Question 3.22: The following data refers to a silty clay that was assumed t...

The following data refers to a silty clay that was assumed to be saturated in the undisturbed condition. On the basis of these data determine the liquidity index, sensitivity, and void ratio of the saturated soil. Classify the soil according to the Unified and AASHTO systems. Assume G_{s}=2.7.

 

Index property Undisturbed Remolded
Unconfmed compressive
strength, q_{u} kN / m ^{2} 244 kN / m ^{2} 144 kN / m ^{2}
Water content, % 22 22
Liquid limit, % 45
Plastic limit, % 20
Shrinkage limit, % 12
% passing no. 200 sieve 90
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Liquidity Index, I_{l}=\frac{w_{n}-w_{p}}{w_{l}-w_{p}}=\frac{22-20}{45-20}=0.08

 

Sensitivity, S=\frac{q_{u} \text { undisturbed }}{q_{u}^{\prime} \text { disturbed }}=\frac{244}{144}=1.7

 

Void ratio, e=\frac{V}{V_{s}}

 

For S=1, e=w G_{s}=0.22 \times 2.7=0.594.

 

Unified Soil Classification

Use the plasticity chart Fig. 3.22 w_{l}=45, I_{p}=25. The point falls above the A-line in the CL-zone, that is the soil is inorganic clay of low to medium plasticity.

AASHTO System

Group Index G I=0.2 a+0.005 a c+0.01 b d

a = 90 – 35 = 55

b = 90 – 15 = 75

c = 45 – 40 = 5

d = 25 – 10 = 15 \text { (here } \left.I_{p}=w_{l}-w_{p}=45-20=25\right)

Group index G I=0.2 \times 55+0.005 \times 55 \times 5+0.01 \times 75 \times 15

 

=11+1.315+11.25=23.63 \text { or say } 24

Enter Table 3.15 with the following data

% passing 200 sieve = 90%

Liquid limit = 45%

Plasticity index = 25%

With this, the soil is either A-7-5 or A-7-6. Since \left(w_{l}-30\right)=(45-30)=15<25\left(I_{p}\right) the soil is classified as A-7-6. According to this system the soil is clay, A-7-6 (24).

 

Table 3.15 Soil Fractions as per U.S. Department of Agriculture
Soil fraction Diameter in mm
Gravel >2.00
Sand 2-0.05
Silt 0.05-0.002
Clay <0.002
3.22

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