sieve analysis
DESCRIPTION
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Sieve Analysis
Geotechnical Engineering Laboratory
Submitted by
Group : 1
Sajadur Rahman (1004034)
Safayat Ahmed(1004035)
Sadia Rifat Shanta (1004036)
Nazmul Haque ( 1004037 )
Pritom Paul (1004038)
Sieve Analysis
November 25, 2013
Page 2
Introduction:
Definitions:
Sieve Analysis :
The process of dividing a sample of aggregate into fractions of same particle size is known as
Sieve Analysis.
Coarse Grained Soil Materials :
The coarse grained soil materials are mineral fragments that may be identified primarily on
the basis of particle size. It consists of Gravel and Sand.
Fine Grained Soil Materials :
The fine grained soil materials are mostly microscopic and submicroscopic flake shaped
materials. It consists of Silt and Clay.
Particle Size Limits of Soil Constituents(ASTM Classification )
Gravel Larger than 4.75
Coarse Sand 4.75 to 2.00
Medium Sand 2.00 to 0.425
Fine Sand 0.425 to 0.075
Fines(combined silt and clay ) Smaller than 0.075
Fineness Modulus :
This is Fineness modulus ,defined as the sum of the cumulative percentages of retained on
the sieves of the standard series divided by 100. The standard series consists of the sieves
having the opening size 150µm,300 µm,600 µm,1.18 mm,2.36 mm,4.75mm or according to
ASTM No. 100,50,30,16,8,4.
Fineness Modulus (FM) =Sum of %retained of No. 100,50,30,16,8,4 Sieves
100
Gradation :
Well Graded It contains a good representation of all particle sizes ranging from coarse to fine.
Uniformly Graded The size of the particles are approximately same
Gap Graded It consists of mixture of uniform coarse sized
Sieve Analysis
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and uniformly fine sized particles with a break in gradation between two sizes.
Poorly Graded Any soil not well-graded may be characterized as poorly graded
Effective Particle Size(Diameter), Dx :
Diameter of the particle size corresponding to the ‘X’ percent finer.
Uniformity Coefficient, Cu :
Cu = D60 D10
Coefficient of Curvature, Cc : (D30)2
Cc = (D10 X D60 )
Criteria for SW( Well graded sand ):
If a sand has Cu > 6 and 1 < Cc < 3 , the sand can be defined as well graded else it is poorly
graded.
Sieve Analysis
November 25, 2013
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Scope :
1. Soil permeability and capillarity are related to effective particle Size.
2. The method of designing inverted filters for dams, levees etc. uses the particle size
distribution of the soils involved.
3. The soil identification and classification can be done with this experiment.
Purpose :
1. Getting the particle size distribution of the soil.
2. Getting the gradation curve.
3. Classify the soil.
4. Determine the uniformity coefficient of soil and coefficient of curvature of the soil.
Apparatus :
1. Set of Sieves
2. Brush
3. Balance
4. Large Pan
5. Shaking Machine
Balance Large Pan
Shaking Machine Set of Sieves
Brush
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Test Procedures :
Step 1:
Record the weight of the given oven dry weight
sample
Step 3:
Make sure that all the sieves are clean
and assemble them according to lowest
to highest sieve opening where the
smallest sieve will be set at the bottom.
Pan
#200
#100
#4
#8
#12
#16
#20
#30
#40
#50
Step 2:
Take the weight of the empty sieves with the
weighing machine.
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November 25, 2013
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Step 4:
Now drop down the soil sample through the sieves
Step 5:
Place the set of sieve in the mechanical shaking
machine.
Step 6:
Shake the containers for 10 min with the mechanical
shaking machine.
Step 7:
Remove the stack from the shaking machine.
Step 8:
Then disassemble the sieves and take the weight of
the individual sieve.
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Data Sheet Department of Civil Engineering, BUET
Geotechnical Engineering Laboratory
Sieve Analysis
Wt of Container + Soil = 181.5 gm
Wt of Container = 81.3 gm
Wt of Soil = 100.2 gm
Date : 25/11/13
D10 = 0.17 D30 = 0.29 D60 = 0.52 Cu = 1.71 Cz = 0.95 F.M = 2.06
Sieve No.
Sieve Opening
(mm)
Wt. of Sieve (gm)
Wt of Sieve +
soil (gm)
Wt of soil
retained (gm)
% of soil retained
Cumulative % retained
%Finer
4 4.75 529.7 529.7 0 0 0 100
8 2.36 467.0 468.3 1.3 1.27 1 99
12 1.70 448.1 450.5 2.4 2.34 4 96
16 1.18 498.9 506.6 7.7 7.51 11 89
20 0.850 501.3 516.1 14.8 14.44 26 74
30 0.60 457.8 464.1 6.3 6.14 32 68
40 0.425 402.7 431.2 28.5 27.81 60 40
50 0.300 423.9 432.4 8.5 8.29 68 32
100 0.150 365.3 392.3 27 26.34 94 6
200 0.075 349.7 354.2 4.5 4.39 99 1
Pan 370.9 372.4 1.5 1.46 100 0
Total 102.5
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Discussion :
1. The sand contains 20% fine sand , 48% medium and 31% course sand. As the %Finer
is less than 50% So this is a course grained soil.(S or G )
2. As the coarser part is less than 50%, it is a sample of sand.(S)
3. From the analysis it can be seen that the value of co-efficient of curvature Cz = 0.75
which isn’t in between 1 and 3 , so the sand is not well graded. In the same way the
uniformity coefficient Cu = 1.71 which isn’t also greater than 6 . These Both criteria
indicates that it is a poorly graded sand. It should be denoted as SP.
4. From visual analysis of the graph it can be said that it is a gap graded gradation
curve.
5. The permeability of a soil (cohesion less) is dependent upon its gradation, it is
expressed by k = C X (D10)2 which leads to 2.9 X 10-2 for our sample soil. It represents
that the soil has a good drainage system and it also indicates that it contains clean
sand and gravel.
6. The accuracy of the distribution curve for fine grained soils is more questionable
than the accuracy of the curves for the coarser soils. In the case of residual soils the
term “individual Particle” size depends on the disaggregation that imparted
mechanically prior to testing.