Foundation Engineering applies the soil mechanics principles of the previous module to the design of the structural elements that transfer building loads to the ground — from selecting a foundation type and computing bearing capacity and settlement, through shallow and deep foundation design and field testing, to slope stability, embankments, dams, earth-retaining structures and ground improvement. Every formula, IS code reference, diagram, solved example and exam-pattern table is included.
After studying this chapter you will be able to:
Prerequisite: Soil Mechanics (bearing capacity, settlement and earth pressure formulas here build directly on the shear strength, consolidation and classification concepts covered there). Leads to: Hydraulics, which covers flow and hydraulic structures relevant to dam spillways and seepage analysis introduced here.
| Category | Type | Depth / Description | Best Suited For |
|---|---|---|---|
| Shallow (\(D_f/B \le 1\)) | Strip / Wall footing | Below load-bearing walls; continuous | Masonry / RCC walls on firm soil |
| Isolated / Spread footing | Single column; square, rectangular or circular | Individual columns; good bearing capacity | |
| Combined footing | Two columns on single slab | Closely spaced columns; edge columns near property line | |
| Raft / Mat foundation | Entire building on single slab; \(D_f\) moderate | Weak soils; heavy loads; high water table | |
| Deep (\(D_f/B > 1\)) | Pile foundation | Slender columns driven/cast into soil | Soft/compressible soils; heavy structures |
| Pier / Drilled shaft | Large diameter bored shaft; \(D > 600\ \text{mm}\) | Bridges, tall buildings; hard strata at depth | |
| Well foundation (Caisson) | Hollow sinking structure; used in rivers | Bridge piers in river beds; scour zones | |
| Basement / Buoyancy raft | Deep slab + walls; uses buoyancy principle | Soft clays; skyscrapers; basements |
Gross pressure intensity \(q\) is the total load divided by the base area of the footing. The net pressure intensity subtracts the overburden pressure that already existed before excavation. Safe Bearing Capacity (SBC) divides the ultimate bearing capacity by a factor of safety, and the final allowable bearing capacity is the lesser of the SBC and the bearing capacity based on the settlement criterion — settlement, not shear failure, is often the governing check for wide footings on sand.
| IS Code | Title / Scope |
|---|---|
| IS 1904 : 2006 | Code of Practice for Design and Construction of Foundations in Soils — General Requirements |
| IS 6403 : 1981 | Code of Practice for Determination of Bearing Capacity of Shallow Foundations |
| IS 8009 (Pt 1) : 1976 | Settlement of Shallow Foundations — Granular Soils |
| IS 8009 (Pt 2) : 1980 | Settlement of Foundations on Cohesive Soils |
| IS 2911 (Pt 1–4) | Design and Construction of Pile Foundations |
| IS 3955 : 1967 | Code of Practice for Design and Construction of Well Foundations |
| IS 9527 : 1980 | Instrumentation for Earth and Rockfill Dams |
| IS 7894 : 1975 | Code of Practice for Stability Analysis of Earth Dams |
Gross pressure intensity: \( q = \dfrac{\text{Total load}}{\text{Base area of footing}} \)
Net pressure intensity: \( q_{net} = q - \gamma D_f \)
Safe Bearing Capacity: \( SBC = \dfrac{q_f}{FOS} \) (\(q_f\) = ultimate bearing capacity)
Net Safe Bearing Capacity: \( q_{net,safe} = \dfrac{q_f - \gamma D_f}{FOS} \)
Allowable Bearing Capacity: \( q_{allow} = \min(SBC,\ q_{settlement}) \)
\( D_f \ge 0.5\ \text{m} \) (minimum)
Below the zone of seasonal volume change in expansive clays
Below frost penetration depth (cold regions)
Given: A footing has ultimate bearing capacity \(q_f = 300\ \text{kPa}\), founding depth \(D_f = 1.5\ \text{m}\), soil unit weight \(\gamma = 18\ \text{kN/m}^3\), FOS = 3.
Solution:
Overburden pressure: \( \gamma D_f = 18\times1.5 = 27\ \text{kPa} \)
\( q_{net,safe} = \dfrac{300-27}{3} = \dfrac{273}{3} = 91\ \text{kPa} \)
Answer: Net safe bearing capacity \( = 91\ \text{kPa} \).
Given: A 20-storey building is to be built on soft marine clay 25 m thick over bedrock, with heavy column loads and a high water table.
Solution:
Soft, deep compressible clay rules out shallow footings (excessive settlement); bedrock is too deep for economical piles to bedrock, so friction piles or a pile-supported raft are indicated; the high water table favours a buoyancy-assisted raft or basement combined with piles to control settlement.
Answer: A piled raft (combined pile + raft foundation) is the typical solution for this combination of soft soil, heavy load and high water table.
Fig. 1.1 — Foundation types: Isolated footing · Strip/Wall footing · Raft/Mat · Pile foundation.
Terzaghi's theory applies to shallow foundations (\(D_f/B \le 1\)) undergoing general shear failure. It assumes a rigid, perfectly rough base and neglects shear resistance of the soil above the base level (treating it purely as a surcharge).
| Failure Mode | Soil Type | Characteristics | Identifiable By |
|---|---|---|---|
| General shear failure | Dense sand / stiff clay | Well-defined slip surface extends to ground; sudden collapse; heaving on sides | Clear peak in load–settlement curve; surface heaving |
| Local shear failure | Medium-dense sand / medium clay | Slip surface does not reach ground; progressive failure; slight heaving | No clear peak; moderate settlement before failure |
| Punching shear failure | Loose sand / soft clay | Footing punches into soil; no slip surface; no heaving | Continuous settlement without peak; no surface movement |
Meyerhof extended Terzaghi's equation into a general form (adopted by IS 6403) that adds shape, depth and load-inclination correction factors to each of the three bearing-capacity terms, making it applicable to rectangular footings, footings embedded deeper than \(B\), and footings under inclined loads — none of which Terzaghi's original equation handles well.
For saturated clays under the undrained (short-term) condition immediately after construction, the friction angle \(\phi_u = 0\), which collapses the general equation dramatically: \(N_q = 1\), \(N_\gamma = 0\), and \(N_c = 5.14\) (Prandtl). This φ=0 analysis governs the critical short-term stability of foundations, embankments and excavations in clay.
| Test | Formula / Correlation | Remarks |
|---|---|---|
| Standard Penetration Test (SPT) | \(q_a = 10N\) for \(B \le 1.2\) m (Teng's method); IS 6403 gives tabulated values vs. N | N = corrected blow count; overburden, energy-ratio and dilatancy corrections applied first |
| Plate Load Test (PLT) | \(q_{ult}(\text{footing}) = q_{ult}(\text{plate})\) for clay; settlement scaling relations for sand | Plate 300 mm or 450 mm square; scale effects are significant, especially for sand |
| Cone Penetration Test (CPT) | \(q_c/N \approx 0.4\) (silty clay) to 1.0 (coarse sand); \(S_u = q_c/N_{kt}\) | Continuous profile; no sample recovered; widely used for offshore and soft-soil sites |
A high water table reduces bearing capacity significantly, especially in sands, because the effective (submerged) unit weight \(\gamma'\) is roughly half the bulk unit weight — and bearing capacity is directly proportional to unit weight through the surcharge (\(N_q\)) and self-weight (\(N_\gamma\)) terms. The correction to use depends on whether the water table lies above, at, or below the footing base.
\( q_u = c\,N_c + q\,N_q + 0.5\,\gamma\,B\,N_\gamma \)
where \(q = \gamma D_f\) (surcharge); \(c\)=cohesion; \(B\)=width; \(D_f\)=depth
\( N_q = \dfrac{a^2}{2\cos^2(45+\phi/2)} \), \(a = e^{(0.75\pi-\phi/2)\tan\phi}\)
\( N_c = \dfrac{N_q-1}{\tan\phi} \) (\(\phi>0\)); \(N_c=5.7\) at \(\phi=0°\) (Prandtl)
\( N_\gamma = (N_q-1)\tan(1.4\phi) \) [Terzaghi's original approximation]
Strip: \( q_u = 1.0cN_c + qN_q + 0.5\gamma B N_\gamma \)
Square: \( q_u = 1.3cN_c + qN_q + 0.4\gamma B N_\gamma \)
Circular: \( q_u = 1.3cN_c + qN_q + 0.3\gamma B N_\gamma \)
Factor of Safety: \( FOS = q_u/q_{gross} \) (typically 2.5–3.0)
\( q_u = cN_cF_{cs}F_{cd}F_{ci} + qN_qF_{qs}F_{qd}F_{qi} + 0.5\gamma B N_\gamma F_{\gamma s}F_{\gamma d}F_{\gamma i} \)
\( N_q = e^{\pi\tan\phi}\tan^2(45+\phi/2) \)
\( N_c = (N_q-1)\cot\phi \) [\(\phi=0 \Rightarrow N_c=5.14\)]
\( N_\gamma = 2(N_q+1)\tan\phi \)
Shape: \( F_{cs}=1+\tfrac{B}{L}\tfrac{N_q}{N_c} \), \( F_{qs}=1+\tfrac{B}{L}\tan\phi \), \( F_{\gamma s}=1-0.4\tfrac{B}{L} \)
Depth (\(D_f/B\le1\)): \( F_{qd}=1+2\tan\phi(1-\sin\phi)^2\tfrac{D_f}{B} \), \( F_{cd}=F_{qd}-\tfrac{1-F_{qd}}{N_q\tan\phi} \), \( F_{\gamma d}=1.0 \)
Inclination (load at angle \(\alpha\) to vertical): \( F_{qi}=F_{ci}=(1-\alpha/90°)^2 \), \( F_{\gamma i}=(1-\alpha/\phi)^2 \)
\( q_u = 5.14\,S_u\,(1+0.2\tfrac{D_f}{B})(1+0.2\tfrac{B}{L}) + \gamma D_f \)
SBC (net) \( = q_{net}/FOS \); \(FOS=3.0\) for clays (IS 6403)
WT at ground surface: \( q_u = cN_c + \gamma' D_f N_q + 0.5\gamma' B N_\gamma \)
WT at base of footing: \( q_u = cN_c + \gamma D_f N_q + 0.5\gamma' B N_\gamma \)
WT at depth \(z\) below base (\(0\le z\le B\)): \( \gamma_{avg} = \gamma' + (z/B)(\gamma-\gamma') \) used in base term
Given: A strip footing, \(B=1.5\ \text{m}\), \(D_f=1\ \text{m}\), on sand with \(\phi=30°\) (\(N_q=18.4\), \(N_\gamma=15.1\)), \(\gamma=18\ \text{kN/m}^3\), \(c=0\). Find \(q_u\).
Solution:
\( q = \gamma D_f = 18\times1 = 18\ \text{kPa} \)
\( q_u = 0 + 18\times18.4 + 0.5\times18\times1.5\times15.1 = 331.2 + 203.9 = 535.1\ \text{kPa} \)
Answer: \(q_u \approx 535\ \text{kPa}\).
Given: A square footing (\(B=L=2\ \text{m}\)), \(D_f=1.5\ \text{m}\), on saturated clay with \(S_u=40\ \text{kPa}\), \(\gamma=18\ \text{kN/m}^3\). Find \(q_u\) and the net safe bearing capacity (FOS=3).
Solution:
\( q_u = 5.14\times40\times(1+0.2\times\tfrac{1.5}{2})(1+0.2\times1) + 18\times1.5 \)
\( q_u = 205.6\times(1.15)(1.2) + 27 = 205.6\times1.38 + 27 = 283.7 + 27 = 310.7\ \text{kPa} \)
Net ultimate: \(q_{u,net} = 310.7-27=283.7\ \text{kPa}\); Net safe: \(q_{net,safe}=283.7/3\approx94.6\ \text{kPa}\)
Answer: \(q_u \approx 311\ \text{kPa}\), net safe bearing capacity \(\approx 94.6\ \text{kPa}\).
Given: A strip footing has \(\gamma_{sat}=19\ \text{kN/m}^3\); water table is at the ground surface. If the same footing on dry soil (\(\gamma=18\ \text{kN/m}^3\)) gives \(q_u=535\ \text{kPa}\) (all from surcharge+self-weight terms, \(c=0\)), estimate the reduced \(q_u\).
Solution:
\( \gamma' = \gamma_{sat}-\gamma_w = 19-9.81 = 9.19\ \text{kN/m}^3 \), roughly half of 18 kN/m³
Both surcharge and self-weight terms scale with unit weight, so \(q_u\) roughly halves: \(q_u \approx 535\times(9.19/18) \approx 273\ \text{kPa}\)
Answer: \(q_u\) drops from 535 kPa to roughly 273 kPa — nearly halved by a fully submerged condition.
Fig. 2.1 — Three shear-failure modes for shallow foundations: general (dense/stiff), local (medium) and punching (loose/soft), after Vesic (1973).
Total settlement of a foundation comprises three components: immediate (elastic) settlement \(S_i\), primary consolidation settlement \(S_c\), and secondary compression (creep) settlement \(S_s\). In sands, consolidation is essentially instantaneous, so total settlement is approximately just \(S_i\). In clays, \(S_c\) dominates, \(S_i\) is also significant, and \(S_s\) becomes important for organic soils.
Immediate settlement is computed by treating the soil as an elastic half-space and the footing as a flexible or rigid loaded area. The influence factor \(I_s\) depends on the footing's length-to-width ratio and increases without bound for an infinitely long strip footing, so a practical finite value is used instead. A rigid footing settles somewhat less at its centre than an equivalent flexible footing, since rigidity forces the settlement to be more uniform across the footing.
For a normally consolidated clay layer (OCR = 1), settlement follows the standard \(C_c\)-based logarithmic formula. For over-consolidated clay (OCR > 1), the calculation depends on whether the final stress stays below the preconsolidation pressure \(\sigma_p'\) (in which case the flatter swelling index \(C_s\) governs throughout) or crosses it (in which case \(C_s\) governs up to \(\sigma_p'\) and \(C_c\) governs beyond it).
The dimensionless time factor \(T_v\) links elapsed time to the degree of consolidation \(U\%\) through Terzaghi's parabolic approximation for \(U\le60\%\) and a logarithmic approximation for \(U>60\%\). The coefficient of consolidation \(c_v\) is back-calculated from laboratory oedometer data using either Casagrande's log-time method (using \(t_{50}\)) or Taylor's square-root-of-time method (using \(t_{90}\)).
After primary consolidation is essentially complete, void ratio continues to decrease slowly under constant effective stress due to plastic re-adjustment of the soil skeleton — this secondary compression is governed by the secondary compression index \(C_\alpha\), important for highly organic soils (peat, muskeg) and soft marine clays.
Boussinesq's solution gives the increase in vertical stress at depth due to a point load on an elastic half-space; it is integrated to give an influence factor for a uniformly loaded rectangular area (via Newmark's chart or tabulated formulas). The simplified 2:1 approximation assumes the load spreads at a slope of 2 (vertical) to 1 (horizontal) and is convenient for quick hand estimates. Westergaard's equation instead assumes alternating rigid horizontal layers (appropriate for stratified/anisotropic soils) and gives lower stresses than Boussinesq for the same load.
| Foundation Type / Structure | Max Total Settlement | Max Differential Settlement | Angular Distortion |
|---|---|---|---|
| Isolated footing on sand | 50 mm | 25 mm | 1/300 |
| Isolated footing on clay | 65 mm | 40 mm | 1/300 |
| Raft on sand | 75 mm | 25 mm | 1/300 |
| Raft on clay | 100 mm | 40 mm | 1/300 |
| Framed structures (all) | — | — | 1/500 (Skempton & MacDonald) |
| Water tanks, silos | — | — | 1/200–1/500 |
\( S_{total} = S_i + S_c + S_s \)
In sands: \(S_{total} \approx S_i\); in clays: \(S_c\) dominates
\( S_i = q\,B\,\dfrac{1-\mu^2}{E_s}\,I_s \)
\(I_s\): Square \(\approx0.82\); Rectangle (\(L/B=2\)) \(\approx1.20\); Strip \(\approx1.53\) (practical)
Rigid footing: \(S_i \approx 0.8\times\)(flexible centre value)
NC clay: \( S_c = \dfrac{C_cH}{1+e_0}\log\dfrac{\sigma_{v0}'+\Delta\sigma}{\sigma_{v0}'} \)
OC clay, stays OC: \( S_c = \dfrac{C_sH}{1+e_0}\log\dfrac{\sigma_{v0}'+\Delta\sigma}{\sigma_{v0}'} \)
OC clay, crosses \(\sigma_p'\): \( S_c = \dfrac{C_sH}{1+e_0}\log\dfrac{\sigma_p'}{\sigma_{v0}'} + \dfrac{C_cH}{1+e_0}\log\dfrac{\sigma_{v0}'+\Delta\sigma}{\sigma_p'} \)
Skempton: \(C_c\approx0.009(w_L-10)\) (undisturbed); \(\approx0.007(w_L-10)\) (remoulded)
\( T_v = \dfrac{c_v t}{H_{dr}^2} \) (\(H_{dr}=H\) one-way; \(H/2\) two-way)
\(U\le60\%\): \( T_v=\dfrac{\pi}{4}\left(\dfrac{U}{100}\right)^2 \); \(U>60\%\): \( T_v=1.781-0.933\log(100-U) \)
\(U=50\%\to T_v=0.197\); \(U=90\%\to T_v=0.848\); \(U=95\%\to T_v=1.129\)
Casagrande: \( c_v=0.197H_{dr}^2/t_{50} \); Taylor: \( c_v=0.848H_{dr}^2/t_{90} \)
\( S_s = C_\alpha H\log(t_2/t_1) \)
\(C_\alpha \approx 0.04\text{–}0.06\,C_c\) (Mesri & Godlewski); 0.04 inorganic, 0.06 organic
Point load: \( \Delta\sigma_z = \dfrac{3P z^3}{2\pi(r^2+z^2)^{5/2}} = \dfrac{P\,I_B}{z^2} \)
Rectangular area: \( \Delta\sigma_z = q\,I_r \) (\(I_r\) from Newmark's chart, \(m=L/z\), \(n=B/z\))
2:1 method: \( \Delta\sigma_z = \dfrac{Q}{(B+z)(L+z)} \)
Westergaard: \( \Delta\sigma_z = \dfrac{PA}{\pi z^2(A+r^2/z^2)^{3/2}} \), \(A=\dfrac{1-2\mu}{2-2\mu}\)
Given: A 4 m thick NC clay layer, \(e_0=1.0\), \(C_c=0.30\), \(\sigma_{v0}'=100\ \text{kPa}\), \(\Delta\sigma=50\ \text{kPa}\). Find \(S_c\).
Solution:
\( S_c=\dfrac{0.30\times4}{1+1.0}\log\left(\dfrac{150}{100}\right)=0.60\times0.176=0.1057\ \text{m} \)
Answer: \(S_c\approx106\ \text{mm}\).
Given: A doubly-drained clay layer 6 m thick, \(c_v=3\ \text{m}^2/\text{yr}\). Find time for 90% consolidation.
Solution:
\(H_{dr}=6/2=3\ \text{m}\); at \(U=90\%\), \(T_v=0.848\)
\( t=\dfrac{T_vH_{dr}^2}{c_v}=\dfrac{0.848\times9}{3}=2.544\ \text{years} \)
Answer: \(t\approx2.54\) years.
Given: A flexible square footing, \(B=2\ \text{m}\), net pressure \(q=150\ \text{kPa}\), \(E_s=15{,}000\ \text{kPa}\), \(\mu=0.3\), \(I_s=0.82\).
Solution:
\( S_i=150\times2\times\dfrac{1-0.3^2}{15000}\times0.82=150\times2\times0.0000607\times0.82 \)
\( S_i\approx0.0149\ \text{m}=14.9\ \text{mm} \)
Answer: \(S_i\approx14.9\ \text{mm}\).
Fig. 3.1 — Total settlement vs. time: instantaneous elastic settlement Si, then primary consolidation Sc as excess pore pressure dissipates, followed by slow secondary compression Ss.
The design proceeds in a fixed sequence: (1) size the base area from the allowable bearing capacity, (2) fix the footing dimensions \(B\times L\), (3) compute the net upward pressure that governs the structural design, (4) check one-way (beam) shear at distance \(d\) from the column face, (5) check two-way (punching) shear at \(d/2\) from the column face — this often governs the required depth, (6) compute the bending moment at the column face for flexural reinforcement, and (7) check the development length of that reinforcement.
A 300 mm or 450 mm square steel plate (minimum 25 mm thick) is loaded in increments and settlement measured, until failure or twice the design load or 25 mm settlement is reached. For clay, the ultimate bearing capacity is essentially independent of the plate size, so \(q_{ult}(\text{footing}) = q_{ult}(\text{plate})\) directly. For sand, however, bearing capacity increases with size, so settlement must be scaled using an empirical relation. The test has significant limitations: it is affected by local heterogeneity and scale effects, cannot capture long-term consolidation settlement, and is unreliable for deposits thicker than about twice the plate width.
A split-spoon sampler (50.8 mm OD, 38.1 mm ID) is driven by a 63.5 kg hammer falling 750 mm; a 150 mm seating drive is discarded, then N is recorded as the blow count for the next 300 mm. Three corrections are applied before using N in design: an overburden correction (accounting for the fact that the same soil resists penetration less at shallow depth), an energy correction to a standard 60% energy ratio, and a dilatancy correction for fine saturated sand when \(N>15\) (dense saturated fine sand dilates during rapid shearing, temporarily raising apparent resistance, and the correction removes this artifact).
Base area: \( A = \dfrac{P+W_f}{q_a} \) (\(W_f\approx0.10P\))
Net pressure: \( q_{net} = P/A \)
One-way shear: \( V_u = q_{net}B(\text{overhang}-d) \le \tau_c B d \)
Two-way (punching) shear: \( \tau_v = \dfrac{V_u}{b_0 d} \le k_s\times0.25\sqrt{f_{ck}} \), \(k_s=0.5+\beta_c\)
Bending moment at column face: \( M_u = q_{net}B\dfrac{(\text{overhang})^2}{2} \) per unit width
\( k_s = q/S \) (pressure per unit settlement)
\( k_s \approx \dfrac{E_s}{B(1-\mu^2)I} \)
Sand: \( k_s(B) = k_s(0.3)/B \); Clay: \(k_s\) independent of \(B\) (approx.)
Compensated raft: \( D_f = \dfrac{P}{\gamma_{soil}A} \)
\( q_{ult}(\text{footing}) = q_{ult}(\text{plate}) \) — clay only
Clay settlement: \( S_f = S_p\times(B_f/B_p) \) — linear
Sand settlement: \( S_f = S_p\left[\dfrac{B_f(B_p+0.3)}{B_p(B_f+0.3)}\right]^2 \) (Terzaghi & Peck)
Overburden: \( N_1 = N\sqrt{95.76/\sigma_v'} \) or \( C_N=0.77\log(20/\sigma_v') \)
Energy: \( N_{60} = N\times E_r/60 \) (India \(E_r\approx60\%\))
Dilatancy (\(N>15\), fine saturated sand): \( N_{cor} = 15+0.5(N-15) \)
\(\phi \approx 28°+0.4°N\) (sands, approximate)
Teng's SBC: \( q_s = 34(N-3)\left[\dfrac{B+0.3}{2B}\right]^2 + 5.4(100+N^2)\dfrac{D_f}{B} \) (kPa, \(B\ge1.2\)m)
Friction ratio: \( F_r = f_s/q_c \) (clay > 4%; sand < 1%)
Undrained strength: \( S_u = \dfrac{q_c-\sigma_{v0}}{N_{kt}} \), \(N_{kt}=10\text{–}20\)
Given: A column carries \(P=800\ \text{kN}\); allowable bearing capacity \(q_a=200\ \text{kPa}\). Find the required square footing size.
Solution:
\( W_f\approx0.10\times800=80\ \text{kN} \); \( A=\dfrac{800+80}{200}=4.4\ \text{m}^2 \)
\( B=\sqrt{4.4}\approx2.10\ \text{m} \)
Answer: Provide a \(2.1\ \text{m}\times2.1\ \text{m}\) square footing.
Given: A field SPT test in fine saturated sand gives \(N=28\). Find the dilatancy-corrected N-value.
Solution:
Since \(N=28>15\): \( N_{cor}=15+0.5(28-15)=15+6.5=21.5\approx22 \)
Answer: \(N_{cor}\approx22\) — used in place of the raw field value of 28 for design.
Given: A 300 mm plate settles 8 mm at the design pressure on sand. Find the expected settlement of a 2 m wide footing at the same pressure.
Solution:
\( S_f = 8\times\left[\dfrac{2(0.3+0.3)}{0.3(2+0.3)}\right]^2 = 8\times\left[\dfrac{1.2}{0.69}\right]^2 = 8\times(1.739)^2 = 8\times3.02 \)
Answer: \(S_f \approx 24.2\ \text{mm}\) — settlement of the full-size footing is roughly 3 times the plate's settlement.
Fig. 4.1 — Critical sections for isolated footing design: one-way (beam) shear at distance d, two-way (punching) shear at d/2, bending checked at the column face.
| Classification | Types | Details |
|---|---|---|
| By material | Timber, Steel, Concrete (precast/cast-in-situ), Composite | RCC precast most common in India; steel H-piles for industrial structures |
| By load transfer | End-bearing pile, Friction (skin-friction) pile, Combined | End-bearing transfers load to a hard stratum; friction piles transfer load via skin resistance along the shaft |
| By installation | Driven (displacement), Bored (non-displacement), Driven-cast-in-situ (Franki pile) | Driven piles densify sand around the pile; bored piles are preferred in urban areas to reduce vibration |
| By function | Compressive, Tension (anchor), Lateral resistance, Batter piles | Batter (raker) piles resist horizontal/inclined loads; tension piles resist uplift |
The ultimate capacity of a single pile combines end-bearing resistance at the tip and skin friction along the shaft. In clay, skin friction is usually computed by the α-method (total stress, using an empirical adhesion factor that decreases as undrained strength increases). In both sand and clay, the more general β-method (effective stress) can be used, relating skin friction to the effective overburden stress through a lateral earth pressure coefficient and the pile-soil interface friction angle.
Piles in a closely-spaced group interact through overlapping stress zones, so the group capacity is generally less than the sum of individual pile capacities — captured by the group efficiency \(\eta\), most commonly estimated by the Converse-Labarre formula. For pile groups in clay, a block failure check (treating the entire group plus enclosed soil as a single large pier) must also be performed, and the design capacity is the lesser of the two. Settlement of a pile group is typically estimated using Meyerhof's equivalent-raft concept, placing a fictitious raft at some depth within the pile length and analysing settlement of that raft.
Negative skin friction develops when the soil surrounding a pile settles more than the pile itself — typically where a pile passes through a recently placed fill, a consolidating soft clay, or a layer subject to groundwater lowering. Instead of resisting the load, the downward-moving soil drags the pile down, so the skin friction over that zone acts as an additional downward load rather than a supporting resistance.
Dynamic formulae relate pile capacity to the observed "set" (penetration per hammer blow) during driving. The Engineering News Record (ENR) formula is simple but crude (implicit FOS ≈ 6); the Hiley formula is more refined, accounting for hammer efficiency and elastic compression losses. Modern practice (and IS 2911) prefers static analysis or an actual pile load test over dynamic formulae, and for large projects, wave equation analysis (based on stress-wave propagation theory) gives the most accurate prediction.
A well foundation consists of a cutting edge, well curb, steining (the main hollow shaft wall), bottom plug, top plug and well cap, sunk under its own weight (reduced by buoyancy) with skin friction resisting sinking. The grip length below the scour level provides lateral stability and increases with the proportion of live/dynamic load carried. Tilt and shift during sinking are critical construction risks, tightly limited by IRC 78.
\( Q_u = Q_p + Q_s \) (point + skin friction)
Point resistance (sand, Meyerhof): \( Q_p = A_p\,\sigma_v'\,N_q^{*} \) (limited to \(q_l=50\tan\phi\) kPa)
Point resistance (clay): \( Q_p = A_p\times9\,S_u \)
Safe capacity: \( Q_a = Q_u/FOS \); \(FOS=2.5\) (IS 2911, single pile)
α-method (clay, total stress): \( Q_s = \sum(\alpha_i\,S_{ui}\,A_{si}) \)
\(\alpha=1.0\) for \(S_u<25\) kPa; tapers to 0.5 for \(S_u>75\) kPa (API)
β-method (sand/clay, effective stress): \( Q_s = \sum(\beta_i\,\sigma_{vi}'\,A_{si}) \), \(\beta=K\tan\delta\)
\(K=0.5\text{–}1.0\,K_0\) (bored); \(K=K_0\) to \(2K_0\) (driven)
\( \eta = 1 - \theta\dfrac{(m-1)n+(n-1)m}{90\,m\,n} \), \(\theta=\arctan(D/s)\) in degrees
Block failure (clay): \( Q_{block}=2(m+n-2)sLS_u + N_c S_u(ms)(ns) \)
\( Q_{ug} = \min(\eta\,n\,Q_{u,single},\ Q_{block}) \)
Min. spacing (IS 2911): friction piles \(3D\); end-bearing piles \(2.5D\)
ENR: \( Q_a = \dfrac{WH}{S+C} \) (\(C=25.4\) mm drop hammer; \(2.54\) mm steam/hydraulic)
Hiley: \( Q_u = \dfrac{\eta_h WH e_h}{S+c/2} \)
Steining thickness (approx.): \( t = D_{outer}/(H/D)^{1/3} \)
Grip length: \( L_g = 0.5D \) (dead load) to \(1.0D\) (live+dynamic); min = 1.5 m
IRC 78 limits: max tilt 1 in 80; max shift = 150 mm
Given: A driven pile, diameter 0.4 m, length 12 m, in uniform clay with \(S_u=50\ \text{kPa}\), \(\alpha=0.75\). Find the safe pile capacity (\(FOS=2.5\)), ignoring point resistance.
Solution:
Shaft area: \(A_s=\pi\times0.4\times12=15.08\ \text{m}^2\)
\( Q_s = 0.75\times50\times15.08 = 565.5\ \text{kN} \)
\( Q_p = A_p\times9S_u = (\pi/4\times0.4^2)\times9\times50=0.1257\times450=56.5\ \text{kN} \)
\( Q_u=565.5+56.5=622\ \text{kN} \); \( Q_a=622/2.5=248.8\ \text{kN} \)
Answer: Safe pile capacity \(\approx249\ \text{kN}\).
Given: A 3×3 pile group, pile diameter \(D=0.4\ \text{m}\), spacing \(s=1.2\ \text{m}\) (i.e. \(3D\)). Find the group efficiency.
Solution:
\( \theta=\arctan(0.4/1.2)=\arctan(0.333)=18.43° \)
\( \eta=1-18.43\times\dfrac{(3-1)3+(3-1)3}{90\times3\times3}=1-18.43\times\dfrac{12}{810}=1-0.273=0.727 \)
Answer: Group efficiency \(\eta\approx0.73\) (73%).
Given: A drop hammer of weight \(W=30\ \text{kN}\) falls \(H=1000\ \text{mm}\), giving a final set \(S=8\ \text{mm}\) per blow. Find the allowable pile capacity.
Solution:
\( Q_a=\dfrac{WH}{S+C}=\dfrac{30\times1000}{8+25.4}=\dfrac{30000}{33.4}\approx898\ \text{kN} \)
Answer: \(Q_a\approx898\ \text{kN}\) (already includes an implicit FOS \(\approx6\)).
Fig. 5.1 — Pile group: individual pile capacities overlap in soft soil, so group capacity uses efficiency η and is checked against block failure of the entire enclosed soil mass.
| Failure Type | Description | Common In |
|---|---|---|
| Plane (translational) failure | Sliding along a planar surface parallel to slope; shallow | Stratified soils; rock slopes with weak planes |
| Rotational (circular) failure | Curved slip surface; toe / slope / base circle | Homogeneous clay slopes; embankments |
| Wedge failure | Sliding of triangular wedge; planar or non-planar | Rock cuts; highway slopes |
| Compound failure | Part planar + part curved | Non-homogeneous soils |
| Flow failure | Rapid shear of liquefied or sensitive soil | Quick clay; loose saturated sand (liquefaction) |
| Progressive failure | Gradual strain softening; peak → residual strength | Stiff fissured clays; OC clays |
Slope stability FOS is defined as the ratio of resisting moment to driving (overturning) moment, or equivalently the ratio of the available shear strength to the shear stress actually mobilised on the trial slip surface. IS 7894 specifies different minimum FOS values depending on the loading condition — the required FOS is lowest for the transient rapid-drawdown condition (1.2), since it involves the least certainty about pore pressures, and highest for long-term steady seepage (1.5).
For a slope of essentially infinite lateral extent (a long, uniform natural or cut slope with a slip surface parallel to the ground surface), the FOS reduces to a simple closed-form expression. For dry cohesionless soil, FOS depends only on the ratio of \(\tan\phi'\) to \(\tan\beta\) — failure is imminent exactly when the slope angle equals the friction angle. Seepage flowing parallel to the slope roughly halves this FOS for typical soils, since the destabilising pore pressure reduces the effective normal stress on the slip plane.
The slip mass above a trial circular surface is divided into vertical slices; the FOS is computed as the ratio of the sum of resisting forces (from cohesion and friction on each slice base) to the sum of driving forces (the tangential component of each slice's weight). Fellenius's ordinary method of slices ignores the forces between adjacent slices entirely — this makes the method conservative (it underestimates FOS), with errors of up to 60% possible when pore pressures are high, but it remains useful as a quick, hand-calculable check.
Bishop's simplified method improves on Fellenius by accounting for the normal (though not shear) forces between adjacent slices, giving a significantly more accurate FOS at the cost of requiring iteration — the FOS appears on both sides of the governing equation (through the term \(m_{\alpha i}\)) and converges after typically 3–4 iterations. The pore pressure ratio \(r_u\) conveniently expresses pore pressure as a fraction of the total overburden stress, ranging from 0 (dry slope) to about 0.5 (fully saturated, the worst case). Bishop's simplified method is standard for routine slope design; Spencer's and Morgenstern-Price's methods (which satisfy full force and moment equilibrium) are reserved for critical projects.
Taylor's method uses a dimensionless stability number to relate the critical height of a slope (the height at which FOS = 1) to cohesion, unit weight, slope angle and friction angle, all pre-computed into charts. For the special case of a vertical cut in saturated clay under undrained conditions (\(\phi=0\)), the stability number takes the well-known value 3.83, giving a simple direct formula for the critical height.
\( FOS = \dfrac{\text{Resisting moment}}{\text{Driving moment}} = \dfrac{c'/F_{c,mob}}{\tan\phi'/F_{\phi,mob}} \)
End of construction (undrained): \(FOS \ge 1.4\)
Steady seepage / long-term: \(FOS \ge 1.5\)
Rapid drawdown: \(FOS \ge 1.2\) (most critical)
Dry cohesionless: \( FOS = \dfrac{\tan\phi'}{\tan\beta} \); critical at \(\beta=\phi'\)
With seepage parallel to slope: \( FOS = \dfrac{\gamma'}{\gamma_{sat}}\cdot\dfrac{\tan\phi'}{\tan\beta} \)
Cohesive, no seepage: \( FOS = \dfrac{c'+(\gamma z\cos^2\beta-u)\tan\phi'}{\gamma z\sin\beta\cos\beta} \)
Taylor's stability number: \( S_n = \dfrac{c}{\gamma H F} \)
\( FOS = \dfrac{\sum[c'l_i+(W_i\cos\alpha_i-u_il_i)\tan\phi']}{\sum(W_i\sin\alpha_i)} \)
Ignores inter-slice forces — conservative (underestimates FOS).
\( FOS = \dfrac{\sum\left[\dfrac{c'b+(W_i-u_ib)\tan\phi'}{m_{\alpha i}}\right]}{\sum(W_i\sin\alpha_i)} \)
\( m_{\alpha i} = \cos\alpha_i + \dfrac{\sin\alpha_i\tan\phi'}{FOS} \) (iterate to converge)
Pore pressure ratio: \( r_u = \dfrac{u}{\gamma h_{total}} \) (0=dry, 0.5=saturated)
Critical height (Taylor): \( H_c = \dfrac{4S_u}{\gamma}\left\{\dfrac{\sin\beta\cos\phi}{1-\cos(\beta-\phi)}\right\} \)
For \(\phi=0\) (undrained clay): \( H_c = \dfrac{N_s S_u}{\gamma} \)
Vertical cut, saturated clay: \(N_s=3.83\), so \( H_c = \dfrac{4S_u}{\gamma} \)
Given: A long sand slope has \(\phi'=32°\), slope angle \(\beta=24°\). Find the FOS.
Solution:
\( FOS = \dfrac{\tan32°}{\tan24°} = \dfrac{0.625}{0.445} \approx 1.40 \)
Answer: \(FOS\approx1.40\) — stable, but not by a large margin.
Given: A saturated clay has \(S_u=45\ \text{kPa}\), \(\gamma=18\ \text{kN/m}^3\). Find the critical height of an unsupported vertical cut.
Solution:
\( H_c = \dfrac{4\times45}{18} = \dfrac{180}{18} = 10\ \text{m} \)
Answer: \(H_c=10\ \text{m}\) — beyond this depth, the vertical cut is theoretically unstable without support.
Given: Same slope as Example 1 (\(\phi'=32°, \beta=24°\)), but now with steady seepage parallel to the slope, \(\gamma_{sat}=19.5\ \text{kN/m}^3\), \(\gamma'=9.7\ \text{kN/m}^3\).
Solution:
\( FOS = \dfrac{9.7}{19.5}\times1.40 = 0.497\times1.40 \approx 0.70 \)
Answer: \(FOS\approx0.70 < 1.0\) — the slope becomes unstable under seepage, illustrating why seepage-parallel-to-slope roughly halves the dry FOS.
Fig. 6.1 — Swedish circle / Bishop's method: slip mass divided into vertical slices; FOS = Σ(resisting) / Σ(driving).
| Component | Function | Material Requirements |
|---|---|---|
| Core (Impervious zone) | Reduces seepage; watertight | Silty clay / clayey silt; \(k<10^{-7}\ \text{m/s}\); placed at optimum moisture content |
| Shell (Pervious zone) | Structural stability; carries loads | Sand / gravel / rockfill; \(k>10^{-4}\ \text{m/s}\); free draining |
| Filter / Transition zone | Prevents piping; allows drainage | Graded filter (Terzaghi's filter criteria); placed between core and shell |
| Toe drain / drainage blanket | Intercepts seepage; prevents softening at toe | Gravel; connects to outlet pipe |
| Rip-rap / pitching | Wave action protection on upstream face | Rock / stone; size depends on wave height and fetch |
| Freeboard | Safety margin above design flood level | Minimum 1.5–2.0 m (IS 8237) above FRL; more for wave action |
A filter placed between fine core material and coarse shell material must satisfy two competing requirements simultaneously: it must be fine enough to physically retain the base soil's particles (preventing piping), yet coarse enough to remain far more permeable than the base soil (so it doesn't itself impede drainage). Where a single filter layer cannot bridge a very large size gap, a two-layer (graded) filter is used, applying the same criteria progressively between each pair of adjacent layers.
The phreatic line is the top flow line (free water surface) within an embankment dam, along which pore pressure is atmospheric. Casagrande's parabola construction gives an approximate but very useful geometric method for locating this line in a homogeneous dam, treating the downstream toe as the focus of a parabola. Where the downstream face is sloped, a correction (Schaffernak/Pavlovsky) accounts for the fact that the phreatic line exits partway up the downstream slope, above the toe, creating a visible "seepage face".
Fill material is compacted near its optimum moisture content (OMC) to achieve maximum dry density (MDD), using either the Standard or Modified Proctor test as the laboratory reference. Field compaction is controlled by comparing the achieved field dry density to the laboratory MDD as a percentage — dam cores typically require 95–98% of Modified Proctor MDD, while the sand shell is instead controlled through relative density (targeting \(D_r \ge 70\%\), since the Proctor concept applies poorly to purely cohesionless material).
Piping control: \( \dfrac{D_{15}(\text{filter})}{D_{85}(\text{base})} < 4\text{–}5 \)
Permeability: \( \dfrac{D_{15}(\text{filter})}{D_{15}(\text{base})} > 4\text{–}5 \)
Uniformity: \( \dfrac{D_{50}(\text{filter})}{D_{50}(\text{base})} < 25 \)
Geotextile (IS 14716): AOS \(\le D_{85}\) of protected soil; permittivity \(\psi\ge0.1\) s⁻¹ (typical), \(\ge0.5\) s⁻¹ (critical drainage)
Dupuit's formula (rectangular section): \( q = \dfrac{k(H^2-h^2)}{2L} \)
Schaffernak/Pavlovsky (sloping downstream face): \( q = k\,a\sin\alpha\tan\alpha \)
\( a = \dfrac{d}{\cos\alpha} - \sqrt{\left(\dfrac{d}{\cos\alpha}\right)^2 - \dfrac{H^2}{\sin^2\alpha}} \)
Standard Proctor (IS 2720 Pt 7): 2.5 kg hammer, 300 mm drop, 3 layers × 25 blows
Modified Proctor (IS 2720 Pt 8): 4.5 kg hammer, 450 mm drop, 5 layers × 25 blows
Degree of compaction: \( DOC = \dfrac{\gamma_{d,field}}{\gamma_{d,MDD}}\times100\% \) (require ≥95–98% for dam core)
Relative density (shell sand): \( D_r = \dfrac{e_{max}-e}{e_{max}-e_{min}}\times100\% \ge 70\% \)
Zero air voids line: \( \gamma_d = \dfrac{G_s\gamma_w}{1+wG_s} \)
Given: A base soil has \(D_{85}=0.5\ \text{mm}\), \(D_{15}=0.05\ \text{mm}\). A candidate filter has \(D_{15}=1.8\ \text{mm}\). Check both filter criteria.
Solution:
Piping: \( \dfrac{1.8}{0.5}=3.6 < 4\text{–}5 \) ✓
Permeability: \( \dfrac{1.8}{0.05}=36 > 4\text{–}5 \) ✓
Answer: The filter satisfies both criteria — safe against piping and sufficiently permeable.
Given: A rectangular seepage section, \(k=2\times10^{-5}\ \text{m/s}\), upstream head \(H=6\ \text{m}\), downstream head \(h=1\ \text{m}\), base length \(L=40\ \text{m}\). Find the seepage rate per unit length.
Solution:
\( q = \dfrac{2\times10^{-5}\times(6^2-1^2)}{2\times40} = \dfrac{2\times10^{-5}\times35}{80} = 8.75\times10^{-6}\ \text{m}^3/\text{s per m} \)
Answer: \(q\approx8.75\times10^{-6}\ \text{m}^3/\text{s}\) per metre length \(\approx0.756\ \text{m}^3/\text{day per m}\).
Given: Field dry density achieved = 17.5 kN/m³; laboratory Modified Proctor MDD = 18.2 kN/m³. Check whether the specification of 96% MDD (for a dam core) is satisfied.
Solution:
\( DOC = \dfrac{17.5}{18.2}\times100 = 96.2\% \)
Answer: \(DOC\approx96.2\% \ge 96\%\) — the specification is satisfied.
Fig. 7.1 — Zoned earth embankment: impervious core, pervious shell, and phreatic (seepage) line from Casagrande's parabola, exiting above the downstream toe.
| Basis | Types | Key Features |
|---|---|---|
| By material | Earth dam, Rock-fill dam, Concrete gravity dam, Arch dam, Buttress dam, Masonry dam | Earth/rockfill = embankment (flexible) dams; concrete = rigid dams |
| By purpose | Storage, Detention (flood control), Diversion (weir/barrage), Coffer dam | Storage is most common; a barrage is a low dam with gates for irrigation diversion |
| By hydraulics | Overflow (spillway), Non-overflow | Overflow sections are usually concrete/masonry; non-overflow may be earth/rockfill |
| Earth dam by section | Homogeneous, Zoned (central core), Diaphragm type | Zoned with a central impervious core is most common (e.g., Hirakud, Tehri) |
A gravity dam must resist self-weight, hydrostatic pressure, uplift, silt pressure, wave pressure, earthquake forces, and (where applicable) ice pressure and thermal effects. Uplift pressure deserves special attention: it varies roughly linearly from full hydrostatic head at the heel toward zero (or tailwater head) at the toe, and a drainage gallery is deliberately provided to cut this uplift distribution short, reducing the total uplift force substantially. Hydrodynamic (earthquake) pressure on the submerged upstream face is estimated by Westergaard's added-mass approach, on top of the ordinary pseudo-static horizontal seismic force on the dam body itself.
Every gravity dam section is checked against overturning about the toe, sliding along its base (and along weak planes in the foundation), and excessive or unbalanced base stress. The middle-third rule — requiring the resultant of all forces to fall within the middle third of the base — ensures no tension develops at the heel, which is critical because concrete and most rock/soil interfaces cannot reliably sustain tension.
The spillway design flood is selected based on dam size and hazard classification — large dams are designed for the Probable Maximum Flood (PMF), while medium dams may use the Standard Project Flood (50–75% of PMF). Seepage through the dam body and foundation is governed by Darcy's law, and the exit gradient at the downstream toe must stay well below the critical hydraulic gradient (which triggers piping) — an FOS against piping of 4–5 is typically required, higher than the 3–4 used for ordinary seepage problems because dam failure by piping is catastrophic. Lane's weighted creep theory offers an alternative, purely empirical seepage-path-length design method.
Hydrostatic pressure: \( P_h = 0.5\gamma_w H^2 \) (at \(H/3\) from base)
Uplift (with drainage gallery, IS 6512): \( U_{mid}=\gamma_w H/3 \)
Silt pressure: \( P_s = 0.5\gamma_{sub}K_a h_s^2 \), \(\gamma_{sub}\approx9.8\ \text{kN/m}^3\)
Wave pressure: \( P_w = 2\gamma_w h_w^2 \), \(h_w=0.032\sqrt{VF}+0.763-0.271F^{1/4}\)
Hydrodynamic (Westergaard): \( P_e=0.726\,p_e H \), \(p_e=0.555\,\alpha_h\gamma_w\sqrt{Hy}\)
Ice pressure: 250 kPa (IS 6512, where present)
Overturning: \( FOS = \dfrac{\sum M_{stab}}{\sum M_{OT}} \ge 1.5 \) (normal), \(\ge1.2\) (seismic)
Sliding: \( FOS = \dfrac{\mu\sum V}{\sum H} \ge 1.5 \) (normal), \(\ge1.2\) (seismic); with cohesion: \( FOS=\dfrac{cA+\mu\sum V}{\sum H} \)
Resultant location: \( \bar{x}=\dfrac{\sum M_v}{\sum V} \); Eccentricity: \( e=B/2-\bar{x} \)
No tension: \( e\le B/6 \) (middle third rule)
Base stress: \( p_{max/min} = \dfrac{\sum V}{B}\left(1\pm\dfrac{6e}{B}\right) \)
Darcy: \( q = kiA \), \(i=\Delta h/L\)
Critical exit gradient: \( i_c=\dfrac{G_s-1}{1+e}\approx1.0 \)
FOS against piping: \( FOS = i_c/i_e \ge 4\text{–}5 \)
Lane's weighted creep ratio: \( C_w = \dfrac{L_h}{3}+L_v \), require \( C_w/H \ge \) safe ratio (8–18)
Given: A gravity dam has base width \(B=12\ \text{m}\); the resultant of all forces acts at \(\bar{x}=5.5\ \text{m}\) from the heel. Check the no-tension condition.
Solution:
\( e = B/2-\bar{x} = 6-5.5 = 0.5\ \text{m} \)
\( B/6 = 12/6 = 2\ \text{m} \)
Answer: \(e=0.5\ \text{m} < B/6=2\ \text{m}\) — resultant lies within the middle third; no tension develops at the base.
Given: A dam retains water to a height \(H=25\ \text{m}\). Find the total horizontal hydrostatic force per metre length and its point of application.
Solution:
\( P_h = 0.5\times9.81\times25^2 = 0.5\times9.81\times625 = 3065.6\ \text{kN/m} \)
Acts at \(H/3=8.33\ \text{m}\) above the base.
Answer: \(P_h\approx3066\ \text{kN/m}\), acting 8.33 m above the base.
Given: Foundation soil below a dam has \(G_s=2.66\), \(e=0.66\). The actual exit gradient computed from a flow net is 0.22. Check the FOS against piping.
Solution:
\( i_c = \dfrac{2.66-1}{1+0.66}=\dfrac{1.66}{1.66}=1.0 \)
\( FOS = 1.0/0.22 \approx 4.5 \)
Answer: \(FOS\approx4.5\) — satisfies the required 4–5 minimum for dam foundations.
Fig. 8.1 — Forces on a gravity dam: hydrostatic pressure Ph (triangular, at H/3), self-weight W, and uplift U (reduced by a drainage gallery), checked against overturning, sliding and the middle-third rule.
| Type | Description / Height | Mechanism | Common Use |
|---|---|---|---|
| Gravity retaining wall | Mass concrete/masonry; height < 3 m | Resists by self-weight; no steel required | Small cuts, garden walls |
| Cantilever retaining wall | RCC; height 3–8 m; stem + base | Moment resisted by cantilever action; soil on base provides stability | Highway embankments, bridges |
| Counterfort retaining wall | RCC; height > 8 m; stem + counterforts behind | Triangular counterforts reduce stem moment; efficient for tall walls | Large retaining walls; highway cuts |
| Buttress retaining wall | RCC; similar to counterfort but bracing in front | Buttresses in compression (more efficient material use) | Walls where exposed face is acceptable |
| Sheet pile wall | Steel/RCC sheet piles driven into ground | Cantilever or anchored; embedded depth provides passive resistance | Excavations, cofferdams, waterfront |
| Diaphragm wall | RCC cast in slurry trench | Anchored or propped; very stiff; minimal settlement to adjacent structures | Deep urban excavations; metro stations |
| Crib wall / Gabion | Interlocked boxes filled with stone | Gravity; permeable; flexible | Stream banks, green walls, low-cost applications |
| Reinforced earth wall | Metallic/geosynthetic strips + granular fill + facing panels | Reinforcement increases soil tensile resistance; self-stable mass | Highway embankment walls; ≤ 20 m height |
Rankine's theory assumes a smooth, vertical wall with a horizontal backfill, giving simple closed-form active and passive pressure coefficients. For cohesive backfill, a tension zone develops near the top of the wall, down to a depth where the net active pressure becomes zero. Coulomb's theory is more general — it accounts for wall inclination, backfill slope and wall friction through a limit-equilibrium wedge analysis, and gives a resultant force direction inclined at the wall friction angle to the wall's normal, rather than the purely horizontal/vertical resultant of Rankine's theory.
Every retaining wall must be checked for four failure modes: overturning about the toe (resisting moment from self-weight and passive resistance vs. overturning moment from active thrust), sliding along the base (base friction plus base cohesion vs. the horizontal thrust, with the weight of soil on the heel included for cantilever walls), bearing capacity of the foundation soil (no tension requires the eccentricity to stay within \(B/6\)), and overall slope stability of the wall, backfill and foundation treated together as a single slope.
For a cantilever sheet pile in cohesionless soil, the required embedment depth is found by the free earth support method: setting net driving moment equal to net resisting moment about the point of zero net pressure, solving the resulting cubic equation for depth, then adding 20–40% extra depth as a practical safety margin. An anchored sheet pile additionally has a tie-rod providing horizontal support, reducing the required embedment depth compared to a pure cantilever wall of the same height. Rowe's moment reduction accounts for the flexibility of the sheet pile section itself, which redistributes bending moment away from the theoretical (rigid) value.
\( K_a = \tan^2(45-\phi/2) = \dfrac{1-\sin\phi}{1+\sin\phi} \); \( K_p = 1/K_a \)
Active: \( \sigma_a = K_a\sigma_v' - 2c\sqrt{K_a} \); Passive: \( \sigma_p = K_p\sigma_v' + 2c\sqrt{K_p} \)
Tension crack depth: \( z_c = \dfrac{2c}{\gamma\sqrt{K_a}} \)
Total active force: \( P_a = 0.5K_a\gamma H^2 \) (cohesionless) at \(H/3\)
c-φ soil: \( P_a = 0.5K_a\gamma H^2 - 2cH\sqrt{K_a}+2c^2/\gamma \)
\( K_a = \dfrac{\sin^2(\alpha+\phi)}{\sin^2\alpha\sin(\alpha-\delta)\left[1+\sqrt{\dfrac{\sin(\phi+\delta)\sin(\phi-\beta)}{\sin(\alpha-\delta)\sin(\alpha+\beta)}}\right]^2} \)
\(\alpha\)=wall inclination; \(\beta\)=backfill slope; \(\delta\)=wall friction angle
Overturning: \( FOS = \dfrac{\sum M_R}{\sum M_O} \ge 2.0 \) (typically); min 1.5
Sliding: \( FOS = \dfrac{\mu\sum V+c_bB}{\sum H} \ge 1.5 \)
Bearing: \( q_{toe} = \dfrac{\sum V}{B}\left(1+\dfrac{6e}{B}\right) \le SBC \); no tension: \(e\le B/6\)
FOS on passive resistance: \( K_{p,design}=K_p/FOS_p \), \(FOS_p=1.5\text{–}2.0\)
Anchored: tie-rod force \( T = P_a-P_p-R_{base} \)
Section modulus: \( Z=M_{max}/f_b \)
Rowe's reduction: \( M_{actual}=M_{computed}\times M_r \)
Tension in reinforcement layer: \( T = \sigma_h\times(\text{tributary area}) \)
Min. pullout length: \( L_e \ge 1.0\ \text{m} \)
Typical spacing: \(S_v=0.5\text{–}1.0\ \text{m}\)
Given: A cantilever wall, \(H=5\ \text{m}\), retains sand \(\gamma=18\ \text{kN/m}^3\), \(\phi=32°\) (\(K_a=0.307\)). Total resisting moment about toe = 220 kN·m/m. Find \(P_a\), the overturning moment, and the FOS.
Solution:
\( P_a = 0.5\times0.307\times18\times25 = 69.1\ \text{kN/m} \), acting at \(H/3=1.67\ \text{m}\)
\( M_O = 69.1\times1.67 = 115.4\ \text{kN·m/m} \)
\( FOS = 220/115.4 \approx 1.91 \)
Answer: \(FOS\approx1.91\), just short of the preferred 2.0, but above the minimum 1.5.
Given: A cantilever sheet pile in sand has \(K_p=3.25\), \(FOS_p=2.0\). Find the design passive coefficient.
Solution:
\( K_{p,design} = 3.25/2.0 = 1.625 \)
Answer: Use \(K_{p,design}=1.625\) in the free earth support embedment calculation.
Given: A wall retains clay backfill, \(\gamma=19\ \text{kN/m}^3\), \(c=18\ \text{kPa}\), \(\phi=0\) (\(K_a=1\)). Find the tension crack depth.
Solution:
\( z_c = \dfrac{2\times18}{19\times\sqrt1} = \dfrac{36}{19} \approx1.89\ \text{m} \)
Answer: \(z_c\approx1.89\ \text{m}\) — tension cracks may form to this depth; often neglected in the resultant thrust calculation.
Fig. 9.1 — Cantilever retaining wall: active earth pressure Pa (triangular, acting at H/3 above the base), checked for overturning about the toe, sliding along the base, and bearing pressure.
Ground improvement is needed when natural soil is inadequate for the intended load — low bearing capacity, excessive settlement, liquefaction potential, or expansive behaviour. Methods are broadly classified into mechanical (densification), hydraulic (drainage/consolidation acceleration), chemical/admixture (stabilisation, grouting), and reinforcement-based techniques (stone columns, soil nailing, geosynthetics).
| Category | Techniques | Suitable Soils | Primary Effect |
|---|---|---|---|
| Densification | Vibro-compaction, dynamic compaction, compaction piles, blasting | Loose sands, fills | Increases relative density; reduces settlement and liquefaction potential |
| Preloading / consolidation acceleration | Surcharge fill, vacuum consolidation, prefabricated vertical drains (PVD) | Soft clays, silts | Accelerates consolidation; increases undrained shear strength |
| Chemical stabilisation | Lime treatment, cement stabilisation, deep soil mixing (DSM), grouting | Clays, sands | Increases stiffness/strength; reduces permeability; treats expansive soils |
| Drainage | Vertical sand drains, PVD (band drains), stone columns, dewatering | Soft clays | Reduces drainage path; accelerates consolidation |
| Reinforcement | Stone columns, soil nailing, anchors, geosynthetic reinforcement, micropiles | Clays, fills, slopes | Increases composite shear strength; prevents failure |
| Thermal | Heat treatment (vitrification), freezing | Any | Temporary or permanent strengthening; containment |
Where drainage is primarily radial (toward closely-spaced vertical drains rather than only vertically toward the top/bottom of the layer), Barron's radial consolidation theory replaces Terzaghi's 1D theory. The spacing between drains, converted to an equivalent circular "unit cell" diameter, and the horizontal coefficient of consolidation (typically several times larger than the vertical \(c_v\) for soft clays, due to natural soil anisotropy and macro-fabric) together control how fast radial consolidation proceeds. The combined effect of simultaneous vertical and radial drainage is found by treating the two as independent, multiplicative processes.
Stone columns are installed by vibro-replacement — a vibrating probe penetrates the soft clay and compacted gravel backfill is added in lifts to form a dense granular column. Because the stone column is much stiffer than the surrounding clay, it attracts a disproportionate share of the applied load (expressed by the stress concentration ratio), which both improves overall bearing capacity and reduces settlement (expressed by Priebe's settlement improvement factor). The typical failure mode is bulging of the column near the top, where lateral confinement from the surrounding soft clay is weakest.
| Stabiliser | Mechanism | Effect on Soil | Optimal Content |
|---|---|---|---|
| Lime (CaO or Ca(OH)₂) | Ion exchange (immediate) + pozzolanic reaction (long-term) | Reduces plasticity, increases OMC, reduces MDD, increases UCS with curing | 3–8% by dry weight; depends on soil PI |
| Cement (OPC) | Hydration + cementation bonds; pozzolanic reaction with soil minerals | Increases UCS rapidly; reduces permeability; effective in sands + gravels | 4–12% by dry weight; per IRC SP 89 |
| Fly ash | Pozzolanic; reacts with Ca(OH)₂; self-cementing class C fly ash | Reduces shrinkage; improves workability; lower cost | 10–30% as blended with lime/cement |
| Bitumen stabilisation | Coating particles; reduces capillary action; waterproofs | Increases cohesion; effective for sandy soils; used in roads | 4–6% for granular soils |
A heavy tamper (10–30 tonne) is repeatedly dropped from a great height (10–40 m) in a grid pattern, transmitting enough energy to densify loose fills, mine tailings, and other problematic granular deposits to significant depth. It is unsuitable for saturated fine-grained soils, since there is no time for excess pore pressure to dissipate between drops, and it must be kept well away from vibration-sensitive structures.
| Method | Process | Soil Type | Outcome |
|---|---|---|---|
| Vibro-compaction | Probe vibrated into soil; horizontal vibrations densify granular soil | Loose sands (fines content < 10%) | Relative density increases to 70–80%; liquefaction potential reduced |
| Vibro-replacement (Stone columns) | Probe penetrates; gravel backfill added and compacted in lifts | Soft clays, silts (fines content > 15%) | Composite foundation; settlement reduced; shear strength and drainage improved |
\( U_r = 1-\exp\left(-\dfrac{8T_h}{\mu}\right) \)
\( T_h = \dfrac{c_ht}{d_e^2} \); Triangular grid: \(d_e=1.05s\); Square grid: \(d_e=1.13s\)
\( \mu = \dfrac{n^2}{n^2-1}\ln n - \dfrac{3n^2-1}{4n^2} \), \(n=d_e/d_w\); for \(n>5\): \(\mu\approx\ln n-0.75\)
Combined: \( U_{total} = 1-(1-U_v)(1-U_r) \)
Stress concentration ratio: \( n_c = \sigma_c/\sigma_s \) (typically 2–5)
Area replacement ratio: \( a_s = A_c/A \)
Priebe's settlement improvement factor: \( \beta = 1+a_s(n_c-1) \)
Composite bearing capacity: \( q_c = (1-a_s)q_s + a_s q_{column} \)
Ultimate pressure (Brauns): \( q_{ult} = K_p(\sigma_{r3}+4S_u) \)
Significant improvement depth: \( D_i \approx n\sqrt{WH} \)
\(n\approx0.5\) (loose fills); 0.3–0.4 (silty soils); \(W\)=tamper weight, \(H\)=drop height
Energy per blow: \( E = Wg H \)
Given: PVDs installed on a square grid, spacing \(s=1.5\ \text{m}\), band drain equivalent diameter \(d_w=0.05\ \text{m}\). Find \(d_e\) and \(\mu\) (using the simplified formula for \(n>5\)).
Solution:
\( d_e = 1.13\times1.5 = 1.695\ \text{m} \)
\( n = 1.695/0.05 = 33.9 \)
\( \mu \approx \ln(33.9)-0.75 = 3.524-0.75 = 2.774 \)
Answer: \(d_e\approx1.70\ \text{m}\), \(\mu\approx2.77\).
Given: A stone column installation has area replacement ratio \(a_s=0.25\), stress concentration ratio \(n_c=4\). Find the settlement improvement factor.
Solution:
\( \beta = 1+0.25(4-1) = 1+0.75 = 1.75 \)
Answer: \(\beta=1.75\) — settlement is reduced to about \(1/1.75\approx0.57\) of the untreated value.
Given: A 15-tonne tamper is dropped from 20 m in a loose sand fill (\(n=0.5\)). Estimate the significant improvement depth.
Solution:
\( D_i = 0.5\sqrt{15\times20} = 0.5\sqrt{300} = 0.5\times17.32 = 8.66\ \text{m} \)
Answer: \(D_i\approx8.7\ \text{m}\) of significant densification depth.
Fig. 10.1 — Stone column ground improvement: gravel columns installed in soft clay, carrying a disproportionate share of load (stress concentration ratio nc), with bulging failure typically occurring near the top.
| Parameter | Value / Formula | Notes |
|---|---|---|
| \(N_c\) (Terzaghi, φ=0) | 5.7 | Strip footing; Meyerhof/Prandtl gives 5.14 |
| Terzaghi shape factors (square) | \(1.3cN_c+qN_q+0.4\gamma BN_\gamma\) | Circle same as square |
| FOS for shallow foundation | 2.5–3.0 (IS 6403) | 3.0 for clays (φ=0) |
| FOS for pile (IS 2911) | 2.5 | Single pile; 2.0 for pile group |
| Min. pile spacing (friction) | 3D | D = pile diameter |
| \(T_v\) at \(U=50\%\) | 0.197 | \(T_v=(\pi/4)(U/100)^2\) |
| \(T_v\) at \(U=90\%\) | 0.848 | \(T_v=1.781-0.933\log(100-U)\) |
| \(S_c\) (NC clay) | \([C_cH/(1+e_0)]\log[(\sigma_0+\Delta\sigma)/\sigma_0]\) | \(C_c\approx0.009(w_L-10)\) |
| \(K_a\) (Rankine) | \((1-\sin\phi)/(1+\sin\phi)=\tan^2(45-\phi/2)\) | \(K_p=1/K_a\) |
| Tension crack depth | \(z_c=2c/(\gamma\sqrt{K_a})\) | Net active pressure = 0 here |
| FOS slope (infinite, dry) | \(\tan\phi/\tan\beta\) | With seepage: \((\gamma'/\gamma_{sat})\tan\phi/\tan\beta\) |
| Critical height (vertical cut) | \(H_c=4S_u/\gamma\) | φ=0 analysis |
| Dam: middle third rule | \(e\le B/6\) | For no tension at base |
| FOS overturning (dam) | ≥ 1.5 (IS 6512) | 1.2 for seismic condition |
| Filter criteria (piping) | \(D_{15}(f)/D_{85}(b)<4\text{–}5\) | Also \(D_{15}(f)/D_{15}(b)>4\text{–}5\) |
| Dynamic compaction depth | \(D=n\sqrt{WH}\); \(n\approx0.5\) | W in tonnes, H in metres |
| PVD \(d_e\) (triangular grid) | 1.05·s | Square: 1.13·s |
| Topic | GATE Focus | ESE Focus | SSC JE Focus |
|---|---|---|---|
| Bearing capacity | Terzaghi/Meyerhof numericals; effect of water table; SPT correlation | Complete Meyerhof with shape/depth/inclination factors; IS 6403 | Definition of SBC, FOS, types of shear failure |
| Settlement | Consolidation Sc formula; Tv; time for U%; Hdr trick | Full settlement components; Boussinesq stress; IS 8009 limits | Types of settlement; consolidation vs immediate |
| Piles | Static capacity; group efficiency; PLT failure criterion | Complete IS 2911 provisions; wave equation; well foundation | Types of piles; material types; spacing rules |
| Slope stability | Fellenius; Bishop's; infinite slope FOS | All methods; stability charts; pore pressure ratio; dam conditions | Types of failures; FOS ≥ 1.5; Taylor's chart concept |
| Retaining walls | Ka, Kp; Pa computation; sliding/overturning FOS | Coulomb vs Rankine; sheet pile analysis; Rowe's reduction | Types of walls; Rankine Ka formula; stability concept |
| Ground improvement | PVD spacing; dynamic compaction depth formula | Stone columns; jet grouting; DSM; selection criteria | Types of ground improvement; OMC/MDD concept |
| IS Code | Scope |
|---|---|
| IS 1904 | General requirements for foundation design & construction |
| IS 6403 | Bearing capacity of shallow foundations (formulae) |
| IS 8009 | Settlement of foundations (Pt 1: sands; Pt 2: clays) |
| IS 2911 | Pile foundations design & construction (Pt 1–4) |
| IS 3955 | Well foundations design & construction |
| IS 2131 | Standard Penetration Test procedure |
| IS 1888 | Plate Load Test procedure |
| IS 4968 | Static Cone Penetration Test (CPT) |
| IS 7894 | Stability analysis of earth dams |
| IS 6512 | Criteria for design of solid gravity dams |
| IS 8237 | Protection of slopes of earth dams and embankments |
| IS 14716 | Geosynthetics in geotechnical applications |
| IS 14458 | Retaining walls for hill area, soil nailing |
| IRC 78 | Standard specifications for road bridges — foundations & substructure (well foundations) |
1. A strip footing, B=2m, Df=1.2m, on sand with φ=28° (Nq=14.7, Nγ=11.2), γ=17 kN/m³, c=0. Find qu (Terzaghi).
2. A clay layer 8 m thick, doubly drained, cv = 5 m²/year. Find the time for 50% consolidation.
3. A pile group of 2×3 piles, D=0.35 m, spacing 1.05 m (3D). Find the group efficiency by Converse-Labarre.
4. A retaining wall retains sand, φ=30°, γ=18 kN/m³, H=4.5 m. Find the total active thrust per metre using Rankine's theory.
5. A stone column installation has area replacement ratio as=0.3, stress concentration ratio nc=3. Find Priebe's settlement improvement factor.