Soil Mechanics is the study of the engineering behaviour of soil — from site investigation and index properties, through classification systems, permeability and seepage, consolidation settlement, shear strength, lateral earth pressure theories and stress distribution, to modern geosynthetics. Every formula, IS code reference, diagram, solved example and exam-pattern table is included.
After studying this chapter you will be able to:
Prerequisite: RCC & Prestressed Concrete (footing/foundation design relies directly on the soil bearing capacity and settlement computed here). Leads to: Foundation Engineering, which applies these soil properties to bearing capacity, pile and deep foundation design.
Soil exploration (site investigation) is the process of determining the sub-surface profile, stratification, engineering properties of soil and rock, and groundwater conditions at a site, so that a safe and economical foundation can be designed. It also identifies potential construction problems (soft clays, expansive soils, fill, cavities) before construction begins.
Planning steps:
Depth and spacing of boreholes depend on the type of structure, subsoil variability, and the zone of influence of the foundation loads:
| Structure Type | Typical Boring Spacing | Typical Depth |
|---|---|---|
| Multi-storey building | 15 – 30 m grid | 1.5 × width of building or up to hard stratum |
| Highway / Runway | 250 – 500 m along alignment | 2 – 3 m below formation level |
| Bridge (per pier) | One boring per pier/abutment | Up to hard stratum or 1.5 × pier width below founding level |
| Earth dam | 30 – 60 m grid | Up to impervious stratum |
| Isolated footing (light) | One boring may suffice for small area | 1.5 × width of footing |
| Method | Suitable Soil | Max. Depth | Remarks |
|---|---|---|---|
| Trial Pits / Test Pits | All soils above water table | 3 – 4 m | Allows direct visual inspection and undisturbed block sampling; cheap but shallow |
| Auger Boring | Soft to stiff clay, silt, loose sand above water table | 3 – 8 m | Hand or power auger; disturbed samples only |
| Wash Boring | Soft to medium soils, no boulders | Up to 30 m | Water jet loosens soil, cuttings washed out; cheap, but disturbs strata for sampling |
| Rotary Drilling | Rock and hard soil, all depths | > 30 m | Rotating bit with drilling mud/water circulation; used for core recovery in rock |
| Percussion Drilling | Boulders, dense gravel, rock | Any depth | Repeated blows of a heavy chisel bit break the material; slow, disturbs soil |
| Shell & Auger Boring | All soil types including gravel | Up to 40–50 m | Common in Indian practice; shell (bailer) used in cohesionless soil below water table |
Samples are classified as disturbed (structure altered but soil composition/moisture roughly preserved — used for classification, index tests) and undisturbed (structure, density, moisture and stress history preserved as far as practicable — used for shear strength, consolidation and permeability tests).
| Sampler | Type | Use |
|---|---|---|
| Split-Spoon Sampler | Disturbed (thick-walled) | Used with SPT; recovers disturbed sample for classification & index tests |
| Thin-Walled (Shelby) Tube | Undisturbed | Pushed hydraulically into soft-to-stiff clay; minimal disturbance |
| Piston Sampler | Undisturbed | Very soft, sensitive clays; internal piston prevents soil flowing in ahead of the tube |
The quality of an undisturbed sample from a thin-walled tube is governed by:
Standard Penetration Test (SPT): A split-spoon sampler is driven 450 mm into the soil at the bottom of a borehole by a 63.5 kg hammer falling freely through 750 mm. The number of blows for the last 300 mm penetration is the SPT N-value (IS 2131:1981).
| N-Value | Relative Density / Consistency |
|---|---|
| 0 – 4 | Very loose sand / Very soft clay |
| 4 – 10 | Loose sand / Soft clay |
| 10 – 30 | Medium dense sand / Medium clay |
| 30 – 50 | Dense sand / Stiff clay |
| > 50 | Very dense sand / Hard clay |
Static/Static Cone Penetration Test (CPT): A cone is pushed at a steady rate (20 mm/s) and the cone resistance q_c and, in a piezocone, the friction ratio R_f are measured continuously with depth — gives a continuous profile rather than discrete points, useful in soft clays and loose sands where SPT sampling is unreliable.
Vane Shear Test: A four-bladed vane is pushed into soft clay and rotated; the torque at failure gives the in-situ undrained shear strength — ideal for very soft, sensitive clays where sampling disturbance is severe.
Plate Load Test (IS 1888, IS 8009): A rigid steel plate (300–750 mm) is loaded incrementally in a test pit and settlement recorded, giving the modulus of subgrade reaction and allowing extrapolation of settlement for the actual footing size — subject to a scale effect since the plate's zone of influence is much smaller than a full footing.
Pressuremeter & Dilatometer Tests: A cylindrical probe is expanded radially against the borehole wall (pressuremeter) or a flat blade is pushed in and a membrane expanded (dilatometer); both give the in-situ stress-strain response and hence modulus and strength parameters with minimal sample disturbance.
| Method | Principle | Application | Limitation |
|---|---|---|---|
| Seismic Refraction | Travel time of refracted shock waves at layer interfaces | Depth to bedrock, layer thickness, dynamic modulus | Requires velocity to increase with depth (fails for inverted layers) |
| Electrical Resistivity | Resistivity of soil/rock to an induced current | Groundwater table, soil/rock boundaries, contamination plumes | Non-unique interpretation; affected by salinity |
| Ground Penetrating Radar (GPR) | Reflection of electromagnetic pulses at material boundaries | Shallow utilities, voids, pavement layers | Poor penetration in clayey/saline soils |
| Cross-hole / Down-hole Seismic | Direct measurement of shear/compression wave velocity between boreholes | Dynamic soil properties (G_max) for earthquake engineering | Needs two or more boreholes; costly |
\( \text{Area Ratio, } A_R = \dfrac{D_o^2 - D_i^2}{D_i^2} \times 100 \)
Good quality sampler: \( A_R \le 10\text{–}15\% \)
\( \text{Inside Clearance Ratio, ICR} = \dfrac{D_s - D_i}{D_i} \times 100 \)
\( \text{Outside Clearance Ratio, OCR} = \dfrac{D_c - D_o}{D_o} \times 100 \)
\(D_o\) = outer dia of cutting edge, \(D_i\) = inner dia of cutting edge, \(D_s\) = inner dia of sample tube, \(D_c\) = outer dia of sample tube.
Overburden correction (Peck, Hanson & Thornburn): \( N_1 = C_N \cdot N_{60}, \quad C_N = 0.77\log_{10}\!\left(\dfrac{2000}{\sigma_v'}\right) \)
Energy correction to 60%: \( N_{60} = \dfrac{N \cdot E_m \cdot C_B \cdot C_S \cdot C_R}{0.60} \)
Dilatancy correction (Terzaghi, for saturated fine sand/silt when \(N > 15\)):
\( N_{corrected} = 15 + \dfrac{1}{2}(N - 15) \)
\( \text{Friction Ratio, } R_f = \dfrac{f_s}{q_c} \times 100\% \)
Correlations: \( \phi' \approx 0.1\,q_c(\text{MPa}) + 30° \) (clean sand, approximate)
\( c_u \approx \dfrac{q_c - \sigma_{v0}}{N_k}, \quad N_k \approx 15\text{–}20 \)
\( c_u = \dfrac{T}{\pi D^2\left(\dfrac{H}{2} + \dfrac{D}{6}\right)} \)
\(T\) = torque at failure, \(D\) = vane diameter, \(H\) = vane height (usually \(H = 2D\)).
Modulus of subgrade reaction: \( k_s = \dfrac{p}{\delta} \)
Settlement scale effect for footing width \(B_f\) from plate width \(B_p\) (IS 8009):
Clay: \( S_f = S_p \cdot \dfrac{B_f}{B_p} \) Sand: \( S_f = S_p\left(\dfrac{2B_f}{B_f + B_p}\right)^2 \)
Given: Field SPT value \(N = 24\), effective overburden pressure \(\sigma_v' = 120\ \text{kPa}\). Find the corrected N-value.
Solution:
\( C_N = 0.77 \log_{10}\!\left(\dfrac{2000}{120}\right) = 0.77 \times 1.222 = 0.94 \)
\( N_1 = 0.94 \times 24 \approx 22.6 \approx 23 \)
Answer: Corrected SPT value \( N_1 \approx 23 \).
Given: A vane of diameter 50 mm and height 100 mm is rotated in soft clay and fails at a torque of 35 N·m. Find the undrained shear strength.
Solution:
\( D = 0.05\ \text{m}, \ H = 0.10\ \text{m}, \ T = 35\ \text{N·m} \)
\( c_u = \dfrac{35}{\pi (0.05)^2\left(\dfrac{0.10}{2} + \dfrac{0.05}{6}\right)} = \dfrac{35}{\pi \times 0.0025 \times 0.0583} \)
\( c_u = \dfrac{35}{4.58\times10^{-4}} \approx 76.4\ \text{kPa} \)
Answer: \( c_u \approx 76.4\ \text{kPa} \) (soft to medium clay).
Given: A Shelby tube has cutting-edge outer diameter 76 mm and inner diameter 73 mm. Check whether it qualifies as a good-quality undisturbed sampler.
Solution:
\( A_R = \dfrac{76^2 - 73^2}{73^2}\times100 = \dfrac{5776 - 5329}{5329}\times100 = \dfrac{447}{5329}\times100 \approx 8.4\% \)
Answer: \( A_R \approx 8.4\% < 10\text{–}15\% \) — acceptable good-quality undisturbed sampler.
Fig. 1.1 — Standard Penetration Test (SPT) set-up: a 63.5 kg hammer falls freely 750 mm to drive a split-spoon sampler; N = blow count for the last 300 mm of a 450 mm drive.
Natural soil is a three-phase system consisting of solid particles, water and air. The volumetric and gravimetric (weight) relationships between these three phases are the foundation of nearly all quantitative soil mechanics — they connect index properties measured in the laboratory to the in-situ state of the soil.
Void ratio, porosity, degree of saturation and air content describe how the total volume is divided between solids, water and air.
Water content, bulk/dry/saturated/submerged unit weights, and specific gravity of solids describe the mass side of the three-phase system; \( e \cdot S = w \cdot G_s \) links the volumetric and gravimetric relationships and is one of the most-used identities in the subject.
| Soil Type | \(G_s\) | Void Ratio \(e\) | Dry Unit Weight \(\gamma_d\) (kN/m³) | Porosity \(n\) |
|---|---|---|---|---|
| Gravel | 2.65 – 2.68 | 0.30 – 0.75 | 15 – 20 | 0.25 – 0.43 |
| Sand (loose) | 2.65 – 2.68 | 0.65 – 0.90 | 13.5 – 16 | 0.40 – 0.47 |
| Sand (dense) | 2.65 – 2.68 | 0.35 – 0.55 | 17 – 19.5 | 0.26 – 0.35 |
| Silt | 2.66 – 2.70 | 0.40 – 1.10 | 13 – 19 | 0.29 – 0.52 |
| Soft clay | 2.68 – 2.75 | 0.90 – 2.50 | 8 – 14 | 0.47 – 0.71 |
| Peat / organic soil | 1.30 – 1.90 | 3 – 12 | 1 – 6 | 0.75 – 0.92 |
Defined by Albert Atterberg and standardised by Arthur Casagrande, these limits (IS 2720 Part 5) mark the moisture-content boundaries between the liquid, plastic, semi-solid and solid states of a fine-grained soil: Liquid Limit (LL), Plastic Limit (PL) and Shrinkage Limit (SL). The Plasticity Index PI = LL − PL is the range of water content over which the soil is plastic.
Liquidity Index (LI) and Consistency Index (CI) locate the natural water content within this range; Activity relates plasticity to clay content.
| LI Range | Consistency State |
|---|---|
| LI < 0 | Semi-solid / solid (below PL) — brittle on remoulding |
| 0 < LI < 0.25 | Stiff |
| 0.25 < LI < 0.50 | Medium stiff |
| 0.50 < LI < 0.75 | Soft |
| 0.75 < LI < 1.0 | Very soft |
| LI > 1.0 | Liquid — flows like a viscous fluid on remoulding |
Coarse fractions (> 75 µm) are analysed by sieve analysis; fine fractions (< 75 µm) by hydrometer/sedimentation analysis based on Stokes' Law, since particles this small no longer settle under a simple sieve. The grading curve (percentage finer vs. log particle size) gives \(D_{10}\), \(D_{30}\), \(D_{60}\) — the sizes at which 10%, 30% and 60% of the sample (by weight) is finer — from which the Coefficient of Uniformity \(C_u\) and Coefficient of Curvature \(C_c\) are computed to judge gradation.
For cohesionless soils, relative density \(D_r\) compares the in-situ void ratio to the loosest (\(e_{max}\)) and densest (\(e_{min}\)) states attainable in the laboratory — a far more meaningful indicator of behaviour than void ratio alone, since two sands of very different gradation can have the same \(e\) but very different relative densities.
| \(D_r\) (%) | State |
|---|---|
| 0 – 15 | Very loose |
| 15 – 35 | Loose |
| 35 – 65 | Medium dense |
| 65 – 85 | Dense |
| 85 – 100 | Very dense |
Void ratio: \( e = \dfrac{V_v}{V_s} \) Porosity: \( n = \dfrac{V_v}{V} = \dfrac{e}{1+e} \)
Degree of saturation: \( S = \dfrac{V_w}{V_v} \times 100\% \)
Air content: \( a_c = \dfrac{V_a}{V_v} \) Percentage air voids: \( n_a = \dfrac{V_a}{V} \)
Water content: \( w = \dfrac{M_w}{M_s} \times 100\% \)
Bulk unit weight: \( \gamma = \dfrac{W}{V} \) Dry unit weight: \( \gamma_d = \dfrac{\gamma}{1+w} \)
Saturated unit weight: \( \gamma_{sat} = \dfrac{(G_s + e)\,\gamma_w}{1+e} \)
Submerged unit weight: \( \gamma' = \gamma_{sat} - \gamma_w \)
\( \gamma_w \approx 9.81\ \text{kN/m}^3 \approx 10\ \text{kN/m}^3 \) (approx., for hand calc)
Master identity: \( e \cdot S = w \cdot G_s \)
\( PI = LL - PL \)
\( LI = \dfrac{w_n - PL}{PI} \) \( CI = \dfrac{LL - w_n}{PI} = 1 - LI \)
Activity: \( A = \dfrac{PI}{\%\ \text{clay-size fraction} \;(< 2\ \mu m)} \)
\(A < 0.75\) inactive, \(0.75 \le A \le 1.25\) normal, \(A > 1.25\) active
Stokes' Law (sedimentation): \( v = \dfrac{(G_s - 1)\gamma_w D^2}{18\mu} \)
Coefficient of Uniformity: \( C_u = \dfrac{D_{60}}{D_{10}} \)
Coefficient of Curvature: \( C_c = \dfrac{D_{30}^2}{D_{10}\times D_{60}} \)
Well-graded sand/gravel: \(C_u > 6\) (sand) or \(>4\) (gravel), and \(1 \le C_c \le 3\).
\( D_r = \dfrac{e_{max} - e}{e_{max} - e_{min}} \times 100\% \)
In terms of dry unit weight: \( D_r = \dfrac{\gamma_{d,max}}{\gamma_d} \cdot \dfrac{\gamma_d - \gamma_{d,min}}{\gamma_{d,max} - \gamma_{d,min}} \times 100\% \)
Given: A soil sample has void ratio \(e = 0.70\), \(G_s = 2.68\), and degree of saturation \(S = 80\%\). Find the water content and bulk unit weight.
Solution:
\( w = \dfrac{e \cdot S}{G_s} = \dfrac{0.70 \times 0.80}{2.68} = 0.2090 = 20.9\% \)
\( \gamma = \dfrac{(G_s + e\,S)\gamma_w}{1+e} = \dfrac{(2.68 + 0.70\times0.80)\times9.81}{1.70} = \dfrac{(2.68+0.56)\times9.81}{1.70} \)
\( \gamma = \dfrac{3.24 \times 9.81}{1.70} \approx 18.70\ \text{kN/m}^3 \)
Answer: \(w \approx 20.9\%\), \(\gamma \approx 18.70\ \text{kN/m}^3\).
Given: LL = 55%, PL = 25%, natural water content \(w_n\) = 40%, clay fraction (< 2 µm) = 30%. Find PI, LI, and Activity; comment on soil state and clay type.
Solution:
\( PI = 55 - 25 = 30\% \)
\( LI = \dfrac{40 - 25}{30} = 0.50 \) → medium stiff consistency
\( A = \dfrac{30}{30} = 1.0 \) → normal activity clay
Answer: \(PI = 30\%\), \(LI = 0.50\) (medium stiff), \(A = 1.0\) (normal-activity clay, e.g. kaolinitic/illitic).
Given: From a sieve analysis, \(D_{10} = 0.12\ \text{mm}\), \(D_{30} = 0.35\ \text{mm}\), \(D_{60} = 0.90\ \text{mm}\). Classify the gradation.
Solution:
\( C_u = \dfrac{0.90}{0.12} = 7.5 \)
\( C_c = \dfrac{0.35^2}{0.12 \times 0.90} = \dfrac{0.1225}{0.108} \approx 1.13 \)
Answer: \(C_u = 7.5 > 6\) and \(1 \le C_c = 1.13 \le 3\) → well-graded sand (SW).
Fig. 2.1 — Three-phase diagram of soil: air, water and solid phases separated by volume (left) and mass (right); \(V = V_a+V_w+V_s\), \(M = M_a+M_w+M_s \approx M_w+M_s\).
The Indian Standard Soil Classification System is essentially the Unified Soil Classification System (USCS) adapted for Indian conditions, using a two-letter symbol: the first letter denotes the primary soil type and the second a qualifier.
| Fraction | Particle Size Range |
|---|---|
| Boulder | > 300 mm |
| Cobble | 80 – 300 mm |
| Gravel (G) | 4.75 – 80 mm |
| Sand (S) | 0.075 – 4.75 mm |
| Silt/Clay (fines, M/C) | < 0.075 mm |
| Symbol | Description |
|---|---|
| GW | Well-graded gravel |
| GP | Poorly-graded gravel |
| GM | Silty gravel |
| GC | Clayey gravel |
| SW | Well-graded sand |
| SP | Poorly-graded sand |
| SM | Silty sand |
| SC | Clayey sand |
| ML | Inorganic silt, low plasticity (LL < 50) |
| CL | Inorganic clay, low plasticity (LL < 50) |
| OL | Organic silt/clay, low plasticity |
| MH | Inorganic silt, high plasticity (LL ≥ 50) |
| CH | Inorganic clay, high plasticity (LL ≥ 50) |
| OH | Organic silt/clay, high plasticity |
| Pt | Peat and highly organic soils |
For fine-grained soils (≥ 50% passing 75 µm), Casagrande's plasticity chart plots Plasticity Index against Liquid Limit; the A-line separates clays (above) from silts (below), and the U-line is an empirical upper bound beyond which no natural soil plots.
Used primarily for highway subgrade evaluation, AASHTO groups soils from A-1 (best, granular) to A-7 (worst, fine-grained clay) and appends a Group Index (GI) — a numerical rating of sub-grade quality within each group, where a higher GI indicates a poorer subgrade.
| Test | Procedure | Interpretation |
|---|---|---|
| Dilatancy Test | Shake a wet pat of soil in the palm, then squeeze | Silt: water appears on shaking, disappears on squeezing (fast reaction). Clay: little to no reaction. |
| Dry Strength Test | Break an air-dried lump of soil between fingers | High strength → clay (CH/CL); low/none → silt (ML) or fine sand. |
| Toughness Test | Roll soil into a 3 mm thread near PL and observe stiffness | Stiff, tough thread near PL → highly plastic clay (CH); weak, crumbly thread → low plasticity (ML/CL). |
| Shine (Lustre) Test | Cut a dried soil pat with a knife and observe the cut surface | Shiny surface → plastic clay; dull surface → silt or low-plasticity soil. |
A-line: \( PI = 0.73\,(LL - 20) \)
U-line (upper bound, empirical): \( PI = 0.9\,(LL - 8) \)
The hatched zone \(4 \le PI \le 7\), below the A-line, is the CL–ML dual-symbol zone (silty clay of low plasticity).
\( GI = (F_{200} - 35)\big[0.2 + 0.005(LL-40)\big] + 0.01(F_{200}-15)(PI-10) \)
\(F_{200}\) = % passing 75 µm (No. 200) sieve, expressed as a whole number 0–100. GI is rounded to the nearest whole number and reported in parentheses, e.g. A-6(9). Negative GI is reported as 0.
Given: A fine-grained soil has LL = 45%, PI = 22%. Classify using the plasticity chart.
Solution:
A-line value at LL = 45: \( PI_{A} = 0.73(45-20) = 0.73\times25 = 18.25\% \)
Actual \(PI = 22\% > 18.25\%\) → soil plots above the A-line, and LL = 45 < 50 → low-plasticity clay
Answer: Classified as CL (inorganic clay of low plasticity).
Given: A subgrade soil has 60% passing the No. 200 sieve, LL = 38%, PI = 16%. Compute the Group Index.
Solution:
\( GI = (60-35)[0.2+0.005(38-40)] + 0.01(60-15)(16-10) \)
\( GI = 25[0.2 - 0.01] + 0.01(45)(6) = 25(0.19) + 2.70 = 4.75 + 2.70 = 7.45 \)
Answer: \(GI \approx 7\) → soil classifies approximately as A-6(7), a fair-to-poor subgrade.
Fig. 3.1 — Casagrande's plasticity chart: A-line (\(PI=0.73(LL-20)\)) separates clays (above) from silts (below); U-line is the empirical upper bound; LL = 50 divides low- and high-plasticity groups.
| Test | IS 2720 Part | Property Determined |
|---|---|---|
| Water content (oven-drying) | Part 2 | Natural water content \(w\) |
| Specific gravity | Part 3 | \(G_s\) |
| Sieve analysis (grain size) | Part 4 | Coarse-fraction gradation |
| Liquid & plastic limit | Part 5 | LL, PL, PI |
| Shrinkage factors | Part 6 | SL, shrinkage ratio |
| Relative density (min./max. dry density) | Part 14 | \(e_{max}, e_{min}, D_r\) |
| Standard Proctor compaction | Part 7 | OMC, \(\gamma_{d,max}\) (light compaction) |
| Modified Proctor compaction | Part 8 | OMC, \(\gamma_{d,max}\) (heavy compaction) |
| Permeability (constant/falling head) | Part 17/36 | Coefficient of permeability \(k\) |
| One-dimensional consolidation | Part 15 | \(C_c, C_s, c_v, m_v\) |
| Direct shear test | Part 13 | \(c, \phi\) |
| Triaxial shear test (UU/CU/CD) | Part 11/12 | \(c, \phi\) under controlled drainage |
Compaction densifies soil by mechanical energy, expelling air (not water) from the voids. A plot of dry density against water content for a fixed compactive effort shows a peak — the Optimum Moisture Content (OMC) gives the Maximum Dry Density (MDD). The Zero Air Voids (ZAV) line represents the theoretical maximum dry density at 100% saturation (zero air voids) and always lies above and to the right of the compaction curve, which it approaches but never crosses.
| Test | Compactive Effort | Hammer | Layers × Blows |
|---|---|---|---|
| Standard Proctor (Light) | ≈ 605 kJ/m³ | 2.6 kg, 310 mm drop | 3 layers × 25 blows |
| Modified Proctor (Heavy) | ≈ 2700 kJ/m³ | 4.9 kg, 450 mm drop | 5 layers × 25 blows |
Increasing compactive effort raises the MDD and lowers the OMC; the peak of the curve shifts up and to the left. Degree of compaction in the field is expressed as a percentage of the laboratory MDD.
A compact set of identities links every quantity introduced so far and is worth memorising as a group: \(e\cdot S = w\cdot G_s\), \(\gamma_d = \dfrac{\gamma}{1+w}\), \(n = \dfrac{e}{1+e}\), and the coefficient of earth pressure at rest \(K_0\) (Jaky's empirical formula for normally consolidated soil, and the Mayne & Kulhawy correction for over-consolidated soil).
Sensitivity measures the loss of shear strength of a clay when it is remoulded at constant water content — it reflects the destruction of the soil's natural bonded/flocculated structure.
| Sensitivity \(S_t\) | Classification |
|---|---|
| 1 | Insensitive |
| 1 – 2 | Low sensitive |
| 2 – 4 | Medium sensitive |
| 4 – 8 | Sensitive |
| 8 – 16 | Extra sensitive |
| > 16 | Quick clay |
Thixotropy is the partial or complete regain of strength with time after remoulding, at constant water content and without any change in volume — purely a structural re-arrangement effect, distinct from consolidation-driven strength gain.
Dry density from bulk density and water content: \( \gamma_d = \dfrac{\gamma}{1+w} \)
Zero Air Voids line: \( \gamma_{d(ZAV)} = \dfrac{G_s\,\gamma_w}{1+w\,G_s} \)
Degree of compaction: \( \text{DOC} = \dfrac{\gamma_{d(field)}}{\gamma_{d,max(lab)}} \times 100\% \)
\( e\cdot S = w\cdot G_s \) \( \gamma_d = \dfrac{\gamma}{1+w} \) \( n = \dfrac{e}{1+e} \)
Jaky's formula (NC soil): \( K_0 = 1 - \sin\phi' \)
Mayne & Kulhawy (OC soil): \( K_{0(OC)} = (1-\sin\phi')\,OCR^{\sin\phi'} \)
\( S_t = \dfrac{q_u\,(\text{undisturbed})}{q_u\,(\text{remoulded})} \)
Both tested at the same water content, typically via unconfined compression test.
Given: \(G_s = 2.70\), \(w = 15\%\). Find the theoretical maximum dry density at zero air voids.
Solution:
\( \gamma_{d(ZAV)} = \dfrac{2.70\times9.81}{1+0.15\times2.70} = \dfrac{26.49}{1.405} \approx 18.85\ \text{kN/m}^3 \)
Answer: \( \gamma_{d(ZAV)} \approx 18.85\ \text{kN/m}^3 \) — the achieved compaction MDD must always be below this value.
Given: An unconfined compression test on undisturbed clay gives \(q_u = 120\ \text{kPa}\); on remoulded clay (same water content) it gives \(q_u = 20\ \text{kPa}\). Classify the clay.
Solution:
\( S_t = \dfrac{120}{20} = 6 \)
Answer: \(S_t = 6\) → sensitive clay.
Given: A normally consolidated clay has \(\phi' = 28°\). Find \(K_0\).
Solution:
\( K_0 = 1 - \sin 28° = 1 - 0.469 = 0.531 \)
Answer: \(K_0 \approx 0.53\).
Fig. 4.1 — Standard Proctor compaction curve: peak gives OMC & MDD; the Zero-Air-Voids (S=100%) line bounds the curve from above and is never crossed.
Henry Darcy (1856) established that the rate of laminar flow of water through a saturated soil is directly proportional to the hydraulic gradient. This is valid for laminar flow only — the seepage velocity must remain low enough that the Reynolds number stays below about 1–10, which holds for essentially all soils except very coarse, clean gravel.
Constant Head Test — suitable for coarse-grained soils (sand, gravel) with relatively high permeability, where enough water flows through the sample in a measurable time that a steady head can be maintained.
Falling Head Test — suitable for fine-grained soils (silt, clay) with low permeability; the head in a standpipe is allowed to fall and the rate of fall is used to back-calculate \(k\), since a constant-head test would take impractically long to yield a measurable outflow.
Hazen's formula gives a quick estimate of \(k\) for clean, uniform sand directly from the effective grain size \(D_{10}\). The Kozeny-Carman equation is a more rigorous, physically-based relation involving void ratio, specific surface and viscosity. Permeability increases with temperature (decreasing viscosity), so lab values are conventionally corrected to 27°C.
Natural deposits are almost always stratified. For flow parallel to the bedding (horizontal), each layer conducts flow like a resistor in parallel, so the equivalent \(k\) is a thickness-weighted arithmetic average. For flow perpendicular to the bedding (vertical), layers act like resistors in series, giving a thickness-weighted harmonic average. This is why \(k_H\) is always greater than \(k_V\) in a layered deposit — the most permeable layer dominates horizontal flow while the least permeable layer dominates and controls vertical flow.
| Soil Type | Typical \(k\) (m/s) |
|---|---|
| Clean gravel | 1 to \(10^{-2}\) |
| Clean sand | \(10^{-2}\) to \(10^{-5}\) |
| Silty sand | \(10^{-5}\) to \(10^{-7}\) |
| Silt | \(10^{-7}\) to \(10^{-9}\) |
| Clay | \(10^{-9}\) to \(10^{-12}\) |
Two-dimensional seepage through soil under a hydraulic structure (dam, sheet pile) is governed by Laplace's equation and is solved graphically by drawing a flow net — a mesh of flow lines (paths followed by water) and equipotential lines (lines of equal total head) intersecting at right angles, forming "curvilinear squares". The flow net directly gives the seepage discharge, the seepage pressure at any point, and the exit gradient at the downstream toe.
When the upward seepage (exit) gradient at the ground surface equals the critical hydraulic gradient, the effective stress drops to zero and the soil loses all shear strength — this is the quicksand condition. If it initiates at a point and progressively erodes a channel back toward the source, it is called piping, one of the principal causes of earth dam and sheet-pile failure. A Factor of Safety against piping of at least 3–4 is conventionally required.
Lane's Weighted Creep Theory is an alternative, empirical design method for the seepage path length below hydraulic structures, weighting vertical creep paths three times more effective than horizontal paths in resisting piping.
Discharge velocity: \( v = k\,i \) Flow rate: \( Q = k\,i\,A \)
Seepage velocity: \( v_s = \dfrac{v}{n} = \dfrac{Q}{n\,A} \)
Reynolds number (validity check): \( R_e = \dfrac{\rho\,v\,D}{\mu} < 1\text{–}10 \)
\( k = \dfrac{Q\,L}{A\,h\,t} \)
\(Q\)=volume collected in time \(t\), \(L\)=sample length, \(A\)=c/s area, \(h\)=constant head.
\( k = \dfrac{a\,L}{A\,t}\ \ln\!\left(\dfrac{h_1}{h_2}\right) = 2.303\,\dfrac{a\,L}{A\,t}\log_{10}\!\left(\dfrac{h_1}{h_2}\right) \)
\(a\)=standpipe c/s area, \(h_1,h_2\)=initial & final heads at times \(t_1,t_2\); \(t=t_2-t_1\).
Hazen's formula: \( k = C\,D_{10}^{2} \), \(C \approx 100\) (k in cm/s, \(D_{10}\) in cm)
Temperature correction: \( k_{27} = k_T\,\dfrac{\mu_T}{\mu_{27}} \)
Horizontal (parallel flow): \( k_H = \dfrac{\sum k_i H_i}{\sum H_i} \)
Vertical (series flow): \( k_V = \dfrac{\sum H_i}{\sum (H_i/k_i)} \)
Always \(k_H \ge k_V\).
Seepage discharge: \( Q = k\,H\,\dfrac{N_f}{N_d} \) (per unit length)
\(N_f\)=number of flow channels, \(N_d\)=number of equipotential drops, \(H\)=total head loss.
Seepage pressure at a point: \( P_s = h\,\gamma_w \) (\(h\)=head dissipated up to that point)
\( i_{cr} = \dfrac{\gamma'}{\gamma_w} = \dfrac{G_s - 1}{1+e} \)
Factor of safety against piping: \( FS = \dfrac{i_{cr}}{i_{exit}} \ge 3\text{–}4 \)
Lane's weighted creep ratio: \( C_L = \dfrac{L_V + \tfrac{1}{3}L_H}{H_{water}} \)
Given: Standpipe area \(a = 4\ \text{cm}^2\), sample area \(A = 80\ \text{cm}^2\), sample length \(L = 15\ \text{cm}\). Head falls from \(h_1 = 60\) cm to \(h_2 = 30\) cm in \(t = 300\) s. Find \(k\).
Solution:
\( k = 2.303\dfrac{4\times15}{80\times300}\log_{10}\!\left(\dfrac{60}{30}\right) = 2.303\times\dfrac{60}{24000}\times0.301 \)
\( k = 2.303\times0.0025\times0.301 \approx 1.73\times10^{-3}\ \text{cm/s} \)
Answer: \( k \approx 1.73\times10^{-3}\ \text{cm/s} \).
Given: A sand deposit has \(G_s = 2.65\), \(e = 0.65\). Estimate \(i_{cr}\), and check the FS against piping if the actual exit gradient is 0.25.
Solution:
\( i_{cr} = \dfrac{2.65-1}{1+0.65} = \dfrac{1.65}{1.65} = 1.0 \)
\( FS = \dfrac{1.0}{0.25} = 4.0 \)
Answer: \(i_{cr} = 1.0\), \(FS = 4.0\) — adequate, at the conventional minimum limit.
Given: Two horizontal layers: Layer 1 (\(H_1=2\ \text{m}, k_1=4\times10^{-4}\ \text{m/s}\)), Layer 2 (\(H_2=3\ \text{m}, k_2=1\times10^{-6}\ \text{m/s}\)). Find \(k_H\) and \(k_V\).
Solution:
\( k_H = \dfrac{4\times10^{-4}\times2 + 1\times10^{-6}\times3}{5} = \dfrac{8\times10^{-4} + 3\times10^{-6}}{5} \approx 1.61\times10^{-4}\ \text{m/s} \)
\( k_V = \dfrac{5}{\dfrac{2}{4\times10^{-4}} + \dfrac{3}{1\times10^{-6}}} = \dfrac{5}{5000 + 3{,}000{,}000} \approx 1.66\times10^{-6}\ \text{m/s} \)
Answer: \(k_H \approx 1.61\times10^{-4}\ \text{m/s}\), \(k_V \approx 1.66\times10^{-6}\ \text{m/s}\) — confirms \(k_H \gg k_V\), dominated respectively by the sand and clay layers.
Fig. 5.1 — Flow net beneath a sheet pile wall: flow lines and equipotential lines cross at right angles forming curvilinear squares; \(Q = k H(N_f/N_d)\).
Total settlement of a foundation on clay comprises three components: immediate (elastic) settlement \(S_i\) (occurs instantly on loading, no volume change), primary consolidation settlement \(S_c\) (occurs over months to years as pore water is squeezed out and effective stress increases), and secondary compression \(S_s\) (continues after excess pore pressure has fully dissipated, due to plastic re-adjustment of the soil skeleton).
Terzaghi's one-dimensional consolidation theory idealises the clay layer as a spring-and-dashpot-in-water analogy: as excess pore water pressure dissipates through drainage, the "spring" (soil skeleton) picks up the load and effective stress rises, causing settlement. The governing partial differential equation is solved using the coefficient of consolidation \(c_v\), which combines permeability, compressibility and unit weight of water.
The compression index \(C_c\) (slope of the virgin compression line on the e vs. \(\log p\) plot) governs settlement for normally consolidated clay beyond the preconsolidation pressure; Skempton's empirical correlation \(C_c = 0.009(LL-10)\) is widely used for a first estimate from the liquid limit alone. The swelling/recompression index \(C_s\) is much flatter, typically \(\dfrac{C_c}{5}\) to \(\dfrac{C_c}{10}\), and governs settlement below the preconsolidation pressure (recompression of previously loaded soil).
The time factor \(T_v\) is a dimensionless parameter linking elapsed time to the coefficient of consolidation and the maximum drainage path length; it is related to the degree of consolidation \(U\) (percentage of ultimate settlement achieved) through Terzaghi's solution — approximated by a parabolic relation for \(U < 60\%\) and a logarithmic relation for \(U > 60\%\). Two values are worth memorising exactly: at \(U=50\%\), \(T_v = 0.197\); at \(U=90\%\), \(T_v = 0.848\).
The Over-Consolidation Ratio (OCR) is the ratio of the maximum past effective stress (preconsolidation pressure) to the current effective overburden pressure — \(OCR = 1\) for normally consolidated clay, \(OCR > 1\) for over-consolidated clay.
After primary consolidation is essentially complete (excess pore pressure ≈ 0), void ratio continues to decrease slowly under constant effective stress — this creep-like behaviour is secondary compression, governed by the secondary compression index \(C_\alpha\). It is most significant in highly organic soils and peat, where \(C_\dfrac{\alpha}{C_c}\) can be as high as 0.05–0.10, versus typically 0.02–0.04 for inorganic clays.
\( \dfrac{\partial u}{\partial t} = c_v\,\dfrac{\partial^2 u}{\partial z^2} \)
\( c_v = \dfrac{k}{m_v\,\gamma_w} \) \( m_v = \dfrac{a_v}{1+e_0} \) \( a_v = \dfrac{\Delta e}{\Delta p} \)
\( C_c = \dfrac{e_1 - e_2}{\log_{10}(p_2/p_1)} \)
Skempton's correlation: \( C_c = 0.009\,(LL - 10) \)
\( C_s \approx \dfrac{C_c}{5}\ \text{to}\ \dfrac{C_c}{10} \)
Normally consolidated clay (\(p_0 = p_c\)):
\( S_c = \dfrac{C_c\,H}{1+e_0}\ \log_{10}\!\left(\dfrac{p_0+\Delta p}{p_0}\right) \)
Over-consolidated clay, case \(p_0+\Delta p \le p_c\):
\( S_c = \dfrac{C_s\,H}{1+e_0}\ \log_{10}\!\left(\dfrac{p_0+\Delta p}{p_0}\right) \)
Over-consolidated clay, case \(p_0+\Delta p > p_c\):
\( S_c = \dfrac{C_s\,H}{1+e_0}\log_{10}\!\dfrac{p_c}{p_0} + \dfrac{C_c\,H}{1+e_0}\log_{10}\!\dfrac{p_0+\Delta p}{p_c} \)
\( T_v = \dfrac{c_v\,t}{H_{dr}^2} \)
\(U < 60\%\): \( T_v = \dfrac{\pi}{4}\left(\dfrac{U\%}{100}\right)^2 \)
\(U > 60\%\): \( T_v = 1.781 - 0.933\log_{10}(100-U\%) \)
Key values: \(U=50\% \Rightarrow T_v=0.197\); \(U=90\% \Rightarrow T_v=0.848\).
\( OCR = \dfrac{p_c}{p_0} \)
\( S_s = C_\alpha\,H\,\log_{10}\!\left(\dfrac{t_2}{t_1}\right) \)
\( C_\alpha = \dfrac{\Delta e}{\log_{10}(t_2/t_1)} \)
Given: A 4 m thick NC clay layer has \(e_0 = 1.0\), \(C_c = 0.30\), initial effective stress \(p_0 = 100\ \text{kPa}\), stress increase \(\Delta p = 50\ \text{kPa}\). Find \(S_c\).
Solution:
\( S_c = \dfrac{0.30\times4}{1+1.0}\log_{10}\!\left(\dfrac{150}{100}\right) = 0.60\times\log_{10}(1.5) = 0.60\times0.1761 \)
\( S_c \approx 0.1057\ \text{m} \approx 105.7\ \text{mm} \)
Answer: \(S_c \approx 106\ \text{mm}\).
Given: A doubly-drained clay layer 6 m thick has \(c_v = 3\ \dfrac{\text{m}^2}{\text{year}}\). Find the time for 50% consolidation.
Solution:
Doubly drained: \(H_{dr} = \dfrac{6}{2} = 3\ \text{m}\). At \(U=50\%\), \(T_v = 0.197\).
\( t = \dfrac{T_v\,H_{dr}^2}{c_v} = \dfrac{0.197\times3^2}{3} = \dfrac{1.773}{3} = 0.591\ \text{years} \)
Answer: \(t \approx 0.59\) years \(\approx 7.1\) months.
Given: A remoulded clay sample has \(LL = 60\%\). Estimate \(C_c\) using Skempton's correlation, and the swelling index \(C_s\) (assume \(C_s = \dfrac{C_c}{6}\)).
Solution:
\( C_c = 0.009(60-10) = 0.009\times50 = 0.45 \)
\( C_s = \dfrac{0.45}{6} = 0.075 \)
Answer: \(C_c \approx 0.45\), \(C_s \approx 0.075\).
Fig. 6.1 — e-log p consolidation curve: recompression branch (\(C_s\)) up to the preconsolidation pressure \(p_c\), then the steeper virgin compression line (\(C_c\)); unloading follows a swelling/rebound branch.
Soil fails in shear, not in tension or compression, when the shear stress on some plane reaches the shear strength on that plane. The Mohr-Coulomb criterion expresses shear strength as a linear function of effective normal stress, characterised by two parameters: effective cohesion \(c'\) and effective angle of internal friction \(\phi'\). All strength must be expressed in terms of effective stress (\(\sigma' = \sigma - u\)), since it is the effective stress, not total stress, that governs soil behaviour (Terzaghi's principle of effective stress).
At failure, the Mohr circle of stress is tangent to the failure envelope; the plane of failure is inclined at \(\theta = 45° + \dfrac{\phi'}{2}\) to the major principal plane.
| Test | Consolidation | Drainage during shear | Parameters obtained |
|---|---|---|---|
| UU (Unconsolidated Undrained) | No | No drainage | Total stress \(c_u\) (\(\phi_u \approx 0\) for saturated clay) |
| CU (Consolidated Undrained) | Yes | No drainage, pore pressure measured | Total stress \(c_{cu}, \phi_{cu}\) and effective stress \(c', \phi'\) |
| CD (Consolidated Drained) | Yes | Full drainage, very slow rate | Effective stress \(c', \phi'\) directly |
| Soil | \(\phi'\) (degrees) | \(c'\) (kPa) |
|---|---|---|
| Loose sand | 28 – 32 | 0 |
| Dense sand | 35 – 45 | 0 |
| Gravel | 35 – 45 | 0 |
| Soft normally-consolidated clay | 20 – 30 | 0 – 10 |
| Stiff over-consolidated clay | 18 – 28 | 10 – 50 |
| Residual (drained) clay | 10 – 20 | 0 – 5 |
Skempton's pore pressure parameters A and B relate the change in pore water pressure to the change in applied stress in an undrained triaxial test — the B parameter depends on the degree of saturation (B ≈ 1 for fully saturated soil, B < 1 otherwise) and the A parameter depends on stress history and soil type, typically negative for heavily over-consolidated clay, near zero for lightly over-consolidated soil, and up to ~1 for normally consolidated (contractive) clay, and can exceed 1 for very sensitive/loose soils.
A special case of the UU triaxial test with zero confining pressure (\(\sigma_3 = 0\)) — a simple, fast test performed on saturated cohesive soil to obtain the undrained shear strength directly, without needing a triaxial cell.
The Critical State framework (Roscoe, Schofield & Wroth) describes the state at which a soil continues to deform in shear at constant volume and constant effective stress — the Critical State Line (CSL). Loose/normally consolidated (contractive) soils compress toward the CSL during shear; dense/over-consolidated (dilative) soils expand toward it. This unifies the drained and undrained, and loose and dense, behaviour of soil within a single framework, plotted either in e-log p' space or q-p' (stress) space.
\( \tau_f = c' + \sigma'\tan\phi' \)
Effective stress: \( \sigma' = \sigma - u \)
Failure plane angle: \( \theta = 45° + \dfrac{\phi'}{2} \)
Mohr circle relation at failure: \( \sigma_1' = \sigma_3'\tan^2\!\left(45°+\dfrac{\phi'}{2}\right) + 2c'\tan\!\left(45°+\dfrac{\phi'}{2}\right) \)
\( \Delta u = B\big[\Delta\sigma_3 + A(\Delta\sigma_1 - \Delta\sigma_3)\big] \)
\(B \approx 1\) for saturated soil. \(A\): NC clay ≈ 0.5–1, OC clay ≈ −0.5–0, loose sand ≈ 0.5–1, dense sand ≈ −0.5–0.
\( \sigma_3 = 0, \quad c_u = \dfrac{q_u}{2} \)
Corrected area (assuming constant volume): \( A_c = \dfrac{A_0}{1-\varepsilon} \)
CSL in e-log p' space: \( e = e_{cs} - \lambda\,\ln p' \)
CSL in q-p' space: \( q = M\,p' \)
\( M = \dfrac{6\sin\phi'}{3-\sin\phi'} \) (triaxial compression)
Given: A CD triaxial test on sand has \(c' = 0\), \(\phi' = 34°\), confining pressure \(\sigma_3' = 100\ \text{kPa}\). Find \(\sigma_1'\) at failure.
Solution:
\( \tan^2(45+17) = \tan^2(62°) = (1.881)^2 = 3.538 \)
\( \sigma_1' = 100 \times 3.538 + 0 = 353.8\ \text{kPa} \)
Answer: \(\sigma_1' \approx 354\ \text{kPa}\).
Given: An unconfined compression test on saturated clay gives \(q_u = 96\ \text{kPa}\) at failure. Find the undrained shear strength.
Solution:
\( c_u = \dfrac{q_u}{2} = \dfrac{96}{2} = 48\ \text{kPa} \)
Answer: \(c_u = 48\ \text{kPa}\).
Given: A dense sand has \(\phi' = 40°\). Find the angle of the failure plane to the major principal plane.
Solution:
\( \theta = 45° + \dfrac{40°}{2} = 45° + 20° = 65° \)
Answer: \(\theta = 65°\) to the major principal plane (or \(25°\) to the minor principal plane).
Fig. 7.1 — Mohr circle tangent to the Mohr-Coulomb failure envelope at failure; the tangent point defines the stresses on the failure plane, inclined at \(\theta=45°+\dfrac{\phi'}{2}\) to the major principal plane.
| Condition | Coefficient | Wall Movement |
|---|---|---|
| At-rest | \(K_0\) (Jaky's formula) | No lateral movement (e.g. basement wall braced on both sides) |
| Active | \(K_a\) (minimum pressure) | Wall moves away from backfill |
| Passive | \(K_p\) (maximum pressure) | Wall pushed into backfill |
Rankine (1857) assumed a cohesionless, homogeneous backfill with a vertical, frictionless wall and horizontal ground surface, so that the wall friction is zero and the failure planes are straight, giving simple closed-form expressions for \(K_a\) and \(K_p\). For cohesive (c-φ) soil, the active pressure distribution shows tension near the top of the wall down to a tension crack depth \(z_c\), below which the net pressure is compressive.
Coulomb (1776) is more general than Rankine — it accounts for wall friction \(\delta\), a sloping wall back \(\alpha\), and a sloping backfill surface \(\beta\), using a limit-equilibrium wedge analysis. Coulomb's expression reduces exactly to Rankine's when the wall is vertical (\(\alpha=90°\)), the backfill is horizontal (\(\beta=0°\)) and wall friction is neglected (\(\delta=0°\)).
Submerged backfill: below the water table, use the submerged unit weight \(\gamma'\) for the soil skeleton's contribution and add full hydrostatic pressure separately — the water pressure itself is not multiplied by \(K_a\). Layered soils: pressure is computed layer-by-layer using each layer's own \(K_a\) and unit weight, with the overburden from layers above carried down as a surcharge on the layer below. Surcharge: a uniform surcharge \(q\) on the ground surface adds a uniform (rectangular) increment \(K_a q\) to the pressure diagram at every depth.
A gravity or cantilever retaining wall must be checked against four modes of failure: overturning about the toe, sliding along the base, bearing capacity failure of the foundation soil, and excessive settlement. The resultant of all forces should ideally fall within the middle third of the base to avoid tension (uplift) at the heel.
Geostatic stress is the stress due to the soil's own self-weight, increasing linearly with depth (with a change in slope at the water table due to buoyancy). Boussinesq's theory (1885) gives the increase in vertical stress at depth due to a point load on the surface of a homogeneous, isotropic, elastic, semi-infinite half-space, and is integrated to give solutions for circular, rectangular (via influence charts such as Fadum's) and line loads. Westergaard's theory assumes alternating rigid horizontal layers (models layered/stratified soil with negligible lateral strain) and gives lower stresses than Boussinesq at the same point for the same load — a mnemonic worth remembering is "Boussinesq is Bigger". The 2:1 method is a simple approximate rule assuming the load spreads out at a slope of 2 vertical to 1 horizontal, useful for quick hand estimates.
Newmark's chart is a graphical method to find the vertical stress increase at any depth beneath a loaded area of any irregular shape — the loaded area is drawn to a matching scale on the chart, and the number of chart units covered is counted and multiplied by an influence factor to give the stress, making it far more general than the point/circular/rectangular formulas which only apply to regular shapes.
\( K_a = \dfrac{1-\sin\phi}{1+\sin\phi} = \tan^2\!\left(45°-\dfrac{\phi}{2}\right) \)
\( K_p = \dfrac{1+\sin\phi}{1-\sin\phi} = \tan^2\!\left(45°+\dfrac{\phi}{2}\right) = \dfrac{1}{K_a} \)
Total active force: \( P_a = \tfrac{1}{2}K_a\gamma H^2 \) acting at \(\dfrac{H}{3}\) above the base
\( \sigma_a = K_a\gamma z - 2c\sqrt{K_a} \) \( \sigma_p = K_p\gamma z + 2c\sqrt{K_p} \)
Tension crack depth: \( z_c = \dfrac{2c}{\gamma\sqrt{K_a}} \)
With surcharge \(q\): \( \sigma_a = K_a(\gamma z + q) - 2c\sqrt{K_a} \)
\( K_a = \dfrac{\sin^2(\alpha+\phi)}{\sin^2\alpha\,\sin(\alpha-\delta)\left[1+\sqrt{\dfrac{\sin(\phi+\delta)\sin(\phi-\beta)}{\sin(\alpha-\delta)\sin(\alpha+\beta)}}\right]^2} \)
Reduces to Rankine's \(K_a\) when \(\alpha=90°, \beta=0°, \delta=0°\).
Overturning: \( FS_{OT} = \dfrac{\sum M_{resisting}}{\sum M_{overturning}} \ge 1.5\text{–}2.0 \)
Sliding: \( FS_{sliding} = \dfrac{\mu\,\sum V}{\sum P_h} \ge 1.5 \)
Bearing (middle-third): \( q_{max,min} = \dfrac{\sum V}{B}\left(1 \pm \dfrac{6e}{B}\right) \)
Above water table: \( \sigma_v = \gamma\,z \) Below water table: \( \sigma_v' = \gamma_{sat}z_1 + \gamma'(z-z_1) \)
2:1 method: \( \Delta\sigma_z = \dfrac{Q}{(L+z)(B+z)} \)
Boussinesq: \( \Delta\sigma_z = \dfrac{3Q}{2\pi z^2}\left[\dfrac{1}{1+(r/z)^2}\right]^{5/2} \)
Westergaard (\(\mu_s=0\)): \( \Delta\sigma_z = \dfrac{Q}{\pi z^2}\left[\dfrac{1}{1+2(r/z)^2}\right]^{3/2} \)
Given: A 5 m high retaining wall retains dry sand, \(\gamma = 18\ \text{kN/m}^3\), \(\phi = 30°\), \(c = 0\). Find \(K_a\), the pressure at the base, and the total active thrust per metre length.
Solution:
\( K_a = \tan^2(45-15) = \tan^2(30°) = (0.577)^2 = 0.333 \)
\( \sigma_a(\text{base}) = K_a\gamma H = 0.333\times18\times5 = 30.0\ \text{kPa} \)
\( P_a = \tfrac{1}{2}\times0.333\times18\times5^2 = \tfrac{1}{2}\times0.333\times18\times25 = 75.0\ \text{kN/m} \)
Answer: \(K_a = 0.333\), base pressure = 30 kPa, \(P_a = 75\ \text{kN/m}\) acting at \(\dfrac{5}{3} = 1.67\ \text{m}\) above the base.
Given: A wall retains clay backfill with \(\gamma=19\ \text{kN/m}^3\), \(c=15\ \text{kPa}\), \(\phi=0\) (so \(K_a=1\)). Find the tension crack depth.
Solution:
\( z_c = \dfrac{2\times15}{19\times\sqrt{1}} = \dfrac{30}{19} \approx 1.58\ \text{m} \)
Answer: \(z_c \approx 1.58\ \text{m}\) — tension cracks may develop to this depth; this zone is often neglected in computing the resultant thrust.
Given: A point load \(Q = 500\ \text{kN}\) acts on the ground surface. Find the vertical stress increase at a depth \(z=3\ \text{m}\) directly below the load (\(r=0\)).
Solution:
At \(r=0\): \( \Delta\sigma_z = \dfrac{3Q}{2\pi z^2} = \dfrac{3\times500}{2\pi\times9} = \dfrac{1500}{56.55} \approx 26.5\ \text{kPa} \)
Answer: \(\Delta\sigma_z \approx 26.5\ \text{kPa}\) directly below the load at 3 m depth.
Fig. 8.1 — Rankine active earth pressure distribution (triangular, cohesionless backfill): resultant \(P_a = \tfrac12 K_a\gamma H^2\) acts at \(\dfrac{H}{3}\) above the wall base.
| Type | Description |
|---|---|
| Geotextile | Permeable, woven/non-woven fabric (usually polypropylene/polyester) |
| Geogrid | Grid-like polymer sheet with large apertures, high tensile stiffness for soil reinforcement |
| Geomembrane | Impermeable polymer sheet (HDPE/LLDPE/PVC) used as a barrier |
| Geocomposite | Combination of two or more geosynthetic types (e.g. geotextile + geonet) |
| Geonet | Open, net-like structure providing high in-plane drainage capacity |
| Geosynthetic Clay Liner (GCL) | Thin bentonite clay layer sandwiched between geotextiles, low-permeability barrier |
| Geofoam | Lightweight expanded polystyrene blocks used as compressible/lightweight fill |
| Geocell | Three-dimensional honeycomb cellular confinement system |
| Function | Description | Typical Application |
|---|---|---|
| Separation | Prevents mixing of two dissimilar soil/aggregate layers | Road base over soft subgrade |
| Filtration | Allows water to pass while retaining soil particles | Drainage trenches, behind retaining walls |
| Drainage | Conveys water/fluid within the plane of the geosynthetic | Geonet/geocomposite drains |
| Reinforcement | Provides tensile resistance the soil itself lacks | MSE walls, reinforced embankments over soft soil |
| Containment / Barrier | Prevents fluid or gas migration | Landfill liners, pond liners |
| Erosion Control | Protects soil surface from erosive forces | Channel linings, slope protection |
| Protection | Cushions a geomembrane from puncture | Above/below geomembrane in landfill liners |
For filtration, the geotextile's apparent opening size (AOS/O95) must be small enough to retain the protected soil, while its permeability must remain higher than that of the soil so it does not itself become the controlling (slowest) layer in the drainage system. For strength design (e.g. as a separator under a road embankment), the required tensile strength is reduced from the ultimate wide-width strength by several reduction factors accounting for installation damage, creep, chemical/biological degradation, and seam strength.
Mechanically Stabilised Earth (MSE) walls use horizontal layers of geogrid (or geotextile/steel strip) reinforcement within a compacted granular backfill; the reinforcement must have adequate tensile capacity to resist the outward-pulling active earth pressure at each level, and adequate embedment length beyond the theoretical failure plane to develop the required pull-out (frictional/interlock) resistance without slipping.
A GCL sandwiches a thin layer (≈ 5 mm) of sodium bentonite between two geotextile layers (needle-punched or stitch-bonded together). On hydration the bentonite swells to form a very low-permeability barrier (\(k \sim 10^{-11}\ \text{m/s}\)), often used as a composite liner together with a geomembrane, or as a stand-alone liner in less critical applications, since it is thinner and easier to install than compacted clay.
| IS Code | Scope |
|---|---|
| IS 13162 (Parts 1–19) | Geotextiles — test methods |
| IS 14714 | Geotextile used as filter — specification |
| IS 16391 | Geogrid — specification |
| IS 17373 | Geomembrane — specification |
| IS 15784 | Geosynthetics used in flexible pavements — guidelines |
| IRC SP-59 | Guidelines for use of geosynthetics in road pavements |
Retention (piping) criterion: \( O_{95} \le B\,D_{85}(\text{soil}) \)
Permeability criterion: \( k_{geotextile} \ge k_{soil} \) (typically \(\ge 10\times\))
Permittivity: \( \Psi = \dfrac{k_n}{t} \) Transmissivity: \( \theta = k_p\,t \)
\( T_{allow} = \dfrac{T_{ult}}{RF_{ID}\times RF_{CR}\times RF_{CBD}\times RF_{W}} \)
\(RF_{ID}\)=installation damage, \(RF_{CR}\)=creep, \(RF_{CBD}\)=chemical/biological degradation, \(RF_W\)=seam/weathering.
Max. tension at each level: \( T_{max} = K_a\,\gamma\,z\,S_v \)
Embedment (pull-out) length: \( L_e = \dfrac{T_{max}\times FS}{2\,\sigma_v'\tan\phi_{sg}} \)
Total reinforcement length: \( L = L_e + L_a \) (\(L_a\)=length within active zone)
Given: A protected soil has \(D_{85} = 0.30\ \text{mm}\); the candidate geotextile has \(O_{95} = 0.20\ \text{mm}\). Using \(B=1\) for the retention criterion, check suitability.
Solution:
Criterion: \(O_{95} \le B\,D_{85} = 1\times0.30 = 0.30\ \text{mm}\)
Actual \(O_{95} = 0.20\ \text{mm} \le 0.30\ \text{mm}\) ✓
Answer: The geotextile satisfies the retention criterion and will not allow excessive soil loss through its openings.
Given: An MSE wall backfill has \(\gamma = 19\ \text{kN/m}^3\), \(\phi = 34°\) (\(K_a = 0.283\)), reinforcement at depth \(z = 3\ \text{m}\), vertical spacing \(S_v = 0.6\ \text{m}\). Find \(T_{max}\).
Solution:
\( T_{max} = 0.283\times19\times3\times0.6 = 9.68\ \text{kN/m} \)
Answer: \(T_{max} \approx 9.68\ \text{kN/m}\) run of wall at that reinforcement level.
Fig. 9.1 — Typical composite landfill liner cross-section: primary geomembrane over GCL, geonet leak-detection layer, secondary geomembrane, and a compacted clay liner over the natural subgrade.
\( e\cdot S = w\cdot G_s \), \( \gamma_d=\dfrac{\gamma}{1+w} \), \( n=\dfrac{e}{1+e} \), \( D_r=\dfrac{e_{max}-e}{e_{max}-e_{min}}\times100 \)
\( PI=LL-PL \), \( LI=\dfrac{w_n-PL}{PI} \), \( A=\dfrac{PI}{\%\,\text{clay}} \), A-line: \( PI=0.73(LL-20) \)
\( v=ki \), \( k=\dfrac{aL}{At}\ln\dfrac{h_1}{h_2} \) (falling head), \( i_{cr}=\dfrac{G_s-1}{1+e} \)
\( S_c=\dfrac{C_c H}{1+e_0}\log\dfrac{p_0+\Delta p}{p_0} \), \( T_v=\dfrac{c_vt}{H_{dr}^2} \); \(T_v=0.197\) at U=50%, \(0.848\) at U=90%
\( \tau_f=c'+\sigma'\tan\phi' \), \( c_u=\dfrac{q_u}{2} \), \( \Delta u=B[\Delta\sigma_3+A(\Delta\sigma_1-\Delta\sigma_3)] \)
\( K_a=\tan^2(45-\dfrac{\phi}{2}) \), \( P_a=\tfrac12K_a\gamma H^2 \), \( \Delta\sigma_z=\dfrac{3Q}{2\pi z^2} \) (Boussinesq, r=0)
| Chapter | GATE Focus | ESE Focus | SSC-JE Focus |
|---|---|---|---|
| Soil Exploration | SPT corrections, N-value interpretation | Boring methods, sampler design ratios | Basic terminology, IS 1892 |
| Properties of Soil | Numerical (e, w, γ, e·S=wGs) | Atterberg limits, activity classification | Direct formula-based numericals |
| Soil Classification | Casagrande chart, USCS symbols | AASHTO Group Index, field ID tests | Symbol identification |
| Tests & Interrelationships | Compaction curve, ZAV line | Sensitivity, IS 2720 parts | OMC/MDD concept questions |
| Permeability & Seepage | Flow net, equivalent k | Piping FS, Lane's creep ratio | Darcy's law numericals |
| Consolidation | Tv-U relation, settlement numericals | Secondary compression, OCR | Basic Terzaghi theory concepts |
| Shearing Resistance | Mohr circle numericals, triaxial types | Skempton's parameters, critical state | UU/CU/CD differences |
| Earth Pressure & Stress | Ka/Kp numericals, Boussinesq stress | Coulomb theory, retaining wall stability | Rankine formula application |
| Geosynthetics | Function identification, filtration criteria | MSE wall design, IS codes | Basic type/function matching |
| IS Code | Title / Scope |
|---|---|
| IS 1498:1970 | Classification and identification of soils for general engineering purposes |
| IS 1892:1979 | Code of practice for subsurface investigation for foundations |
| IS 2131:1981 | Method for Standard Penetration Test for soils |
| IS 4434:1978 | Code of practice for in-situ vane shear test |
| IS 4968:1976 | Method for subsurface sounding for soils (dynamic and static cone penetration) |
| IS 2720 (multi-part) | Methods of test for soils — index & engineering property determination |
| IS 6403:1981 | Code of practice for determination of bearing capacity of shallow foundations |
| IS 8009:1976 | Code of practice for calculation of settlement of foundations (plate load test) |
| IS 15462:2004 | Geosynthetics — terms and definitions |
| IRC SP-59:2019 | Guidelines for use of geosynthetics in road pavements |
1. A soil has void ratio 0.8 and water content 22% with Gs = 2.70. Find the degree of saturation.
2. Two clay layers of thickness 2 m and 4 m have permeabilities \(2\times10^{-8}\) and \(5\times10^{-9}\ \text{m/s}\) respectively. Find the equivalent vertical permeability.
3. A retaining wall retains sand with \(\phi=32°\), \(\gamma=17.5\ \text{kN/m}^3\), height 4 m. Find the total active thrust per metre length using Rankine's theory.
4. A clay layer 5 m thick (single drainage) has \(c_v = 4\ \dfrac{\text{m}^2}{\text{year}}\). Find time to reach 90% consolidation.
5. A CU triaxial test on saturated clay gives B = 0.98. What does this indicate about the sample?