Soil Properties & Classification

5% of exam

Phase relationships and index properties, USCS and AASHTO classification, shear strength, permeability, compressibility and consolidation, lateral earth-pressure coefficients, and bearing-capacity parameters.

4 concepts

Exam areas

A. Soil classification and properties

Soil Phase Relationships & Classification

The weight-volume phase diagram (e, w, S, unit weights) tied to USCS and AASHTO classification with the group index — the foundation every geotech calc rests on.

Almost every geotechnical number on the PE — a bearing capacity, an active thrust, a settlement, an effective stress — starts from a unit weight and a void ratio that you back out of a few measured masses and volumes. Get the phase relationships wrong and the entire downstream chain is wrong, so this is the highest-leverage skill in the whole geotechnical area. The NCEES PE Civil Reference Handbook collects these in §3.7 Soil Classification (the three-phase block and the volume-weight table) and pairs them with the USCS and AASHTO classification charts; this concept teaches you to move fluently across that block and to name a soil two different ways.

The three-phase block

Real soil is three phases — mineral solids, water, and air — and the entire subject is bookkeeping on a single diagram that stacks volumes on one side and weights on the other. By convention you draw solids at the bottom, water in the middle, air on top; the voids are water plus air. Every relationship below is just a ratio read off this block, so when a problem stalls, redraw the block and label what you know. A favorite shortcut is to set the volume of solids

V_s = 1

(or

V_v = e

), which turns ratios into direct quantities.

V = V_s + V_w + V_a, \qquad V_v = V_w + V_a, \qquad W = W_s + W_w \;\; (W_a \approx 0)

Void ratio, porosity, water content, saturation

Four ratios describe the state of the voids. Void ratio

e

compares void volume to solid volume and is the workhorse in consolidation; porosity

n

compares void volume to total volume. Water content (gravimetric)

w

is the ratio of water weight to solid weight, the single most measured soil property. Degree of saturation

S

is the fraction of the voids that is water;

S=1

(100%) below the water table,

S<1

above it. Memorize the bridge identity

S\,e = w\,G_s

— it ties the moisture state to the void state and shows up constantly.

e = \dfrac{V_v}{V_s}, \quad n = \dfrac{V_v}{V} = \dfrac{e}{1+e}, \quad w = \dfrac{W_w}{W_s}, \quad S = \dfrac{V_w}{V_v}, \qquad \boxed{S\,e = w\,G_s}

Unit weights

Specific gravity

G_s

(typically 2.65 for quartz sands, 2.70-2.75 for clays) converts solid volume to solid weight via

\gamma_w = 62.4\ \text{pcf}

. From the block you get a family of unit weights that differ only in which phases you count. Dry unit weight ignores water; total (moist) unit weight is what a field sample weighs; saturated assumes the voids are full; and submerged (buoyant) is what governs effective stress below the water table — note

\gamma_b = \gamma_{sat}-\gamma_w

, not

\gamma_{sat}/2

\gamma_d = \dfrac{G_s\,\gamma_w}{1+e}, \quad \gamma_t = \dfrac{(G_s + S e)\gamma_w}{1+e} = \gamma_d(1+w), \quad \gamma_{sat} = \dfrac{(G_s+e)\gamma_w}{1+e}, \quad \gamma_b = \gamma_{sat}-\gamma_w

Particle size and Atterberg limits

Classification needs two pieces of information: how coarse the soil is, and how the fines behave. A sieve analysis splits the sample at the No. 200 (0.075 mm) sieve — coarse-grained if more than half is retained, fine-grained if half or more passes. For coarse soils, the grading is summarized by the uniformity coefficient

C_u = D_{60}/D_{10}

and the coefficient of curvature

C_c = D_{30}^2/(D_{10}D_{60})

. For the fines, the Atterberg limits — liquid limit

LL

and plastic limit

PL

— give the plasticity index

PI = LL-PL

, the measure of how much water a clay can take before it flows.

C_u = \dfrac{D_{60}}{D_{10}}, \qquad C_c = \dfrac{D_{30}^{\,2}}{D_{10}\,D_{60}}, \qquad PI = LL - PL

USCS: the two-letter symbol

The Unified Soil Classification System (ASTM D2487, reproduced in handbook §3.7.2) builds a two-letter group symbol. The first letter is the dominant grain type — G gravel, S sand, M silt, C clay, O organic, Pt peat. The second is a qualifier: for clean coarse soils W (well-graded) or P (poorly graded); for fine soils L (low plasticity,

LL<50

) or H (high plasticity,

LL\ge50

). A clean sand is well-graded (SW) when

C_u\ge6

and

1\le C_c\le3

. Fines are split by the A-line on the plasticity chart:

PI = 0.73(LL-20)

. Plotting on or above the A-line with

PI>7

makes a clay (CL/CH); below it makes a silt (ML/MH).

\text{A-line:}\quad PI = 0.73\,(LL - 20)

AASHTO and the group index

For pavement subgrades, AASHTO M145 sorts soils A-1 through A-7, with smaller numbers being better subgrade. The group index

GI

then rates the material — larger is poorer — and is reported in parentheses, e.g. A-6(10). It is a pure formula from the percent passing the No. 200 sieve

F

and the Atterberg limits; round to the nearest integer and report 0 if the formula returns a negative number. For the A-2-6 and A-2-7 subgroups, use only the second (PI) term.

GI = (F-35)\big[0.2 + 0.005(LL-40)\big] + 0.01\,(F-15)(PI-10)

Exam strategy

When a problem gives you masses and volumes, draw the block, set

V_s=1

V_v=e

, and fill in the rest — do not hunt for a single formula. Keep

w

and

S

as decimals inside

Se=wG_s

even though they are quoted as percents. For unit weight, decide first which phases count: below the water table you almost always want

\gamma_b

. For classification, run the No. 200 split first, then the second test (grading for coarse, A-line for fine); for AASHTO, compute

GI

, round to an integer, and never report it negative.

Key equations

Void ratio and porosity

e = \dfrac{V_v}{V_s}, \qquad n = \dfrac{e}{1+e}, \qquad e = \dfrac{n}{1-n}

$e$ = void ratio (used in consolidation), $n$ = porosity. Both dimensionless; voids = water + air.

Bridge identity

S\,e = w\,G_s

Ties saturation $S$ and void ratio $e$ to water content $w$ and specific gravity $G_s$ . Use $S$ and $w$ as decimals.

Dry unit weight

\gamma_d = \dfrac{G_s\,\gamma_w}{1+e}

Weight of solids per total volume. $\gamma_w = 62.4\,\text{pcf}$ ( $9.81\,\text{kN/m}^3$ ).

Total (moist) unit weight

\gamma_t = \gamma_d(1+w) = \dfrac{(G_s + S e)\gamma_w}{1+e}

What a field sample weighs at water content $w$ and saturation $S$ .

Saturated and buoyant unit weight

\gamma_{sat} = \dfrac{(G_s+e)\gamma_w}{1+e}, \qquad \gamma_b = \gamma_{sat} - \gamma_w

$\gamma_{sat}$ at $S=1$ ; $\gamma_b$ (submerged/effective) governs effective stress below the water table.

Grading coefficients

C_u = \dfrac{D_{60}}{D_{10}}, \qquad C_c = \dfrac{D_{30}^{\,2}}{D_{10}\,D_{60}}

$D_{xx}$ = grain size with $xx\%$ passing. Well-graded sand needs $C_u\ge6$ , gravel $C_u\ge4$ , both with $1\le C_c\le3$ .

Plasticity index and A-line

PI = LL - PL, \qquad PI_{A} = 0.73\,(LL-20)

On/above A-line with $PI>7\Rightarrow$ clay (C); below $\Rightarrow$ silt (M). $LL<50$ low (L), $\ge50$ high (H).

Liquidity index

LI = \dfrac{w - PL}{PI}

Locates in-situ water content between the plastic ( $LI=0$ ) and liquid ( $LI=1$ ) limits; $LI>1$ flags sensitive clay.

AASHTO group index

GI = (F-35)\big[0.2+0.005(LL-40)\big] + 0.01(F-15)(PI-10)

$F$ = % passing No. 200. Round to nearest integer; report 0 if negative. A-2-6/A-2-7 use only the last term.

Relative density

D_r = \dfrac{e_{max}-e}{e_{max}-e_{min}} \times 100\%

Density state of a coarse soil between its loosest ( $e_{max}$ ) and densest ( $e_{min}$ ) states.

Worked examples

Phase relationships from a moist unit weight

Problem. A moist soil sample has a total unit weight of $120\,\text{pcf}$ , a water content of $15\%$ , and a specific gravity of solids $G_s = 2.70$ . Find the dry unit weight, void ratio, porosity, and degree of saturation.

Solution. Dry unit weight:

\gamma_d = \gamma_t/(1+w) = 120/1.15 = 104.3\,\text{pcf}

. Void ratio from

\gamma_d = G_s\gamma_w/(1+e)

1+e = G_s\gamma_w/\gamma_d = (2.70)(62.4)/104.3 = 1.615

, so

e = 0.615

. Porosity:

n = e/(1+e) = 0.615/1.615 = 0.381

(i.e.

38.1\%

). Saturation from

Se = wG_s

S = wG_s/e = (0.15)(2.70)/0.615 = 0.659

, i.e.

\mathbf{65.9\%}

. Sanity check: at

S=1

this soil would weigh

\gamma_{sat} = (G_s+e)\gamma_w/(1+e) = (3.315)(62.4)/1.615 = 128.1\,\text{pcf} > 120\,\text{pcf}

— consistent with the sample being only partly saturated.

S = \dfrac{w G_s}{e} = \dfrac{(0.15)(2.70)}{0.615} = 0.659

Classify a fine-grained soil (USCS and AASHTO)

Problem. A soil has $65\%$ passing the No. 200 sieve, a liquid limit $LL = 42$ , and a plastic limit $PL = 24$ . Give the USCS group symbol and the AASHTO classification with its group index.

Solution. Plasticity index:

PI = LL-PL = 42-24 = 18

. USCS: more than

50\%

passes No. 200, so the soil is fine-grained. Since

LL = 42 < 50

, it is low plasticity. A-line value at

LL=42

PI_A = 0.73(42-20) = 16.1

. The point

PI = 18 > 16.1

plots above the A-line with

PI>7

, so it is a lean clay:

\mathbf{CL}

. AASHTO:

F = 65\% > 35\%

is a silt-clay material;

LL\ge41

and

PI\ge11

place it in A-7. Test

LL-30 = 12

; since

PI = 18 > 12

, it is the

\mathbf{A\text{-}7\text{-}6}

subgroup. Group index:

GI = (65-35)[0.2+0.005(42-40)] + 0.01(65-15)(18-10) = 30(0.21) + 0.01(50)(8) = 6.30 + 4.00 = 10.3 \Rightarrow \mathbf{A\text{-}7\text{-}6\,(10)}

. Sanity check: a high-plasticity-ish clayey subgrade should classify as a poor subgrade —

GI=10

is firmly in the poor range, consistent with CL clay.

GI = 30\,(0.21) + 0.01(50)(8) = 6.30 + 4.00 = 10.3 \to 10

Classify a coarse-grained soil (grading coefficients)

Problem. A sand has only $4\%$ passing the No. 200 sieve (and none retained on the No. 4), with grain sizes $D_{10} = 0.12\,\text{mm}$ , $D_{30} = 0.55\,\text{mm}$ , and $D_{60} = 1.8\,\text{mm}$ . Give the USCS group symbol.

Solution. No. 200 split: only

4\% < 50\%

passes, so the soil is coarse-grained; with nothing retained on the No. 4 it is a sand (first letter

S

). Fines content:

4\% < 5\%

fines, so the sand is clean and the second letter comes from grading (W or P), not plasticity. Uniformity coefficient:

C_u = D_{60}/D_{10} = 1.8/0.12 = 15.0

. Coefficient of curvature:

C_c = D_{30}^2/(D_{10}D_{60}) = (0.55)^2/[(0.12)(1.8)] = 0.3025/0.216 = 1.40

. Well-graded sand requires BOTH

C_u \ge 6

AND

1 \le C_c \le 3

: here

C_u = 15.0 \ge 6

and

C_c = 1.40

lies in

[1,3]

, so both pass. The soil is

\mathbf{SW}

(well-graded sand). Sanity check: a uniform (single-size) sand would have

C_u

near

1

–

2

and fail the

C_u \ge 6

test, classifying as SP — the large

C_u = 15

here confirms a broad, well-distributed gradation.

C_u = \dfrac{1.8}{0.12} = 15.0, \qquad C_c = \dfrac{(0.55)^2}{(0.12)(1.8)} = 1.40 \;\Rightarrow\; \textbf{SW}

Common pitfalls

•Reporting porosity when the formula calls for void ratio (or vice-versa). Consolidation uses $e$ ; $n = e/(1+e)$ and $e = n/(1-n)$ — never interchangeable.
•Plugging $w$ or $S$ as percents into $Se = wG_s$ . They must be decimals: $15\%\to0.15$ , $66\%\to0.66$ .
•Computing buoyant unit weight as $\gamma_{sat}/2$ . The correct relation is $\gamma_b = \gamma_{sat}-\gamma_w$ (subtract $62.4\,\text{pcf}$ ), which is roughly half but never exactly.
•Using $\gamma_t$ below the water table for effective stress. Below the table you need $\gamma_b$ (or total stress minus pore pressure) — using moist unit weight overestimates effective stress.
•Mixing up the USCS second letter: W/P apply to clean coarse soils (grading), L/H to fine soils (the $LL=50$ split). A clay is never SW.
•Forgetting that the AASHTO group index is reported as a non-negative integer — a negative computed value is reported as 0, and A-2-6/A-2-7 use only the $0.01(F-15)(PI-10)$ term.
•Treating well-graded as a single criterion: it requires BOTH the $C_u$ threshold AND $1\le C_c\le3$ ; failing either makes the soil poorly graded.

References

NCEES PE Civil Reference Handbook — §3.7 Soil Classification and Boring Log Interpretation
ASTM D2487 — Unified Soil Classification System (USCS)
AASHTO M145 — Classification of Soils and Soil-Aggregate Mixtures (group index)
FHWA-NHI-06-088 Soils and Foundations, Vol. I — Source of the volume-weight relationships table reproduced in the handbook.

Shear Strength & Effective Stress

Mohr-Coulomb strength with c and phi, total vs effective stress through pore pressure, and the drained/undrained distinction that decides which strength to use.

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Consolidation & Compressibility Parameters

Compression and recompression indices Cc/Cr, preconsolidation pressure, the NC vs OC distinction, and the primary-consolidation settlement equations.

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I. Earth pressure and surcharge

Lateral Earth Pressure (Rankine, At-Rest, Surcharge)

Active/passive Rankine coefficients, the at-rest K0, the triangular thrust and its location, and the lateral pressure a surcharge adds to a wall.

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