Question

The following results are obtained from particle size analysis:

Uniformity coefficient = 8 Coefficient of curvature = 2.8

Percentage of soil passing through 75 μ IS sieve = 10%

Percentage of soil passing   through 4·75 mm IS sieve = 70%

If liquid limit and plastic limit for the soil are 38% and 34.2% respectively, the soil can be classified as per IS soil classification system as

1. SP-SM
2. SW-SM
3. SP-SC
4. SW-SC

Answer 1

Correct Answer - Option 2 : SW-SM

Concept

Note: As data given of soil is passing through 75 microns and 4.75 mm sieve we go for classification of coarse-grained soil.

Classification of Coarse-Grained soil

i) Particle size

ii) Gradation characteristics (C_u and C_c)

iii) % fineness (% fraction passing through 75 μ sieve)

→ Coarse-grained soils are those having 50 % or more retaining on 75 micron (0.075 mm) sieve

→ Further, if coarse-grained soils are designated as gravel (G) if 50 % or more of the coarse fraction retained on 4.75 mm sieve: Otherwise they are designates as Sand (S).

Calculation

Given,

C_u = 8, C_c = 2.8

Percentage of soil passing through 75 μ IS sieve = 10%

Percentage of soil passing   through 4·75 mm IS sieve = 70%

W_L = 38 %, W_P = 34.2 %

i) % of fineness = 10% (i.e. finesses is between 5 - 12 %)

Soil retained on 75 micron sieve = 100 – 10 = 90 % (> 50 %) (∴ Coarse grained)

ii) Soil retained on 4.75 mm sieve = 100 – 70 = 30 % (< 50 %) (∴ Sand)

For well graded sand

C_u > 6, 1 < C_c < 3

Here C_u = 8, C_c = 2.8

∴ Sand is well graded (SW)

W- Well graded, S - Sand.

Plasticity Index (I_P)

\({I_P} = {W_L} - {W_P}\)

I_P = 38 – 34.2

= 3.8 %

\({I_p}of\;A - line = 0.73 \times \left( {{W_L} - 20} \right)\)

\({I_{P,\;\;A}} = 0.73 \times \;\left( {38 - 20} \right)\)

= 13.4 %

To determine whether sand is clay or silt

If

I_P < I_{P, A} sand is silt

I_P > I_{P, A} sand is a clay

∴ I_P < I_{P, A}

∴ Silty soil (M)

∴ Soil is SW – SM well-graded sand containing silt as fine