In civil engineering, several studies showed that organic soils may show considerably low bearing capacities, and considered them as soft soils due to their high settlement values indeed under relatively applied loads. Consequently, there’s a tendency in the field of construction to avoid executing any projects above this type of problematic soils.
when exposed to structure’s loads, especially that wide areas of the area are covered with fine granulated soils containing different rates of organic matter. It is very necessary to know whether soil is organic or not. However, organic matter can also decrease the efficiency of compaction because the organic matter absorbs some of the energy transmitted to the sample.As per MORT& H specification , organic soil can not be used in road construction. Here is procedure how to find out the organic percentage in soil.
Determination of total Organic matter in soils.
Sample Preparation:
Total weight of original soil sample (Oven dried) – W1
Sieve the sample on 10 mm IS sieve and weighed the passing material – W2
Then sieve the sample on 425 micron IS sieve and take the soil sample for the test – Approximately 5 grams – W3 grams.
Preparation of Reagents:
Potassium Dichromate Normal solution: Dissolve 49.035 grams of Potassium Dichromate in one liter of distilled water.
Ferrous Sulphate 0.5 N solution: Dissolve 140 grams of Ferrous Sulphate in 0.5 N Sulphuric acid to make one liter of solution (Add 14 ml of concentrated Sulphuric acid to distilled water to make one liter of solution for 0.5 N Sulphuric acid).
Concentrated Sulphuric acid: Gr 1.83.
Ortho Phosphoric acid: Gr 1.70 to 1.75.
Indicator: 25 grams of Sodium Diphemylamine-Sulphonate dissolved in 100 ml of distilled water.
Standardization of Ferrous Sulphate Solution:
Take 10 ml of Normal Potassium Dichromate solution in to 500 ml conical
Add 20 ml concentrated Sulphuric acid and swirled and allowed to cool for some
Add 200 ml of distilled water, 10 ml of Ortho Phosphoric acid and 1ml of the Indicator and the mixture shall be shaking thoroughly.
Ferrous Sulphate solution added through burette in 0.5 ml increments, up to the solution changes from blue to
Add 5 ml Potassium Dichromate, then solution changing the color back to blue.
And then Ferrous Sulphate added drop by drop until the color of the solution changes from blue to Measure the total volume of Ferrous Sulphate solution in ml and recorded it as ‘X’.
Procedure:
Take 5 grams of soil sample of 425 micron IS sieve passing in 500 ml conical
Add 10 ml of Potassium Dichromate
Add 20 ml of concentrated Sulphuric acid and allowed to 30 minutes on a heat insulating surface like asbestos
Add 200 ml distilled water, 10 ml of Ortho Phosphoric acid, 1 ml of the Indicator and the mixture shall be shake
Ferrous Sulphate solution added through burette in 0.5 ml increments, up to the solution changes from blue to
Add 5 ml Potassium Dichromate, then solution changing the color back to blue.
And then Ferrous Sulphate added drop by drop until the color of the solution changes from blue to Measure the total volume of Ferrous Sulphate solution in ml and recorded it as ‘Y’.
Calculations:
The Volume of Potassium Dichromate used to oxidize organic, V= 5(1 – Y/X).
Percentage of Organic Matter in soil = (0.67W2V) / (W1W3)
The expansion of soil containing high sulphates occurs during ambient temperature drops from daytime temperatures of roughly 30 degree to below 5 degree at night. This expansion caused structural damage to single storey house in particular and to interior concrete bottoms and asphalt driveways. This phenomena, herein appertained to as” salt heave,” . Only by early recognition of the presence of water-soluble sulphate in the soil to control structural damage building or pavement. The final test procedure described herein provides a system to estimate soils containing sodium sulfate. In this session we will learn how to determine total soluble sulphates by volumetric system as per( IS 2720 – Part – 27).
Object: Determine total soluble sulphates by volumetric method as per (IS : 2720 – Part – 27)
Sample Preparation: Take about 100 grams of oven dried soil sample and sieve it in a 425 micron IS Sieve.
Preparation of Reagents:
Barium Chloride Solution (N/4): Dissolve 5 grams Barium chloride in one liter of distilled water.
Potassium Chromate Solution (N/4): Dissolve 275 grams of Potassium chromate in a small amount of distilled water. Add few drops of Silver Nitrate solution to it to remove any Chloride, filter and dilute to 250 ml.
Silver Nitrate Indicator: Dissolve 500 mg of Silver Nitrate in 100 ml of distilled water.
Dilute Solution of Ammonium Hydroxide: (Sp.Gr 0.888) Mix Ammonium Hydroxide and distilled water in the ratio of 1:2.
Concentrated Hydrochloric acid: Gr 1.11.
Test Procedure:
10 grams of soil sample taken in to
Add 50 ml water, stir well, allow decanting and
Take 10 ml filtrate sample by pipette in a conical
Make it slightly acidic by adding concentrated hydrochloric acid (i.e.: few drops) and heat to
Add Barium Chloride solution (N/4) from the burette till the precipitation is complete, measure the volume in ml and recorded it as ‘X’.
Neutralize the solution with Ammonium hydroxide (i.e.: few drops).
Titrate the excess of Barium chloride against Potassium Chromate solution(N/4), measure the volume in ml and recorded it as ‘Y’. the end point may be confirmed, if considered necessary, by using Silver nitrate solution as an external indicator.
Calculations:
Sulphates as Sodium Sulphate in Soil, percent by mass = 0.0177 x (X – Y)
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
Small sand pouring cylinder for depth up to 150 mm.
Large sand pouring cylinder for depth more than 150 mm and not exceeding 250 mm.
Tools for excavating holes such as a scraper tool for leveling the surface, bent spoon or dibber for digging holes.
Cylindrical calibrating container with material diameter of 100 mm and an internal depth of 150 mm.
Balance of capacity 15 kg and sensitivity 1 gram.
A glass plate o about 600 mm square area and atleast 10 mm or more thicker.
Metal containers.
Metal tray having area 300 mm square , 40 mm deep with a 100 mm hole in the center of the tray.
1 mm and 600 microns IS sieves..
Clean, uniformly graded natural sand passing 1 mm sieve and retained on 600 microns sieve.
PROCEDURE
Calibration of Sand Pouring Cylinder
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.
Remove the cap of the pouring cylinder.
Weigh the empty pouring cylinder (W).
Close the shutter of the cone.
Fill the sand in to the pouring cylinder about 10 mm below from the top.
Determine the net weight of sand in the cylinder (W1).
Now place the pouring cylinder on a clean plane surface, open the shutter and allow the sand to flow in to the cone.
Close the shutter when the flow stops.
Carefully collect and weigh (W2) the sand discharged from the pouring cylinder.
Refill the pouring cylinder with sand such that the initial weight is W1.
Place the pouring cylinder on the top of the calibration cylinder concentrically.
Open the shutter and allow the sand to flow in to the calibrating cylinder.
Close the shutter when the flow stops or no further movement of sand takes place in the cylinder.
Determine the weight (W3) of the pouring cylinder.
Repeat the above procedure for at least three times and determine the mean values of W2 and W3.
Determine the volume (V) of the calibrating cylinder either by measuring the dimensions
(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
Prepare a flat approximately 450 mm square area with the aid of a scraper tool.
Place the metal tray just above the central hole on the prepared surface of the soil to be tested.
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.
Carefully collect the excavated soil from the hole and weigh (Ww).
Determine the water content (W) of the excavated soil as per IS: 2720 (Part 2) 1973.
Fill the pouring cylinder to the constant weight (W1) i.e. weight equal to the initial weight during calibration.
Remove the metal tray before the pouring cylinder is placed in position over the excavated hole.
Place the cylinder such that the base of the cylinder covers the hole concentrically.
Open the shutter and allow the sand to run out in to the hole.
Close the shutter of the pouring cylinder when no further movement of sand takes place in the cylinder.
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
Care shall be taken to see that the test sand used is clean, dry and uniformly graded.
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.
Care shall be taken to see that the same initial weight of sand is taken during calibration and during density measurement in the field.
Example: A soil in the borrow pit is at a dry density of 17 kN/m³ 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/m³ 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 m³ 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
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.
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
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%.
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 Ccbetween 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
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
Take 20 gm soil sample passing from 425 micron IS sieve .
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.
Soil should be allowed to season for sufficient time for allowing water to permeate throughout the soil mass.
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.
Continued the rolling till the thread becomes 3 mm. in diameter and see the crumbling.
If not soil is then kneaded together to a uniform mass and rolled again.
The process is to be continued until the thread crumbled and offering shear when rolled into 3 mm diameter.
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:
