CivilSolve

Geotechnical

AASHTO Soil Classification (M 145)

AASHTO group classification (A-1-a through A-7-6) with group index, from percent passing the No. 10, No. 40 and No. 200 sieves plus liquid limit and plasticity index, by left-to-right elimination per AASHTO M 145. Reports GI = (F−35)[0.2+0.005(LL−40)] + 0.01(F−15)(PI−10) rounded and clamped, with the partial (PI-term) GI for A-2-6/A-2-7. Use for: 'classify by AASHTO', highway subgrade group and group index. Not for: USCS symbols (separate solver).

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.)