Gravel 35%, fines 4%: sand is 61%. The soil is

sand since gravel is less than half the coarse fraction

With Cu=5 and Cc=2, sand is

poorly graded (SP) since Cu < 6

2079 Asoj · 5 marks

USCS classification, 75 micron 4%

Short answer

SP, poorly graded sand with gravel. Fines 4%, coarse 96%; gravel is 36% of coarse fraction, so sand dominates. Well-graded sand requires Cu>=6, but Cu=5 (Cc=2 ok), so poorly graded.

The sieve analysis of a soil gave the following results: % passing 75 micron sieve = 4 % retained on 4.75 mm sieve = 35 Coefficient of curvature = 2 Coefficient of uniformity = 5 Classify the soil according to the USCS system.

Given

Passing 75 micron $= 4\%$ ; retained on 4.75 mm $= 35\%$ ; $C_c = 2$ ; $C_u = 5$ .

Step 1: Coarse or fine?

Only 4% passes 75 micron, so more than 50% is retained: coarse-grained.

Step 2: Gravel or sand?

Gravel (retained on 4.75 mm) $= 35\%$
Sand $= 100 - 35 - 4 = 61\%$

Coarse fraction $= 96\%$ ; gravel is $35/96 = 36\%$ of it, less than half. So the soil is a sand (S).

Step 3: Fines content

4% is less than 5%, so only the grading is needed (symbols SW or SP).

Step 4: Grading

For well-graded sand: $C_u \ge 6$ and $1 \le C_c \le 3$ .

$C_c = 2$ satisfies the condition.
$C_u = 5 < 6$ fails.

The soil is not well graded.

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

✓ Answer checked by Claude Opus 5.5 (high effort)

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