SAND ML SILT CL CLAY OL ORGANIC SILT, ORGANIC CLAY MH SILT OF HIGH PLASTICITY, ELASTIC SILT CH CLAY OF HIGH PLASTICITY, FAT CLAY OH ORGANIC CLAY, ORGANIC SILT PT PEAT

CE 240 Soil Mechanics & Foundations Lecture 3.2. Engineering Classification of Soil (AASHTO and USCS) (Das, Ch. 4)

CE 240 Soil Mechanics & Foundations Lecture 3.2 Engineering Classification of Soil (AASHTO and USCS) (Das, Ch. 4) Outline of this Lecture 1. Particle distribution and Atterberg Limits 2. Soil classification