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Chapter 3 · 4 hours

Soil Identifications and Classification

IOE past exam questions

Past questions and answers

24 questions set from this chapter, 6 of them more than once; 3 are most repeated (set, or a close variant set, in 3 or more exams). Most repeated first.

  • Most repeated · 4 of 12 exams
  • Asked 4 times
  • 2073 Magh · 5 marks
  • 2074 Bhadra · 1 mark
  • 2077 Chaitra · 4 marks
  • 2078 Poush · 2 marks

Describe the field identification tests used to identify fine-grained soils and to distinguish between clay and silt.

Answer

Field identification tests are quick hand tests on a small moist sample, used to tell fine-grained soils apart without laboratory equipment. They use the dry strength, dilatancy, toughness, plasticity, and appearance of the soil. They are carried out on the fraction finer than 0.425 mm (No. 40 sieve).

1. Dilatancy (shaking) test

Make a pat of moist soil (soft putty consistency) and shake it in the palm. Then squeeze it.

  • Silt: water rises quickly to the surface giving a shiny look; it disappears on squeezing (rapid reaction).
  • Clay: no change (no reaction).

2. Dry strength (crushing) test

Dry a moulded pat and crush it between the fingers.

  • Clay: high dry strength, cannot be crushed easily.
  • Silt: low dry strength, crumbles easily to powder.

3. Toughness (thread) test

Roll the soil into a thread about 3 mm diameter, fold and re-roll repeatedly until it crumbles at the plastic limit.

  • Clay: strong, stiff thread; tough lump after crumbling.
  • Silt: weak thread, soft lump, loses coherence quickly.

4. Plasticity / sticking

Clay is plastic and sticky in hand and cannot be easily washed off; silt feels gritty or floury and washes off easily.

5. Other tests

  • Visual and touch: clay feels smooth, silt feels slightly gritty on the teeth.
  • Shine test: a cut surface of clay shows a shiny surface when scratched; silt stays dull.
  • Sedimentation test: shaken in water, silt settles in 30 - 60 seconds, clay stays in suspension for hours.
TestSiltClay
DilatancyRapidNone
Dry strengthLowHigh to very high
ToughnessLowMedium to high
FeelGritty/flourySmooth, sticky
  • Most repeated · 3 of 12 exams
  • Asked 3 times
  • 2079 Jestha · 2 marks
  • 2078 Chaitra · 3 marks
  • 2078 Poush · 2 marks

What is the purpose (importance) of soil classification?

Answer

Soil classification groups soils of similar engineering behaviour into the same class, using simple index properties (grain size and plasticity).

Purposes and importance

  1. Common language: engineers share a standard name and symbol for a soil (e.g. SC, CH), so reports are understood by everyone.
  2. Prediction of behaviour: from the group, the probable strength, compressibility, permeability, compaction and swelling can be estimated.
  3. Preliminary design: gives approximate design values (bearing capacity, subgrade suitability, frost susceptibility) before costly tests are done.
  4. Selection of construction material: helps to select borrow soil for embankments, dam cores, and road subgrades.
  5. Site investigation planning: helps decide what tests are needed.
  6. Use of past experience: data from similar soils elsewhere can be applied.
  7. Economy: avoids costly lab tests for small or simple projects.
  8. Correlation: index properties correlate with engineering properties (e.g. wLw_L with compression index).
  • Most repeated · 3 of 12 exams
  • Asked 3 times
  • 2074 Bhadra · 2 marks
  • 2076 Baisakh · 1 mark
  • 2078 Poush · 1 mark

What are the common (important engineering) soil classification systems? Write down the types of soil classification.

Answer

Types of soil classification

  1. Textural (particle size) classification: based on percent of sand, silt and clay, e.g. USDA/triangular chart, MIT system.
  2. Classification based on plasticity and size (engineering classification): Unified Soil Classification System (USCS), Indian Standard Soil Classification System (ISSCS, IS 1498), AASHTO, Casagrande's airfield system.
  3. Classification based on origin and genesis: residual, transported, alluvial, etc.
  4. Classification for specific purposes: AASHTO for highways, Federal Aviation Administration (FAA) for airfields, and agricultural (pedological) classification.

Important engineering systems

SystemMain use
Unified (USCS)General engineering, foundations, dams
IS Soil Classification (IS 1498)Indian/Nepali practice
AASHTOHighways, subgrade
Textural (MIT, USDA)Based on grain size only
  • Asked 2 times
  • 2079 Asoj · 3 marks
  • 2073 Magh · 3 marks

Describe the soil classification according to the MIT classification system, giving the grain size ranges of the different soil types.

Answer

The MIT (Massachusetts Institute of Technology) classification is a textural system that names a soil only by its particle size. It was proposed by Gilboy and Casagrande and later adopted by the British Standards. The soil is divided into gravel, sand, silt and clay.

Soil typeParticle size (mm)
Gravel> 2.0
Coarse sand2.0 - 0.6
Medium sand0.6 - 0.2
Fine sand0.2 - 0.06
Coarse silt0.06 - 0.02
Medium silt0.02 - 0.006
Fine silt0.006 - 0.002
Clay< 0.002
  • The system has the same limits for sand, silt, and clay as the International classification, except for the sand boundaries.
  • The mixture is named by its major fraction, for example "silty sand" or "sandy clay".
  • Limitation: it ignores plasticity, so soils with the same size distribution can behave very differently.
  • Asked 2 times
  • 2078 Poush · 2 marks
  • 2076 Baisakh · 1 mark

What are the basic requirements of soil classification?

Answer

A good soil classification system should:

  1. Be simple, easy to learn and to use, with a few clear groups.
  2. Use standard index tests that are quick and cheap (grain size, Atterberg limits).
  3. Place soils of similar engineering behaviour in the same group.
  4. Use clear boundaries so that different engineers get the same classification.
  5. Use a systematic nomenclature (symbols) that conveys properties of the soil.
  6. Be accepted widely and linked to experience, so that behaviour can be predicted.
  7. Cover all types of soil, coarse and fine, organic and inorganic.
  • Asked 2 times
  • 2077 Chaitra · 2 marks
  • 2074 Bhadra · 3 marks

Draw neatly the plasticity chart (IS / USCS) and label the group symbols of the various soil regions in the chart.

Answer

The plasticity chart (Casagrande) plots plasticity index IpI_p (vertical) against liquid limit wLw_L (horizontal). It is used for the fine-grained fraction (passing 75 micron).

Lines

  • A-line: Ip=0.73(wL−20)I_p = 0.73(w_L - 20), separating clays (above) from silts and organic soils (below).
  • U-line: Ip=0.9(wL−8)I_p = 0.9(w_L - 8), the upper limit of natural soils.
  • Vertical line at wL=50%w_L = 50\% (USCS) separates low and high plasticity. IS adds wL=35%w_L = 35\% for low (L), medium (I), and high (H) plasticity.
 Ip
 60|                       /U
 50|                  CH  /
 40|                    /
 30|        CI    CH  /  A
 20|    CL   /       /
 10|  CL-ML /     MH & OH
  7|--/--------/
  4|ML & OL    MI    MH & OH
  0+---+---+---+---+---+--- wL
    0  10 20 35 50 60 70 80

Group symbols

RegionSymbolName
Above A-line, wL<35w_L < 35 (IS) / 50 (USCS)CLInorganic clay of low plasticity
Above A-line, 35<wL<5035 < w_L < 50 (IS)CIClay of medium plasticity
Above A-line, wL>50w_L > 50CHInorganic clay of high plasticity
Below A-line, wL<35w_L < 35 / 50ML (OL)Silt (organic silt) of low plasticity
Below A-line, 35<wL<5035 < w_L < 50MI (OI)Silt of medium plasticity
Below A-line, wL>50w_L > 50MH (OH)Silt of high plasticity / organic clay
Hatched zone, IpI_p 4 - 7 above A-lineCL-MLSilty clay
  • 2078 Baisakh · 2 marks

Name the tests generally done to identify sandy soil and clayey soil in the field.

Answer

Sandy soil

  • Visual and touch: coarse, gritty, grains visible to the naked eye.
  • Dry strength test: crumbles easily; no strength.
  • Dilatancy test: quick reaction with water rising to the surface (for fine sand).
  • Sedimentation test: settles in less than a minute.
  • Does not form a thread (no plasticity).

Clayey soil

  • Dry strength test: high strength; hard to crush.
  • Dilatancy test: no or very slow reaction.
  • Toughness (thread) test: rolls into 3 mm thread; tough and cohesive.
  • Stickiness/shine test: sticky, smooth, with shiny cut surface.
  • Sedimentation test: stays in suspension for a long time.
  • 2078 Baisakh · 2 marks

Name the soil classification systems which use both particle size and plasticity characteristics of soil.

Answer

Soil classification systems that use both particle size and plasticity are:

  1. Unified Soil Classification System (USCS) - Casagrande.
  2. Indian Standard Soil Classification System (ISSCS) - IS 1498.
  3. AASHTO classification system - uses sieve analysis and Atterberg limits (also gives a group index).
  4. Casagrande's airfield classification (the origin of the USCS).
  5. British Soil Classification System (BSCS).
  • 2075 Bhadra · 3 marks

Write down the names of soil classification systems based on particle size and plasticity of soil. Define the plasticity chart of a soil based on ISSCS (Indian Standard Soil Classification System).

Answer

Systems based on particle size and plasticity

  • Unified Soil Classification System (USCS)
  • Indian Standard Soil Classification System (ISSCS, IS 1498)
  • AASHTO classification system
  • British Soil Classification System (BSCS)

Plasticity chart of ISSCS

The plasticity chart is a graph of plasticity index IpI_p (y-axis) against liquid limit wLw_L (x-axis), used for classifying the fine-grained soils (passing 75 micron).

  • A-line: Ip=0.73(wL−20)I_p = 0.73(w_L - 20). Inorganic clays (C) lie above it; silts (M) and organic soils (O) lie below it.
  • Vertical lines at wL=35%w_L = 35\% and 50%50\% divide the soils into low (L, wL<35w_L < 35), intermediate (I, 35 - 50) and high plasticity (H, wL>50w_L > 50).
  • The hatched area above the A-line with IpI_p between 4 and 7 gives dual symbol CL-ML.

The resulting groups are CL, CI, CH (clays), ML, MI, MH (silts), and OL, OI, OH (organic soils).

  • 2078 Poush · 3 marks

Describe in detail one important engineering soil classification system, clearly bringing out its limitations.

Answer

Unified Soil Classification System (USCS)

Proposed by A. Casagrande (1948) and adopted by the US Army Corps of Engineers and Bureau of Reclamation; it is the most widely used engineering system. Soil is classified by grain size distribution and plasticity.

Procedure

  1. Coarse-grained if more than 50% is retained on the 75 micron (No. 200) sieve; fine-grained if 50% or more passes.
  2. Coarse-grained: G (gravel) if more than half of the coarse fraction is retained on 4.75 mm sieve, else S (sand).
    • Fines < 5%: GW, GP, SW, SP. Well graded if Cu≥4C_u \ge 4 (gravel) or ≥6\ge 6 (sand) and 1≤Cc≤31 \le C_c \le 3.
    • Fines > 12%: GM, GC, SM, SC using Atterberg limits (A-line).
    • Fines 5 - 12%: dual symbols (e.g. GW-GC).
  3. Fine-grained: use the plasticity chart. Below A-line: M (silt); above: C (clay); organic: O; wL<50%w_L < 50\% gives L, wL≥50%w_L \ge 50\% gives H: ML, CL, OL, MH, CH, OH. Pt is peat.
Major divisionGroups
GravelsGW, GP, GM, GC
SandsSW, SP, SM, SC
Silts and clays, wL<50w_L < 50ML, CL, OL
Silts and clays, wL>50w_L > 50MH, CH, OH
Highly organicPt

Limitations

  • Only the fraction smaller than 75 mm is considered; cobbles and boulders are not classified.
  • Gives no direct measure of engineering properties such as strength or compressibility.
  • Dual symbols and borderline cases cause ambiguity.
  • Particle shape, density and structure (in-situ condition) are not considered.
  • Single boundary at wL=50%w_L = 50\% gives no medium plasticity class (the IS system corrects this).
  • Based on the remoulded soil, so natural structure effects are ignored.
  • 2078 Poush · 2 marks

Point out similarities and differences between the USCS system and the AASHTO system of soil classification.

Answer

Both classify soil using grain size distribution and Atterberg limits.

Similarities

  • Both use sieve analysis and liquid and plastic limits.
  • Both are based on the fraction of soil finer than 75 mm.
  • Both separate coarse-grained from fine-grained soil using the 75 micron sieve, though at different percentages.
  • Both are used widely in engineering practice.

Differences

BasisUSCSAASHTO
PurposeGeneral engineeringHighway subgrade
Coarse/fine boundary50% passing 75 micron35% passing 75 micron
SymbolsLetters (GW, SC, CH)A-1 to A-7 with group index
Plasticity chartUsed (A-line)Not used
Organic soilSeparate OL, OH, PtNo separate group
Group indexNot usedUsed to rate subgrade
RatingOrder of groups not numericalSubgrade quality worsens from A-1 to A-7
  • 2074 Bhadra · 2 marks

For finding the suitability of soils as subgrade for highways, which soil classification is generally used? Write down the name of each group according to that classification. Show the general rating of those groups as a suitability of subgrade.

Answer

The AASHTO classification system (American Association of State Highway and Transportation Officials) is generally used for rating soil as a highway subgrade.

Groups

Granular materials (35% or less passing 75 micron):

  • A-1-a: stone fragments, gravel and sand
  • A-1-b: stone fragments, gravel and sand
  • A-3: fine sand
  • A-2-4, A-2-5, A-2-6, A-2-7: silty or clayey gravel and sand

Silt-clay materials (more than 35% passing 75 micron):

  • A-4: silty soils
  • A-5: elastic (micaceous or diatomaceous) silty soils
  • A-6: clayey soils
  • A-7-5, A-7-6: elastic clayey soils (A-7-5 moderate plasticity; A-7-6 high change in volume)

General rating as subgrade

GroupRating
A-1-a, A-1-bExcellent to good
A-3Excellent to good
A-2Excellent to good (A-2-6, A-2-7 fair)
A-4, A-5Fair to poor
A-6Fair to poor
A-7-5, A-7-6Fair to poor (poor, very poor in swelling)

Within each group, a lower group index (0 for best soils) indicates a better subgrade.

  • 2075 Baisakh · 2 marks

How is the plasticity chart useful for classifying fine-grained soils?

Answer

The plasticity chart plots plasticity index (IpI_p) against liquid limit (wLw_L) and has the A-line (Ip=0.73(wL−20)I_p = 0.73(w_L - 20)) and a vertical line at wL=50%w_L = 50\%.

It is useful because:

  1. It classifies fine-grained soils (more than 50% passing 75 micron) into clay (C), silt (M) or organic (O) by position relative to the A-line (above = clay; below = silt/organic).
  2. It divides them into low (wL<50w_L<50) and high (wL>50w_L>50) plasticity (L or H), giving CL, CH, ML, MH, OL, OH.
  3. The hatched zone above A-line with Ip=4I_p = 4 to 7 gives the borderline class CL-ML.
  4. It is also used to classify the fines of coarse-grained soils with more than 12% fines (GM, GC, SM, SC).
  5. It indicates relative engineering behaviour: compressibility, swelling and strength rise with position upward and right.
  • 2079 Asoj · 5 marks

The sieve analysis of a soil gave the following results: % passing 75 micron sieve = 4 % retained on 4.75 mm sieve = 35 Coefficient of curvature = 2 Coefficient of uniformity = 5 Classify the soil according to the USCS system.

Answer

Given

Passing 75 micron =4%= 4\%; retained on 4.75 mm =35%= 35\%; Cc=2C_c = 2; Cu=5C_u = 5.

Step 1: Coarse or fine?

Only 4% passes 75 micron, so more than 50% is retained: coarse-grained.

Step 2: Gravel or sand?

  • Gravel (retained on 4.75 mm) =35%= 35\%
  • Sand =100−35−4=61%= 100 - 35 - 4 = 61\%

Coarse fraction =96%= 96\%; gravel is 35/96=36%35/96 = 36\% of it, less than half. So the soil is a sand (S).

Step 3: Fines content

4% is less than 5%, so only the grading is needed (symbols SW or SP).

Step 4: Grading

For well-graded sand: Cu≥6C_u \ge 6 and 1≤Cc≤31 \le C_c \le 3.

  • Cc=2C_c = 2 satisfies the condition.
  • Cu=5<6C_u = 5 < 6 fails.

The soil is not well graded.

Answer: SP, poorly graded sand (with gravel).

  • 2079 Jestha · 6 marks

A soil sample on laboratory test gives the following results. Classify the soil and give its symbol as per the USCS classification system. Passing through 75-micron sieve = 8% Passing through 4.75 mm sieve = 42% Coefficient of uniformity = 6 Coefficient of curvature = 4 Plasticity index = 4

Answer

Given

Passing 75 micron =8%= 8\%; passing 4.75 mm =42%= 42\%; Cu=6C_u = 6; Cc=4C_c = 4; Ip=4I_p = 4.

Step 1: Coarse or fine?

92% is retained on the 75 micron sieve, so the soil is coarse-grained.

Step 2: Gravel or sand?

  • Retained on 4.75 mm (gravel) =100−42=58%= 100 - 42 = 58\%
  • Sand =42−8=34%= 42 - 8 = 34\%

Gravel is more than half of the coarse fraction (58/92=63%58/92 = 63\%), so the soil is gravel (G).

Step 3: Fines content

8% is between 5% and 12%, so a dual symbol is required.

Step 4: Grading

For well-graded gravel: Cu≥4C_u \ge 4 and 1≤Cc≤31 \le C_c \le 3.

  • Cu=6≥4C_u = 6 \ge 4: OK
  • Cc=4C_c = 4 is greater than 3: fails

So the gravel is poorly graded (GP).

Step 5: Nature of fines

Ip=4I_p = 4 is at the lower limit of the hatched zone. Since IpI_p is not above 4 (the fines plot as silt, ML or CL-ML), the fines are taken as silty (M).

Answer: GP-GM, poorly graded gravel with silt and sand.

  • 2078 Chaitra · 5 marks

Classify the soils A and B with the properties shown below according to the unified soil classification system.
SoilwLw_L (%)IpI_p (%)% passing through 4.75 mm sieve% passing through 75 µ sieve
A452910059
B551510085

Answer

Both soils have 59% and 85% passing the 75 micron sieve (more than 50%), so both are fine-grained. Use the plasticity chart with the A-line Ip=0.73(wL−20)I_p = 0.73(w_L - 20).

Soil A: wL=45%w_L = 45\%, Ip=29%I_p = 29\%

Ip,A-line=0.73(45−20)=18.25%I_{p,A\text{-line}} = 0.73(45 - 20) = 18.25\%

Ip=29>18.25I_p = 29 > 18.25: the soil lies above the A-line (clay), and wL=45<50w_L = 45 < 50 (low plasticity).

Soil A: CL, inorganic clay of low plasticity (lean clay). It is also sandy, since 41% is coarse.

Soil B: wL=55%w_L = 55\%, Ip=15%I_p = 15\%

Ip,A-line=0.73(55−20)=25.55%I_{p,A\text{-line}} = 0.73(55 - 20) = 25.55\%

Ip=15<25.55I_p = 15 < 25.55: the soil lies below the A-line (silt), and wL=55>50w_L = 55 > 50 (high plasticity).

Soil B: MH, inorganic silt of high plasticity (elastic silt).

(Both are taken as inorganic, since no data on organic content is given.)

  • 2078 Poush · 4 marks

Classify the given soil according to the USCS classification system. % of soil passing through sieve no. 200 (0.075 mm) = 40% % of soil retained in sieve no. 4 (4.75 mm sieve) = 55% The grading characteristics of soil were: D10D_{10} = 1.2 mm, D60D_{60} = 3.8 mm, D30D_{30} = 2.6 mm

Answer

Given

Passing 75 micron (No. 200) =40%= 40\%; retained on 4.75 mm (No. 4) =55%= 55\%; D10=1.2D_{10} = 1.2 mm, D30=2.6D_{30} = 2.6 mm, D60=3.8D_{60} = 3.8 mm.

Step 1: Coarse or fine?

60% is retained on the 75 micron sieve, so the soil is coarse-grained.

Step 2: Gravel or sand?

  • Gravel =55%= 55\%
  • Sand =100−55−40=5%= 100 - 55 - 40 = 5\%

Gravel is more than half of the coarse fraction, so the soil is gravel (G).

Step 3: Grading parameters

Cu=D60D10=3.81.2=3.17,Cc=D302D10D60=2.621.2×3.8=1.48C_u = \frac{D_{60}}{D_{10}} = \frac{3.8}{1.2} = 3.17, \qquad C_c = \frac{D_{30}^2}{D_{10}D_{60}} = \frac{2.6^2}{1.2 \times 3.8} = 1.48

The gravel fraction is poorly graded (Cu<4C_u < 4), but CcC_c is in the range 1 - 3.

Step 4: Fines content

The fines are 40%, which is more than 12%. So the grading is not used; the symbol depends on the type of fines (GM or GC) found from wLw_L and IpI_p using the A-line. The plasticity data are not given.

Answer: a gravel with fines, GM or GC (silty or clayey gravel). If the fines are silty the symbol is GM; if the fines plot above the A-line with Ip>7I_p > 7 it is GC. The grading (Cu=3.17C_u = 3.17, Cc=1.48C_c = 1.48) of the coarse fraction is poor.

  • 2078 Baisakh · 4 marks

Classify the following soil if the test results obtained from sieve analysis and consistency tests are given below: Percentage passing No. 4 sieve (4.75 mm) = 70%, percentage passing No. 200 sieve (0.075 mm) = 30%; liquid limit = 33% and plastic limit = 11%.

Answer

Given

Passing No. 4 =70%= 70\%; passing No. 200 =30%= 30\%; wL=33%w_L = 33\%; wP=11%w_P = 11\%.

Ip=wL−wP=33−11=22%I_p = w_L - w_P = 33 - 11 = 22\%

Step 1: Coarse or fine?

70% is retained on the No. 200 sieve, so the soil is coarse-grained.

Step 2: Gravel or sand?

  • Gravel (retained on No. 4) =100−70=30%= 100 - 70 = 30\%
  • Sand =70−30=40%= 70 - 30 = 40\%

Gravel is 30/70 = 43% of the coarse fraction (less than half), so the soil is sand (S).

Step 3: Fines

Fines are 30% (more than 12%), so classify the fines with the plasticity chart.

Ip,A-line=0.73(33−20)=9.5%I_{p,A\text{-line}} = 0.73(33 - 20) = 9.5\%

Ip=22I_p = 22 is above the A-line and greater than 7: clayey fines (C).

Answer: SC, clayey sand (with gravel).

  • 2075 Bhadra · 5 marks

Particle size distribution curves for two types of soil, Soil A and Soil B, are shown in the figure [Figure: percentage passing versus particle size (mm), log scale; labelled points on the curves: 99% and 55% for soil A, 63% and 13% for soil B, with the 0.075 mm and 4.75 mm sizes marked on the axis]. Water contents measured at the boundaries between the liquid state-plastic state and plastic state-semi solid for soil A are 45% and 15% respectively. Similarly, for Soil B, they are 25% and 10% respectively. Classify these soils based on the Unified Soil Classification System. Draw the plasticity chart if required.

Answer

Reading the curves

For each soil the larger percent is taken as the percent passing the 4.75 mm sieve and the smaller as the percent passing the 0.075 mm sieve.

SoilPassing 4.75 mmPassing 0.075 mmwLw_LwPw_PIpI_p
A99%55%451530
B63%13%251015

Soil A

Fines =55%>50%= 55\% > 50\%, so it is fine-grained.

Ip,A-line=0.73(45−20)=18.25%I_{p,A\text{-line}} = 0.73(45 - 20) = 18.25\%

Ip=30>18.25I_p = 30 >18.25: above the A-line (clay); wL=45<50w_L = 45 < 50 (low plasticity).

Soil A: CL, lean clay (sandy).

Soil B

Fines =13%>12%= 13\% > 12\% and less than 50%, so it is coarse-grained with fines.

  • Gravel =100−63=37%= 100 - 63 = 37\%
  • Sand =63−13=50%= 63 - 13 = 50\%

Sand is more than half of the coarse fraction (50/87=57%50/87 = 57\%), so it is S.

Ip,A-line=0.73(25−20)=3.65%I_{p,A\text{-line}} = 0.73(25 - 20) = 3.65\%

Ip=15>7I_p = 15 > 7 and above the A-line: clayey fines.

Soil B: SC, clayey sand.

Plasticity chart

 Ip
 30|  * A
 20|        / A-line
 15|   * B /
  7|----/-----------
  4| /
  0+----+----+----+---- wL
    0   25   45   50

Point A (45, 30) lies above the A-line; B (25, 15) lies above it as well.

  • 2075 Baisakh · 6 marks

A soil has the following characteristics: a) Percentage of soil passing No. 200 sieve = 55 b) Percentage of coarse fraction passing No. 4 sieve = 60 c) Liquid limit = 68% d) Plastic limit = 22% Classify the given soil according to ISSCS.

Answer

Given

Passing No. 200 =55%= 55\%; wL=68%w_L = 68\%; wP=22%w_P = 22\% (the percentage of coarse fraction passing No. 4 is not needed).

Ip=68−22=46%I_p = 68 - 22 = 46\%

Step 1: Coarse or fine?

Since 55% passes the 75 micron sieve (more than 50%), the soil is fine-grained.

Step 2: Plasticity group (IS 1498)

  • wL<35%w_L < 35\%: low (L); 35−50%35 - 50\%: medium (I); >50%> 50\%: high (H).
  • wL=68%>50%w_L = 68\% > 50\%: high plasticity (H).

Step 3: Position relative to the A-line

Ip,A-line=0.73(wL−20)=0.73(68−20)=35.0%I_{p,A\text{-line}} = 0.73(w_L - 20) = 0.73(68 - 20) = 35.0\%

Ip=46>35.0I_p = 46 > 35.0: the soil plots above the A-line, so it is a clay (C) (inorganic).

Answer: CH, inorganic clay of high plasticity (fat clay). Because 45% is coarse, it may be described as a sandy or gravelly clay.

  • 2076 Baisakh · 6 marks

A sample of inorganic soil has the following grain size characteristics:
Size (mm)Percent passing
0.075 (No. 200)58
0.425 (No. 40)80
2 mm (No. 10)100
The liquid limit is 30% and PI is 10%. Classify the soil according to the AASHTO classification system.

Answer

Given

Passing No. 200 =58%= 58\%; passing No. 40 =80%= 80\%; passing No. 10 =100%= 100\%; wL=30%w_L = 30\%; Ip=10%I_p = 10\%.

Step 1: Granular or silt-clay?

58% passes the No. 200 sieve, which is more than 35%, so the soil is in the silt-clay group (A-4 to A-7).

Step 2: Check groups from the left of the table (AASHTO)

GroupwLw_L maxIpI_pResult
A-44010 maxSatisfied: wL=30≤40w_L = 30 \le 40, Ip=10≤10I_p = 10 \le 10
A-541 min10 maxNot satisfied (wL<41w_L < 41)
A-640 max11 minNot satisfied (Ip<11I_p < 11)
A-741 min11 minNot satisfied

So the soil is A-4.

Group index

GI=(F−35)[0.2+0.005(wL−40)]+0.01(F−15)(Ip−10)=(58−35)[0.2+0.005(30−40)]+0.01(58−15)(10−10)=23×0.15+0=3.45≈3\begin{aligned} GI &= (F - 35)[0.2 + 0.005(w_L - 40)] + 0.01(F - 15)(I_p - 10) \\ &= (58 - 35)[0.2 + 0.005(30 - 40)] + 0.01(58 - 15)(10 - 10) \\ &= 23 \times 0.15 + 0 = 3.45 \approx 3 \end{aligned}

Answer: A-4(3), silty soil; fair to poor subgrade.

  • 2073 Bhadra · 3 marks

Classify the following soil as per the unified soil classification system: soil passing from 75 µ sieve = 4%, soil passing from 4.75 mm sieve (coarse fraction) = 62%, coefficient of uniformity = 5, coefficient of curvature = 2.6.

Answer

Given

Passing 75 micron =4%= 4\%; passing 4.75 mm =62%= 62\%; Cu=5C_u = 5; Cc=2.6C_c = 2.6.

  • Fines are only 4%, so the soil is coarse-grained.
  • Gravel (retained on 4.75 mm) =100−62=38%= 100 - 62 = 38\%; sand =62−4=58%= 62 - 4 = 58\%. Sand is more than half of the coarse fraction (58/96 = 60%), so it is sand (S).
  • Fines are < 5%, so grading decides: well-graded sand needs Cu≥6C_u \ge 6 and 1≤Cc≤31 \le C_c \le 3. Here Cc=2.6C_c = 2.6 passes, but Cu=5<6C_u = 5 < 6 fails.

Answer: SP, poorly graded sand (with gravel).

  • 2073 Bhadra · 3 marks

Classify the following soil as per the unified soil classification system: soil passing from 75 µ sieve = 62%, liquid limit = 54%, plastic limit = 23%.

Similar questions: USCS classification, 75 micron 39% (2073 Bhadra)

Answer

Given

Passing 75 micron =62%= 62\%; wL=54%w_L = 54\%; wP=23%w_P = 23\%.

Ip=54−23=31%I_p = 54 - 23 = 31\%
  • 62% passes the 75 micron sieve (more than 50%), so the soil is fine-grained.
  • wL=54>50w_L = 54 > 50: high plasticity (H).
  • A-line at wL=54w_L = 54: Ip=0.73(54−20)=24.8%I_p = 0.73(54 - 20) = 24.8\%. Since Ip=31>24.8I_p = 31 > 24.8, the soil is above the A-line: clay (C).

Answer: CH, inorganic clay of high plasticity (fat clay).

  • 2073 Bhadra · 2 marks

Classify the following soil as per the unified soil classification system: soil passing from 75 µ sieve = 39%, liquid limit = 33%, plastic limit = 18%.

Similar questions: USCS classification, 75 micron 62% (2073 Bhadra)

Answer

Given

Passing 75 micron =39%= 39\%; wL=33%w_L = 33\%; wP=18%w_P = 18\%.

Ip=33−18=15%I_p = 33 - 18 = 15\%
  • Only 39% is finer than 75 micron (less than 50%), so the soil is coarse-grained, with more than 12% fines.
  • Fines classification: A-line at wL=33w_L = 33: Ip=0.73(33−20)=9.5%I_p = 0.73(33-20) = 9.5\%. Ip=15I_p = 15 lies above the A-line and is greater than 7, so the fines are clayey (C).
  • The gravel/sand split is not given. Assuming the coarse fraction is mainly sand (the usual case), the symbol is SC; if gravel dominates it would be GC.

Answer: SC, clayey sand (GC if gravel exceeds sand).

Questions from Old Question Collection (CE 552) (IOE BCE Soil Mechanics (CE552) papers from 2073 Bhadra to 2079 Asoj). Answers are written for this site; check them against your class notes.

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