DETERMINATION OF DRY DENSITY OF SOILS IN PLACE (SAND REPLACEMENT METHOD)

Soil compaction is a artificial method in which expulsion of air from soil is done by mechanical means thereby increasing the density of soil In construction.
It is important to know and control the soil density during compaction process .To determine the proper soil compaction of embankment or subgrade in highway project , several methods were developed.The most prominent method is Sand Replacement method to check the relative density at site.In this article ,we will discuss in detail “how to perform sand replacement test at site”.

STANDARD USED 

IS: 2720 (Part 28) 1974.

OBJECTIVE

To determine the in place dry density of natural or compacted fine and medium grained soils by sand replacement method.

APPARATUS

  1. Small sand pouring cylinder for depth up to 150 mm.
  2. Large sand pouring cylinder for depth more than 150 mm and not exceeding 250 mm.
  3. Tools for excavating holes such as a scraper tool for leveling the surface, bent spoon or dibber for digging holes.
  4. Cylindrical calibrating container with material diameter of 100 mm and an internal depth of 150 mm.
  5. Balance of capacity 15 kg and sensitivity 1 gram.
  6. A glass plate o about 600 mm square area and atleast 10 mm or more thicker.
  7. Metal containers.
  8. Metal tray  having area 300 mm square , 40 mm deep with a 100 mm hole in the center of the tray.
  9. 1 mm and 600 microns IS sieves..
  10. Clean, uniformly graded natural sand passing 1 mm sieve and retained on 600 microns sieve.

PROCEDURE

Calibration of Sand Pouring Cylinder

  1. Clean and dry, sand passing 1 mm sieve and retained on 600 microns sieve approximately 5 to 6 kg for small pouring cylinder and 23 to 24 kg of sand for large pouring cylinder.
  2. Remove the cap of the pouring cylinder.
  3. Weigh the empty pouring cylinder (W).
  4. Close the shutter of the cone.
  5. Fill the sand in to the pouring cylinder about 10 mm below from the top.
  6. Determine the net weight of sand in the cylinder (W1).
  7. Now place the pouring cylinder on a clean plane surface, open the shutter and allow the sand to flow in to the cone.
  8. Close the shutter when the flow stops.
  9. Carefully collect and weigh (W2) the sand discharged from the pouring cylinder.
  10. Refill the pouring cylinder with sand such that the initial weight is W1.
  11. Place the pouring cylinder on the top of the calibration cylinder concentrically.
  12. Open the shutter and allow the sand to flow in to the calibrating cylinder.
  13. Close the shutter when the flow stops or no further movement of sand takes place in the cylinder.
  14. Determine the weight (W3) of the pouring cylinder.
  15. Repeat the above procedure for at least three times and determine the mean values of W2 and W3.
  16. Determine the volume (V) of the calibrating cylinder either by measuring the dimensions
  17. (Diameter and height) or by filling with water until the brim.

CALCULATIONS

Weight of sand (WS) in the calibration container up to level top

WS = W1 – W3 – W2 grams

Volume of calibrating container = V cc

DETERMINATION OF SOIL DENSITY

  1. Prepare a flat approximately 450 mm square area with the aid of a scraper tool.
  2. Place the metal tray just above the central hole on the prepared surface of the soil to be tested.
  3. Excavate the hole in the soil with a chisel & hammer using the hole in the tray as a pattern to the depth of the layer to be tested.
  4. Carefully collect the excavated soil from the hole and weigh (Ww).
  5. Determine the water content (W) of the excavated soil as per IS: 2720 (Part 2) 1973.
  6. Fill the pouring cylinder to the constant weight (W1) i.e. weight equal to the initial weight during calibration.
  7. Remove the metal tray before the pouring cylinder is placed in position over the excavated hole.
  8. Place the cylinder such that the base of the cylinder covers the hole concentrically.
  9. Open the shutter and allow the sand to run out in to the hole.
  10. Close the shutter of the pouring cylinder when no further movement of sand takes place in the cylinder.
  11. Remove the cylinder and determine the net weight of sand (W4).

CALCULATIONS

REPORT

  • Report the bulk density and dry density of soil to the nearest second decimal.

PRECAUTIONS

  1. Care shall be taken to see that the test sand used is clean, dry and uniformly graded.
  2. Care shall be taken in excavating the hole to see that the hole is not enlarged  against the side of the hole, as this will result in lower densities.
  3. Care shall be taken to see that the same initial weight of sand is taken during calibration and during density measurement in the field.

 

 

 

 

 

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DETERMINATION OF QUANTITY OF WATER ADDED , FOR REQUIRED COMPACTION OF SUB GRADE & EMBANKMENT

Example: A soil in the borrow pit is at a dry density of 17 kN/  with a moisture content of 10%. The soil is excavated from this pit and compacted in a embankment to a dry density of 18 kN/  with a moisture content of 15%. Compute the quantity of soil to be excavated from the borrow pit and the amount of water to be added for 100   of compacted soil in the embankment. 

Ans :Volume of compacted soil = 100  m³ & Dry density of compacted soil = 18 kN/m3

Weight of compacted dry soil = 100 × 18 = 1800 kN. This is the weight of dry soil to be excavated  from the borrow pit.

Weight of wet soil to be excavated = 1800 (1 + w) = 1800 (1 + 0.10) = 1980 kN.

 Wet density of soil in the borrow pit = 17 (1 + 0.10) = 18.7 kN/ m³

Volume of wet soil to be excavated = 1980 / 18.7  = 105.9  m³

Moisture present in the wet soil, in the borrow pit for every 100  m³ of compacted soil

 = 1800 × 0.10 = 180 kN

 Moisture present in the compacted soil of 100  m³ = 1800 × 0.15 = 270 kN

 Weight of water to be added for 100  m³ of compacted soil = (270 – 180) kN = 90 kN

= 90/ 9.81   m³  = 9.18 kl

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WHAT IS 90th PERCENTILE CBR

The kth percentile is a value in a data set that splits the data into two pieces: If lower piece contains k percent of the data, then upper piece contains the rest of the data means (100 – k) percent, because the total amount of data percentage is 100% where  k is any number between 0 and 100 & median will the 50th percentile.

PROCEDURE
1. First of all arrange all values from smallest to largest.
2. Mulitiply the 90 percent with total number of data if multiplied product is not whole number, round them.
3. Rounding number will be the test data.(Smallest to largest) suppose it is 5
4. Then you go until you find the 5th value in the data set . This value will be 90th percentile. Lets take an example ,suppose 4 days soaked CBR for 10 test in percentage is 15.86,10.32,19.08,11.87,23.84,18.73,16.74,18.15,17.27 & 19.32
then simplified solution can be determined as below :

step 1. First arrange with smallest to largest value it will be
10.32, 11.87, 15.86, 16.74,17.27,18.15,18.73,19.08,19.32,23.84
Step 2. Multiply 10 by 90% i.e 10 x .09 = 9
Step 3. Counting from left to right (from the smallest to the largest value in the above CBR set), you go until you find the 9th value in the step 2 data sheet & this value is 19.32, and it’s the 90th percentile for this data set.

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DETERMINATION OF SAFE BEARING CAPACITY OF SOIL WITHOUT PLATE LOAD TEST

In this article we will discuss with example on ‘How to calculate safe bearing capacity of soil on site ,when unit weight of soil , depth of foundation , width of foundation C & ɸ (Phi) value is given ? So what is bearing capacity of soil ? in simple way , bearing capacity is the load carrying capacity of the soil & bearing capacity after applying the factor of safety (FS) is called safe bearing capacity of soil. This vale is used for design of foundation. We can understand it with following example.

Q. A foundation in a sand is 5 metres wide & 1.5 metres deep. Considering factor of safety 2.50 what will be safe bearing capacity if the unit weight of sand is 1.9 gm/cc and angle of internal friction is 30 ˚. How does it compare with safe loading capacity for surface loading.

  • Ans: From chart ɸ = 30 ˚ Nc =37.2 , Nq = 22.5 , N ү = 19.7 , C=0 , Y= 1.9 t/m³ , b= 5 m , d=1.5 m
  • Safe bearing capacity =1 /F [ c Nc + ү. d (Nq – 1) + ½ ү b Nү] + ү .d

                      = 1/2.5 [ 0 x 37.2+1.9 x 1.5 (22.5-1) + 1/2 x 1.9 x 5 x 19.7] + 1.9 x 1.5

                      = 1/2.5  [ 0 + 1.9 x 1.5 x 21.5 + 0.5 x 1.9 x 5  x 19.7) + 2.85

                      = 1/2.5 [ 0 + 61.28 + 93.58] + 2.85

                      =  1/2.5 X 154.86 + 2.85

                      =  64.79 t/m²

If surface loading is there then d ( depth) will be zero then Safe Bearing Capacity

 = 1/2.5 [½ үb Nү]

 = 1/2.5 [ ½ x 1.9 x 5 x 19.7] = 1/2.5 x 92.63  = 37.052 t/m² hence we see that loading capacity of the foundation is 1.75 times more than surface loading capacity

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VARIATION IN MOISTURE CONTENT DURING COMPACTION TEST

SCOPE

For the determination of dry density of soil when water is added into the sample of soil , it becomes easier for soil particles to move over another particle after applying external forces or compactive force . Owing to that the soil particles come closer & closer hence voids are reduced ultimately which causes the dry density to increase. As we go on increasing the water content , the soil particles creates larger water films around them.

Owing to that dry density goes on increasing till a stage is reached where water starts occupying the space which have been occupied by the soil particle. At this stage  achieved density is called maximum dry density. For any compacted soil there is certain moisture content at which soil can be compacted at maximum instant.
In MORT&H specification following tolerances has been given while carrying out the compaction test.

1.EMBANKMENT: As per MORT&H 5th Revision compaction requirement is 95% of dry density & moisture requirement is 1% above and 2% below of OMC. For example if OMC is 11% range will be 9 % to 12% •

2.SUBGRADE: As per MORT&H 5th Revision compaction requirement is 97% of dry density & moisture requirement is 1% above and 2% below of OMC. For example if OMC is 11% range will be  9 % to 12% •

3.GRANULAR SUB BASE(GSB): As per MORT&H 5th Revision compaction requirement is 98% of dry density & moisture requirement is 1% to 2% below of OMC. For example if OMC is 11% ; range will be   11 % to 9% •

4.WET MIX MACADAM: As per IRC 109 – 2015 compaction requirement is 100% of dry density & moisture requirement is 2% above and 2% below of OMC. For example if OMC is 11% ; range will be 9% to 13% 0f OMC.

5.CEMENT TREATED BASE & SUBBASE: As per MORT&H 5th Revision compaction requirement is 98 % of dry density & moisture requirement should not be more 2% OMC. For example if OMC is 11% range will be 11 % to 13%.

Reference :

1.MORT&H 5th Revision

2.IRC 109 -2015

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FREE SWELL INDEX OF SOIL

OBJECTIVE

Finding Out Free Swell Index Of Soil

REFERENCE TAKEN

IS: 2720(Part 40)-1985

EQUIPMENT

1.Oven(1000C to 2200C minimum)

2.Balance 500 grams with 0.01 g accuracy

3.Sieve 425 micron

4.Two glass cylinder having 100 ml capacity

PREPARATION OF SAMPLE

The soil passing through 425 micron IS sieve is used in this test.

PROCEDURE

1.Take 2 nos sample of 10 g oven dried soil sample passing through 425 micron IS sieve .

2.Each soil sample is poured into each of the two glass graduated cylinders having 100 ml. capacity.

3.Fill kerosene oil into one glass graduated cylinder and fill distilled water up to the 100 ml mark into another cylinder.

4. Remove entrapped air, if any , by stirring with suitable means.

5.Allow  about 24 hours time to soil sample to attain equilibrium state of volume without any further change in the volume of the soils.

6. Take the reading of each cylinder , after completion of 24 hours the final volume of soils .

CALCULATION


Formula for free swell index of the soil is :

Free swell index of soil, percent = ((Vd-Vk) / Vk)*100

Where

Vd = The volume of soil specimen reading from the graduated cylinder containing distilled water.

Vk = The volume of soil specimen reading from the graduated cylinder containing kerosene oil

REPORT

The free swell index of soil is reported to the nearest whole number.

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INDIAN SOIL CLASSIFICATION SYSTEM

Introduction

Soil classification is like a language between engineers .Soil classification for engineering purposes should be based mainly on the mechanical properties, permeability & strength.

The Unified Soil Classification System (USCS) , the American Association of State Highway and Transportation Officials (AASHTO) and Indian soil classification system are the common classification system in the present scenario in civil engineering practice. Here we will discuss pertaining to soil classification in the following order.

1.Classification System

2. Symbolization System

3.Finding Out Cc & Cv

4.Finding Out Clay & Silt From A Line

5.Coarse Grain Soil Classification System

6.Fine Grain Soil Classification System

7. Example

1. Classification System:

The aim of a classification system is to differentiate between different soils. The system must be simple.Classifying soils into groups with similar behavior can provide geotechnical engineers a general guidance,

2. Symbolization System

Symbols and other soil properties used for soil classification which are beneficial are given below :

3.Finding Out Cc & Cv

What is D10 , D30 & D60 ?

Practical Definition Of D10: The size of the sieve from which 10 % material are passing. (10% finer than size size)

Practical Definition Of D30: The size of the sieve from which 30 % material are passing. (30% finer than size) •

Practical Definition Of D60: The size of the sieve from which 60 % material are passing. (60% finer than size)

Coefficient Of Curvature

Coefficient Of Uniformity

Both Cuand Cc will be 1 for a single-sized soil.

If Cu > 5 means a well-graded soil means a soil which having particles over a wide size range.

If Cc between 1 and 3 it indicates a well-graded soil.

If Cu < 3  it indicates a uniform soil

3.1.Border line (Dual Symbol)

For the below given conditions, a dual symbol will be used.

1.For Coarse-grained soils with PI between 5% – 12%  and LL between about 10 and 30). –For  Sand it is denoted as SW-SM and for gravel it is denoted as GW-GM.

2.For Fine-grained soils with limits within the shaded zone. (PI between 4 and 7 and LL between 10 and 30  and more clay type materials. CL-ML means Silty clay

3.2 Organic soil

Organic soils -A sample having decay vegetable tissue in various stages of decomposition and looks like a dark-brown to black color, and smells like organic odor will be designated as organic soil and will be classified as peat, PT.

Organic clay or silt: -“If soil’s liquid limit (LL) after oven drying is less than 75 % of its liquid limit before drying.” it will be organic soil & the first symbol shall be O. -The second symbol can be obtained by locating the values of PI and LL   (as usual not oven dried) in the plasticity chart

4.Finding Out Clay(C), Silt( M) & Organic Soil(O) From A Line

We need grain analysis table & 3 sieves are very important  i.e 4.75 mm,75 micron and 425 micron(for LL & PL).

For determining A Line formula A line IP =.73(WL-20) ,The IP obtain from this will be compare from original IP.

Suppose Original IP given is 9.03% and WL is 25.86%.

Find out the A line , A line=.73(WL-20)=.73*5.86=4.28.

Compare it with original IP which is 9.03% which is greater than 4.28% So Sample comes above A line. Above A line will be denoted by C and below A line will be denoted by M or O.

5. Coarse Soil Identification

If 50 % 0r less material is passing from 0.075 mm soil it will be treated as coarse soil  & they can be further divided into either gravels (G) or sands (S).According to gradation, they are further symbolized  as well-graded (W) or poorly graded (P). If fine soils are present, they can be grouped as  silt fines (M) or  clay fines (C).

6. Fine Grain Soil Classification

Fine-grained soils are those which passes more than  50% of the material from IS sieve 0.075 mm. A plasticity chart , based on the values of liquid limit (WL) and plasticity index (IP), is provided in IS 1498 to aid classification. The ‘A’ line in this chart has been given by as IP = 0.73 (WL – 20).Any soil which is above A line ; will always be denoted as Clay(C). In the same manner , if soil is below A line ; will be denoted as Silt(M) as discussed earlier in para 4.

Now depending on the point in the chart, fine soils are divided into clays (C)silts (M), or organic soils (O).Three divisions of plasticity are also defined as follows.

If Liquid Limit of the soil is less than 35% ; soil will be classified as CL/ML/OL.If Liquid Limit of the soil is in between 35% & 50% soil will be classified as CI/MI/OI.In the same manner if Liquid Limit of the soil is more than 50% soil will be classified as CH/MH/OH.

Low plasticity means liquid Limit is less than 35

WL< 35%
Intermediate plasticity means liquid Limit is between
35 % & 50 %

35% < WL< 50%
High plasticity means liquid Limit is more than 50%

WL> 50%

Example  1

1.Percentage passing  from sieve  4.75mm=38.66%

2.Percentage passing from sieve 0.425mm=37.47%

3 .Percentage passing from sieve 0.075mm=33.47%

4.Liquid Limit                                                =16.80%

5.Plastic Index                                                =.16%

Classify the soil

ANSWER: Less % passing from .075mm sieve so it is coarse grain soil. Less % passing from 4.75mm sieve, hence it is GRAVEL. Its A line=.73*-3.2=-2.336 Sample comes above A line and at .075mm sieve passing more than 12%; so it is classified as GC

Example 2  

1.Percentage passing  from sieve  4.75mm=68.12% 

2.Percentage passing from sieve 0.425mm=56.23%

3 .Percentage passing from sieve 0.075mm=34.62%

4.Liquid Limit                                                =24.5%

5.Plastic Index                                                =N.P Classify the soil

ANSWER: Less % passing from .075mm sieve so it is coarse grain soil. higher % passing from 4.75mm sieve, hence it is SAND. Its A line=.73*4.5=3.285 Sample comes below A line and at .075mm sieve passing more than 12%; so it is classified as SM


Example  3 

1.Percentage passing  from sieve  4.75mm=99.96% 

2.Percentage passing from sieve 0.425mm=97%

3 .Percentage passing from sieve 0.075mm=74.91%

4.Liquid Limit                                                =25.86%

5.Plastic Index                                                =9.03%

Classify the soil

ANSWER:

More % passing from .075mm sieve so it is fine grain soil. Liquid Limit is less than 35% so it is low plastic. Its A line=.73(WL-20)=.73*5.86=4.28.Sample comes above A line  so it is classified as CL

Example  4 

1.Percentage passing  from sieve  4.75mm=99.80% 

2.Percentage passing from sieve 0.425mm=99.55%

3 .Percentage passing from sieve 0.075mm=36.10%

4.Liquid Limit                                                =17.54%

5.Plastic Index                                                =N.P

Classify the soil ?

ANSWER:

Less  % passing from .075mm sieve so it is coarse grain soil.its % passing from .075mm sieve is more than 12% .Its A line=.73(WL-220)=.73*-2.46= -1.80 Sample comes  above A line  so it is classified as SC.

Example  5

1.Percentage passing  from sieve  4.75mm=99.96% 

2.Percentage passing from sieve 0.425mm=93.87%

3 .Percentage passing from sieve 0.075mm=58.99%

4.Liquid Limit                                                =22%

5.Plastic Index                                                =3.7%

Classify the soil

ANSWER:

More % passing from .075mm sieve so it is fine grain soil. Liquid Limit is less than 35% so it is low plastic.Its A line=.73(WL-20)=.73*2=1.46 Sample comes above A line  so it is classified as CL

Example  6 

1.Percentage passing  from sieve  4.75mm=100% 

2.Percentage passing from sieve 0.425mm=94.30%

3 .Percentage passing from sieve 0.075mm=51.54%

4.Liquid Limit                                                =23.0%

5.Plastic Index                                                =4.54%

Classify the soil

ANSWER:

More % passing from .075mm sieve so it is fine grain soil. Liquid Limit is

 less than 35% but PI is 4.54 and it comes on hatched line   so it is classified  ML-CL

Example  7 

1.Percentage passing  from sieve  4.75mm=82.3% 

2.Percentage passing from sieve 0.425mm=73.11%

 3 .Percentage passing from sieve 0.075mm=4.82%

4.Liquid Limit                                                =25.42%

5.Plastic Index                                                =8.64%

6.Coefficient of  Uniformity                      =6.66

7.Coefficient of curvature                           =1.35

Classify the soil

ANSWER:  Less % passing from .075mm sieve so it is course grain soil. Greater% passing from 4.75 so it is SAND. Less than 5%  passing from .075mm sieve and well graded so it is classified as SW

Example  8

1.Percentage passing  from sieve  4.75mm=99.31% 

2.Percentage passing from sieve 0.425mm=88.57%

3 .Percentage passing from sieve 0.075mm=50.38

4.Liquid Limit                                                =21.50%

5.Plastic Index                                                =4.84%

Classify the soil

ANSWER:

 More % passing from .075mm sieve so it is fine grain soil. Liquid Limit is less than 35% but PI is 4.54 and it comes on hatched line   so it is classified  ML-CL

 

 

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DETERMINATION OF LIQUID LIMIT & PLASTIC LIMIT

DETERMINATION OF LIQUID LIMIT OF SOIL

DETERMINATION OF LIQUID LIMIT OF SOIL

The moisture content  at which the cohesive soil passes from a liquid state into a plastic state is called the liquid limit of the soil. Similarly, the moisture contents at which the soil changes its behavior from a plastic to a semisolid state is called as the plastic limit of soil.

Objective

This test is done to find out  the liquid limit of soil as per IS: 2720 (Part 5) – 1985. The liquid limit of fine-grained soil is the water content at which soil behaves practically just like a liquid, but have small shear strength. It’s flow closes the groove of 12.7 mm in just 25 blows in Casagrande’s liquid limit device.

Apparatus

The apparatus used :-
i) Casagrande’s liquid limit device
ii) Grooving tools both standard and  ASTM types
iii) Oven
iv) Evaporating dish
v) Spatula
vi) IS Sieve of size 425µm
vii) Weighing balance, with 0.01 /1 gm accuracy
viii) Wash bottle
ix) Air-tight container for determination of moisture content

Preparation Of Sample

i) Dry the soil sample in and break the clods with wooden hammer. Remove any organic matter like tree roots& pieces of bark, etc.
ii) Take about 120 g of the specimen passing through 425µm IS Sieve and  mixed thoroughly  it with distilled water in the evaporating dish and left it for 24 hrs. for soaking.

Procedure

i) Place a portion of the paste into the cup of the liquid limit device.

ii) Level the mix from top so as to have a maximum depth of 1 cm.

iii) Draw the groove from special design tool through the sample along the symmetrical axis of the cup & holding the tool perpendicular to the cup.

iv) For the normal fine grained soil: The Casagrande’s tool is used to cut a groove of 2 mm wide at the bottom, 11 mm wide at the top and 8 mm deep.

v) For sandy soil: The ASTM tool is used to cut a groove of  2 mm wide at the bottom, 13.6 mm wide at the top and 10 mm deep.

vi) After the soil paste has been cut by a suitable grooving tool, the handle of the device is rotated at the rate of about 2 revolutions per second and the no. of blows counted, till the two parts of the sample come into contact for about 10 mm length.

vii) Take about 10 g of soil near the closed groove as a sample for  determing its water content.

viii) The soil of the cup is transferred into the dish containing the soil paste and mixed thoroughly after adding some water. Repeat the test as earlier.

ix) By changing  the water content of the soil and repeating the foregoing operations & obtain at least 5 readings in the range of  between 15 to 35 blows.  Keep in mind don’t mix dry soil to change its consistency.

x) Now Liquid limit is determined by plotting a ‘flow curve’ on a semi-log graph, with no. of blows as abscissa X -axis (log scale) and the water content as ordinate on Y axis and drawing the best possible straight line through the plotted points.

Reporting Of Results

Report the water content corresponding to 25 number blows by reading from the flow Curve as the Liquid Limit. A sample of Flow curve is given below for your reference.

Safety: .1 Use hand gloves while opening the door of oven

DETERMINATION OF PLASTIC LIMIT

Objective

For determination of the plastic limit of soil.

Reference Standard

IS : 2720(Part 5)-1985    Determination of Plastic limit.

Equipment & Apparatus
  • Oven
  • Balance with 0.01 g accuracy
  • IS Sieve of 425 micron
  • Flat surface glass for rolling
Preparation Of Sample

After receiving the soil sample from the site it is dried in air or in oven ( by maintaining a temperature of 600C). If clods are there in the soil sample it is broken with the help of wooden mallet. The soil passing through 425 micron sieve is used for this test.

Procedure
  1. Take 20 gm soil sample passing from 425 micron IS sieve .
  2. It is then mixed with distilled water thoroughly in the evaporating dish till the soil mass becomes plastic enough to be easily molded with the fingers itself.
  3. Soil should be allowed to season for sufficient time for allowing water to permeate throughout the soil mass.
  4. The 10 gms. of the sample is taken and rolled between fingers and glass plate with just sufficient pressure to roll the mass into a thread of uniform diameter throughout its length. The rate of rolling shall be kept between 60 and 90 stokes per minute.
  5. Continued the rolling till the thread becomes 3 mm. in diameter and see the crumbling.
  6. If not soil is then kneaded together to a uniform mass and rolled again.
  7. The process is to be continued until the thread crumbled and offering shear when rolled into 3 mm diameter.
  8. The pieces of the crumbled thread are collected in a air tight container for determination of moisture content .For more practical see the video given below:

Report

Sample Format



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