Soil Properties & Classification · Study · PE Civil: Structural · FE → PE Prep
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
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 Vs=1
(or
Vv=e
), which turns ratios into direct quantities.
V=Vs+Vw+Va,Vv=Vw+Va,W=Ws+Ww(Wa≈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 Se=wGs — it ties the moisture state to the void state and shows up constantly.
Specific gravity Gs (typically 2.65 for quartz sands, 2.70-2.75 for clays) converts solid volume to solid weight via γw=62.4pcf. 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 γb=γsat−γw, not γsat/2.
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 Cu=D60/D10 and the coefficient of curvature Cc=D302/(D10D60). 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.
Cu=D10D60,Cc=D10D60D302,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≥50). A clean sand is well-graded (SW) when Cu≥6 and 1≤Cc≤3. 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).
A-line: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)[0.2+0.005(LL−40)]+0.01(F−15)(PI−10)
Exam strategy
When a problem gives you masses and volumes, draw the block, set Vs=1 or Vv=e, and fill in the rest — do not hunt for a single formula. Keep w and S as decimals inside Se=wGs even though they are quoted as percents. For unit weight, decide first which phases count: below the water table you almost always want γb. 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 porositye=VsVv,n=1+ee,e=1−nn
e = void ratio (used in consolidation), n = porosity. Both dimensionless; voids = water + air.
Bridge identitySe=wGs
Ties saturation S and void ratio e to water content w
Dry unit weightγd=1+eGsγw
Total (moist) unit weightγt=γd(1+w)=1+e(Gs+Se)γw
Saturated and buoyant unit weightγsat=1+e(Gs+e)γw,γb=γsat−γw
Plasticity index and A-linePI=LL−PL,PIA=0.73(LL−20)
Liquidity indexLI=PIw−PL
Locates in-situ water content between the plastic (LI=0
AASHTO group indexGI=(F−35)[0.2+0.005(LL−40)]+0.01(F−15)(PI−10)
Relative densityDr=emax−eminemax−e×100%
Worked examples
Phase relationships from a moist unit weight
Problem. A moist soil sample has a total unit weight of 120pcf, a water content of 15%, and a specific gravity of solids Gs=2.70. Find the dry unit weight, void ratio, porosity, and degree of saturation.
Solution. Dry unit weight: γd=γt/(1+w)=120/1.15=104.3pcf
S=ewGs=0.615(0.15)(2.70)=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
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 D10=0.12mm, D30=0.55mm
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=wGs. They must be decimals: 15%→0.15
•Computing buoyant unit weight as γsat/2. The correct relation is γb=γsat−γw
•Using γt below the water table for effective stress. Below the table you need γb (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 Cu threshold AND 1≤Cc≤3; 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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.
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→10
, and
D60=1.8mm
. 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: Cu=D60/D10=1.8/0.12=15.0.
Coefficient of curvature: Cc=D302/(D10D60)=(0.55)2/[(0.12)(1.8)]=0.3025/0.216=1.40.
Well-graded sand requires BOTH Cu≥6 AND 1≤Cc≤3: here Cu=15.0≥6 and Cc=1.40 lies in [1,3], so both pass. The soil is SW (well-graded sand).
Sanity check: a uniform (single-size) sand would have Cu near 1–2 and fail the Cu≥6 test, classifying as SP — the large Cu=15 here confirms a broad, well-distributed gradation.