10.4 Shallow and Deep Foundation Bearing Capacity and Slope Stability

Key Takeaways

  • Terzaghi's ultimate bearing capacity equation for continuous strip footings is qult = c Nc + q Nq + 0.5 γ B Nγ, modified by specific shape factors for square (1.3c Nc + q Nq + 0.4 γ B Nγ) and circular footings (1.3c Nc + q Nq + 0.3 γ B Nγ).

  • Groundwater table proximity reduces bearing capacity: a water table within depth Df modifies surcharge q, while a water table within depth B below the footing base reduces the third unit-weight term toward buoyant unit weight γ'.

  • Deep foundation pile capacity combines end-bearing and shaft friction (Qult = qp Ap + ∑ fs As); for cohesive soils, end bearing is qp = 9 cu, while skin friction is governed by the α-method (fs = α cu).

  • In pile group design, capacity is governed by the smaller of individual pile capacity summation (with Converse-Labarre efficiency) and monolithic block failure capacity.

  • Slope stability evaluation requires distinct models: infinite slopes depend on friction and seepage ratios (FS = [γ'/γsat] * [tan φ'/tan β]), while finite slopes are analyzed via Swedish circle (φ = 0), Ordinary Method of Slices, or Bishop's Simplified method (FS ≥ 1.5).

Last updated: October 2026

10.4 Shallow and Deep Foundation Bearing Capacity and Slope Stability

Foundation design bridges structural engineering and soil mechanics. Structural engineers must proportion structural footings and deep piles to transmit superstructure column loads into underlying soil strata without exceeding the shear strength of the supporting ground or causing excessive differential settlement. Furthermore, natural and engineered slopes must be stabilized against catastrophic gravitational mass wasting.


Terzaghi's Bearing Capacity Theory for Shallow Foundations

A shallow foundation is defined as one where the embedment depth is less than or equal to its width (Df≤BD_f \le B). In 1943, Karl Terzaghi published the classical ultimate bearing capacity equation for a continuous strip footing based on general shear failure:

                    Column Load P
                         |
                     +---+---+
     Ground Surface  |       |        Ground Surface
    -----------------+   Df  +-----------------
                     |       |
                     +-------+ Footing Base (Width B)
                    /|   I   |\
                   / |       | \
                  /  +-------+  \
                 /   /       \   \
                / II/         \II \
               /   /           \   \
              +---+             +---+
              <------- Zone III ------->

General shear failure divides the underlying soil mass into three distinct behavioral zones:

  • Zone I (Triangular Elastic Wedge): Directly under the base, pushes downward into the soil as a rigid body.
  • Zone II (Radial Shear Zone): Plastic shear zones along log-spiral boundaries.
  • Zone III (Rankine Passive Linear Wedge): Pushed upward against the surcharge overburden.

General Form for Continuous Strip Footings

qult=cNc+qNq+12γBNγq_{\text{ult}} = c N_c + q N_q + \frac{1}{2} \gamma B N_\gamma Where:

  • cc = soil cohesion (kPa\text{kPa})
  • q=γDfq = \gamma D_f = effective surcharge at foundation base level (kPa\text{kPa})
  • γ\gamma = unit weight of soil supporting the base wedge (kN/m3\text{kN/m}^3)
  • BB = footing width (shortest lateral dimension, m\text{m})
  • Nc,Nq,NγN_c, N_q, N_\gamma = dimensionless Terzaghi bearing capacity factors, which depend purely on the internal friction angle ϕ\phi: Nq=e2(3π/4−ϕ/2)tan⁡ϕ2cos⁡2(45∘+ϕ/2)N_q = \frac{e^{2(3\pi/4 - \phi/2)\tan \phi}}{2 \cos^2(45^\circ + \phi/2)} Nc=(Nq−1)cot⁡ϕ(for ϕ=0∘,Nc=5.7)N_c = (N_q - 1) \cot \phi \quad (\text{for } \phi = 0^\circ, N_c = 5.7) Nγ≈12(Kpγcos⁡2ϕ−1)tan⁡ϕN_\gamma \approx \frac{1}{2} \left( \frac{K_{p\gamma}}{\cos^2\phi} - 1 \right) \tan \phi

For saturated clays under undrained conditions (ϕ=0∘\phi = 0^\circ): Nc=5.7,Nq=1.0,Nγ=0N_c = 5.7, \quad N_q = 1.0, \quad N_\gamma = 0 qult=5.7cu+qq_{\text{ult}} = 5.7 c_u + q

Footing Shape Modifications

To adapt the 2D strip footing equation to 3D isolated footings, empirical shape coefficients are applied:

Footing GeometryTerzaghi Ultimate Bearing Capacity Equation (qultq_{\text{ult}})
Continuous Stripqult=cNc+qNq+0.5γBNγq_{\text{ult}} = c N_c + q N_q + 0.5 \gamma B N_\gamma
Square (B×BB \times B)qult=1.3cNc+qNq+0.4γBNγq_{\text{ult}} = 1.3 c N_c + q N_q + 0.4 \gamma B N_\gamma
Circular (Diameter BB)qult=1.3cNc+qNq+0.3γBNγq_{\text{ult}} = 1.3 c N_c + q N_q + 0.3 \gamma B N_\gamma
Rectangular (B×LB \times L)qult=cNc(1+0.3BL)+qNq+0.5γBNγ(1−0.2BL)q_{\text{ult}} = c N_c \left(1 + 0.3 \frac{B}{L}\right) + q N_q + 0.5 \gamma B N_\gamma \left(1 - 0.2 \frac{B}{L}\right)

General Bearing Capacity Equation (Meyerhof, Hansen, Vesic)

For footings subjected to inclined loads, deep embedment, or eccentricities, the General Bearing Capacity Equation incorporates shape (ss), depth (dd), and load inclination (ii) factors: qult=cNcscdcic+qNqsqdqiq+12γBNγsγdγiγq_{\text{ult}} = c N_c s_c d_c i_c + q N_q s_q d_q i_q + \frac{1}{2} \gamma B N_\gamma s_\gamma d_\gamma i_\gamma

Groundwater Table Modifications

The presence of a high water table reduces effective stresses and diminishes bearing capacity:

   Case 1: Water at depth d_w <= D_f
   -------------------------- GWT (d_w)
   /////////// Submerged Overburden
   ----------- Footing Base (D_f)
   
   Case 2: Water at depth D_f < d_w <= D_f + B
   ----------- Footing Base (D_f)
   ........... Moist Wedge
   -------------------------- GWT (d_w)
   /////////// Submerged Wedge
   -------------------------- Depth D_f + B
  1. Case 1: Groundwater Table Above Footing Base (dw≤Dfd_w \le D_f):
    • Surcharge qq becomes: q=γdw+γ′(Df−dw)q = \gamma d_w + \gamma' (D_f - d_w)
    • Unit weight in third term becomes buoyant: γ=γ′=γsat−γw\gamma = \gamma' = \gamma_{\text{sat}} - \gamma_w
  2. Case 2: Groundwater Table Within Wedge Depth (Df<dw≤Df+BD_f < d_w \le D_f + B):
    • Surcharge is unaffected: q=γDfq = \gamma D_f
    • Unit weight in third term is an effective weighted average: γˉ=γ′+(dw−DfB)(γ−γ′)\bar{\gamma} = \gamma' + \left( \frac{d_w - D_f}{B} \right) (\gamma - \gamma')
  3. Case 3: Groundwater Table Deep Below Base (dw>Df+Bd_w > D_f + B):
    • Water table has no physical effect on bearing capacity; use moist/dry γ\gamma in both terms.

Allowable Bearing Capacity & Factor of Safety

qall=qultFS(standard FS=3.0)q_{\text{all}} = \frac{q_{\text{ult}}}{FS} \quad (\text{standard } FS = 3.0) Net Allowable Bearing Capacity: qall,net=qult−qFS\text{Net Allowable Bearing Capacity: } q_{\text{all,net}} = \frac{q_{\text{ult}} - q}{FS} Allowable Column Load: Pall=qall×Afooting\text{Allowable Column Load: } P_{\text{all}} = q_{\text{all}} \times A_{\text{footing}}


Deep Foundations: Piles and Drilled Shafts

When surficial soils are too weak or compressible to support shallow footings within tolerable settlement limits, structural loads are transferred to deeper, competent strata via deep foundations.

Ultimate Axial Pile Capacity

The ultimate compressive load capacity QultQ_{\text{ult}} of an isolated single pile is the sum of its base end-bearing resistance (QpQ_p) and shaft skin friction resistance (QsQ_s): Qult=Qp+Qs=qpAp+∑fsAsQ_{\text{ult}} = Q_p + Q_s = q_p A_p + \sum f_s A_s Where:

  • ApA_p = cross-sectional area of the pile tip
  • qpq_p = unit point bearing capacity
  • As=pΔLA_s = p \Delta L = surface area of pile shaft perimeter over length segment ΔL\Delta L
  • fsf_s = unit skin friction (shaft resistance)

Cohesive Soils (Clays)

  1. End-Bearing (QpQ_p): For saturated clays under undrained conditions, Nc∗=9.0N_c^* = 9.0: qp=9cu  ⟹  Qp=9cuApq_p = 9 c_u \implies Q_p = 9 c_u A_p
  2. Skin Friction (QsQ_s) — The α\alpha-Method: fs=αcu  ⟹  Qs=∑αcupΔLf_s = \alpha c_u \implies Q_s = \sum \alpha c_u p \Delta L Where α\alpha is the empirical adhesion factor (ranging from ≈1.0\approx 1.0 for soft clays down to 0.40.4 for stiff clays).

Cohesionless Soils (Sands)

  1. End-Bearing (QpQ_p): qp=q′Nq∗≤qlimitq_p = q' N_q^* \le q_{\text{limit}} Where q′q' is the effective vertical stress at the pile tip (often capped at a critical depth Lc≈10D to 20DL_c \approx 10D \text{ to } 20D).
  2. Skin Friction (QsQ_s) — The β\beta-Method: fs=βσv′=Ktan⁡δσv′f_s = \beta \sigma'_v = K \tan \delta \sigma'_v Where KK is the lateral earth pressure coefficient and δ\delta is the interface friction angle.

Pile Group Efficiency & Block Failure

Piles are typically installed in clusters connected by a reinforced concrete pile cap. The ultimate capacity of a pile group containing nn piles is governed by the lesser of:

  1. Sum of Individual Capacities (with efficiency η\eta): Qgroup=η⋅n⋅Qult,singleQ_{\text{group}} = \eta \cdot n \cdot Q_{\text{ult,single}} Where the Converse-Labarre efficiency equation models group interaction: η=1−θ90∘[(n−1)m+(m−1)nm⋅n]\eta = 1 - \frac{\theta}{90^\circ} \left[ \frac{(n-1)m + (m-1)n}{m \cdot n} \right] Where mm is number of rows, nn is number of columns, and θ=arctan⁡(d/s)\theta = \arctan(d/s) in degrees (d=diameter,s=spacingd = \text{diameter}, s = \text{spacing}).
  2. Block Failure Capacity (QblockQ_{\text{block}}): The pile cluster and enclosed soil mass act as a single large monolithic pier of dimensions Bg×LgB_g \times L_g and depth LL: Qblock=9cu(BgLg)+2(Bg+Lg)LcuQ_{\text{block}} = 9 c_u (B_g L_g) + 2(B_g + L_g) L c_u (Note: For block failure, soil shears along soil, so α=1.0\alpha = 1.0.)

Slope Stability Analysis

Slope failures occur when gravity-induced shear stresses along a potential sliding surface exceed the available shear strength of the soil.

1. Infinite Slopes (Translational Failure)

Applies to long, planar natural slopes where the depth of the active sliding layer HH is small compared to slope length.

  • Cohesionless Soil (c′=0c' = 0), Dry or No Seepage: FS=Resisting Shear StressDriving Shear Stress=σ′tan⁡ϕ′τ=γHcos⁡βtan⁡ϕ′γHsin⁡β=tan⁡ϕ′tan⁡βFS = \frac{\text{Resisting Shear Stress}}{\text{Driving Shear Stress}} = \frac{\sigma' \tan \phi'}{\tau} = \frac{\gamma H \cos \beta \tan \phi'}{\gamma H \sin \beta} = \frac{\tan \phi'}{\tan \beta} The slope is stable against failure as long as slope angle β≤ϕ′\beta \le \phi'.

  • Cohesionless Soil with Steady Seepage Parallel to Slope (Water at Surface): FS=γ′γsattan⁡ϕ′tan⁡β≈12tan⁡ϕ′tan⁡βFS = \frac{\gamma'}{\gamma_{\text{sat}}} \frac{\tan \phi'}{\tan \beta} \approx \frac{1}{2} \frac{\tan \phi'}{\tan \beta}

    Important

    Steady seepage parallel to the slope face with water at the surface cuts the factor of safety roughly in half because buoyant unit weight γ′\gamma' is approximately half of saturated unit weight γsat\gamma_{\text{sat}}.

  • Cohesive-Frictional Soil (c′−ϕ′c' - \phi'): FS=c′γHsin⁡βcos⁡β+tan⁡ϕ′tan⁡βFS = \frac{c'}{\gamma H \sin \beta \cos \beta} + \frac{\tan \phi'}{\tan \beta}

2. Finite Slopes (Rotational Circular Arc Failure)

Applies to engineered embankments, cuts, and retaining berms.

  • Swedish Circle / Mass Method (ϕ=0∘\phi = 0^\circ Saturated Clay): FS=Resisting MomentOverturning Moment=cuLaRW⋅d=cu(Rθ)RW⋅dFS = \frac{\text{Resisting Moment}}{\text{Overturning Moment}} = \frac{c_u L_a R}{W \cdot d} = \frac{c_u (R \theta) R}{W \cdot d} Where La=RθL_a = R \theta is the circular arc length, RR is radius of the slip circle, WW is total weight of sliding mass, and dd is horizontal distance from circle center to mass centroid.

  • Ordinary Method of Slices (Fellenius): The sliding soil mass is subdivided into vertical slices. Assumes interslice forces are zero or equal and opposite: FS=∑[c′Δl+(Wicos⁡αi−uiΔl)tan⁡ϕ′]∑Wisin⁡αiFS = \frac{\sum [ c' \Delta l + (W_i \cos \alpha_i - u_i \Delta l) \tan \phi' ]}{\sum W_i \sin \alpha_i}

  • Bishop's Simplified Method: Considers normal interslice forces while assuming interslice shear forces are zero. Solved iteratively: FS=∑c′bi+(Wi−uibi)tan⁡ϕ′mα∑Wisin⁡αi,mα=cos⁡αi(1+tan⁡αitan⁡ϕ′FS)FS = \frac{\sum \frac{c' b_i + (W_i - u_i b_i) \tan \phi'}{m_\alpha}}{\sum W_i \sin \alpha_i}, \quad m_\alpha = \cos \alpha_i \left( 1 + \frac{\tan \alpha_i \tan \phi'}{FS} \right) Bishop's method is standard practice, typically providing 5% to 15%5\% \text{ to } 15\% higher and more realistic factors of safety than Fellenius.


Step-by-Step Worked Problem Examples

Worked Example: Square Footing with Shallow Water Table

Problem: A reinforced concrete square column footing B×B=2.0 m×2.0 mB \times B = 2.0\text{ m} \times 2.0\text{ m} is founded at depth Df=1.5 mD_f = 1.5\text{ m} below the ground surface. Soil properties are: moist unit weight γmoist=18.5 kN/m3\gamma_{\text{moist}} = 18.5\text{ kN/m}^3, saturated unit weight γsat=20.0 kN/m3\gamma_{\text{sat}} = 20.0\text{ kN/m}^3 (giving γ′=20.0−9.81=10.19 kN/m3\gamma' = 20.0 - 9.81 = 10.19\text{ kN/m}^3), cohesion c=20.0 kPac = 20.0\text{ kPa}, and ϕ=25∘\phi = 25^\circ. Terzaghi bearing capacity factors for ϕ=25∘\phi = 25^\circ are Nc=25.1N_c = 25.1, Nq=12.7N_q = 12.7, Nγ=9.7N_\gamma = 9.7. The groundwater table is located at depth dw=2.5 md_w = 2.5\text{ m} below the ground surface. Utilizing a factor of safety of FS=3.0FS = 3.0:

  1. Compute the modified effective unit weight γˉ\bar{\gamma} for the soil wedge below the base.
  2. Determine the ultimate bearing capacity qultq_{\text{ult}}.
  3. Calculate the maximum allowable column load PallP_{\text{all}}.

Solution:

Step 1: Water Table Position Analysis Df=1.5 m,dw=2.5 m,Df+B=1.5+2.0=3.5 mD_f = 1.5\text{ m}, \quad d_w = 2.5\text{ m}, \quad D_f + B = 1.5 + 2.0 = 3.5\text{ m} Since Df<dw≤Df+BD_f < d_w \le D_f + B (1.5<2.5≤3.5 m1.5 < 2.5 \le 3.5\text{ m}), this is Case 2: the water table lies within the failure wedge depth BB below the base.

  • Surcharge qq (above footing base) is entirely dry/moist: q=γmoistDf=18.5 kN/m3×1.5 m=27.75 kPaq = \gamma_{\text{moist}} D_f = 18.5\text{ kN/m}^3 \times 1.5\text{ m} = 27.75\text{ kPa}
  • Effective unit weight γˉ\bar{\gamma} for the third term is weighted over depth BB: γˉ=γ′+(dw−DfB)(γmoist−γ′)\bar{\gamma} = \gamma' + \left( \frac{d_w - D_f}{B} \right) (\gamma_{\text{moist}} - \gamma') γˉ=10.19+(2.5−1.52.0)(18.5−10.19)=10.19+(0.50)(8.31)=10.19+4.155=14.345 kN/m3\bar{\gamma} = 10.19 + \left( \frac{2.5 - 1.5}{2.0} \right) (18.5 - 10.19) = 10.19 + (0.50)(8.31) = 10.19 + 4.155 = 14.345\text{ kN/m}^3

Step 2: Calculate Ultimate Bearing Capacity (qultq_{\text{ult}}) Using Terzaghi's square footing equation: qult=1.3cNc+qNq+0.4γˉBNγq_{\text{ult}} = 1.3 c N_c + q N_q + 0.4 \bar{\gamma} B N_\gamma Evaluate each term:

  • Cohesion term: 1.3×20.0×25.1=652.60 kPa1.3 \times 20.0 \times 25.1 = 652.60\text{ kPa}
  • Surcharge term: 27.75×12.7=352.43 kPa27.75 \times 12.7 = 352.43\text{ kPa}
  • Soil wedge term: 0.4×14.345×2.0×9.7=111.32 kPa0.4 \times 14.345 \times 2.0 \times 9.7 = 111.32\text{ kPa} qult=652.60+352.43+111.32=1,116.35 kPa≈1,116.4 kPaq_{\text{ult}} = 652.60 + 352.43 + 111.32 = 1,116.35\text{ kPa} \approx 1,116.4\text{ kPa}

Step 3: Calculate Allowable Column Load (PallP_{\text{all}}) qall=qultFS=1,116.353.0=372.12 kPaq_{\text{all}} = \frac{q_{\text{ult}}}{FS} = \frac{1,116.35}{3.0} = 372.12\text{ kPa} Pall=qall×B2=372.12 kPa×(2.0 m)2=372.12×4.0=1,488.48 kN≈1,488.5 kNP_{\text{all}} = q_{\text{all}} \times B^2 = 372.12\text{ kPa} \times (2.0\text{ m})^2 = 372.12 \times 4.0 = 1,488.48\text{ kN} \approx 1,488.5\text{ kN}


CELE Board Exam Traps & Strategic Checklists

Warning

Footing Shape Factor Mix-Up: Pay close attention to the third term coefficient in Terzaghi's equation. Strip footings use 0.5γBNγ0.5 \gamma B N_\gamma; square footings use 0.4γBNγ0.4 \gamma B N_\gamma; circular footings use 0.3γBNγ0.3 \gamma B N_\gamma. Confusing 0.40.4 and 0.50.5 is one of the most common point losses on the CELE.

Clay Bearing Capacity Trap (ϕ=0∘\phi = 0^\circ): For saturated clay, Nc=5.7N_c = 5.7, Nq=1.0N_q = 1.0, and Nγ=0N_\gamma = 0. For a square footing, qult=1.3(5.7)cu+q=7.41cu+qq_{\text{ult}} = 1.3(5.7)c_u + q = 7.41 c_u + q. Examinees often forget the 1.31.3 shape multiplier or mistakenly add a third term.

Pile Block Failure Check: Never assume individual pile capacity governs group capacity. In soft sensitive clays with close pile spacing (s<3Ds < 3D), block failure almost always governs. You must evaluate both ∑Qsingle\sum Q_{\text{single}} and QblockQ_{\text{block}}.

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Foundation Capacity and Slope Stability Assessment Architecture
Test Your Knowledge

A square footing (2.5 m × 2.5 m) and a circular footing (diameter D = 2.5 m) are both founded at depth Df = 1.2 m in a deep saturated cohesive clay layer with undrained shear strength cu = 60.0 kPa, φ = 0° (Terzaghi factors: Nc = 5.7, Nq = 1.0, Nγ = 0), and saturated unit weight γsat = 18.0 kN/m³. What are the gross ultimate bearing capacities (qult) of the square footing and circular footing, respectively?

A

Square: 512.4 kPa; Circular: 485.6 kPa

B

Square: 466.2 kPa; Circular: 444.6 kPa

C

Square: 363.6 kPa; Circular: 363.6 kPa

D

Square: 466.2 kPa; Circular: 466.2 kPa

Test Your Knowledge

A 16-pile group (arranged in a 4 × 4 square grid) is driven into a deep saturated clay stratum with uniform undrained shear strength cu = 40.0 kPa and unit weight γ = 18.0 kN/m³. Each precast square pile has a width of 0.40 m × 0.40 m, embedment length L = 15.0 m, center-to-center spacing s = 1.20 m, and adhesion factor α = 0.70. Taking Nc* = 9.0 for end bearing, what is the single pile capacity (Qult,single), the block failure capacity of the group (Qblock), and the governing ultimate pile group capacity?

A

Qult,single = 672.0 kN, Qblock = 10,240 kN, Qgroup = 10,240 kN

B

Qult,single = 729.6 kN, Qblock = 11,674 kN, Qgroup = 15,360 kN

C

Qult,single = 729.6 kN, Qblock = 15,360 kN, Qgroup = 11,674 kN

D

Qult,single = 810.0 kN, Qblock = 18,200 kN, Qgroup = 12,960 kN

Test Your Knowledge

An infinite granular slope inclined at β = 18° is composed of clean sand with an effective internal friction angle φ' = 32°, dry unit weight γd = 16.5 kN/m³, and saturated unit weight γsat = 19.5 kN/m³. What is the factor of safety (FS) against translational failure under dry conditions, and what does the factor of safety become during severe monsoon conditions when steady seepage occurs parallel to the slope face with the groundwater table at the ground surface?

A

Dry FS = 1.25; Seepage FS = 0.62

B

Dry FS = 1.92; Seepage FS = 0.96

C

Dry FS = 1.92; Seepage FS = 1.45

D

Dry FS = 1.50; Seepage FS = 1.15

Sections you finish are checked off in the contents.