Skip to main content

USCS classification, coarse fraction 62%

Short answer

Only 4% fines, so coarse-grained. Gravel=38%, sand=58%, so sand dominates; with Cu=5 (<6) and Cc=2.6, it is poorly graded sand with gravel, SP.

Classify the following soil as per the unified soil classification system: soil passing from 75 µ sieve = 4%, soil passing from 4.75 mm sieve (coarse fraction) = 62%, coefficient of uniformity = 5, coefficient of curvature = 2.6.

Given

Passing 75 micron =4%= 4\%; passing 4.75 mm =62%= 62\%; Cu=5C_u = 5; Cc=2.6C_c = 2.6.

  • Fines are only 4%, so the soil is coarse-grained.
  • Gravel (retained on 4.75 mm) =100−62=38%= 100 - 62 = 38\%; sand =62−4=58%= 62 - 4 = 58\%. Sand is more than half of the coarse fraction (58/96 = 60%), so it is sand (S).
  • Fines are < 5%, so grading decides: well-graded sand needs Cu≥6C_u \ge 6 and 1≤Cc≤31 \le C_c \le 3. Here Cc=2.6C_c = 2.6 passes, but Cu=5<6C_u = 5 < 6 fails.

Answer: SP, poorly graded sand (with gravel).

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Soil Mechanics chapter 3 questions →