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    SOIL CLASSIFICATION

    CLASSIFICATION OF SOIL

    Classification systems are used to group soils according to their order of performance

    under given set of physical conditions. Soils that are grouped in order of performance forone set of physical conditions will not necessarily have the same order of performance

    under some other physical conditions. Therefore, number of classification systems have

    been developed depending on the intended purpose of the system. Soilclassification hasproved to be a very useful tool to the soil engineer. It gives general guidelines in an

    empirical manner for making use of the field experience of others. Soil may be broadly

    classified as follows:

    1. Classification based on grain size2. Textural classification

    3. AASHTO classification system

    4. Unified soil classification system

    (i) Grain size classification system:

    Grain size classification systems were based on grain size. In this system the terms clay,

    silt, sand and gravel are used to indicate only particle size and not to signify nature ofsoil

    type. There are several classification systems fin use, but commonly used systems areshown here.

    (ii) Textural Classification:

    The classification of soil exclusively based on particle size and their percentage

    distribution is known as textural classification system. This system specifically names thesoil depending on the percentage of sand, silt and clay. The triangular charts are used to

    classify soil by this system. Figure 1 shows the typical textural classification system.

    Fig-1: Textural Classification of U.S. Public Roads Administration

    (iii) AASHTO classification system:

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    AASHTO classification, (table-2) is otherwise known as PRA classification system. It

    was originally developed in 1920 by the U.S. Bureau of Public Roads for the

    classification of soil for highway subgrade use. This system is developed based onparticle size and plasticity characteristics of soil mass. After some revision, this system

    was adopted by the AASHTO in 1945. In this system the soils are divided into seven

    major groups. Some of the major groups further divided into subgroups. A soil isclassified by proceeding from left to right on the classification chart to find first the group

    into which the soil test data will fill. Soil having fine fractions are further classified based

    on their group index. The group index is defined by the following equation.

    Group index = (F 35)[0.2 + 0.005 (LL 40)] + 0.01(F 15)(PI 10)

    F Percentage passing 0.075mm size

    LL Liquid limit

    PI Plasticity index

    When the group index value is higher, the quantity of the material is poorer.

    Click Here to View AASHTO Classification Chart

    (iv) Unified soil classification system:

    Unified soil classification system was originally developed by Casagrande (1948) andwas known as airfield classification system. It was adopted with some modification by

    the U.S. Bureau of Reclamation and the U.S. Corps of Engineers. This system is based on

    both grain size and plasticity characteristics of soil. The same system with minormodification was adopted by ISI for general engineering purpose (IS1498 1970). IS

    system divides soil into three major groups, coarse grained, fine grained and organic soils

    and other miscellaneous soil materials.

    Coarse grained soils are those with more than 50% of the material larger than 0.075mmsize. Coarse grained soils are further classified into gravels (G) and sands (S). The

    gravels and sands are further divided into four categories according to gradation, silt or

    clay content.

    Fine grained soils are those for which more than 50% of soil finer than 0.075 mm sieve

    size. They are divided into three sub-divisions as silt (M), clay , and organic salts andclays (O). based on their plasticity nature they are added with L, M and H symbol to

    indicate low plastic, medium plastic and high plastic respectively.

    Examples:

    GW well graded gravel

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    GP poorly graded gravel

    GM silty gravel

    SW well graded sand

    SP poorly graded sand

    SM silty sand

    SC clayey sand

    CL clay of low plastic

    CI clay of medium plastic

    CH clay of higher plastic

    ML silt of medium plastic

    MI silt of medium plastic

    MH silt of higher plastic

    OL organic silt and clays of low plastic

    OI organic silt and clays of medium plastic

    OH organic silt and clays of high plastic.

    Fine grained soils have been sub-divided into three subdivisions of low, medium and high

    compressibility instead of two sub-divisions of the original Unified Soil Classification

    System.

    Table-3 below shows the classification system. Table 2 lists group symbols for soils of

    table-3.

    Table-2: Significance of letters for group symbol in table-3.

    Soil Soil Component Symbol

    Coarse Grained

    Boulder None

    Cobble None

    Gravel G

    Sand S

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    Fine Grained

    Silt M

    Clay C

    Organic Matter O

    Table 3

    Soil Soil Component Symbol

    Peat Peat Pt

    Applicable to

    Coarse grained Soils

    Well graded W

    Poorly Graded P

    Applicable to

    Fine grained soils

    Low compressibility

    WL50)

    H

    The standard recommends that when a soil possesses characteristics of two groups either

    in particle size distribution or in plasticity, it is designed by combination of groupsymbols.

    Click Here to View Unified Soil Classification Chart

    Field identification is recommended through the following tests:

    For fine grained soils

    a) Visual examination

    b) Dilatancy test

    c) Toughness test

    d) Dry strength test

    e) Organic content and colour

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    f) Other identification test

    INDIAN STANDARD CLASSIFICATION SYSTEM:

    Indian Standard Classification System (ISC) was adopted by Bureau of Indian Standards

    is in many respect similar to the Unified Soil Classification (USC) system.

    Soils are divided into three broad divisions:

    1. Coarse grained soils, when 50% or more of the total material by weight is retained

    on 75 micro IS sieve.

    2. For fine grained soils, when more than 50% of the total material passes through75 micron IS sieve.

    3. If the soil is highly organic and contains a large percentage of organic matter and

    particles of decomposed vegetation, it is kept in a separate category marked as

    peat (Pt).

    In all there are 18 groups of soils: 8 groups of coarse grained, 9 groups of fine grained

    and one of peat.

    Fig.2: Indian Standard Classification Plasticity Chart

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    Soils are the fundamental resource supporting agriculture and forestry,as well as contributing to the aesthetics of a green planet. They are

    also a base from which minerals are extracted and to which solidwastes are disposed. In addition, soils act as a medium and filter for

    collection and movement of water. By supporting plant growth, soil

    becomes a major determinant of atmospheric composition andtherefore earths climate.

    ORIGIN OF SOILS

    Soils are formed by weathering of rocks due to mechanicaldisintegration or chemical decomposition. When a rock surface gets

    exposed to atmosphere for an appreciable time, it disintegrates ordecomposes into small particles and thus the soils are formed.

    FORMATION OF SOILS

    Soils are formed either by (A) Physical Disintegration or (B) Chemical

    decomposition of rocks.

    A. PHYSICAL DISINTEGRATION

    Physical disintegration or mechanical weathering of rocks occurs due

    to the following physical processes:

    1. Temperature changes

    Different minerals of rocks have different coefficients of thermal

    expansion. Unequal expansion and contraction of these minerals occurdue to temperature changes. When the stresses induced due to such

    changes are repeated many times, the particles get detached from the

    rocks and the soils are formed.

    2. Wedging action of ice

    Water in the pores and minute cracks of rocks gets frozen in very cold

    climates. As the volume of ice formed is more than that of water,

    expansion occurs. Rocks get broken into pieces when large stressesdevelop in the cracks due to wedging action of the ice formed.

    3. Spreading of roots of plants

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    As the roots of trees and shrubs grow in the cracks and fissures of therocks, forces act on the rocks. The segments of the rock are forced

    apart and disintegration of rocks occurs.

    4. Abrasion

    As water, wind and glaciers move over the surface of rock, abrasion

    and scouring takes place. It results in the formation of soils.

    Note: In all the processes of physical disintegration, there is nochange in the chemical composition. The soil formed has the

    properties of the parent rock. Coarse grained soils, such as gravel andsand, are formed by the process of physical disintegration.

    B. CHEMICAL DECOMPOSITION

    When chemical decomposition or chemical weathering of rocks takesplace, original rock minerals are transformed into new minerals bychemical reactions. The soils formed do not have the properties of the

    parent rock. The following chemical processes generally occur innature:

    1. Hydration

    In hydration, water combines with rock minerals and results in theformation of a new chemical compound. The chemical reaction causes

    a change in volume and decomposition of rock into small particles.

    An example of hydration reaction that is taking place in soils is thehydrolysis of SiO2

    SiO2+ 2H2O Si(OH)4

    2. Carbonation

    It is a type of chemical decomposition in which carbon dioxide in theatmosphere combines with water to form carbonic acid. The carbonicacid reacts chemically with rocks and causes their decomposition.

    The example for this type of is, that is taking place in sedimentary

    rocks which contain calcium carbonate.

    3. Oxidation

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    Oxidation occurs when oxygen ions combine with minerals in rock.Oxidation results in decomposition of rocks. Oxidation of rocks is

    somewhat similar to rusting of steel.

    4. Solution

    Some of the rock minerals form a solution with water when they get

    dissolved in water. Chemical reaction takes place in the solution and

    the soils are formed.

    5. Hydrolysis

    It is a chemical process in which water gets dissociated into H+ and OH-

    ions. The hydrogen cations replace the metallic ions such as calcium,sodium and potassium in rock minerals and soils are formed with a

    new chemical composition.

    Note: Chemical decomposition of rocks result in the formation of clayminerals. The clay minerals impart plastic properties of soils. Clayey

    soils are formed by chemical decomposition.

    TRANSPORTATION OF SOILS

    The soils formed at a place may be transported to other places by

    agents of transportation, such as water, ice, wind and gravity.

    1. Water transported soils

    Flowing water is one of the most important agents of transportation of

    soils. the size of the soil particles carried by water depends upon the

    velocity. The swift water can carry the particles of large size such asboulders and gravels. With a decrease in velocity, the coarser particles

    get deposited. The finer particles are carried further downstream anddeposited when the velocity reduces. A delta is formed when the

    velocity slows down to almost zero at the confluence with a receiving

    body of still water such as lake, a sea or an ocean.

    All types of soils carried and deposited by water are known as alluvialdeposits. Deposits made in lakes are called lacustrine deposits. Marinedeposits are formed when the following water carries soils to ocean or

    sea.

    2. Wind transported soils

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    Soil particles are transported by winds. the particle size of the soildepends on the velocity of wind. The finer particles are carried far

    away from the place of the formation. Soil deposits by wind are knownas Aeolian deposits.

    Large sand dunes are formed by winds. Sand dunes occur in aridregions and on the lee ward side of the sea with sandy beaches.

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

    low. The permeability in vertical direction is large.

    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 sizefrom fine grained to huge boulders. Soils get mixed with ice and are

    transported far away from their original position.

    AIM

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    To determine the particle size distribution of the given fine aggregate and to determine

    ,the fineness modulus,the effective size and uniformly coefficent .

    THEORY

    Fine aggregate is the sand used in mortars. Coarse aggregate is the broken stone used inconcrete .The coarse aggregate unless mixed with fine aggregate serves no purpose in

    cement works .the size of fine aggregate is limited to a maximum of 4.75 mm gauge

    beyond which it is known as coarse aggregate.

    FINENESS MODULUS.

    Fineness modulus is only a numerical index of fineness, giving some idea of the mean

    size of the particle s in the entire body of the aggregate. To a certain extent it is a method

    of standardization of the grading of the aggregate. It is obtained by adding the percentage

    weight of material retained in each of the standard sieves and dividing it by 100.

    To object of finding the fineness modulus is to grade a given aggregate for the most

    economical mix and workability with minimum quantity of cement .Certain limits of

    fineness modulus for fine coarse aggregates are given in the table below and a sample

    under test should satisfy these results so that the aggregate may give good workabilityunder economical conditions.

    LIMIT OF FINENESS MODULUS

    Maximum size ofaggregate

    Fineness modulus

    Minimum Maximum

    Fine Aggregate 2 3.5

    Coarse aggregate

    20mm

    6 6.9

    Coarse aggregate

    40mm

    6.9 7.5

    Coarse aggregate75mm

    7.5 8.0

    If the test aggregate gives higher fineness modulus the mix will be harsh and if on theother hand gives a lower fineness modulus it gives uneconomical mix .

    EFFECTIVE SIZE

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    Effective size (in microns) is the maximum particle size of the smallest 10% of the

    aggregate or it is the sieve opening corresponding to 10% finer and is designated by the

    symbol D10.

    UNIFORMITY COEFFICIENT

    This is the ratio of the maximum size of the smallest 60% to the effective size .

    Uniformity coefficent = D60/D10

    APPARATUES

    Indian standard test sieves, weighing balance ,sieve shaker etc .

    Size of sieves to be used.

    (I) for fine agggregate- 4.75mm, 2.36mm, 1.18mm, 600 microns, 300microns ,150microns .

    (II) For coarse aggregate-25mm,20mm 12.5mm, 10mm, 4.75mm.

    PROCEDURE

    FINE AGGREGATE

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    (i)Take one kg of sand from the laboratory sample

    (ii)Arrange the sieves in order of IS sieves nos 480, 240, 120, 60, 30 and 15, Keeping

    sieve no.480 at the top and 15 at the bottom and cover the top.

    (iii)Keep the sample in the top sieve no.480.

    (iv) Carry out the sieving in the set of sieves for not less than 10 minutes .

    (v) find the weight of sample retained in each sieve.

    (vi) Tabulate the values in given tabular column .

    COARS AGGREGATES

    (i)Take one kg of coarse aggregate

    (ii)Arrange the sieves one over the other in relation to their size of opening.

    (25mm,20mm,12.5mm10mm,4.75mm)

    (iii) Carry out the sieving for the specified time

    (iv) Find the weight of agggregate retained on each sieve taken in order and tabulate in

    table.

    GRAPH

    Draw a graph with sieve opening to log scale on the X-axis and % finer on Y-axis .Thecurve iscalled a grading curve.

    CALCULATION

    FINE AGGREGATE

    1. Effective size in microns(Di0 sieve opening corresponding to 10%finer in the graph ) =

    2. Uniformity coefficient (D60/ D10),D to be obtained from the graph ) =

    3. Fineness modulus (Sum of cumulative % wt retained /100) =

    COARSEAGGREGATE

    1. Effective size in microns (D10,sieve opening corresponding to 10% finer in the graph)

    2. Uniformity coefficient (D60/ D10),D to be obtained from the graph ) =

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    3. Fineness modulus (Sum of cumulative % wt retained /100) =

    RESULT S

    FINE AGGREGATE

    1. Effective size =..micron

    2. Uniformity coefficent =

    3. Fineness modulus =

    COARS AGGREGATES

    4. Effective size =..micron

    5. Uniformity coefficient =

    6. Fineness modulus =

    O DETERMINE THE PARTICLE SIZE DISTRIBUTION BY HYDROMETER

    METHOD

    Theory:

    Hydrometer method is used to determine the particle size distribution of fine-grained

    soils passing 75 sieve. The hydrometer measures the specific gravity of the soilsuspension at the centre of its bulb. The specific gravity depends upon the mass of solids

    present, which in turn depends upon the particle size. The particle size (D) is given by:

    Where

    In which, = viscosity of water in poise, G= specific gravity of solids,

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    = density of water (gm/ml); , = effective depth, t= time in

    minutes at which observation is taken, reckoned with respect to the beginning of

    sedimentation.

    The percentage finer than the size D is given by

    Where R= corrected hydrometer reading, Ms= mass of dry soil in 1000ml suspension.

    Equipment:

    1. Hydrometer

    2. Glass measuring cylinder (jar), 1000ml

    3. Rubber bung for the cylinder (jar)

    4. Mechanical stirrer

    5. Weighing balance, accuracy 0.01g

    6. Oven

    7. Desiccator

    8. Evaporating dish

    9. Conical flask or beaker, 1000ml

    10. Stop watch

    11. Wash bottle

    12. Thermometer

    13. Glass rod

    14. Water bath

    15. 75 sieve

    16. Scale

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    17. Deflocculating agent.

    Procedure:

    Part-I: Calibration of hydrometer

    1. Take about 800ml of water in one measuring cylinder. Place the cylinder on a table and

    observe the initial reading.

    2. Immerse the hydrometer in the cylinder. Take the reading after the immersion.

    3. Determine the volume of the hydrometer ( ) which is equal to the difference

    between the final and initial readings. Alternatively weigh the hydrometer to the nearest

    0.1g. The volume of the hydrometer in ml is approximately equal to its mass in grams.

    4. Determine the area of cross section (A) of the cylinder. It is equal to the volume

    indicated between any two graduations divided by the distance between them. Thedistance is measured with an accurate scale.

    5. Measure the distance (H) between the neck and the bottom of the bulb. Record it as the

    height of the bulb (h).

    6. Measure the distance (H) between the neck to each marks on the hydrometer ( ).

    7. Determine the effective depth ( ), corresponding to each of the mark ( ) as

    [Note: the factor should not be considered when the hydrometer is not taken out

    when taking readings after the start of the sedimentation at , 1, 2, and 4 minutes.]

    8. Draw a calibration curve between and . Alternatively, prepare a table between

    and . The curve may be used for finding the effective depth corresponding to

    reading .

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    Fig: Hydrometer Method

    Fig: Hydrometer Calibration Chart

    Part-II: Meniscus Correction

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    1. Insert the hydrometer in the measuring cylinder containing about 700ml of water.

    2. Take the readings of the hydrometer at the top and at the bottom of the meniscus.

    3. Determine the meniscus correction, which is equal to the difference between the two

    readings.

    4. The meniscus correction is positive and is constant for the hydrometer.

    5. The observed hydrometer reading is corrected to obtain the corrected hydrometer

    reading as

    Part-III: Pretreatment and Dispersion

    1. Weigh accurately, to the nearest 0.01g about 50g air-dried soil sample passing 2mm IS

    sieve, obtained by riffling from the air-dried sample passing 4.75mm IS sieve.

    Place the sample in a wide mouthed conical flask.

    2. Add about 150ml of hydrogen peroxide to the soil sample in the flask. Stir it gently

    with a glass rod for a few minutes.

    3. Cover the flask with a glass plate and leave it to stand overnight.

    4. Heat the mixture in the conical flask gently after keeping it in an evaporating dish. Stirthe contents periodically. When vigorous frothing subsides, the reaction is complete.

    Reduce the volume to 50ml by boiling. Stop heating and cool the contents.

    5. If the soil contains insoluble calcium compounds, add about 50ml of hydrochloric acid

    to the cooled mixture. Stir the solution with a glass rod for a few minutes. Allow it tostand for one hour or so. The solution would have an acid reaction to litmus when the

    treatment is complete.

    6. Filter the mixture and wash it with warm water until the filtrate shows no acid reaction.

    7. Transfer the damp soilon the filter and funnel to an evaporating dish using a jet ofdistilled water. Use the minimum quantity of distilled water.

    8. Place he evaporating dish and its contents in an oven and dry it at 105 to 110 degree C.

    transfer the dish to a desiccator and allow it to cool.

    9. Take the mass of the oven dried soil after pretreatment and find the loss of mass due to

    pretreatment.

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    10. Add 100ml of sodium hexa-metaphosphate solution to the oven dried soil in the

    evaporating dish after pretreatment.

    11. Warm the mixture gently for about 10minutes.

    12. Transfer the mixture to the cup of a mechanical mixture. Use a jet of distilled water towash all traces of the soil out of the evaporating dish. Use about 150ml of water. Stir the

    mixture for about 15minutes.

    13. Transfer the soil suspension to a 75 IS sieve placed on a receiver (pan). Wash the

    soil on this sieve using a jet of distilled water. Use about 500ml of water.

    14. Transfer the soil suspension passing 75 sieve to a 1000ml measuring cylinder. Add

    more water to make the volume exactly equal to 1000ml.

    15. Collect the material retained on 75 sieve. Dry it in an oven. Determine its mass. If

    required, do the sieve analysis of this fraction.

    Part-IV: Sedimentation Test

    1. Place the rubber bung on the open end of the measuring cylinder containing the soil

    suspension. Shake it vigorously end-over-end to mix the suspension thoroughly.

    2. Remove the bung after the shaking is complete. Place the measuring cylinder on thetable and start the stop watch.

    3. Immerse the hydrometer gently to a depth slightly below the floating depth, and then

    allow it to float freely.

    4. Take hydrometer reading ( ) after 1/2, 1, 2 and 4 minutes without removing thehydrometer from the cylinder.

    5. Take out the hydrometer from the cylinder, rinse it with distilled water.

    6. Float the hydrometer in another cylinder containing only distilled water at the same

    temperature as that of the test cylinder.

    7. Take out the hydrometer from the distilled water cylinder and clean its stem. Insert it in

    the cylinder containing suspension to take the reading at the total elapsed time interval of8minutes. About 10 seconds should be taken while taking the reading. Remove the

    hydrometer, rinse it and place it in the distilled water after reading.

    8. Repeat the step (7) to take readings at 15, 30, 60, 120 and 240minutes elapsed timeinterval.

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    9. After 240 minutes (4 hours) reading, take readings twice within 24 hours. Exact time

    of reading should be noted.

    10. Record the temperature of the suspension once during the first 15minutes andthereafter at the time of every subsequent reading.

    11. After the final reading, pour the suspension in an evaporating dish, dry it in an oven

    and find its dry mass.

    12. Determine the composite correction before the start of the test and also at 30min, 1, 2

    and 4 hours. Thereafter just after each reading, composite correction is determined.

    13. For the determination of composite correction (C), insert the hydrometer in the

    comparison cylinder containing 100ml of dispersing agent solution in 1000 ml of distilled

    water at the same temperature. Take the reading corresponding to the top of meniscus.

    The negative of the reading is the composite correction.

    Fig: Downward Movement of Hydrometer

    DATA SHEET FOR HYDROMETER TEST

    Mass of dry soil (Ms)=_______g

    Meniscus correction (Cm)= +_______

    Specific gravity of solids (G)= ______

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    Sl.

    No.

    OBSERVATIONS CALCULATIONS

    Elaspsed

    time

    Hydrometer

    reading

    Temp-

    erature

    Composite

    correction

    Corrected

    reading

    Rh=Rh+Cm

    Height

    (cm) He

    Reading

    R= Rh+C

    Factor

    M

    Particle

    size

    D

    %

    Finer

    1 1/2min

    2 1

    3 2

    4 4

    5 8

    6 15

    7 30

    8 1 hr

    9 2 hr

    10 4 hr

    11 8 hr

    12 12 hr

    13 24 hr

    Result:

    Particle Size distribution curve can be plotted using the last two columns.

    O DETERMINE SPECIFIC GRAVITY OF THE SOLIDS BY DENSITY BOTTLE

    METHOD

    Theory;

    The specific gravity of solid particles is the ratio of the mass density of solids to thatwater. It is determined in the laboratory using the relation:

    Where

    M1 = mass of empty bottle

    M2 = mass of the bottle and dry soil

    M3 = mass of bottle, soil and water

    M4 = mass of bottle filled with water only.

    http://theconstructor.org/category/practical-guide/material-testing/soil/http://theconstructor.org/category/practical-guide/material-testing/soil/http://theconstructor.org/category/practical-guide/material-testing/soil/http://theconstructor.org/category/practical-guide/material-testing/soil/
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    Equipment:

    1. 50ml density bottle with stopper

    2. Oven (105 0 to 110 0C)

    3. Constant temperature water bath (27 0C)

    4. Vacuum desiccator

    5. Vacuumpump

    6. Weighing balance accuracy 0.001g

    7. Spatula

    Procedure:

    1. Wash the density bottle and dry it in an oven at 105 0C to 100 0C. Cool it in the

    desiccator.

    2. Weigh the bottle, with stopper to the nearest 0.001g (M1).

    3. Take 5 to 10g of the oven dried soilsample and transfer it the density bottle. Weigh the

    bootle with the stopper and the dry sample (M2).

    4. Add de-aired distilled water to the density bottle just enough to cove thesoil. Shake

    gently to mix the soil and water.

    5. Place the bottle containing the soil and water after removing the stopper in the vacuumdesiccator.

    6. Evacuate the desiccator gradually by operating the vacuumpump. Reduce the pressure

    to about 20 mm of mercury. Keep the bottle in the desiccator for at least 1 hour or until

    no further movement of air is noticed.

    7. Replace the vacuum and remove the lid of the desiccator. Stir the soil in the bottle

    carefully with a spatula. Before removing the spatula from the bottle, the particles of soil

    adhering to it should be washed off with a few drops of air free water. Replace the lid ofthe desiccator and again apply vacuum. Repeat the procedure until no more air is evolved

    from the specimen.

    Note: In case of vacuum desiccator is not available, the entrapped air can be removed by

    heating the density bottle on a water bath or a sand bath.

    http://theconstructor.org/category/practical-guide/hydraulics-lab/turbines-pumps/http://theconstructor.org/category/practical-guide/material-testing/soil/http://theconstructor.org/category/practical-guide/material-testing/soil/http://theconstructor.org/category/practical-guide/material-testing/soil/http://theconstructor.org/category/practical-guide/material-testing/soil/http://theconstructor.org/category/practical-guide/material-testing/soil/http://theconstructor.org/category/practical-guide/material-testing/soil/http://theconstructor.org/category/practical-guide/hydraulics-lab/turbines-pumps/http://theconstructor.org/category/practical-guide/hydraulics-lab/turbines-pumps/http://theconstructor.org/category/practical-guide/hydraulics-lab/turbines-pumps/http://theconstructor.org/category/practical-guide/material-testing/soil/http://theconstructor.org/category/practical-guide/material-testing/soil/http://theconstructor.org/category/practical-guide/material-testing/soil/http://theconstructor.org/category/practical-guide/material-testing/soil/http://theconstructor.org/category/practical-guide/hydraulics-lab/turbines-pumps/
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    8. Remove the bottle from the desiccator. Add air-free water until the bottle is full. Insert

    the stopper.

    9. Immerse the bottle upto the neck in a constant-temperature bath for approximately 1hour or until it has attained the constant temperature.

    If there is an apparent decrease in the volume of the liquid in the bottle, remove the

    stopper and add more water to the bottle and replace the stopper. Again place the bottle in

    the water bath. Allow sufficient time to ensure that the bottle and its content attain theconstant temperature.

    10. Take out the bottle from the water bath. Wipe it clean and dry it from outside. Fill the

    capillary in the stopper with drops of distilled water, if necessary.

    11. Determine the mass of the bottle and its contents (M3).

    12. Empty the bottle and clean it thoroughly. Fill it with distilled water. Insert thestopper.

    13. Immerse the bottle in the constant-temperature bath for 1 hour or until it has attained

    the constant temperature of the bath.

    If there is an apparent decrease in the volume of the liquid, remove the stopper and add

    more water. Again keep it in the water bath.

    14. Take out the bottle from the water bath. Wipe it dry and take the mass (M 4).

    Observations and calculations:

    Sl.

    No.

    Observations an Calculations Determination No.

    1 2 3

    Observation

    1 Density bottle No.

    2 Mass of empty density bottle (M1)

    3 Mass of Mass of bottle dry soil (M2)

    4 Mass of bottle, soil and water (M3)

    5 Mass of bottle filled with water (M4)

    Calculations

    6 M2 M1

    7 M3 M4

    8 Calculate G using formula

    Result:

    Specific gravity of solids = _____.