Skip to content
CivilSolve

Geotechnical

AASHTO Soil Classification (M 145)

Find the AASHTO soil group and group index from sieve passing percentages, liquid limit, and plasticity index, with classification steps.

Direct solving is free and needs no account. Have a word problem instead? Submit it as text.

Inputs

Load a sample problem:

Worked examples

These are the solver's own reference problems — the answers come from published sources or independent hand computation, never from the solver itself. The solution below is the live solver output for each.

Example 1

A-1-a gravel: No10=40, No40=20, No200=5, PI=3 (GI = 0 by definition)

Given

passing no10
40
passing no40
20
passing no200
5
plasticity index
3

Assumptions

  • Classification per AASHTO M 145 by left-to-right elimination: the FIRST group whose limits all pass is the classification.
  • Result: A-1-a (0) — stone fragments, gravel and sand, general subgrade rating excellent to good.

Solution steps

  1. Granular screen

    With 35% or less passing the No. 200 sieve the soil is a granular material (A-1, A-3, A-2 families); more would make it a silt-clay material (A-4 to A-7).

    F = 5 % ≤ 35 % → granular

  2. Test A-1-a

    Stone fragments/gravel: tight limits on all three sieves and PI ≤ 6.

    passes: passing No. 10 = 40 % ≤ 50 %; passing No. 40 = 20 % ≤ 30 %; passing No. 200 = 5 % ≤ 15 %; PI = 3 % ≤ 6 %

  3. Group index for A-1-a

    By definition the group index of A-1, A-3, A-2-4 and A-2-5 soils is always taken as zero.

    GI = 0

    = 0

  4. Classification: A-1-a (0)

    Group A-1-a — stone fragments, gravel and sand; general subgrade rating excellent to good. Reported with the group index in parentheses.

Results

Group index GI (whole number)

0

Percent passing the No. 200 sieve (F)

5

Where this answer was checked
source
AASHTO M 145 classification table (as tabulated in the NCEES FE Reference Handbook)A-1-a limits: No10 ≤ 50, No40 ≤ 30, No200 ≤ 15, PI ≤ 6; GI ≡ 0
verified by
hand-recomputed
derivation
40 ≤ 50 ✓, 20 ≤ 30 ✓, 5 ≤ 15 ✓, PI 3 ≤ 6 ✓ → first group passes → A-1-a. GI of A-1 soils is always 0.

Example 2

A-2-6 with partial GI: No200=30, LL=35, PI=15 → GI = round(0.75) = 1

Given

passing no10
80
passing no40
60
passing no200
30
liquid limit
35
plasticity index
15

Assumptions

  • Classification per AASHTO M 145 by left-to-right elimination: the FIRST group whose limits all pass is the classification.
  • Result: A-2-6 (1) — silty or clayey gravel and sand, general subgrade rating excellent to good.

Solution steps

  1. Granular screen

    With 35% or less passing the No. 200 sieve the soil is a granular material (A-1, A-3, A-2 families); more would make it a silt-clay material (A-4 to A-7).

    F = 30 % ≤ 35 % → granular

  2. Test A-1-a

    Stone fragments/gravel: tight limits on all three sieves and PI ≤ 6.

    fails: passing No. 10 = 80 % ≤ 50 % is violated

  3. Test A-1-b

    Coarse sand mixture: No. 40 and No. 200 limits with PI ≤ 6.

    fails: passing No. 40 = 60 % ≤ 50 % is violated

  4. Test A-3

    Clean fine sand: mostly finer than No. 40, few fines, nonplastic.

    fails: passing No. 200 = 30 % ≤ 10 % is violated

  5. A-2 family (silty or clayey gravel and sand)

    Granular soils that fail A-1/A-3 fall into A-2; the suffix comes from the LL split at 40 and the PI split at 10.

    LL = 35 % ≤ 40 %; PI = 15 % > 10 %

  6. Partial group index for A-2-6

    A-2-6 and A-2-7 use only the PI term of the group index formula; a negative value is reported as zero and the result is rounded to the nearest whole number.

    GI = 0.01·(F − 15)·(PI − 10)

    GI = 0.01 × 15 × 5 = 0.75 → 1

    = 1

  7. Classification: A-2-6 (1)

    Group A-2-6 — silty or clayey gravel and sand; general subgrade rating excellent to good. Reported with the group index in parentheses.

Results

Group index GI (whole number)

1

Percent passing the No. 200 sieve (F)

30

Where this answer was checked
source
AASHTO M 145 classification table and group index formulaA-2-6: F ≤ 35, LL ≤ 40, PI ≥ 11; partial GI = 0.01(F−15)(PI−10)
verified by
hand-recomputed
derivation
Elimination: A-1-a fails (F 30 > 15), A-1-b fails (F 30 > 25), A-3 fails (PI ≠ 0); F 30 ≤ 35 → A-2; LL 35 ≤ 40 and PI 15 ≥ 11 → A-2-6. Partial GI = 0.01·(30−15)·(15−10) = 0.01·15·5 = 0.75 → rounds to 1.

Example 3

A-6 clay: No200=60, LL=32, PI=15 → GI = 6.25 → 6

Given

passing no10
95
passing no40
80
passing no200
60
liquid limit
32
plasticity index
15

Assumptions

  • Classification per AASHTO M 145 by left-to-right elimination: the FIRST group whose limits all pass is the classification.
  • Result: A-6 (6) — clayey soil, general subgrade rating fair to poor.

Solution steps

  1. Silt-clay screen

    With more than 35% passing the No. 200 sieve the soil is a silt-clay material (A-4 to A-7).

    F = 60 % > 35 % → silt-clay

  2. LL and PI grid

    The silt-clay groups split on the liquid limit at 40 (A-4/A-6 low vs A-5/A-7 high) and the plasticity index at 10 (silty ≤ 10 vs clayey ≥ 11).

    LL = 32 % ≤ 40 %; PI = 15 % > 10 %

  3. Group index

    The group index penalizes fines content and plasticity; a negative value is reported as zero and the result is rounded to the nearest whole number.

    GI = (F − 35)·[0.2 + 0.005·(LL − 40)] + 0.01·(F − 15)·(PI − 10)

    GI = 25 × 0.16 + 0.01 × 45 × 5 = 4 + 2.25 = 6.25 → 6

    = 6

  4. Classification: A-6 (6)

    Group A-6 — clayey soil; general subgrade rating fair to poor. Reported with the group index in parentheses.

Results

Group index GI (whole number)

6

Percent passing the No. 200 sieve (F)

60

Where this answer was checked
source
AASHTO M 145 classification table and group index formulaF > 35, LL ≤ 40, PI ≥ 11 → A-6; full GI formula
verified by
hand-recomputed
derivation
GI = (60−35)[0.2+0.005(32−40)] + 0.01(60−15)(15−10) = 25·(0.2−0.04) + 0.01·45·5 = 25·0.16 + 2.25 = 4 + 2.25 = 6.25 → rounds to 6. (The negative LL−40 term legitimately reduces the first bracket.)