Soil Mechanics and Foundation Engineering · 12 min read

Soil properties and laboratory tests

Soil is a three-phase material of solid grains, water and air. Before designing a foundation, embankment or slope, the engineer must quantify how much of each phase is present, identify the soil type and measure its strength, permeability and compressibility. These index and engineering properties come from standard laboratory tests (IS 2720 series, widely used in Nepal). This note covers phase relationships, classification, sieve analysis, Atterberg limits, the main tests and how to read a boring log.

Key concepts

Phase relationships

TermDefinition
Void ratio eeVv/VsV_v/V_s
Porosity nnVv/VV_v/V; n=e/(1+e)n = e/(1+e)
Degree of saturation SSVw/VvV_w/V_v (0 dry, 1 saturated)
Water content wwWw/WsW_w/W_s
Specific gravity GGUnit weight of solids ÷ unit weight of water (2.6–2.8 for most soils)
Air content aca_cVa/VvV_a/V_v; na=n acn_a = n\,a_c
Density index IDI_Demax−eemax−emin\dfrac{e_{max}-e}{e_{max}-e_{min}} (relative density, for sands)

Index property tests

  • Water content: oven drying at 105–110 °C for 24 h (standard); also sand bath, calcium carbide (rapid moisture meter), pycnometer.
  • Specific gravity: density bottle (fine soils) or pycnometer (coarse).
  • Field density: core cutter (soft cohesive), sand replacement (gravelly/hard soils).
  • Grain size: sieve analysis (> 75 µm) and hydrometer/pipette (sedimentation using Stokes' law, < 75 µm).

Sieve analysis and interpretation

  • Plot % finer (y-axis) vs particle size on log scale (x-axis) → grain size distribution curve.
  • D10D_{10} = effective size; coefficient of uniformity Cu=D60/D10C_u = D_{60}/D_{10}; coefficient of curvature Cc=D302D10D60C_c = \dfrac{D_{30}^2}{D_{10}D_{60}}.
  • Well-graded: gravel Cu>4C_u > 4, sand Cu>6C_u > 6, and 1≤Cc≤31 \le C_c \le 3. Otherwise poorly graded (uniform or gap-graded).
  • Steep curve → uniform soil; flat S-curve → well graded; horizontal step → gap graded.

Atterberg (consistency) limits

  • Liquid limit (LL, wLw_L): water content at which soil just begins to flow; Casagrande apparatus (groove closes 13 mm at 25 blows) or cone penetrometer (20 mm penetration).
  • Plastic limit (PL, wPw_P): water content at which a soil thread just crumbles at 3 mm diameter.
  • Shrinkage limit (SL, wSw_S): water content below which further drying causes no volume decrease.
  • Plasticity index IP=wL−wPI_P = w_L - w_P; liquidity index IL=w−wPIPI_L = \dfrac{w - w_P}{I_P}; consistency index IC=wL−wIPI_C = \dfrac{w_L - w}{I_P} (IL+IC=1I_L + I_C = 1); flow index, toughness index IT=IP/IFI_T = I_P/I_F; activity A=IP/(% clay<2μm)A = I_P/(\%\text{ clay} < 2\mu m).

Soil classification

SystemBasis
DescriptiveField description: colour, texture, gravel/sand/silt/clay, consistency
Textural (triangular chart, US Bureau of Public Roads)% sand, silt, clay plotted on triangle
MITGravel > 2 mm, sand 2–0.06 mm, silt 0.06–0.002 mm, clay < 0.002 mm
ISI (IS 1498)Gravel 80–4.75 mm, sand 4.75–0.075 mm, silt 0.075–0.002 mm, clay < 0.002 mm; groups like GW, SP, CL, MH, CI (intermediate)
USCSCoarse-grained if > 50% retained on 75 µm (No. 200) sieve; fine-grained classified on plasticity chart using A-line IP=0.73(wL−20)I_P = 0.73(w_L - 20)

Symbols: G gravel, S sand, M silt, C clay, O organic, Pt peat; W well graded, P poorly graded, L low plasticity (wLw_L < 50), H high plasticity (wLw_L > 50). IS adds I (intermediate, 35–50).

Engineering property tests

  • Strength: direct shear, triaxial (UU, CU, CD), unconfined compression (UCS, cu=qu/2c_u = q_u/2), vane shear (soft clays), SPT/CPT in field.
  • Permeability: constant-head test (coarse soils, k=QLAhtk = \dfrac{QL}{Aht}) and falling-head test (fine soils, k=2.303aLAtlog⁡10h1h2k = 2.303\dfrac{aL}{At}\log_{10}\dfrac{h_1}{h_2}). Typical kk: gravel > 10⁻² m/s, clay < 10⁻⁹ m/s.
  • Compressibility: one-dimensional oedometer (consolidation) test giving CcC_c, CvC_v, mvm_v.
  • Compaction: Proctor (standard/modified) giving OMC and MDD.

Boring log interpretation

A borehole log shows depth, soil description and layers, groundwater table, sample types (disturbed/undisturbed, SPT split spoon, Shelby tube), SPT N-values with depth, core recovery and RQD in rock, and lab results. Engineers use it to identify weak layers, the bearing stratum, water level and suitable foundation depth.

High-yield points

  • Se=wGSe = wG; γd=γ/(1+w)\gamma_d = \gamma/(1+w); n=e/(1+e)n = e/(1+e).
  • Liquid limit at 25 blows (Casagrande) or 20 mm cone penetration; plastic limit thread 3 mm.
  • Well graded: Cu>4C_u > 4 (gravel) or >6> 6 (sand) and 1≤Cc≤31 \le C_c \le 3.
  • IS 1498: sand 4.75–0.075 mm, clay < 0.002 mm; plasticity L < 35, I 35–50, H > 50.
  • Constant-head test for coarse soils, falling-head for fine soils.
  • A-line: IP=0.73(wL−20)I_P = 0.73(w_L - 20).

Formulas

Se=wG,γ=(G+Se)γw1+e,γd=Gγw1+e=γ1+wSe = wG,\qquad \gamma = \frac{(G + Se)\gamma_w}{1+e},\qquad \gamma_d = \frac{G\gamma_w}{1+e} = \frac{\gamma}{1+w} Submerged unit weight: γ′=γsat−γw=(G−1)γw1+e\gamma' = \gamma_{sat} - \gamma_w = \dfrac{(G-1)\gamma_w}{1+e}; Cu=D60/D10C_u = D_{60}/D_{10}; IP=wL−wPI_P = w_L - w_P.

Worked example

A soil has γ=19\gamma = 19 kN/m³, w=15%w = 15\%, G=2.70G = 2.70. Find γd\gamma_d, ee and SS (γw=9.81\gamma_w = 9.81 kN/m³).

  • γd=19/1.15=16.52\gamma_d = 19/1.15 = 16.52 kN/m³
  • e=Gγw/γd−1=2.70×9.81/16.52−1=26.49/16.52−1=0.603e = G\gamma_w/\gamma_d - 1 = 2.70\times9.81/16.52 - 1 = 26.49/16.52 - 1 = 0.603
  • S=wG/e=0.15×2.70/0.603=0.672S = wG/e = 0.15\times2.70/0.603 = 0.672 → 67.2%

Exam tips

  • Se=wGSe = wG is the most-used relation; for saturated soil e=wGe = wG.
  • Liquid limit at 25 blows; plastic limit thread 3 mm.
  • Well-graded sand: Cu>6C_u > 6 and 1≤Cc≤31 \le C_c \le 3.
  • Constant-head test for sands, falling-head for silts/clays.
  • A-line: IP=0.73(wL−20)I_P = 0.73(w_L - 20); above A-line = clay, below = silt/organic.

Exam-style questions

From Calmon's question bank, which has 25 questions on this topic.

1. A clay has a natural water content of 30%, liquid limit of 40% and plastic limit of 20%. Its liquidity index is

  • A. 0.50
  • B. 0.25
  • C. 0.75
  • D. 1.00

Answer: A. IL=w−wPwL−wP=30−2040−20=0.5I_L = \dfrac{w - w_P}{w_L - w_P} = \dfrac{30 - 20}{40 - 20} = 0.5.

2. In a constant head permeability test, a sample 15 cm long with cross-sectional area 50 cm² discharges 300 cm³ of water in 5 minutes under a constant head of 30 cm. The coefficient of permeability is

  • A. 0.050.05 cm/s
  • B. 1×10−31 \times 10^{-3} cm/s
  • C. 0.10.1 cm/s
  • D. 0.010.01 cm/s

Answer: D. k=QLAht=300×1550×30×300=0.01k = \dfrac{QL}{Aht} = \dfrac{300 \times 15}{50 \times 30 \times 300} = 0.01 cm/s.

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