Question
Download Solution PDFSieve analysis on a dry soil sample of mass 1000 g showed that 980 g & 270 g of soil pass through a 4.75 mm and 0.075 mm sieve respectively. The LL = 40% & PL = 18%. The soil classification as per IS system is:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
The soil classification according to the Indian Standard Soil Classification System (IS 1498) depends on the grain size distribution and Atterberg limits (Liquid Limit, Plastic Limit, and Plasticity Index).
Given Data:
- Total soil mass = 1000 g
- Mass passing through 4.75 mm sieve = 980 g
- Mass passing through 0.075 mm sieve = 270 g
- Liquid Limit (LL) = 40%
- Plastic Limit (PL) = 18%
Grain Size Distribution:
From the sieve analysis:
- Percentage passing through 4.75 mm sieve = \( \frac{980}{1000} \times 100 = 98\% \)
- Percentage passing through 0.075 mm sieve = \( \frac{270}{1000} \times 100 = 27\% \)
Since 27% of the soil passes through the 0.075 mm sieve, the soil contains significant fines.
Plasticity Index (PI):
The Plasticity Index (PI) is calculated as:
\( PI = LL - PL = 40\% - 18\% = 22\% \)
Soil Classification:
Given that 27% of the soil passes through the 0.075 mm sieve (fines), and the Plasticity Index (PI = 22%) indicates significant plasticity, the soil is classified as Clayey Sand (SC) according to the IS classification system.
Last updated on Jun 4, 2025
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