Soil properties, investigation and testing
Soil is made of solids, water and air. The empty gaps (water + air) are voids.
- Water content w = weight of water ÷ weight of dry soil × 100.
- Void ratio e = volume of voids ÷ volume of solids.
- Porosity n = volume of voids ÷ total volume × 100.
- Degree of saturation Sr = volume of water ÷ volume of voids. Sr = 100% means all voids hold water.
- Unit weight γ = weight ÷ volume (soil is about 16 to 20 kN/m³).
Investigation: engineers bore holes, take samples and note the layers. Tests: oven drying for water content, sieving for grain size, permeability test, consolidation test and shear test.
Water flow in soil
Water moves from a high water level to a low one through the voids. This is seepage.
- Hydraulic gradient i = head difference Δh ÷ length of path L.
- Darcy law: v = k × i. Here k is the permeability (how easily water passes).
- Flow per second q = k × i × A.
- k is about 10⁻² m/s for gravel, 10⁻⁴ for sand and 10⁻⁹ m/s or less for clay.
Fast seepage under a dam can carry soil grains away and weaken the dam, so engineers check it.
Stress in soil and consolidation
Stress is force per area (kPa). Soil weight gives total stress σ = γ × z at depth z.
- Water in the voids pushes back with pore water pressure u = γw × (depth below water table). γw ≈ 10 kN/m³.
- The grains carry only the effective stress: σ′ = σ − u. Effective stress decides strength and settlement.
- A load on top spreads out, so the added stress gets smaller with depth. A simple rule is the 2:1 spread: Δσ = Q ÷ ((B + z)(L + z)).
Consolidation: when a clay layer is loaded, the water is squeezed out slowly through tiny pores. Pore pressure falls, effective stress rises and the ground settles. In sand this takes seconds; in clay it can take years. A longer drainage path means a slower process.
Soil strength
Soil fails by sliding along a surface, not by crushing. The sliding resistance is the shear strength.
- Coulomb equation: τ = c + σ′ × tan φ.
- c = cohesion (sticking of clay particles; about zero for dry sand).
- φ = angle of internal friction (sand about 30° to 40°).
- More pressing weight σ′ gives more friction and so more strength.
A direct shear test pushes the top half of a soil box sideways under a fixed weight and measures the force at failure. Slopes, foundations and walls are all designed so that stresses stay below this strength.
Earth pressure
Soil behind a wall pushes it sideways. This is earth pressure.
- Active pressure (wall moves away a little): coefficient Ka = (1 − sin φ) ÷ (1 + sin φ). For φ = 30°, Ka = 1/3.
- Passive pressure (wall pushes into soil): Kp = 1 ÷ Ka. It is much bigger.
- At rest (wall cannot move): K₀ ≈ 1 − sin φ.
- Pressure at depth z = K × γ × z. It is zero at the top and biggest at the bottom: a triangle.
- Total push per metre of wall: P = ½ K γ H². It acts at H/3 above the base.
Because of H², doubling the height makes the push four times larger.
Try it
In the 3D: in the last step set the soil height to 2 m and note the push. Then set it to 4 m. Is the push 4 times bigger?
At home: pile dry sand into a cone on a tray. The steepest angle where it stays is the angle of repose, close to φ. Wet it a little and see if it can stand steeper (cohesion).
Key formulas and definitions
- w = Ww ÷ Ws × 100; e = Vv ÷ Vs; n = Vv ÷ V × 100; Sr = Vw ÷ Vv
- Darcy: v = k × i, i = Δh ÷ L, q = k × i × A
- σ = γ × z; u = γw × zw; σ′ = σ − u
- Shear strength: τ = c + σ′ tan φ
- Ka = (1 − sin φ) ÷ (1 + sin φ); Kp = 1 ÷ Ka
- Active push: P = ½ Ka γ H² (acts at H/3 from the base)
- Key terms: voids, permeability, effective stress, consolidation, cohesion, friction angle
Worked examples
1. A wet soil sample weighs 200 g. After oven drying it weighs 160 g. Find the water content.
Water = 40 g. w = 40 ÷ 160 × 100 = 25%.
2. A soil has solids volume 0.6 m³ and voids volume 0.4 m³. Find void ratio and porosity.
e = 0.4 ÷ 0.6 = 0.67. Total volume = 1.0 m³, so n = 0.4 ÷ 1.0 × 100 = 40%.
3. Sand with k = 1 × 10⁻⁴ m/s. Water head difference 0.5 m over a 2 m path. Find the seepage velocity v.
i = 0.5 ÷ 2 = 0.25. v = k × i = 1 × 10⁻⁴ × 0.25 = 2.5 × 10⁻⁵ m/s.
4. Soil weight γ = 18 kN/m³. The water table is at the ground surface. Find the effective stress at 4 m depth.
σ = 18 × 4 = 72 kPa. u = 10 × 4 = 40 kPa. σ′ = 72 − 40 = 32 kPa.
5. Clay has c = 10 kPa and φ = 30°. Find its shear strength under σ′ = 50 kPa. (tan 30° = 0.577)
τ = 10 + 50 × 0.577 = 10 + 28.9 = 38.9 kPa.
6. A wall 4 m high holds soil with γ = 18 kN/m³ and φ = 30°. Find the active push per metre and where it acts.
Ka = 1/3. P = ½ × (1/3) × 18 × 4² = ½ × 6 × 16 = 48 kN/m. It acts at 4 ÷ 3 = 1.33 m above the base.
Common mistakes
- Using total stress for strength. Strength depends on effective stress σ′ = σ − u.
- Thinking clay with many pores drains fast. Its pores are tiny, so k is very small.
- Forgetting that earth pressure grows with H², not H. Doubling height gives four times the push.
- Placing the push at mid-height. For a triangle it acts at H/3 from the base.