Question bank of soil mechanics

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DEPARTMENT OF
CIVIL ENGINEERING
Question Bank on
on
CE603PC / SOIL MECHANICS
(Regulation - R16)
Prepared By
Prashantha T. R.
Assistant Professor/Civil
Approved by
G. Nagamanidevi
HoD/Civil

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TABLE OF CONTENT
Sl.No TOPIC Page
No
UNIT – I 5
1. Introduction Soil Mechanics 5
1.1. Question of 2 marks. 5
1.2. Questions of 3 marks. 9
1.3. Questions of 5marks and above: 9
1.4. JNTU previous year selective questions. 11
UNIT – II 13
2. Soil Water, Permeability, Seepage 13
2.1. Question of 2 marks. 13
2.2. Question of 3 marks 16
2.3. Questions of 5marks and above: 17
2.4. JNTU previous papers: 19
UNIT – III 21
3. Stress Distribution in soil and Compaction 21
3.1. Questions for 2 marks, 21
3.2. Questions for 3 Marks 23
3.3. Questions for 5 and above marks. 24
3.4. Previous year papers JNTU Questions 25
UNIT – IV 27
4. Consolidation 27
4.1. Two marks questions 27

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4.2. Question with mor than 5 marks 30
4.3. JNTU Previous year Exam questions 34
UNIT - V 35
5. Shear Strength of Soils 35
5.1. Questions for 2 marks. 35
5.2. Questions for 5 and above marks. 37
5.3. Previous year papers JNTU Questions 40
Previous Year Model Question Papers 42
Geotechnical Engineering R13 JNTUH May -2018 42
Geotechnical Engineering -1 R09 -JNTUH May 2018 44
Remarks:
In Final Examination following pattern will be followed
Part-A
Five question with two marks
Five question with three marks
Part-A
Five sets of questions will be there in which each set will have two questions. From each set
one question bearing ten marks should be answered.
Two Model papers at 6 have been given at the end of this question Bank.

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SOIL MECHANICS JNTUH R-16 SYLLABUS
B.Tech. III Year II Sem. L T/P/D C Course Code: CE603PC 4 0/0/0 4
Pre-Requisites: Engineering Geology, Applied Mechanics, Fluid Mechanics Course
Objectives: To enable the student to study the properties of soil and to determine the behaviour
soil under various conditions and loads.
Course Outcomes: At the end of the course, the student will be able to:: Understand the
mechanism Behaviour of Soil for different loads : and from Soil Condition will be able to
determine properties of soil.
UNIT – I
Introduction: Soil formation and structure – moisture content – Mass- volume relationship –
Relative density. Index Properties of Soils: Grain size analysis – Sieve–
UNIT – II
Permeability: Soil water – capillary rise – flow of water through soils – Darcy’s law
permeability – Factors affecting permeability – laboratory determination of coefficient of
permeability –Permeability of layered soils – In-situ permeability tests (Pumping in & Pumping
out test).
Effective Stress & Seepage Through Soils: Total, neutral and effective stress – principle of
effective stress - quick sand condition – Seepage through soils – Flownets: Characteristics and
Uses.
UNIT – III
Stress Distribution In Soils: Boussinesq’s and Westergaard’s theories for point load, uniformly
loaded circular and rectangular areas, pressure bulb, variation of vertical stress under point load
along the vertical and horizontal plane, and Newmark’s influence chart for irregular areas.
Compaction: Mechanism of compaction – factors affecting compaction effects of compaction
on soil properties – Field compaction Equipment – compaction quality control.
UNIT – IV
Consolidation: Types of compressibility – Immediate Settlement, primary consolidation and
secondary consolidation - stress history of clay; e-p and e-log(p) curves – normally consolidated
soil, over consolidated soil and under consolidated soil - preconsolidation pressure and its
determination - Terzaghi’s 1-D consolidation theory – coefficient of consolidation: square root
time and logarithm of time fitting methods - computation of total settlement and time rate of
settlement.
UNIT - V
Shear Strength Of Soils: Importance of shear strength – Mohr’s– Coulomb Failure theories –
Types of laboratory tests for strength parameters – strength tests based on drainage conditions
– strength envelops – Shear strength of sands - dilatancy – critical void ratio.

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UNIT – I
Relative density. Index Properties Of Soils: Grain size analysis – Sieve
1. Introduction Soil Mechanics
1.1. Question of 2 marks.
1.1.1. Define is soil mechanics and soil according to Civil Engineer?
Ans: Soil is an unconsolidated accumulation of solid particles, which are produced by the
mechanical and chemical disintegration of rocks, regardless of whether or not they contain an
admixture of organic constituents.
To an engineer, it is a material that can be:
built on: foundations of buildings, bridges
built in: basements, culverts, tunnels
built with: embankments, roads, dams
supported: retaining walls
1.1.2. Define water content of soil? Write it’s mathematical expression.
Ans: The ratio of the mass of water present to the mass of solid particles is called the water
content (w), or sometimes the moisture content or Water Content.
w =
𝑊 𝑊
𝑊𝑠
Its value is 0% for dry soil and its magnitude can exceed 100%.
1.1.3. What do you mean by degree of saturation? Write it’s mathematical expression.
Ans: The volume of water (Vw) in a soil can vary between zero (i.e. a dry soil) and the volume
of voids. This can be expressed as the degree of saturation (S) in percentage.
𝑆𝑟 =
𝑉 𝑊
𝑉𝑉
For a dry soil, S = 0%, and for a fully saturated soil, S = 100%.
1.1.4. Define bulk density of soil? How it is related with submerged density.
Ans: Bulk unit weight (γt or γ ) is a measure of the amount of solid particles plus water per
unit volume.
γ 𝑡 = γ =
(𝑊𝑠 + 𝑊𝑤)
(𝑉𝑠 + 𝑉𝑣)
submerged unit weight γ = γ 𝑠𝑎𝑡 − γ 𝑤
1.1.5. Draw 3 - phase system of soil with weight and volume.
Ans:

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1.1.6. What is void ratio? Show the relation ship with porosity.
Ans: Void ratio is the ratio of the volume of voids (Vv) to the volume of soil solids (Vs), and
is expressed as a decimal.
𝑒 =
𝑉 𝑉
𝑉𝑠
Void ratio = 𝑒 =
𝑛
(1−𝑛)
1.1.7. What is specific gravity? List different methods to find specific gravity.
Ans: The mass of solid particles is usually expressed in terms of their particle unit
weight (γs) or specific gravity (Gs) of the soil grain solids.
γ 𝑠 =
𝑊 𝑊
𝑉𝑠
= 𝐺𝑆. γ 𝑤
It can be determined by 1. Pycnometer 2. Density bottle
1.1.8. List any five soil forming minerals.
Ans: Kolinite, silica, Illite, Mont morillonite, halloysite.
1.1.9. What is soil profile? Show a neat diagram
of soil profile.
Ans: Soil formation starts first with breakdown of
rock by weathering and soil horizon development
process leads to the development of a soil profile.
A soil profile is the vertical display of soil
horizons. A soil profile is divided into layers called
horizons.
Figure 1: Soil Profile with A,B, C horizonFigure 1
shows an Example soil horizons(Profile). a) top
soil and colluvium b) mature residual soil c) young
residual soil d) weathered rock.
A very general and simplified soil profile can be
described as follows:
Figure 1: Soil Profile with A,B, C horizon

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a. The plough layer (20 to 30 cm thick): is rich in organic matter and contains many live roots.
This layer is subject to land preparation (e.g. ploughing, harrowing etc.) and often has a dark
colour (brown to black).
b. mature residual soil: contains much less organic matter and live roots. This layer is hardly
affected by normal land preparation activities. The colour is lighter, often grey, and sometimes
mottled with yellowish or reddish spots.
c. young residual soil : hardly any organic matter or live roots are to be found. This layer is
not very important for plant growth as only a few roots will reach it.
d. The parent rock layer or weathered rock.: consists of rock, from the degradation of which
the soil was formed. This rock is sometimes called parent material.
The depth of the different layers varies widely: some layers may be missing altogether.
1.1.10. What is the relationship between unit weight & mass density?
Ans: Unit weight (γ) is defined as total unit weight of soil per unit total volume
Unit weight (γ) = Total Weight t /unit total volume
in KN/m3
or N/mm3
Density(ρ) is defined as total unit weight of soil per unit total volume
Mass Density (ρ) = Total Mass /unit total volume
in Kg/m3
or g/cc
They are related by γ = ρ.g
Where standard value g = 9.807m/s2
for all practical purposes
1.1.11. Distinguish between Residual and Transported soil.
Ans: Residual Soils
Residual soils are found at the same location where they have been formed. Generally, the
depth of residual soils varies from 5 to 20 m.
Chemical weathering rate is greater in warm, humid regions than in cold, dry regions causing
a faster breakdown of rocks. Accumulation of residual soils takes place as the rate of rock
decomposition exceeds the rate of erosion or transportation of the weathered material. In
humid regions, the presence of surface vegetation reduces the possibility of soil transportation.
As leaching action due to percolating surface water decreases with depth, there is a
corresponding decrease in the degree of chemical weathering from the ground surface
downwards. This results in a gradual reduction of residual soil formation with depth, until
unaltered rock is found.
Residual soils comprise of a wide range of particle sizes, shapes and composition.
Transported Soils
Weathered rock materials can be moved from their original site to new locations by one or
more of the transportation agencies to form transported soils. Tranported soils are classified
based on the mode of transportation and the final deposition environment.
(a) Soils that are carried and deposited by rivers are called alluvial deposits.
(b) Soils that are deposited by flowing water or surface runoff while entering a lake are called
lacustrine deposits. Atlernate layers are formed in different seasons depending on flow rate.

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(c) If the deposits are made by rivers in sea water, they are called marine deposits. Marine
deposits contain both particulate material brought from the shore as well as organic remnants
of marine life forms.
(d) Melting of a glacier causes the deposition of all the materials scoured by it leading to
formation of glacial deposits.
(e) Soil particles carried by wind and subsequently deposited are known as aeolian deposits.
1.1.12. Draw A- line chart and Brief the function of A-line Chart in soil classification?
Ans: A plasticity chart , based on the values of liquid limit (WL) and plasticity index (IP), is
provided in ISSCS to aid classification. The 'A' line in this chart is expressed as
IP = 0.73 (WL - 20).
Depending on the point in the chart, fine soils are divided into clays (C), silts (M), or organic
soils (O). The organic content is expressed as a percentage of the mass of organic matter in a
given mass of soil to the mass of the dry soil solids. Three divisions of plasticity are also
defined as follows.
Low plasticity WL< 35%
Intermediate plasticity 35% < WL<
50%
High plasticity WL> 50%
The 'A' line and vertical lines at WL equal to 35% and 50% separate the soils into various
classes.
For example, the combined symbol CH refers to clay of high plasticity.
1.1.13. list out different types of soil according to I S Classification with their sizes in
mm.
Ans:
Very coarse soils Boulder size > 300 mm
Cobble size 80 - 300 mm
Coarse soils Gravel size (G) Coarse 20 - 80 mm

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Fine 4.75 - 20 mm
Sand size (S) Coarse 2 - 4.75 mm
Medium 0.425 - 2 mm
Fine 0.075 - 0.425 mm
Fine soils Silt size (M) 0.002 - 0.075 mm
Clay size (C) < 0.002 mm
1.1.14. Recall the geology, list different types of weathering and with their product.
Ans: Physical weathering by wind, frost, temperature, water – Produces mainly Coarse soils
such as Boulder, Gravel, sand, silt.
Chemical weathering by hydration oxidation and carbonation – Produces fine soil particles
such as silt and clay.
Biological weathering by decomposition of rock by plants, animals, microbes – Produces
organic soil, clays and sometimes gravels.
1.2. Questions of 3 marks.
1.2.1. Define plasticity index, flow index and liquidity index.
1.2.2. What are the methods available for determination of in-situ density.
1.2.3. Give the relation between γsat, G, γw and e.
1.2.4. What is consistency index of a soil? Write the equation for consistency Index.
1.2.5. How organic soils are formed?
1.2.6. how organic soil can be identified?
1.2.7. Draw structures of montmorillonite, illite, and kaolinite mineral.
1.2.8. Compare montmorillonite, illite, and kaolinite mineral.
1.2.9. Define stokes law? Write the expression for settling velocity.
1.2.10. List different chemicals used in hydrometer analyses. Why these chemicals are added.
1.2.11. What are the correction to be applied for hydrometer analyses and why?
1.2.12. What is relative density? What is it’s significance?
1.2.13. Show the relation between specific gravity, bulk density, void ratio and degree of
saturation.
1.3. Questions of 5marks and above:

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1.3.1. Write down a neat procedure for determining water content and specific gravity of a
given soil in the laboratory by using a pycnometer.
1.3.2. Explain soil cycle with a neat sketch.
1.3.3. Explain in detail how silt, sand, gravel, clay and organic soil are formed.
1.3.4. Explain all the consistency limits and indices.
1.3.5. Explain in detail the procedure for determination of grain size distribution of soil by
sieve analysis.
1.3.6. Explain field methods of finding water content and dry densities of soil in details.
1.3.7. Explain the procedure of finding liquid limits.
1.3.8. Explain the procedure of finding shrinkage limit of soil.
1.3.9. Explain in details how plastic limit of soil can be found.
1.3.10. List different method to classify the soil. Explain I.S. method of classification of soil.
1.3.11. Sandy soil in a borrow pit has unit weight of solids as 25.8 kN/m3
, water content
equal to 11% and bulk unit weight equal to 16.4 kN/m3
. How many cubic meter of
compacted fill could be constructed of 3500 m3
of sand excavated from borrow pit, if
required value of porosity in the compacted fill is 30%. Also calculate the change in
degree of saturation.
1.3.12. The following data on consistency limits are available for two soils A and B. Find
which soil is (i) More plastic. (ii) Better
foundation material on remoulding. (iii) Better
shear strength as function of water content.
(iv) Better shear strength at plastic limit.
Sl. No. Property Soil a Soil B
1 Plastic limit 16% 19%
2 Liquid limit 30% 52%
3 Flow index 11 06
4 Natural water content 32% 40%

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1.3.13. By three phase soil system, prove that the degree of saturation S (as ratio) in terms of
mass unit weight (γ), void ratio (e), specific gravity of soil grains (G) and unit weight
of water (γW) is given by the expression. mass unit weight (γ) =
(𝑮+𝒆𝑺)𝛄 𝒘
(𝟏+𝒆)
1.3.14. An earthen embankment of 106 m3
volume is to be constructed with a soil having a
void ratio of 0.80 after compaction. There are three borrow pits marked A, B and C
having soils with voids ratios of 0.90, 0.50 and 1.80 respectively. The cost of
excavation and transporting the soil is Rs 0.25, Rs 0.23 and Rs 0.18 per m3
respectively. Calculate the volume of soil to be excavated from each pit. Which borrow
pit is the most economical? (Take G = 2.65).
1.3.15. A laboratory compaction test on soil having specific gravity equal to 2.67 gave a
maximum dry unit weight of 17.8 kN/m3
and a water content of 15%. Determine the
degree of saturation, air content and percentage air voids at the maximum dry unit
weight. What would be theoretical maximum dry unit weight corresponding to zero air
voids at the optimum water content?
1.3.16. A soil sample has a porosity of 40 per cent. The specific gravity of solids is 2.70.
Calculate i) Voids ratio ii) Dry density and iii) Unit weight if the soil is completely
saturated.
1.3.17. A soil has a bulk unit weight of 20.11 KN/m3 and water content of 15 percent.
Calculate the water content of the soil partially dries to a unit weight of 19.42 KN/m3
and the voids ratio remains unchanged.
1.3.18. An earth embankment is compacted at a water content of 18% to a bulk density of
1.92 g/cm3
. If the specific gravity of the sand is 2.7, find the void ratio and degree of
saturation of the compacted embankment.
1.4. JNTU previous year selective questions.
1.4.1. Explain liquid limit, plastic limit and shrinkage limit. (JNTUK-R13-Nov/Dec-
2016)
1.4.2. Define the terms specific gravity of particles, porosity and submerged density.
(JNTUK-R13-Nov/Dec-2017)

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1.4.3. Explain the formation of soil. (JNTUH-R13-March-2017)
1.4.4. Explain any two methods of identify the silts from clay. (JNTUH-R13-March-2017)
1.4.5. Derive the relationship between dry density and bulk density in terms of water
content. (JNTUH-R13-March-2017)
1.4.6. What is the difference between the classification based on particle size and based on
textural? Discuss the limitations of the two.
1.4.7. In a field exploration, a soil sample was collected in a sampling tube of internal
diameter 5.0 cm below the ground water table. The length of the extracted sample was
10.2 cm and its mass was 387 g. If G = 2.7, and the mass of the dried sample is 313 g,
find the porosity, void ratio, degree of saturation, and the dry density of the sample.
1.4.8. Explain laboratory method to determine shrinkage limit of a given soil sample.
1.4.9. The Atterberg limits of a clayey soil are liquid limit = 63%, plastic limit = 40% and
shrinkage limit = 27%. If a sample of this soil has a volume of 10cm3
at the liquid
limit and a volume of 6.4 cm3
at the shrinkage limit, determine specific gravity of
solids, shrinkage ratio and volumetric shrinkage. (JNTUK-R13-
Nov/Dec-2017)
1.4.10. Explain step by step procedure to classify soils as per I.S. Classification of soils.
1.4.11. An oven dry soil sample of volume 300cc weighs 450g. If the specific gravity of
solids is 2.65, what is the water content when the soil becomes fully saturated without
any change in it’s volume? What will be the water content which will fully saturate the
soil sample and also cause an increase in volume equal to 15% of the original dry
volume? (JNTUH-R13-March-2017)
1.4.12. Sketch the phase diagram for a soil and indicate the volumes and weights of the
phases on it. Define ‘Void ratio’, ‘Degree of saturation’, and ‘Water content’.
1.4.13. What is a unit phase diagram?
1.4.14. The volume of an undisturbed clay sample having a natural water content of 40% is
25.6 cm3
and its wet weight is 0.435 N. Calculate the degree of saturation of the
sample if the grain specific gravity is 2.75. (JNTUK-R13-April/May-2017)

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UNIT – II
Permeability: Soil water – capillary rise – flow of water through soils – Darcy’s law
permeability – Factors affecting permeability – laboratory determination of coefficient of
permeability –Permeability of layered soils – In-situ permeability tests (Pumping in & Pumping
out test).
Effective Stress & Seepage Through Soils: Total, neutral and effective stress – principle of
effective stress - quick sand condition – Seepage through soils – Flownets: Characteristics and
Uses.
2. Soil Water, Permeability, Seepage
2.1.1. List different methods to find permeability of soil in laboratory?
Ans: i. Constant head permeability
ii. Falling head permeability test
iii. Capillary - permeability test
2.1.2. List different methods to find permeability of soil in the field, differentiate
between them?
Ans: Pumping in Test
Pumping out test
2.1.3. What do you mean piezometric head? Write it’s mathematical equation in terms
of water head.
Ans: The height of water level in the standpipe above the datum is
the piezometric head (h).
h = hz + hw
i.e., Peizometric head is the sum of datum head and pressure head.
2.1.4. Sate Darcy’s law with its equation.
Ans: Darcy's law states that there is a linear relationship between flow velocity (v) and
hydraulic gradient (i) for any given saturated soil under steady laminar flow conditions.

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If the rate of flow is q (volume/time) through cross-sectional area (A) of the soil mass, Darcy's
Law can be expressed as
𝒗 =
𝒒
𝑨
= 𝒌. 𝒊
where k = permeability of the soil
𝒊 =
∆𝒉
𝑳
∆𝒉 = difference in total heads
L = length of the soil mass
2.1.5. What is surface tension? What is it’s significance in flow of water through soil.
Ans: Surface tension is the elastic tendency of a fluid surface which makes it acquire the least
surface area possible.
Surface is the cause of capillarity in soil. The rise of water in the capillary tubes, or the fine
pores of the soil, is due to the existence of surface tension which pulls the water up against the
gravitational force.
2.1.6. What is meant by capillary rise in soil and how it affects the stress level in soils?
Ans: The pores of soil mass may be looked upon as a series of
capillary tubes, extending vertically above water table.
The rise of water in the capillary tubes, or the fine pores of the
soil, is due to the existence of surface tension which pulls the
water up against the gravitational force.
Above the water table, when the soil is saturated, pore pressure
will be negative (less than atmospheric) because of capillarity.
Thus, the height above the water table to which the soil is
saturated is called the capillary rise, and this depends on the
grain size and the size of pores. In coarse soils, the capillary
rise is very small.
2.1.7. How effective stress varies under submerged
conditiof soil?
Ans: For Submerged Condition
at top 𝜎 = 0, 𝑈 = 0, 𝜎′
= 𝜎 − 𝑈 = 0
At hw
𝜎 = ℎ 𝑤 𝛾 𝑤, 𝑈 = ℎ 𝑤 𝛾 𝑤, 𝜎′
= 0
At z
𝜎 = ℎ 𝑤 𝛾 𝑤 + 𝑧 𝛾𝑧, 𝑈 = ℎ 𝑤 𝛾 𝑤 + 𝑧 𝛾 𝑤, 𝝈′
= 𝒛 𝜸′
For fully saturated soil Condition
𝛾 𝑤
𝛾𝑠𝑎𝑡

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at top
𝜎 = 0, 𝑈 = 0, 𝜎′
= 𝜎 − 𝑈 = 0
At z
𝜎 = 𝑧 𝛾𝑠𝑎𝑡, 𝑈 = 𝑧𝛾 𝑤, 𝝈′
= 𝒛 𝜸′
Where, 𝜎= Norma stress
𝑈 = neutral stress
𝜎′
= effective stress
It indicates that effective stress variation is not affected
irrespective variation of water level above ground
2.1.8. List out the uses of flow net?
Ans: i. Estimation of seepage losses from reservoirs
ii. Determination of uplift pressures below dams
iii. Checking the possibility of piping beneath dams
2.1.9. What do you mean by ‘Capillary rise of water in soil’? when this condition
occurs in soil.
Ans: The groundwater can be sucked upward by the soil through very small pores that are
called capillars. This process is called capillary rise.
In fine textured soil (clay), the upward movement of water is slow but covers a long distance.
On the other hand, in coarse textured soil (sand), the upward movement of the water is quick
but covers only a short distance.
Soil texture Capillary rise (in cm)
coarse (sand) 20 to 50 cm
medium 50 to 80 cm
fine (clay) more than 80 cm up to several metres
2.1.10. Define quick sand? In what kind of soil quick sand condition arises?
Ans: In case of upward seepage or flow, decreases in effective stress can be seen. In such a
situation, effective stress is reduced to zero and the soil behaves like a very viscous liquid.
Such a state is known as quick sand condition.
In nature, this condition is usually observed in coarse silt or fine sand subject to artesian
conditions.
𝛾𝑠𝑎𝑡

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2.1.11. Derive equation for seepage pressure for upward
flow though soil.
Ans:
At the bottom of the soil column,
𝜎 = 𝐿 𝛾𝑠𝑎𝑡, 𝑈 = (𝐿 + ∆𝐻)𝛾 𝑤,
𝝈′
= 𝐿 𝛾𝑠𝑎𝑡 − (𝐿 + ∆𝐻)𝛾 𝑤
Seepage pressure => 𝑃𝑠 = −∆𝐻 𝛾 𝑤
2.2. Question of 3 marks
2.2.1. Define effective stress, neutral stress and total stress?
Solution: 𝝈= Norma stress(Total stress): total Stress in the soil mass is defined as the
total load per unit cross section area at any point of soil mass.
At z
𝜎 = 𝑧 𝛾𝑠𝑎𝑡,
i.e., Stress in the ground due to self-weight of soil, water and external load.
𝑼 = neutral stress: (Pore water pressure): Stress in the ground due to only water.
𝑈 = 𝑧𝛾 𝑤,
i.e., pressure due water in the pores or voids inside the soil mass.
𝝈′
= effective stress : effective stress is the stress obtained by deducting neutral stress from
Normal stress or total stress.
𝜎′
= 𝜎 − 𝑈
𝝈′
= 𝒛 𝜸′
Theoretically, stress in the ground which is subjected on only through soil grains.

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2.2.2. what is the relation between Total stress, neutral stress and effective stress?
2.2.3. What is flow net? What is the difference between potential line and flow lines?
2.2.4. What is the hydraulic gradient and critical hydraulic gradient?
2.2.5. What are the different types of soil water?
2.2.6. Define seepage pressure, seepage velocity and discharge velocity?
2.2.7. List out the factor affecting permeability?
2.2.8. For a homogeneous earth dam 52 m high and 2 m free board, a flow net was
constructed, and following results were obtained: Number of potential drops = 25;
Number of flow channels = 4 Calculate the discharge per metre length of the dam if the
co-efficient of permeability of the dam material is 3 x 10-5
m/sec.
2.2.9. What is phreatic line? Is it different from flow lines? If yes/no how.
2.2.10. What are the remedies to avoid quick sand condition?
2.2.11. What is permeability of soil, intrinsic permeability of soil and hydraulic
permeability of soil.
2.2.12. The following data were obtained when a sample of medium sand was tested in a
constant head permeameter: cross-section area of a sample :100 cm2
, hydraulic
gradient : 10, discharge collected : 10cc/s, Find the coefficient of permeability of the
sand.
2.3. Questions of 5marks and above:

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2.3.1. Explain in detail procedure for determining permeability by constant head test in the
laboratory.
2.3.2. List methods to obtain flow net. Briefly explain any two.
2.3.3. Explain in detail how to find field permeability by pumping out test with neat figure.
2.3.4. Explain in detail how to find field permeability by pumping in test with neat figure.
2.3.5. Explain in detail procedure for determining permeability by variable head test in the
laboratory.
2.3.6. What will be the permeability equivalent for horizontal flow through three layered
soil? Derive expression with neat figure.
2.3.7. Explain properties of flow net with neat figures.
2.3.8. What will be the permeability equivalent for vertical flow through three layered soil?
Derive expression with neat figure.
2.3.9. The water table in a deposit of sand 8 m thick is at a depth of 3 m below the ground
surface. Above the water table, the sand is saturated with capillary water. The bulk
density of sand is 19.62 kN/m3
. Calculate the effective pressure at 1m, 3m and 8m
below the ground surface. Hence plot the variation of total pressure, neutral pressure
and effective pressure over the depth of 8m.
2.3.10. Compute the total, effective and pore pressure at a depth of 20 m below the bottom
of a lake 6m deep. The bottom of lake consists of soft clay with a thickness of more
than 20 m. the average water content of the clay is 35% and specific gravity of the soil
may be assumed to be 2.65.
2.3.11. The subsoil strata at a site consist of fine sand 1.8 m thick overlying a stratum of
clay 1.6 m thick. Under the clay stratum lies a deposit of coarse sand extending to a
considerable depth. The water table is 1.5 m below the ground surface. Assuming the
top fine sand to be saturated by capillary water, calculate the effective pressures at
ground surface and at depths of 1.8 m, 3.4 m and 5.0 m below the ground surface.
Assume for fine sand G = 2.65, e = 0.8 and for coarse sand G = 2.66, e = 0.5. What will
be the change in effective pressure at depth 3.4 m, if no capillary water is assumed to
be present in the fine sand and its bulk unit weight is assumed to be 16.68 kN/m3
. The
unit weight of clay may be assumed as 19.32 kN/m3
.

19
2.3.12. The following data were recorded in a constant head permeability test. Internal
diameter of permeameter = 7.5cm Head lost over a sample length of 18cm = 24.7cm
Quantity of water collected in 60 Sec = 626 ml Porosity of soil sample was 44%
Calculate the coefficient of permeability of the soil. Also determine the discharge
velocity and seepage velocity during the test.
2.3.13. A stratified soil deposit is shown in Fig.
Along with the coefficient of permeability
of the individual strata. Determine the
ratio of KH and KV. Assuming an average
hydraulic gradient of 0.3 in both
horizontal and vertical seepage, Find (i)
Discharge value and discharge velocities in each layer for horizontal flow and (ii)
Hydraulic gradient and loss in head in each layer for vertical flow.
2.3.14. Determine the effective stress at 2m, 4m, 6m, 8m and 10m is a soil mass having γs =
21 KN/m3
. Water table is 2m below ground surface. Above water table there is
capillary rise upto ground surface. Also draw total stress diagram up to 10m.
2.3.15. A foundation trench is to be
excavated in a stratum of stiff
clay, 10m thick, underlain by a
bed of coarse sand(fig.). In a trial
borehole the ground water was
observed to rise to an elevation
of 3.5m below ground surface.
Determine the depth up to which
an excavation can be safely
carried out without the danger of the bottom
becoming unstable under the artesian pressure in the sand stratum. The specific gravity
of clay particles is 2.75 and the void ratio is 0.8. if excavation is to be carried out safely
to a depth of 8m, how much should the water table be lowered in the vicinity of the
trench?
2.4. JNTU previous papers:
JNTUH March-2017
K= 3 X 10-4
cm/s, z = 2.5m
K= 5 X 10-4
cm/s, z = 2m
K= 5 X 10-4
cm/s, z = 2m
10m
Excavation
0m
@1.6m

20
2.4.1. 1. What is adsorbed water?
2.4.2. 2. What is Darcy’s law? and under what conditions it’s valid?
2.4.3. 3. Derive an expression to determine coefficient of permeability of soil by laboratory
falling head permeability test.
2.4.4. 4. In a deposit of silty soil, the water table which was at originally at depth of l m below
ground level was lowered to 3m below ground level. The bulk and saturated unit weight
of silty soil was l 8kN/m3
and 20kN/m3
respectively. What is the change in effective
pressure at a depth of I.0 m and 3.0 m.

21
UNIT – III
Stress Distribution In Soils: Boussinesq’s and Westergaard’s theories for point load, uniformly
loaded circular and rectangular areas, pressure bulb, variation of vertical stress under point load
along the vertical and horizontal plane, and Newmark’s influence chart for irregular areas.
Compaction: Mechanism of compaction – factors affecting compaction effects of compaction
on soil properties – Field compaction Equipment – compaction quality control.
3. Stress Distribution in soil and Compaction
3.1. Questions for 2 marks,
3.1.1. What are the assumption made in Boussinesq’s theory?
Ans: Boussinesq’s assumptions are
1. The soil mass is elastic, isotropic, homogeneous and semi-infinite.
2. The soil is weightless.
3. The load is a point load acting on the surface.
3.1.2. What are the assumption made in Westergaards’s theory
Ans: The soils is a semi-infinite and laterally reinforced by numerous, closely spaced,
horizontal sheets of negligible thickness but of infinite rigidity, which prevent the mass as a
whole from undergoing lateral movement of soil grains.
3.1.3. What do you mean by semi-infinite and homogeneous soil?
Ans: A semi-infinite solid is the one bounded on one side by a horizontal surface, here the
surface of the earth, and infinite in all the other directions.
The soil is said to be homogeneous if there are identical elastic properties at every point of the
mass in identical
directions.
In fig: E= Youngs
modulous
∞ =infinity in the
given direction
3.1.4. Define Pressure bulb with neat figure?
Ans: An isobar is a line which connects all
points of equal stress below the ground
surface. In other words, an isobar is a stress
contour. We may draw any number of isobars as
shown in Fig. for any given load system. Each
isobar represents a fraction of the load applied at
the surface. Since these isobars form closed
figures and resemble the form of a bulb, they are
also termed bulb of pressure or simply the
pressure bulb.
3.1.5. List any three methods to find the
stress distribution under the soil.
Ans: Boussinesq’s method

22
Westergaard's method
Newmark's Influence Chart
Approximate method
3.1.6. What are the equipment used in standard proctor test?
Ans: Compaction mould, capacity 1000ml.
Rammer, mass 2.6 kg
Detachable base plate
Collar, 60mm high
IS sieve, 4.75 mm
Oven
Desiccator
Weighing balance, accuracy 1g
Large mixing pan
Straight edge
Spatula
Graduated jar
Mixing tools, spoons, trowels, etc.
3.1.7. List any four ‘field compacting equipments’.
Ans: Smooth steel drum rollers (static or vibratory)
Pneumatic tyred rollers
Sheepsfoot rollers
Grid rollers
Vibrating plates
Tampers and rammers
3.1.8. How modified proctor test is different from standard proctor test?
Ans: It was found that the Light Compaction Test (Standard Test) could not reproduce the
densities measured in the field under heavier loading conditions, and this led to the
development of the Heavy Compaction Test (Modified Test). The equipment and procedure
are essentially the same as that used for the Standard Test except that the soil is compacted in
5 layers, each layer also receiving 25 blows. The same mould is also used. To provide the
increased compactive effort, a heavier rammer of 4.9 kg and a greater drop height of 450 mm
are used.
3.1.9. List any four effects of compaction on soil properties.
Ans: Increase in soil shear strength
Increase in its bearing capacity.
Reduction subsequent settlement under working loads.
Reduction in soil permeability making it more difficult for water to flow through.
Increase in it’s dry density.
3.1.10. Daw a Comparative graph showing relation between OMC and MDD for sand
and clay.
Ans:

23
3.1.11. What are isobars and pressure bulb? How
they are related.
Ans: An isobar is a line which connects all points of
equal stress below the ground surface. In other
words, an isobar is a stress contour. We may draw
any number of isobars as shown in Fig. for any given
load system. Each isobar represents a fraction of the
load applied at the surface. Since these isobars form
closed figures and resemble the form of a bulb, they
are also termed bulb of pressure or simply the
pressure bulb.
3.1.12. What do you mean by isotropic soil? Show
with figure.
Ans: The soil is said to be isotropic if there are
identical elastic properties throughout the mass and in
every direction through any point of it.
E= Youngs modulous
3.2. Questions for 3 Marks
3.2.1. What is OMC and MDD? Write their expression.
3.2.2. How Newmarks’s influence chart is calculated?
3.2.3. How MDD can be calculated during Proctor test write? Show in mathematical
expressions.
3.2.4. What are the difference between standard and modified proctor test? Writ energy
ratio.
3.2.5. A concentrated load of 50KN acts on the surface of ground. Calculate the increase in
vertical stress directly below the load at a depth of 4m.

24
3.2.6. Draw a neat diagram for variation of vertical stress under point load along the vertical and
horizontal plane.
3.2.7. Write the difference between Boussinesq’s theory and Westergaards theory.
3.2.8. List the factors affecting compaction.
3.2.9. Briefly explain mechanism of compaction.
3.2.10. Write Boussinesq’s equation, Show each term means in a figure for point load.
3.2.11. Write Westergaards’s equation, Show each term means in a figure for point load.
3.2.12. Write a short not on compaction quality control.
3.3. Questions for 5 and above marks.
3.3.1. Explain the assumptions made by Boussinesque and Westergaraard in stress
distribution on soils.
3.3.2. Explain in detail to find the stress under uniformly loaded circular and rectangular area
by Westergaard and Boussinesq theory.
3.3.3. Explain in detail the variation of vertical stress under point load along the vertical and
horizontal plane with neat figure.
3.3.4. Derive Boussinesque equations to find intensity of vertical pressure and tangential
stress when a concentrated load is acting on the soil.
3.3.5. Explain the Newmark’s influence chart in detail.
3.3.6. Derive Westergaards equations to find intensity of vertical pressure and tangential
stress when a concentrated load is acting on the soil.
3.3.7. Explain the procedure for determining the relationship between dry density and
moisture content by proctor compaction test.
3.3.8. What is compaction? What is the Engineering purpose compaction in the field. List out
some cases where compaction can used.
3.3.9. Explain Standard Proctor Compaction test with neat sketches.
3.3.10. Explain in detail factors affecting compaction.
3.3.11. Explain in detail effects of compaction soil properties.
3.3.12. A water tank is supported by a ring foundation having outer diameter of 10 m and
inner diameter of 7.5 m. the ring foundation transmits uniform load intensity of 160
kN/m2
. Compute the vertical stress induced at depth of 4 m, below the center of ring
foundation, using (i) Boussinesque analysis and (ii) Westergaard’s analysis, taking μ = 0
3.3.13. The load from a continuous footing of width 2m, which may be considered to be strip
load of considerable length, is 200 kN/m2
. Determine the maximum principal stress at
1.5m depth below the footing, if the point lies (i) directly below the centre of the footing,
(ii) directly below the edge of the footing and (iii) 0.8m away from the edge of the
footing.
3.3.14. A concentrated point load of 200 kN acts at the ground surface. Find the intensity of
vertical pressure at a depth of 10 m below the ground surface and situated on the axis of
the loading. What will be the vertical pressure at a point at a depth of 5 m and at a radial
distance of 2 m from the axis of loading? Use Boussinesque analysis.

25
3.3.15. A line load of 100 kN/m run extends to a long distance. Determine the intensity of
vertical stress at a point, 2 m below the surface and i) Directly under the line load and ii)
At a distance 2 m perpendicular to the line. Use Boussinesq’s theory.
3.4. Previous year papers JNTU Questions
3.4.1. What do you understand by geostatic stresses? How are these determined? (JNTUK
Dec-2016)
3.4.2. A point load of 3000kN is acting at the ground surface. Determine the vertical stress
at a point which is 5m directly below the load. What will be the vertical stress at a
point which is at a depth of 5m and at a horizontal distance of 3m from the axis of the
load?
3.4.3. State the assumptions made in computing stresses below the ground surface due to
point load acting on it. Discuss their validity in practice. (JNTUK Nov-2017)
3.4.4. Find the intensity of vertical pressure and horizontal shear stress at a point 4m directly
and shear stress at a point 2m horizontally away from the axis of loading but at
the same depth of 4m.
3.4.5. Find the intensity of vertical pressure at a point 4m directly below a 20 kN point
load acting at a horizontal ground surface. What will be the vertical pressure at a
point 2 m horizontally away from the axis of loading but at the same depth of 4m
and directly under the load at a depth of 3.
(JNTUH March-2017)

26
3.4.6. What do you understand by ‘Pressure bulb’? Illustrate with sketches. (JNTUK
April/May -2017)
3.4.7. State the basic requirements to be satisfied for the validity of Boussinesq equation for
stress distribution.
3.4.8. What merits and demerits between wet side and dry side compaction of MDD
compaction
3.4.9. A ring foundation is of 3 m external diameter and 2.00 m internal diameter. It
transmits a uniform pressure of 90 kN/m2
. Calculate the vertical stress at a depth of 1.5
m directly beneath the center of the loaded area.
3.4.10. Describe modified proctor test. (JNTUK NOV/DEC -2016)
3.4.11. The following are the result of compaction test. Find the compaction curve
showing the optimum moisture content and maximum dry density Plot the zero air
void line. Determine the degree of saturation at the maximum dry density. Volume of
mould = 1000ml; Mass of mould = 1000g; specific gravity of solids = 2.70
3.4.12. Describe standard proctor test. (JNTUK Nov-2017)
3.4.13. A laboratory compaction test on soil having specific gravity equal to 2.68 gave
a maximum dry density of 1.82 g/cc and a water content of 17%. Determine the degree
of saturation, air content and percentage air voids at the maximum dry density.
What would be theoretical maximum dry density corresponding to zero air voids
at the optimum water content?
3.4.14. Briefly explain factors affecting-compaction 'or soil (JNTUH March-2017)
3.4.15. Write the differences between standard and modified proctor compaction test:
3.4.16. Derive the relationship between dry density and bulk density in terms of water
content.
3.4.17. Write short notes on field compaction control.(JNTUK April/May -2017)
3.4.18. Derive an expression for ‘zero air-void line’ and draw the line for a specific gravity
of 2.65. The maximum dry density and optimum moisture content of a soil from
standard proctor’s test are 18 kN/m3 and 16% respectively. Compute the degree of
saturation of the sample, assuming the specific gravity of soil grains as 2.70.
Water Content in % 10 12 14.3 16.1 18.2
Mass of Mould + wet soil in g 2925 3095 3150 3125 3070

27
UNIT – IV
Consolidation: Types of compressibility – Immediate Settlement, primary consolidation and
secondary consolidation - stress history of clay; e-p and e-log(p) curves – normally consolidated
soil, over consolidated soil and under consolidated soil - preconsolidation pressure and its
determination - Terzaghi’s 1-D consolidation theory – coefficient of consolidation: square root
time and logarithm of time fitting methods - computation of total settlement and time rate of
settlement.
4. Consolidation
4.1. Two marks questions
4.1.1. What consolidation? At what conditions consolidation takes place?
Ans: Consolidation is the reduction in volume of a soil due to the dissipation of excess pore
water pressures.
This phenomenon only happens in cohesive soils, since granular soils drain rapidly and do not
generate the excess pore water pressures in the first place.
4.1.2. What is compressibility of soil? List out different types compressibility.
Ans: When a soil mass is subjected to a compressive force, its volume decreases. The
property of the soil due to which a decrease in volume occurs under compressive force is
known as the compressibility of soil.
Compression of solid particles and water in the voids
Compression and expulsion of air in the voids
Expulsion of water in the voids
4.1.3. What is coefficient of compressibility and compression index?
Ans: The plot of void ratio versus effective stress the coefficient of compressibility, av.
For a small range of effective stress,
The plot of void ratio versus log of effective stress can be approximated to a straight line, and
the slope of this line is indicated by a parameter termed as compression index, Cc.
4.1.4. Define OCR? List out it’s value for different cases.
Ans: The overconsolidation ratio (OCR) is defined as the ratio of the preconsolidation
stress to the current effective stress.

28
when the soil is normally consolidated, OCR = 1
for overconsolidated soils OCR > 1
4.1.5. Draw the graph showing relation between i) 𝒆 − 𝝈 ii) 𝒆 − 𝒍𝒐𝒈(𝝈)
Ans: Graph 𝑒 − 𝜎 and 𝑒 − log(𝜎) -
4.1.6. Write down Terzagi’s 1-D
consolidation equation with their terms.
Ans:
By introducing a parameter called the coefficient of consolidation, ,
the flow eqn. then becomes
This is Terzaghi's one-dimensional consolidation equation. A solution of this for a set of
boundary conditions will describe how the excess pore water pressure u dissipates with
time t and location z. When all the u has dissipated completely throughout the depth of the
compressible soil layer, consolidation is complete and the transient flow situation ceases to
exist.
4.1.7. What is volume compressibility and coefficient of consolidation?
Ans: the coefficient of volume compressibility, mv, which is expressed as

29
It represents the compression of the soil, per unit original thickness, due to a unit increase of
pressure.
the coefficient of consolidation (cv)
,
1. In a consolidation test void ratio decreases from 0.80 to 0.70 when the load 𝜎 changed from 40
KN/m2
to 80 KN/m2
what is the compression Index?
4.1.8. What is Pore water pressure? How it is different from normal pressure i.e.
Normal stress on the ground?
Ans: Pore water pressure(u) refers to the pressure of groundwater held within a soil or
rock, in gaps between particles (pores).
Pore water pressures below the phreatic level of the groundwater are measured with
piezometers.
Normal stress is from whole soil mass with water i.e., saturated soil mass, where as neutral
or pore water pressure due to only water.
4.1.9. List out any few advantages of consolidation?
Ans: Properties such as effective stress, dry density shear strength will be increased
Properties such as pore water pressure, void ratio, permeability and compressibility will
be decreases.
Above results are desirable for Engineering Purposes.
4.1.10. How do you determine the pre-consolidation pressure?
Ans: It is possible to determine the preconsolidation stress that the soil had experienced. The
soil sample is to be loaded in the laboratory so as to obtain the void ratio - effective stress
relationship. Empirical procedures are used to estimate
the preconsolidation stress, the most widely used
being Casagrande's construction which is illustrated.
The steps in the construction are:
• Draw the graph using an appropriate scale.
• Determine the point of maximum curvature A.
• At A, draw a tangent AB to the curve.
• At A, draw a horizontal line AC.
• Draw the extension ED of the straight line portion
of the curve.
• Where the line ED cuts the bisector AF of
angle CAB, that point corresponds to the preconsolidation stress.
4.1.11. What are the assumption are made in the Terzaghi’s theory of one-dimensional
consolidation.
Ans: Soil homogenous and fully saturated

30
Soil particles and water are incompressible.
Deformation of the soil is due entirely to change in volume
Darcy’s law for the velocity of flow of water thorough soil is perfectly valid.
Coefficient of permeability is constant during consolidation
Load is applied deformation occurs only in direction
The change in thickness of the layer during consolidation is insignificant
4.1.12. Draw a figure showing pore water pressure variation in single drain and double
drained soil layer.
Ans:
𝑢̅ variation for both drainage cases is as shown in the figure.
4.2. Question with mor than 5 marks

31
4.2.1. Discuss Terzaghi’s theory of consolidation, stating the various assumptions and their
validity
4.2.2. Explain the different e-log p curves for the consolidation.
4.2.3. How do you determine the pre-consolidation pressure and its determination in soil
engineering practice?
4.2.4. Explain the significance of pre-consolidation pressure. Describe the Casagrande
method of determining it
4.2.5. Write a brief procedure of consolidation test and to determine the coefficient of
consolidation by both logarithmic time fitting method and square root of time method.
4.2.6. What is over consolidation soil? Explain briefly with an example.
4.2.7. Explain the square root of time fitting method of determining the coefficient of
consolidation of a clay sample.
4.2.8. Explain the phenomena of secondary consolidation. Differentiate
between the secondary consolidation index and the coefficient of secondary
consolidation.
4.2.9. What are the different causes of pre-consolidation of soils? What is the effect of pre-
consolidation on the settlement?

32
4.2.10. Define the following terms: (i) Coefficient of compressibility (ii) Coefficient of
volume change (iii) Compression index (iv)Expansion index (v) Recompression
index
4.2.11. Explain different types of consolidation and their uses.
4.2.12. Differentiate between normally consolidated and over consolidated soils.
How would you determine the over consolidation pressure?
4.2.13. A soil sample 20 mm thick takes 20 minutes to reach 20% consolidation. Find the
time taken for a clay layer 6 m thick to reach 40% consolidation. Assuming double
drainage in both the cases.
4.2.14. A stratum of normally consolidated clay 7m thick is located at a depth 12m below
ground level. The natural moisture content of the clay is 43% and its liquid limit is 48%.
The specific gravity of the solid particles is 2.76. The water table is at a depth of 5m
below ground surface. The soil is sand above the clay stratum. The submerged unit
weight of the sand is 11kN/m-3
and 18 kN/m3
above the water table. The average
increase in pressure at the centre of the clay stratum is 120kN/m3
due to the weight of the
building that will be constructed on the sand above the clay stratum. Estimate the
expected settlement of the structure.
4.2.15. Saturated soil of 5 m thick lies above an impervious stratum and below a pervious
stratum. It has a compression index of 0.25 with k = 3.2×10-10
m/sec. Its void ratio at a
stress of 147 kN/m2
is 1.9. Compute (i) The change in voids ratio due to increase of
stress to 196 kN/m (ii) Coefficient of volume compressibility (iii) Coefficient of
consolidation (iv) Time required for 50% consolidation.
4.2.16. A soil has compression index of 0.28. At a stress of 120 kN/m2
the void ratio is 1.02.
Compute (i) void ratio if the stress on the soil is increased to 180 kN/m2
(ii) total
settlement of the stratum of 6 m thickness.
4.2.17. A 10m thick submerged clay layer which is drained at both the upper and lower
boundaries is subjected to a wide surface pressure of 50kN/m2
. The water table is
coincident with the top of the clay layer at the ground surface. If the coefficient of
consolidation of the clay is 1.16 x 10-2
cm2
/sec .Determine the pore pressure at the mid
depth of the layer 50 days after the surface pressure was applied. Consider the degree of
consolidation= 0.23.
4.2.18. A layer of submerged soil 8m thick is drained at its upper surface but is
underlain by impermeable shale. The sol is subjected to a uniform vertical stress which is
produced by the construction of an extensive embankment on the ground surface. If the
coefficient of consolidation for the soil is 2 x 10-3
cm2
/sec calculate the times when 50%
and 90% respectively of the final settlement will take place. Consider T50 =0.197
4.2.19. A laboratory sample of lay 2cm thick took 15min to attain 60% consolidation under a
double drainage condition. What will be the time required to attain the same degree of
consolidation for a clay layer 3cm thick under the foundation of a building for a similar
loading and drainage condition, What is the value of cv.

33
4.2.20. A oedometer test is performed on a 2cm thick clay sample. After 5min, 50%
consolidation is reached. After how long a time would the same degree of
consolidation is achieved in the field where the clay layer is 3.7m thick? Assume the
sample and the clay layer has the same drainage boundary conditions (double drainage).
4.2.21. A recently completed fill was 10m thick and its initial average void ratio was 1.0.
The fill was loaded on the surface by constructing an embankment covering a large
area of the fill.
Some months after the embankment was constructed, measurements of the fill
indicated an average void ratio of 0.8. Estimate the compaction of the fill.
4.2.22. During a consolidation test, as sample of fully saturated clay 3cm thick is
consolidated under a pressure increment of 200kN/m2
. When equilibrium is reached,
the sample thickness is reduced to 2.6cm. The pressure is then removed and the sample is
allowed to expand and absorb water. The final thickness is
observed as 2.8cm and the final moisture content is determined as 24%. If the specific
gravity of the soil solids is 2.7, find the void ratio of the sample before and after
consolidation.
4.2.23. A soil sample has a compression index of 0.3. If the void ratio at stress of 1.4kg/m2
is
0.5, compute (i) the void ratio if the stress is increased to 2kg/m2
and (ii) the settlement
of a soil stratum 4m thick.
4.2.24. A 2.5cm thick sample of clay was taken from the field for predicting the time of
settlement for a proposed building which exerts pressure of 100kN/m2
over the clay
stratum. The sample
was loaded to 100kN/m2
and proper drainage allowed from top to bottom. It was seen
that 50% of the total settlement occurred in 3minutes. Find the time required for
50% of the total settlement of the building, if it is to be constructed on a 6m thick layer
of clay which extends from the ground surface and is underlain by sand.
4.2.25. Soil investigation at a site gave the following information. Fine sand exists to a depth
of 10.6m and below this lies a soft clay layer 7.6m thick. The water table is at 4.6m
below the ground surface. The submerged unit weight of sand is 10.4 kN/m3
and the unit
weight above the water table is 17.6kN/m3
. The water
4.2.26. content of the normally consolidated clay is 40%, its liquid limit is 45% and the
specific gravity of the solid particles is 2.78. The proposed construction will transmit a
net stress of 120kN/m3 at the centre of the clay layer. Find the average settlement of the
clay layer.
4.2.27. The loading period for a new building extended form May 1995 to May 1997. In May
1960, the average measured settlement was found to be 11.43cm. It is known that
the ultimate settlement will be about 35.56cm. Estimate the settlement in May 1965.
Assume double drainage to occur.
4.2.28. A stratum of normally consolidated clay of thickness 3m is drained on one side
only. It has the hydraulic conductivity of k= 5x 10-8
cm/s and a coefficient of volume
compressibility mv.

34
4.3. JNTU Previous year Exam questions
JNTUK NOV/DEC -2016
4.3.1. What is the time factor? How is it related to the average degree of consolidation?
4.3.2. In a consolidation test the following results have been obtained, when the load was
changed for 62 kN/m2
to 90kN/m2
, the void ratio changed from 0.7 to 0.65, Determine
the coefficient of volume decrease, mv and compression index, Cc.
JNTUK Nov-2017
4.3.3. Explain in detail square root of time fitting method for evaluation of consolidation
from laboratory test data.
4.3.4. Normally consolidated clay settled by 2 cm when the effective stress was
increased from 100kPa to 200kPa. Calculate the settlement when the effective stress is
increased to 400kPa and 800kPa.
4.3.5. A normally consolidated clay layer 2m thick is sandwiched between two sand
layers. The average overburden stress at the middle of clay layer can be taken as
160kN/m2
. Due to construction of a structure there is an increase in effective
vertical stress of 40kN/m2
at the middle of clay layer. The liquid limit of clay layer.
is 60% and the initial Void ratio is 0.9. Estimate the primary settlement.
4.3.6. Expain square root of time fitting method for determination of co-efficent of
coefficent of consolidation
JNTUK April/May -2017
4.3.7. State the assumptions made in Terzaghi’s theory of one-dimensional consolidation
4.3.8. Write short notes on the Log fitting method for evaluation of Cv from laboratory
consolidation test.
4.3.9. A 30 mm thick oedometer sample of clay reached 30% consolidation in 15 minutes
with drainage at top and bottom. How long would it take the clay layer from which this
sample was obtained to reach 60% consolidation? The clay layer had one-way drainage
and was 6 m. thick

35
UNIT - V
Shear Strength Of Soils: Importance of shear strength – Mohr’s– Coulomb Failure theories –
Types of laboratory tests for strength parameters – strength tests based on drainage conditions
– strength envelops – Shear strength of sands - dilatancy – critical void ratio.
5. Shear Strength of Soils
5.1. Questions for 2 marks.
5.1.1. What are the important characteristics of Mohr’s circle?
Ans:
Normal stress
Shear stress
The plane inclined at an angle of to the
horizontal has acting on it the maximum shear
stress equal to , and the normal stress on
this plane is equal to .
The plane with the maximum ratio of shear stress
to normal stress is inclined at an angle
of to the horizontal, where a is the slope of the line tangent to the Mohr circle and
passing through the origin.
5.1.2. What are the merits of direct shear test?
Ans: a) It is easy to test sands and gravels.
b) Large samples can be tested in large shear boxes, as small samples can give
misleading results due to imperfections such as fractures and fissures or may not
be truly representative.
c) Samples can be sheared along predetermined planes, when the shear strength along
fissures or other selected planes are needed.

36
5.1.3. What are the different tests for shear strength?
Ans: a) Direct shear test
b) Tri axial Shear test
c) Unconfined compression test
d) Vane shear test
5.1.4. What are the demerits of direct shear test?
Ans: a) The failure plane is always horizontal in the test, and this may not be the weakest
plane in the sample. Failure of the soil occurs progressively from the edges towards
the centre of the sample.
c) The area under shear gradually decreases as the test progress. But the corrected area
cannot be determined and therefore, the original area is take is for the computation of
stress.
b) There is no provision for measuring pore water pressure in the shear box and so it is
not possible to determine effective stresses from undrained tests.
c) The shear box apparatus cannot give reliable undrained strengths because it is
impossible to prevent localised drainage away from the shear plane.
5.1.5. Define Dilatancy. [1]
Ans: The correct description of volume changes due to imposed stress is fundamental to the
modeling of the stress-strain behavior of soils. The remarkable phenomenon of the coupling
between volume and shape changes observed qualitatively and termed granular dilatancy.
Or
in simpler terms “The phenomenon of increase in volume of soil during shearing is called
dilation’.
5.1.6. Define Critical void ratio.
Ans: The “critical void ratio” by Casagrande defined as void ratio at which drained shear takes
place at constant volume.
5.1.7. Define Shear strength of soil.
Ans: Shear strength of a soil is equal to the maximum value of shear stress that can be
mobilized within a soil mass without failure taking place.
5.1.8. State Mohr- Coulomb failure theories.
Ans: When the soil sample has failed, the shear stress on the failure plane defines the shear
strength of the soil. Thus, it is necessary to identify the failure plane. Is it the plane on which
the maximum shear stress acts, or is it the plane where the ratio of shear stress to normal stress
is the maximum.

37
5.1.9. What are the merits and demerits of vane
shear test.
5.1.10. What is unconfined compression test?
Ans: The unconfined compression test is a special
ease of tri-axial compression test in which
𝜎2 = 𝜎3 = 0. The cell pressure in the tri-axial cell
is also called the confining pressure. Due to the
absence of such a confining pressure, the uni-axial
test is called the unconfined ompression test. The
cylindrical specimen of soil is subjected to major
principal stress 𝜎1 till the specimen fails due to
shearing along a critical plane of failure.
5.1.11. Write Mohr-coulomb equation?
Ans: The Mohr-Coulomb failure criterion can be written as the equation for the line that
represents the failure envelope. The general equation is
Where = shear stress on the failure
plane
c = apparent cohesion
= normal stress on the failure plane
f = angle of internal friction
5.1.12. What are the advantages of tri axial shear test over the Direct
shear test?
Ans: i. The shear test under all the three drainage conditions can be performed with
complete control
ii. The precise measurements of the pore pressure and volume change during the test
are possible.
iii. The stress distribution on the failure plane is uniform
iv. The state of stress with in the specimen during any stage of stress, as well as at
failure is completely determines.
5.2. Questions for 5 and above marks.

38
5.2.1. Explain Mohr-Coulomb theory of shear strength. Sketch typical strength envelope for a
soft clay, clean sand and a silty clay
5.2.2. Classify the shear tests based on drainage conditions. Explain how the pore pressure
variation and volume change take place during these tests. Enumerate the field
conditions which necessitate each of these tests.
5.2.3. What types of field tests are necessary for determining the shear strength parameters of
sensitive clays?
5.2.4. What are the advantages and disadvantages of a triaxial compression test in
comparison with a direct shear test.
5.2.5. What are the advantages and disadvantages of direct shear test over triaxial test?
5.2.6. Explain about triaxial compression test
5.2.7. Discuss the shearcharacteristics of cohesionless and cohesive soils.
5.2.8. Discuss modified failure envelope. What are its advantages and disadvantages over the
standard failure envelope?
5.2.9. Derive the relation between the principle stresses at failure using Mohr-Coulomb failure
criterion.
5.2.10. Explain liquefaction of soils.
5.2.11. What is Coulomb’s equation for shear strength of soil? Discuss the factors that affect
the shear strength parameters of soil.
5.2.12. Enlist the features of a triaxial compression test apparatus and describe them briefly.
5.2.13. What are the advantages and disadvantages of triaxial compression test in
comparison to direct shear test?
5.2.14. What is critical void ratio? How would you determine it in the laboratory? Also
explain the conditions causing liquefaction of sand.
5.2.15. For which types of soils will the unconfined compression test give reliable results?
Draw a Mohr circle for this test. How do you consider the change in the area of the
specimen which takes place during the test in final results?
5.2.16. A soil specimen when tasted in unconfined compression test fails at axial test of
120kN/m2
the same sample tested in tri-axial compression test. The failure occurs at cell
pressure of 40kN/m2
and axial deviator stress of 160kN/m2
. Determine shear strength
parameter.
5.2.17. A UU test is carried out on a saturated normally consolidated clay sample at a
confining pressure of 3 kg/cm2
. .The deviator stress at failure is 1 kg/cm2
.
5.2.18. Two samples were tested in a triaxial machine. The all found pressure maintained
further first sample was 2kg/cm2
and 20kg/cm2 and the failure occurred at
additional axial stress of 7.7 kg/cm2
,while for the second the values were 5.0 kg/cm2
and
13.7 kg/m2
resp. Find c and ¢ of the soil.

39
5.2.19. A cylindrical specimen of a saturated soil fails at an axial stress of 180 kN/m2
in an
unconfined compression test. The failure plane makes an angle of 54º with horizontal.
Calculate the shear strength parameters of soil.
5.2.20. A remoulded specimen of soil prepared by compaction to standard proctor
maximum dry unit weight at optimum moisture content is used for consolidated-
undrained triaxial test with pore
pressure measurements. The test results are given below: Test
5.2.21. Cell pressure Deviator stress Pore pressure
No (kN/m2
) at failures (kN/m2
) (kN/m2
)
1 40 300 05
2 100 443 10
3 165 615 12
Determine the values of effective shear stress parameters by
(i) Drawing Mohr envelope (ii) Drawing modified envelope
5.2.22. A direct shear test was conducted on a soil, whose results are given below:
Normal stress, kN/m2
150 250
Shear stress at failure kN/m2
110 120
Plot the graph and determine the shear strength of parameters of the soil. If a triaxial
test is conducted on the same soil, what would be the deviator stress at failure when the
cell pressure is 150 kN/m2
5.2.23. A vane 11.25 cm long and 7.5 cm in diameter was pressed into soft clay at the bottom
of a borehole. Torque was applied to cause failure of soil. The shear strength of clay was
found to be
37 kN/m2
. Determine the torque that was applied.
5.2.24. A series of shear tests was performed on a soil. Each test was carried out until the soil
sample sheared and the principal stress for each test are as follows:
Test σ3(kN/m2
) σ1(kN/m2
)
1 300 875
2 400 1160
3 500 1460
Plot the Mohr circle of stress and determine strength envelope and angle of internal
friction of the soil.
5.2.25. A direct shear test was performed on a 6cm x 6cm sample of dry sand the normal load
was 360N. The failure occurred at a shear load of 180N.Plot the Mohr strength envelope
and determine ᴓ . Assume c=0 also determine principal stress at failure.
5.2.26. A series of direct shear test was conducted on soil each test was carried out till the
same sample failed. The following results were obtained. Determine cohesion intercept
and angle of shearing resistance and plot the Mohr circle. What is the shear strength of
soil along a horizontal plane at a depth 4m in a deposit of sand having the following
properties: Angle of internal friction =35o
, Dry unit weight =17kN/m3
, Specific gravity
=2.7. Assume the ground water table is at a depth of 2.5m from the ground surface. Also
find the change ins hear strength when the water table rises to ground surface.

40
5.2.27. A consolidated drained triaxial test was conducted on a granular soil. At failure
σ1’/σ3’ =4.0. The effective minor principal stress at failure was 100kN/m2
. Compute ϕ’
and the principal stress difference at failure.
5.2.28. A drained triaxial test on sand with σ3’ =150 kN/m2
gave
(σ1’/σ3’) =3.7. Compute (a) σ1f’ (b) (σ1-σ3)f and ϕ’.
5.2.29. At a depth of 6m below the ground surface at a site, a vane shear tests gave a torque
value of 6040 N-cm. The vane was 10cm high and 7cm across the blades. Estimate the
shear strength of the soil.
5.2.30. A vane 11.25cm long and 7.5cm in diameter was pressed into soft clay at the bottom
of a borehole. Torque was applied to cause failure of soil. The shear strength of clay was
found to be
37kN/m2
. Determine the torque that was applied.
5.3. Previous year papers JNTU Questions
JNTUK NOV/DEC -2016
5.3.1. What is Mohr Circle? Discuss its important characteristics?
5.3.2. A sample of dry sand was subjected to triaxial test, with a confining pressure of
250kN/m2
. The angle of shearing resistance was found to be 360
. At what value of
major principal stress, the sample is likely to fail?
JNTUK Nov-2017
5.3.3. Differentiate between unconsolidated undrained test and drained test.
5.3.4. Under what conditions are these test results used for design purposes? A sample of
dry cohesionless soil was tested in a triaxial machine. If the angle of shearing
resistance was 320 and the confining pressure 100kN/m2, determine the deviator stress
at which the sample failed.
JNTUH March-2017
5.3.5. Differentiate between conventional failure envelope and modified failure envelope
with the neat sketches.
5.3.6. Define stress path, and draw typical stress paths (TSP, TSSP, ESP) consolidatd
clay, and on over- for a drained test and undrained test on normally consolidated
clay.
5.3.7. Discuss Skempton's pore pressure parameters.
5.3.8. In a direct shear test the major and minor principal stresses were found to be 500
kN/m2
and 300 kN/m2
respectively. Determine the normal and shear stresses on a
plane inclined at 30° to the major principal plane in a clock-wise direction
JNTUK April/May -2017

41
5.3.9. Explain the shear characteristics of sand and normally loaded clay. Write short notes
on field compaction control.
5.3.10. State the assumptions made in Terzaghi’s theory of one-dimensional consolidation
5.3.11. Explain the basic differences between a box shear test and a triaxial shear test for
soils
5.3.12. A cylindrical specimen of saturated clay, 4.5 cm in diameter, and 9 cm long, is tested
in an unconfined compression apparatus. Find the cohesion if the specimen fails at an
axial load of450 N. The change in length of the specimen at failure is 9mm.

42
6. Previous Year Question Papers
Geotechnical Engineering-1 R13 JNTUH May -2018

44
Geotechnical Engineering -1 R09 -JNTUH May 2018

46
Bibliography
[1] N. SASIHARAN, "MECHANICS OF DILATANCY AND ITS APPLICATION TO
LIQUEFACTION," WASHINGTON STATE UNIVERSITY, Dec 2006. [Online].
Available:
http://www.dissertations.wsu.edu/Dissertations/Fall2006/n_sasiharan_120706.pdf.
[Accessed 20 Dec 2018].
[2] "Wikipedia," [Online]. Available: https://en.wikipedia.org/wiki/Soil_mechanics.
[3] K. Dept. of Civil Engg. Indian Institute of Technology, "NPTEL- Advanced Geotechnical
Engineering," [Online]. Available:
https://nptel.ac.in/courses/105104132/Module5/lecture30.pdf. [Accessed Dec 2018].
[4] A. P. Dr. Baleshwar Singh, "NPTEL, E learning coarses from IIT IISc," Indian Institute
of Technology Guwahati, [Online]. Available: https://nptel.ac.in/syllabus/105103097/.
[Accessed Sep-Oct 2018].

5
UNIT – I
Relative density. Index Properties Of Soils: Grain size analysis – Sieve
1. Introduction Soil Mechanics
1.1.1. Define is soil mechanics and soil according to Civil Engineer?
Ans: Soil is an unconsolidated accumulation of solid particles, which are produced by the
mechanical and chemical disintegration of rocks, regardless of whether or not they contain an
admixture of organic constituents.
To an engineer, it is a material that can be:
built on: foundations of buildings, bridges
built in: basements, culverts, tunnels
built with: embankments, roads, dams
supported: retaining walls
1.1.2. Define water content of soil? Write it’s mathematical expression.
Ans: The ratio of the mass of water present to the mass of solid particles is called the water
content (w), or sometimes the moisture content or Water Content.
w =
Its value is 0% for dry soil and its magnitude can exceed 100%.
1.1.3. What do you mean by degree of saturation? Write it’s mathematical expression.
Ans: The volume of water (Vw) in a soil can vary between zero (i.e. a dry soil) and the volume
of voids. This can be expressed as the degree of saturation (S) in percentage.
=
For a dry soil, S = 0%, and for a fully saturated soil, S = 100%.
1.1.4. Define bulk density of soil? How it is related with submerged density.
Ans: Bulk unit weight (γt or γ ) is a measure of the amount of solid particles plus water per
unit volume.
γ = γ =
( + )
( + )
submerged unit weight γ = γ − γ
1.1.5. Draw 3 - phase system of soil with weight and volume.
Ans:

6
1.1.6. What is void ratio? Show the relation ship with porosity.
Ans: Void ratio is the ratio of the volume of voids (Vv) to the volume of soil solids (Vs), and
is expressed as a decimal.
=
Void ratio = = ( )
1.1.7. What is specific gravity? List different methods to find specific gravity.
Ans: The mass of solid particles is usually expressed in terms of their particle unit
weight (γs) or specific gravity (Gs) of the soil grain solids.
γ = = . γ
It can be determined by 1. Pycnometer 2. Density bottle
1.1.8. List any five soil forming minerals.
Ans: Kolinite, silica, Illite, Mont morillonite, halloysite.
1.1.9. What is soil profile? Show a neat diagram
of soil profile.
Ans: Soil formation starts first with breakdown of
rock by weathering and soil horizon development
process leads to the development of a soil profile.
A soil profile is the vertical display of soil
horizons. A soil profile is divided into layers called
horizons.
Figure 1: Soil Profile with A,B, C horizonFigure 1
shows an Example soil horizons(Profile). a) top
soil and colluvium b) mature residual soil c) young
residual soil d) weathered rock.
A very general and simplified soil profile can be
described as follows:
Figure 1: Soil Profile with A,B, C horizon

11
1.3.13. By three phase soil system, prove that the degree of saturation S (as ratio) in terms of
mass unit weight (γ), void ratio (e), specific gravity of soil grains (G) and unit weight
of water (γW) is given by the expression. mass unit weight (γ) =
( )
( )
1.3.14. An earthen embankment of 106 m3
volume is to be constructed with a soil having a
void ratio of 0.80 after compaction. There are three borrow pits marked A, B and C
having soils with voids ratios of 0.90, 0.50 and 1.80 respectively. The cost of
excavation and transporting the soil is Rs 0.25, Rs 0.23 and Rs 0.18 per m3
respectively. Calculate the volume of soil to be excavated from each pit. Which borrow
pit is the most economical? (Take G = 2.65).
1.3.16. A soil sample has a porosity of 40 per cent. The specific gravity of solids is 2.70.
Calculate i) Voids ratio ii) Dry density and iii) Unit weight if the soil is completely
saturated.
1.3.17. A soil has a bulk unit weight of 20.11 KN/m3 and water content of 15 percent.
Calculate the water content of the soil partially dries to a unit weight of 19.42 KN/m3
and the voids ratio remains unchanged.
1.3.18. An earth embankment is compacted at a water content of 18% to a bulk density of
1.92 g/cm3
. If the specific gravity of the sand is 2.7, find the void ratio and degree of
saturation of the compacted embankment.
1.4. JNTU previous year selective questions.
1.4.1. Explain liquid limit, plastic limit and shrinkage limit. (JNTUK-R13-Nov/Dec-
2016)
1.4.2. Define the terms specific gravity of particles, porosity and submerged density.

14
If the rate of flow is q (volume/time) through cross-sectional area (A) of the soil mass, Darcy's
Law can be expressed as
= = .
where k = permeability of the soil
=
∆
∆ = difference in total heads
L = length of the soil mass
2.1.5. What is surface tension? What is it’s significance in flow of water through soil.
Ans: Surface tension is the elastic tendency of a fluid surface which makes it acquire the least
surface area possible.
Surface is the cause of capillarity in soil. The rise of water in the capillary tubes, or the fine
pores of the soil, is due to the existence of surface tension which pulls the water up against the
gravitational force.
2.1.6. What is meant by capillary rise in soil and how it affects the stress level in soils?
Ans: The pores of soil mass may be looked upon as a series of
capillary tubes, extending vertically above water table.
The rise of water in the capillary tubes, or the fine pores of the
soil, is due to the existence of surface tension which pulls the
water up against the gravitational force.
Above the water table, when the soil is saturated, pore pressure
will be negative (less than atmospheric) because of capillarity.
Thus, the height above the water table to which the soil is
saturated is called the capillary rise, and this depends on the
grain size and the size of pores. In coarse soils, the capillary
rise is very small.
2.1.7. How effective stress varies under submerged
conditiof soil?
Ans: For Submerged Condition
at top = 0, = 0, = − = 0
At hw
= ℎ , = ℎ , = 0
At z
= ℎ + , = ℎ + , = ′
For fully saturated soil Condition

15
at top
= 0, = 0, = − = 0
At z
= , = , = ′
Where, = Norma stress
= neutral stress
= effective stress
It indicates that effective stress variation is not affected
irrespective variation of water level above ground
2.1.8. List out the uses of flow net?
Ans: i. Estimation of seepage losses from reservoirs
ii. Determination of uplift pressures below dams
iii. Checking the possibility of piping beneath dams
2.1.9. What do you mean by ‘Capillary rise of water in soil’? when this condition
occurs in soil.
Ans: The groundwater can be sucked upward by the soil through very small pores that are
called capillars. This process is called capillary rise.
In fine textured soil (clay), the upward movement of water is slow but covers a long distance.
On the other hand, in coarse textured soil (sand), the upward movement of the water is quick
but covers only a short distance.
Soil texture Capillary rise (in cm)
coarse (sand) 20 to 50 cm
medium 50 to 80 cm
fine (clay) more than 80 cm up to several metres
2.1.10. Define quick sand? In what kind of soil quick sand condition arises?
Ans: In case of upward seepage or flow, decreases in effective stress can be seen. In such a
situation, effective stress is reduced to zero and the soil behaves like a very viscous liquid.
Such a state is known as quick sand condition.
In nature, this condition is usually observed in coarse silt or fine sand subject to artesian
conditions.

16
2.1.11. Derive equation for seepage pressure for upward
flow though soil.
Ans:
At the bottom of the soil column,
= , = ( + ∆ ) ,
= − ( + ∆ )
Seepage pressure => = −∆
2.2. Question of 3 marks
2.2.1. Define effective stress, neutral stress and total stress?
Solution: = Norma stress(Total stress): total Stress in the soil mass is defined as the
total load per unit cross section area at any point of soil mass.
At z
= ,
i.e., Stress in the ground due to self-weight of soil, water and external load.
= neutral stress: (Pore water pressure): Stress in the ground due to only water.
= ,
i.e., pressure due water in the pores or voids inside the soil mass.
= effective stress : effective stress is the stress obtained by deducting neutral stress from
Normal stress or total stress.
= −
=
Theoretically, stress in the ground which is subjected on only through soil grains.

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