Question: 500 gm of dry soil was subjected to a sieve analysis. The weight of soil retained on each sieve is as follows as:-
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I.S sieve size weight of soilg I.S sieve size weight of soilg
4.75 mm 10 425 $\mu$ 85
2.00 mm 165 212 $\mu$ 40
1.00 mm 100 1.50 $\mu$ 30
75 $\mu$ 50

Plot the grain size distribution curve and determine the following:-

  1. Percentage of gravel, coarse and medium sand, fine sand and slit clay fraction in the soil as per IS: 1498 - 1970.
  2. Effective size
  3. Uniformly coefficient
  4. Coefficient of curve
  5. The gradation of soil.
sieve analysis • 312 views
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modified 6 days ago by gravatar for Sanket Shingote Sanket Shingote ♦♦ 280 written 13 months ago by gravatar for Mayank Aggarwal Mayank Aggarwal30
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The diameter of soil grain and the corresponding percent fines than are calculated as below:-

I.S sieve size Diameter of grain mm Weight retained(g) Cumulative weight retained (g) % retained % finer than
4.75 4.75 10 10 2.0 98.0
2.00 2.00 165 175 35.0 65.0
1.00 1.00 100 275 55.0 45.0
425 0.425 85 360 72.0 28.0
212 0.212 60 420 84 18
150 0.150 20 440 88 15
75 0.075 40 480 96 4

The grain size distribution curve is plotted between the diameter of soil as shown in fig.

A) Is grain size classification scale is fitted on the abscissa,

  1. Percentage of gravel = 100 - 98 = 2% > 4.75 mm
  2. Percentage of coarse sand = 98 - 65 = 33% (4.75 mm - 2.0 mm)
  3. Percentage of medium sand = 65 - 28 = 37% (2.00 mm - 0.425 mm)
  4. Percentage of fine sand = 28 - 4 = 24% (0.425 mm - 0.075 mm)
  5. Percentage of fine sand = 4% (<0.075 mm)

B) Effective size $D_{10} = 0.13 mm$

C) Uniformly coefficient = $C_U = \frac{D_{60}}{D_{10}} = 13.8$

D) Coefficient of curvature = $C_C = \frac{(D_{30})^2}{D_{10}*D_{60}} = 1.1$

E) The sand soil is classed as well- graded sand (SW) since $C_U$ > 6 and $C_c$ is between 1 & 3.

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written 13 months ago by gravatar for Mayank Aggarwal Mayank Aggarwal30
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