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Soil Mechanics Test 2
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Soil Mechanics Test 2
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  • Question 1/10
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    In a clay mineral lattice, two silicon ions in tetrahedral unit is substituted by two aluminum ions due to readily availability of aluminum ions in water. Which of the following is true?
    Solutions

    Concept:

    An Aluminium ion has 3 positive charges and one silicon ion has 4 positive charges.

    When one silicon ion replaced by one aluminum ion, there would be deficiency of one unit (4 – 3 = 1) positive charge per substitution since silicon ion has more positive charge then aluminum ion.

    ∴ Net charge deficient = 1 × 2 = 2 unit of positive charge
  • Question 2/10
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    Match the following List-I (Soil) with List-II (Transporting agent) and select an appropriate choice.

    List-I

    List-II

    A. Drift

    1. Water

    B. Talus

    2. Wind

    C. Loess

    3. Glacier

     D. alluvial 

    4. Gravity

    Solutions

    The soils formed at a place may be transported to other places by agents of transportation, such as water, wind, ice and gravity.

    1) Water transported Soils: Flowing water is one of the most important agents of transportation of soils. Swift running water carries a large quantity of soil either in suspension or by rolling along the bed.

    The size of the soil particles carried by water depends upon the velocity.

    All type of soils carried and deposited by water are known as alluvial deposits. Deposits made in lakes are called lacustrine deposits. Such deposits are laminated or varved in layers. Marine deposits are formed when the flowing water carries soils to ocean or sea.

    2) Wind transported Soils: Soil particles are transported by winds. The particle size of the soil depends upon the velocity of wind. Soils deposited by wind are known as aeolian deposits.

    Loess is a silt deposit made by wind. These deposits have low density and high compressibility. The bearing capacity of such soils is very low.

    3) Glacier-Deposited Soils: Glaciers are large masses of ice formed by the compaction of snow. As the glaciers grow and move, they carry with them soils varying in size from fine grained to huge boulders.

    Drift is a general term used for the deposits made by glaciers directly or indirectly. Deposits directly made by melting of glaciers are called till. The soil carried by thw meling water from the frint of a glacier is termed out-wash.

    4) Gravity deposited soil: These are soils transported through short distances under the action of gravity. Colluvial soils such as talus have been deposited by the gravity. Talus consists of irregular, corase particles. It is a good source of broken rock pieces and coarse grained soils for many engineering works.
  • Question 3/10
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    The composite structure formed when bulky grains appear floating in clay without touching is called as
    Solutions

    Clay-Matrix Structure:

    • Clay-matrix structure is a composite structure formed by clay particles is very large as compared with bulky, coarse-grained particles. The clay forms a matrix in which bulky grains appear floating without touching one another.
    • Soils with a clay-matrix structure have almost the same properties as clay. They are more stable, as disturbance has very little effect on the soil formation with a clay matrix structure.

  • Question 4/10
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    The surface area of soil particle is measured using air permeability method and found to be 10 cm2. The same soil when placed in a 10 cc beaker full of water the volume of water remaining inside is found to be 6 cc. The specific surface area of the soil is
    Solutions

    Concept:

    The specific surface area is defined as the total surface area of any material per unit mass or volume.

    Specific surface \(= \frac{{Surface\;Area}}{{Volume}}\)

    Calculation:

    Surface area = 10 cm2, Volume = 10 cc – 6 cc = 4 cc

    Specific surface area = \(\frac{{10}}{4} = 2.5/cc\)

  • Question 5/10
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    Consider the following statements regarding clay particles.

    1. The clay particles carry net negative charge.

    2. The cation concentration increases with the distance from the surface of the particle.

    3. When the spacing between the two clay particles is very small, the force of attraction is greater than the force of repulsion.

    The correct statements are:

    Solutions

    The clay particles carry a net negative charge on their surfaces. This is the result both of isomorphous substitution and of a break in continuity of the structure at its edges. In dry clay, the negative charge is balanced by exchangeable cations like Ca2+, Mg2+, Na+, and K+ surrounding the particles being held by electrostatic attraction. When water is added to clay, these cations and a few anions float around the clay particles. This configuration is referred to as a diffuse double layer.

     

    The cation concentration decreases with the distance from the surface of the particle. The force of attraction between water and clay decreases with distance from the surface of the particles.

    When two clay particles in suspension come close to each other, the tendency for interpenetration of the diffuse double layers results in repulsion between the particles. At the same time, an attractive force exists between the clay particles that is caused by van der Waals forces and is independent of the characteristics of water. Both repulsive and attractive forces increase with decreasing distance between the particles, but at different rates. When the spacing between the particles is very small, the force of attraction is greater than the force of repulsion.

  • Question 6/10
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    Total weight of soil sample = 10 kg

    Soil particles > 4.75 mm = 3 kg

    Soil particles < 75 μ = 2 kg

    The liquid limit of the sample is 40% and the plastic limit is 20%. The soil can be classified as
    Solutions

    Concept:

    Classification of Coarse-grained soil is done on the basis of

    (1) Grain Size

    (2) Gradation characteristics 

    (3) Percentage of fines present in soil by weight (fines means particle size < 75 μ)

    Case – 1: When fines are less than 5%

    GW

    Well graded gravel

    GP

    Poorly graded gravel

    1.  More than half of coarse fraction > 4.75 mm

    2.      \({C_u} = \frac{{{D_{60}}}}{{{D_{10}}}} > 4\)

    3.     \({C_c} = \frac{{D_{10}^2}}{{{D_{60}} \times {D_{10}}}}\;;\;\;\;\;\)1 < Cc < 3

    Otherwise

     

    SW

    Well graded sand

    SP

    Poorly graded sand

    1.       More than ½ of coarse fraction < 4.75 mm

    2.       Cu > 6

    3.       1 < Cc < 3

    Otherwise

     

    Case-2 : When fines are. between 5-12%:-

    • Dual symbols are used. The first part of the dual symbol indicates gradation (SW), where as second part indicates the nature of fines (SC).
    • The soil having dual symbol SW — SC signifies well-graded sand with clay as fines which is plotted above A-line on Plasticity chart.


    The Classification can be done as follows:

    1. GW — GC → Well graded gravel containing clay as fines
    2. GP — GC → Poorly graded gravel containing clay as fines
    3. GW — GM → Well graded gravel containing Silt as fines
    4. GP — GM → Poorly graded gravel containing Silt as fines
    5. SW—SC → Well graded sand containing clay as fines
    6. SP—SC → Poorly graded sand containing clay as fines
    7. SW — SM → Well graded sand containing silt as fines
    8. SP — SM → Poorly graded sand containing silt as fines.


    Case – 3: When fines are greater than 12%:-

    (1) Gravel:

    (a) Fineness > 12% & Ip < 4% e.g GM → Silty Gravel

    (b) Fineness > 12% & Ip > 7% e.g. GC → Clayey Gravel

    (2) Sand:

    (a) Fineness > 12% & Ip  < 4% e.g. SM → Silty Sand

    (b) Fineness > 12%, & Ip > 7% e.g. SC → Clayey Sand

    If Ip is between 4 – 7 then again Dual symbol are used.

    Calculation:

     

    % fines \(= \frac{2}{{10}} \times 100 = 20\%\)

    % fines > 12%

    Gravel fraction (> 4.75 mm) = 3.0 kg

    Sand fraction (75 μ – 4.75 mm) = 5 kg

    Sand fraction > Gravel fraction, Thus the soil can be classified as sand.

    Ip = ωL - ωp = 40 – 20 = 20%

    Ip > 7% and fineness > 12%

    Soil is clayey sand i.e SC.

  • Question 7/10
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    A soil taken from the construction site was subjected to grain size analysis in the laboratory and following results was obtained:

    Sieve Seize

    4.75 mm 

    600 μ 

    425 μ 

    222 μ 

    150 μ 

    75 μ 

    % finer

    100

    60

    45

    30

    10

    0

    The soil sample is classified as ____
    Solutions

    Concept:

    Soil classification as per Indian standards:

    Coarse-grained soils

    (More than half of the material is larger than 75 μ sieve)

    Gravels

    Sands

    More than half of the coarse fraction is larger than 4.75 mm sieve

    More than half of the coarse fraction is smaller than 4.75 mm sieve

    Clean gravel

    Gravel with appreciable fines

    Clean sands

    Sand with appreciable fines

    GW

    GP

    GM

    GC

    SW

    SP

    SM

    SE

    Fine-grained soils

    (More than half of the material is smaller than 75 μ sieve)

    Silts and clay

    Low compressibility

    (wL < 35)

    Medium compressibility

    (50 < wL > 35)

    Higher compressibility

    (wL > 50)

    ML

    CL

    OL

    MI

    CI

    OI

    MH

    CH

    OH

    High organic peat soil → (Pt)

    Computation:
    % retained on 75 μ = 100 – 5 = 95% > 50 > Thus soil is classified as coarse grained soil.

    Since the soil ranges between 4.75 mm to 75 μ, thus it a sandy soil.

    Here fines are less than 5%.

    For classification of finer soil D60, D30, and D10 are required to be computed.

    D60 = 600 μ, D 30 = 222 μ, and D10 = 150 μ

    Coefficient of uniformity \(\left( {{{\rm{C}}_{\rm{u}}}} \right) = \frac{{{{\rm{D}}_{60}}}}{{{{\rm{D}}_{10}}}} = \frac{{600{\rm{\mu }}}}{{150{\rm{\mu }}}} = 4< 6\) 

    Coefficient of curvature \(\left( {{{\rm{C}}_{\rm{c}}}} \right) = \frac{{{\rm{D}}_{30}^2}}{{{{\rm{D}}_{60}} \times {{\rm{D}}_{10}}}} = \frac{{{{222}^2}}}{{600 \times 150}} = 0.547 < 1\)

    For well graded soil: Cu > 6 & 1 < Cc < 3

    Thus, soil is classified as poorly graded sand.
  • Question 8/10
    1 / -0

    Sieve analysis is conducted on two soils A and B. The results of the test are given below:

    Soil A

    Soil B

    61% of the soil passed through 75 μ Is sieve

    27% of the soil passed through 75 μ Is sieve

    100% of the soil is passing through 4.75 mm IS sieve

    70% of the soil is passing through 4.75 mm IS sieve

     

    Soil

    Cu

    Ip(%)

    WL(%)

    A

    -

    11

    45

    B

    7

    9

    -

     
    Identify the soil on the basis of the result obtained?
    Solutions

    For Soil A:

    More than 50% of the soil fraction passed through 0.075 mm Is sieve is known fine Grained soil

    IP(A-line) = 0.73 (WL - 20)

    IP = 0.73 (45 - 20) = 18.25

    IP for soil = 11

    Thus soil lies below A-line of the Plasticity chart. it is known as silt.

    WL for soil lies between 35% to 50% WL = 45% (Given)

    Soil A is silt of Intermediate Compressibility Soil A = MI

    For Soil B:

    More than 50% of the soil is retained through 0.075 mm Is sieve. It is termed as coarse Grained soil.

    Less than 50% of the soil is retained over 4.75 mm Is sieve, so it is sand

    % fineness Given = 27% > 12% and IP = 9% (Given)

    i.e. IP > 7%

    So, it is termed as SC soil

    Soil B = SC

  • Question 9/10
    1 / -0

    The coefficient of uniformity of a soil specimen is 4 and the coefficient of curvature is 1.6. The ratio D30/D10 correct up to 2 decimal places is______
    Solutions

    Concept:

    Coefficient of uniformity \(= \frac{{{D_{60}}}}{{{D_{10}}}}\)

    Coefficient of curvature \(= \frac{{D_{30}^2}}{{{D_{60}}\; \times \;{D_{10}}}}\)

    Calculation:

    CU = 4, Cc = 1.6

    \({C_U} = \frac{{{D_{60}}}}{{{D_{10}}}} \Rightarrow 4 = \frac{{{D_{60}}}}{{{D_{10}}}} \Rightarrow {D_{60}} = 4\;{D_{10}}\)

    \({C_c} = 1.6 \Rightarrow \frac{{D_{30}^2}}{{{D_{60}}\; \times \;{D_{10}}}} = 1.6 \Rightarrow \frac{{D_{30}^2}}{{4\;{D_{10}}\; \times \;{D_{10}}}} = 1.6\)

    \(\Rightarrow \frac{{D_{30}^2}}{{D_{10}^2}} = 6.4 \Rightarrow \frac{{{D_{30}}}}{{{D_{10}}}} = \sqrt {6.4} = 2.53\)
  • Question 10/10
    1 / -0

    Sedimentation analysis was performed on a soil and the results are listed below

    Seive size

    90 μ

    75 μ

    60 μ

    4.72 μ

    % Passing

    95%

    80%

    60%

    20%


    The liquid limit of the soil is 70% and plastic limit is 30%. The group index of the soil is 
    Solutions

    Concept:

    Group Index (GI): It is used to give rating to the quality of soil within its group.
    GI = 0.2a + 0.005ac + 0.01 bd

    Where,

    a = % passing 75 μ seive – 35 ; a ≯ 40

    b = % passing 75 μ seive – 15 ; b 40

    c = wL - 40 ; c ≯ 20

    d = Ip – 10 ; d ≯ 20

    GI values ranges from 0 - 20.
    GI value 0 → Soil is of Superior quality.

    GI value 20 → Soil is of Poor quality.

    Calculation:

    % Passing 75 μ sieve = 80%, wL = 70%, wp = 30%

    a = 80 – 35 = 45

    But a 40 a = 40

    b = 80 – 15 = 65

    But b ≯ 40 ⇒ b = 40

    Also,

    Ip = ωL - ωp = 70 – 30 = 40%

    \(\therefore c = min\left\{ {\begin{array}{*{20}{c}} {{\omega _L} - 40}\\ {20} \end{array}} \right. = min\left\{ {\begin{array}{*{20}{c}} {70 - 40}\\ {20} \end{array}} \right. = 20\) 

    \(d = min\left\{ {\begin{array}{*{20}{c}} {{I_p} - 10}\\ {20} \end{array}} \right.= min\left\{ {\begin{array}{*{20}{c}} {40 - 10}\\ {20} \end{array}} \right. = 20\)   

    G.I. = 0.2 a + 0.005 ac + 0.01 bd

    = 0.2 × 40 + 0.005 × 40 × 20 + 0.01 × 40 × 20 = 20

    The soil is of extremely poor quality as suggested by group index.

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