Phase diagram and weight–volume relations
A lump of soil has three parts: solids (the grains), water and air. We draw them stacked in one block, the phase diagram: volumes on the left, weights on the right. Air has volume but (almost) no weight.
Void ratio, porosity, water content, saturation
The key link
Gs = specific gravity of the solids (about 2.60 to 2.80). Fully saturated: S = 1, so e = wGs.
Unit weights (γw = 9.81 kN/m³)
γ′ is the buoyant (submerged) unit weight, used below the water table.
Relative density of sand
emax goes with γd,min (loosest), emin with γd,max (densest).
| Dr (%) | 0–15 | 15–35 | 35–65 | 65–85 | 85–100 |
|---|---|---|---|---|---|
| Sand is | very loose | loose | medium dense | dense | very dense |
Atterberg limits
Add water to a fine soil and it changes from solid to semisolid to plastic (you can mould it) to liquid. The water contents at these boundaries are the shrinkage limit SL, the plastic limit PL and the liquid limit LL.
Plasticity, liquidity and consistency
LI < 0: stiff (below PL). LI from 0 to 1: plastic. LI > 1: the natural soil is wetter than its LL. LI + CI = 1.
Activity (Skempton)
A < 0.75 inactive, 0.75 to 1.25 normal, more than 1.25 active (swells a lot).
Liquid limit from the flow curve
Casagrande cup: plot water content w against the number of blows N on a log scale. It is a straight line. LL is the water content at 25 blows.
IF = flow index (slope), IT = toughness index.
One-point method
Use one trial with N between 20 and 30 blows.
Shrinkage limit
wi = starting water content (%), Vi, Vf = starting and final (dry) volume, ms = dry mass.
Grain size distribution
A sieve analysis gives the percent of soil passing each sieve. We plot it against grain size on a log scale. D10 is the size that 10% of the soil is finer than (the "effective size"); D30 and D60 the same way.
Sieves used in classification
No. 4 = 4.75 mm (gravel / sand)
No. 10 = 2.00 mm
No. 40 = 0.425 mm
No. 200 = 0.075 mm (sand / fines)
Fines (silt and clay) are the part passing No. 200. A hydrometer test is used for sizes below that.
Uniformity and curvature
Well graded (USCS)
Well graded = a good mix of all sizes. If either rule fails, the soil is poorly graded.
Unified Soil Classification System (USCS)
The group symbol has two letters. First letter: G gravel, S sand, M silt, C clay, O organic. Second letter: W well graded, P poorly graded, L low plasticity, H high plasticity.
Step 1: coarse or fine?
50% or more passing No. 200 → fine-grained (M or C). Otherwise → coarse-grained (G or S).
Coarse: if more than half of the coarse part is retained on No. 4, it is gravel; if not, sand.
Step 2: coarse soils by the fines F
F < 5%: GW, GP, SW, SP (use Cu, Cc).
F > 12%: GM, GC, SM, SC (use the plasticity chart; in the hatched zone, GC-GM or SC-SM).
F = 5 to 12%: dual symbol, e.g. GW-GM, GP-GC, SW-SM, SP-SC.
Step 3: fine soils (plasticity chart)
LL < 50: CL (PI > 7, on or above the A-line), CL-ML (PI 4 to 7, on or above), ML (below, or PI < 4).
LL ≥ 50: CH (on or above the A-line), MH (below).
AASHTO classification
Used for roads. Go through the table from left to right; the first group whose limits the soil meets is the answer. Granular soils have 35% or less passing No. 200; silt-clay soils have more than 35%.
| Group | % passing No. 10 | No. 40 | No. 200 | LL | PI |
|---|---|---|---|---|---|
| A-1-a | ≤ 50 | ≤ 30 | ≤ 15 | — | ≤ 6 |
| A-1-b | — | ≤ 50 | ≤ 25 | — | ≤ 6 |
| A-3 | — | ≥ 51 | ≤ 10 | — | NP |
| A-2-4 | — | — | ≤ 35 | ≤ 40 | ≤ 10 |
| A-2-5 | — | — | ≤ 35 | ≥ 41 | ≤ 10 |
| A-2-6 | — | — | ≤ 35 | ≤ 40 | ≥ 11 |
| A-2-7 | — | — | ≤ 35 | ≥ 41 | ≥ 11 |
| A-4 | — | — | ≥ 36 | ≤ 40 | ≤ 10 |
| A-5 | — | — | ≥ 36 | ≥ 41 | ≤ 10 |
| A-6 | — | — | ≥ 36 | ≤ 40 | ≥ 11 |
| A-7-5 / A-7-6 | — | — | ≥ 36 | ≥ 41 | ≥ 11 |
A-7-5 or A-7-6?
Group index (F = % passing No. 200)
Round to a whole number. A negative GI is taken as 0. A-1-a, A-1-b, A-3, A-2-4 and A-2-5: GI = 0. A-2-6 and A-2-7: use only the second part,
Write the answer as group(GI), e.g. A-7-6(14). A larger GI means a poorer subgrade.
Compaction and field density
Compaction squeezes out air to make the soil denser. In the Proctor test, the soil is compacted in a mold at several water contents. Dry unit weight rises, peaks, then drops. The peak gives the optimum moisture content (OMC) and the maximum dry unit weight.
| Test | Hammer | Drop | Layers × blows | Mold | Energy |
|---|---|---|---|---|---|
| Standard Proctor | 24.4 N (2.5 kg) | 305 mm | 3 × 25 | 943.3 cm³ | about 600 kN·m/m³ |
| Modified Proctor | 44.5 N (4.54 kg) | 457 mm | 5 × 25 | 943.3 cm³ | about 2700 kN·m/m³ |
From one trial
Zero-air-void line and S lines
The ZAV line (S = 100%) is the upper limit: no test point can be above it.
Relative compaction
Specifications usually ask for R of at least 90% to 95%.
Sand-cone test
Mass of sand in the hole = sand used − sand that fills the cone. If the unit weight of the sand γsand (kN/m³) is given instead, Vhole = Wsand/γsand.
Borrow pit
The weight of solids does not change when soil is dug, hauled and compacted:
Truck loads = Vborrow ÷ truck volume, rounded up.
Soil lab
Choose a mode. Phase relations solves every property from three known values. Classification gives the USCS symbol and the AASHTO group. Compaction finds the OMC and maximum dry unit weight, checks a field test and sizes a borrow pit.
Solution
Checks and notes
Worked examples
A moist soil has a unit weight of 18.5 kN/m³ and a water content of 15%. Gs = 2.70. Find the dry unit weight, void ratio, porosity, degree of saturation and saturated unit weight.
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A sand in the field has a dry unit weight of 16.0 kN/m³ and a water content of 12%. Gs = 2.65. In the laboratory, emax = 0.85 and emin = 0.50. Find the relative density and describe the sand.
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Check with dry unit weights: γd,min = 2.65(9.81)/1.85 = 14.052 and γd,max = 2.65(9.81)/1.50 = 17.331 kN/m³.
Dr = 64.35%: medium dense (35 to 65%)Liquid limit test results: N = 35, 28, 21, 16 blows at w = 38.2%, 40.1%, 42.5%, 44.6%. PL = 22.4%. The natural water content is 36% and 32% of the soil is clay finer than 2 µm. Find LL, PI, LI, CI, the flow index and the activity. Also check LL with the one-point method using the 28-blow trial.
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Fit a straight line w = a + b log N through the four points (calculator: linear regression with x = log N):
LL = 41.00%, PI = 18.60, LI = 0.731, CI = 0.269, I_F = 18.86, A = 0.58; one-point LL = 40.65%Sieve analysis: 72% passes No. 4, 60% passes No. 10, 25% passes No. 40 and 4% passes No. 200. D10 = 0.15 mm, D30 = 0.60 mm, D60 = 2.0 mm. The fines are non-plastic. Classify by USCS and AASHTO.
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USCS: 4% passes No. 200 (less than 50%) → coarse. Gravel = 100 − 72 = 28%, sand = 72 − 4 = 68% → sand. Fines 4% (less than 5%) → SW or SP.
AASHTO: A-1-a needs No. 10 ≤ 50, but 60 > 50 → no. A-1-b: No. 40 = 25 ≤ 50, No. 200 = 4 ≤ 25, PI = 0 ≤ 6 → yes. GI = 0 for A-1-b.
USCS: SW (well-graded sand) · AASHTO: A-1-b(0)A soil has 100% passing No. 4, 98% passing No. 10, 90% passing No. 40 and 68% passing No. 200. LL = 45% and PL = 22%. Classify by USCS and AASHTO with the group index.
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USCS: 68% passes No. 200 → fine-grained. LL = 45 < 50 and PI = 23 is above the A-line (and more than 7) → CL.
AASHTO: 68% > 35% → silt-clay group. LL = 45 ≥ 41 and PI = 23 ≥ 11 → A-7. LL − 30 = 15 and PI = 23 > 15 → A-7-6.
USCS: CL (lean clay) · AASHTO: A-7-6(14)Percent passing: No. 4 = 85, No. 10 = 70, No. 40 = 45, No. 200 = 30. LL = 36% and PL = 18%. Classify by USCS and AASHTO.
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USCS: 30% passes No. 200 → coarse. Gravel = 15%, sand = 85 − 30 = 55% → sand. Fines 30% (more than 12%) → SM or SC.
The fines plot above the A-line with PI > 7 → clay fines → SC.
AASHTO: A-1-a fails (No. 10 = 70 > 50). A-1-b fails (No. 200 = 30 > 25). A-3 fails (not NP). A-2-4 and A-2-5 fail (PI > 10). A-2-6: No. 200 ≤ 35, LL ≤ 40, PI ≥ 11 → yes. Use the partial GI:
USCS: SC (clayey sand) · AASHTO: A-2-6(1)A standard Proctor test (mold 943.3 cm³, Gs = 2.70) gave: w = 10, 12, 14, 16, 18% with moist soil masses 1.772, 1.879, 1.957, 1.965, 1.923 kg. Find the OMC and γd,max. A sand-cone test on the compacted fill: 1.795 kg of moist soil came out of the hole, the hole took 1.420 kg of sand (density 1500 kg/m³), and the field water content is 13%. Is 95% relative compaction met?
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| w (%) | 10 | 12 | 14 | 16 | 18 |
|---|---|---|---|---|---|
| γ (kN/m³) | 18.428 | 19.541 | 20.352 | 20.435 | 19.999 |
| γd (kN/m³) | 16.753 | 17.447 | 17.853 | 17.617 | 16.948 |
The highest point is at w = 14%. Fit a parabola through it and its two neighbours (12%, 14%, 16%) and take its top:
Sand cone:
OMC = 14.26%, γd,max = 17.858 kN/m³; R = 92.18%, so the fill FAILS and must be compacted moreUsing the Proctor results of Example 7, a 20,000 m³ embankment is to be compacted to 95% of γd,max. The borrow soil has a moist unit weight of 17.2 kN/m³ at 12% water content. Find the volume of borrow soil needed, the number of 12 m³ truck loads (measured in the borrow-pit state) and the water to add to bring the soil to the OMC.
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