Question

Part a : answer is A-1-B

Not sure part B

ented for the given soil sample, determine the soil classification using Using the oillassification System (USCS) and the AASHTO Soil classification system. Give the group index if needed. Percent Passing 98 85 Sieve No. 19mm k in.) 12.5mm (1/2in.) No. 4 (4.75 mm) No. 10 (2.00 mm) No. 20 (0.850 mm) No. 40 (0.425 mm) No. 100(0.150 mm) No. 200 (0.075 mm) 80 60 45 30 20 10 / レ Atterberg limits: Liquid limit 31 %, Plastic Limit: 25%, Non organic soil

Answer 1

1. USCS Classification:

Since only 10% of the soil sample passes through the No. 200 sieve, 90% gets retained over it or are larger than it. Therefore, the soil is coarse grained. A soil is considered coarse grained if 50% or more than 50% of soil gets retained over No. 200 sieve. Otherwise it's fine grained.

Now, 80% of total sample passed through the No. 4 sieve, meaning only 20% retained over it, which is 20x100/90 = 22.22% of the coarse grained fraction. Therefore, our sample is Sand. A coarse grained soil is considered Gravel if 50% or more than 50% of soil gets retained over the No. 4 sieve. Otherwise it's sand.

Now, percent passing through No. 200 sieve = 10%, therefore, fineness of our coarse grained soil is 10%.

Since fineness is more than 5% but less than 12%, it's a borderline condition. In this case, both gradation and silt/clay content are important and it will require dual representation.

Gradation:

Here, D₆₀ = 2 mm, D₃₀ = 0.425 mm D₁₀ = 0.075 mm.

Now, the coefficient of uniformity, C_u = D₆₀/D₁₀ = 2/0.075 = 26.67, which is greater than 6.

And the coefficient of curvature, C_c = D₃₀(D₁₀ x D₆₀) = 0.425/(2 x 0.075) = 2.83, which is between 1 and 3.

Since both the conditions are satisfied, the given sample is a Well Graded Sand. or SW.

Silt/clay content:

Given, Liquid limit = 31% and Plastic limit = 25%.
Therefore, plasticity index = 31 - 25 = 6%.
We will use the A - line concept to find out whether silt is dominant or clay in our sample.

The A - line: IP% = 0.73 (LL% - 20)

Therefore, for the given liquid limit, plasticity index on the A line will be 0.73 x (31 - 20) = 8.03%, but the given plastic limit is 6% as calculated above, meaning, the soil lies below the A - line and therefore is silty. A soil sample above the A - line is clayey and below the A - line is silty.

Therefore, the dual representation is SW-SM

The soil is classified as Well Graded Silty Sand.

2. AASHTO classification:

Given,
Percent passing through No. 10 sieve = 60%
Percent passing through No. 40 sieve = 30%
Percent passing through No. 200 sieve = 10%
Liquid limit = 31%, Plasticity index = 31 - 25 = 6%

Using the above data, from the AASHTO soil classification system table, we can see that our soil sample falls in group A-1-b.

Since group index is always zero for soils of groups A - 1, A - 3, A - 2 - 4 and A - 2 - 5, group index of our sample is 0.