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USCS classification, 75 micron 62%

Short answer

Since 62% passes 75µm sieve, it's fine-grained; with LL=54% (>50) and PI=31% above A-line (PI>0.73(LL-20)), the soil is inorganic clay of high plasticity, CH.

Classify the following soil as per the unified soil classification system: soil passing from 75 µ sieve = 62%, liquid limit = 54%, plastic limit = 23%.

Given

Passing 75 micron =62%= 62\%; wL=54%w_L = 54\%; wP=23%w_P = 23\%.

Ip=54−23=31%I_p = 54 - 23 = 31\%
  • 62% passes the 75 micron sieve (more than 50%), so the soil is fine-grained.
  • wL=54>50w_L = 54 > 50: high plasticity (H).
  • A-line at wL=54w_L = 54: Ip=0.73(54−20)=24.8%I_p = 0.73(54 - 20) = 24.8\%. Since Ip=31>24.8I_p = 31 > 24.8, the soil is above the A-line: clay (C).

Answer: CH, inorganic clay of high plasticity (fat clay).

✓ Answer checked by Claude Opus 5.5 (high effort)

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