Bearing resistance of shallow foundations

Determining the bearing resistance of strip and pad footings against ground (bearing) failure according to Eurocode 7 — through the worked examples presented in the practical classes, with interactive calculators.

Course Foundation Engineering BSc · BMEEOGMAT45

How to use this material

The top bar lets you switch between the main chapters, and the sub-menu that appears beneath it lets you move between the sections of a given chapter. Beyond the introduction, three content chapters build on one another:

The calculation procedure — the design workflow, then the complete formula for the bearing resistance of a shallow foundation against ground failure for the drained and undrained cases, the treatment of the effective foundation area and the groundwater, the verification of adequacy, and finally the sliding check (drained and undrained).
Worked examples — 11 examples reconstructed from the practical material and derived step by step: drained and undrained strip and pad footings, concentric and eccentric, vertical and inclined loads, the effect and depth of groundwater, the effect of the direction of the horizontal force, the required foundation width, and the sliding check (drained and undrained) with passive earth resistance.
Calculators — three interactive calculators (drained, undrained, sliding): adjustable geometry, loads and soil parameters; the figure updates to scale, and the program computes the design resistance R_d and compares it with the design load V_d.

1. The calculation procedure

Designing a shallow foundation consists of successive, interdependent steps. This chapter details the theory of two of them — bearing resistance (ground failure) and sliding — for both the drained and the undrained case.

The usual design workflow

choosing the founding level (bearing stratum, groundwater, frost depth)
selecting the foundation type
determining the foundation width — by the ground-failure (bearing-resistance) and settlement checks
structural design of the foundation (material, depth, reinforcement)
checking stability — sliding, overturning, uplift (flotation)

Structure of this chapter: the bearing-resistance calculation (drained and undrained case), the treatment of the effective foundation area and the groundwater, the verification of adequacy, and finally the sliding check (drained and undrained).

2. Bearing-resistance calculation

Depending on the rate of loading and the degree of water saturation, the shear strength of the soil can be accounted for in two ways — for the drained and the undrained case.

2.1 Drained case

The drained case is typical of slow loading / granular soils: the shear strength of the soil is described by the φ – c pair, and the favourable effect of water buoyancy may be taken into account. The resistance of the shallow foundation against ground failure, according to EC7 (MSZ EN 1997-1), is:

R_k / (B′·L′) = 0.5·B′·γ′·N_γ·s_γ·i_γ·b_γ + q′·N_q·s_q·i_q·b_q + c·N_c·s_c·i_c·b_c

Width (self-weight) term

0.5·B′·γ′·N_γ·s_γ·i_γ·b_γ
comes from the foundation width and the weight of the soil beneath it

Surcharge (depth) term

q′·N_q·s_q·i_q·b_q
the effect of the overburden pressure q′ beside the foundation base (founding depth)

Cohesion term

c·N_c·s_c·i_c·b_c
the resistance arising from the cohesion of the soil

Input parameters

γ′ — effective unit weight of the soil beneath the foundation
B / L — actual width / length of the foundation; B′ / L′ — dimensions of the effective foundation area
q′ — effective overburden (geostatic) pressure acting beside the foundation at the founding level
c — cohesion of the soil beneath the foundation; φ — its angle of internal friction

Bearing-resistance (bearing capacity) factors

N_q = e^π·tanφ · tan²(45° + φ/2)
N_γ = 2 · (N_q − 1) · tanφ
N_c = (N_q − 1) · cotφ

All three factors depend solely on the angle of internal friction (φ) and rise steeply as φ increases. A few representative values:

φ [°]	0	10	15	20	24	26	30	32	35	40
N_q	1.00	2.47	3.94	6.40	9.63	11.85	18.40	23.18	33.30	64.20
N_c	5.14	8.34	10.98	14.83	19.32	22.25	30.14	35.49	46.12	75.31
N_γ	0.00	0.52	1.58	3.93	7.66	10.59	20.09	27.72	45.23	106.05

Correction factors

Shape factors (s)

s_γ = 1 − 0,3·(B′/L′)
s_q = 1 + (B′/L′)·sinφ
s_c = (s_q·N_q − 1)/(N_q − 1)

For a strip footing (L ≫ B): s_γ = s_q = s_c ≈ 1.

Load-inclination factors (i)

i_γ = (1 − f)^m+1
i_q = (1 − f)^m
i_c = (i_q·N_q − 1)/(N_q − 1)
f = R_h / (R_v + B′·L′·c·cotφ)

For a vertical load only, R_h = 0 → f = 0 → i = 1.
m: for a strip footing m_B ≈ 2, for a pad footing 1.5.

Base-inclination factors (b)

b_q = b_γ = (1 − α·tanφ)²
b_c = b_q − (1 − b_q)/(N_c·tanφ)

For a horizontal foundation base, α = 0° → b_γ = b_q = b_c ≈ 1.

The load-inclination exponent (m)

The m exponent in the load-inclination (i) factors depends on the aspect ratio of the effective area, according to whether the horizontal force is parallel to side B or side L:

m_B = [2 + (B′/L′)] / [1 + (B′/L′)] (if the horizontal force is parallel to B)
m_L = [2 + (L′/B′)] / [1 + (L′/B′)] (if parallel to L)

Foundation type	m_B (H parallel to B)	m_L (H parallel to L)
Strip footing (L ≫ B)	2.0	1.0
Square pad footing (L = B)	1.5	1.5

2.2 Undrained case

The undrained case is typical of rapid loading / fine-grained (cohesive) soils: the shear strength of the soil is described by the undrained shear strength c_u, with φ = 0. Water buoyancy must not be taken into account — we work with total stresses.

σ_t = R_v / (B′·L′) = (2 + π)·c_u·b_c·s_c·i_c + q

Factors (undrained)

s_c = 1 + 0,2·(B′/L′)
b_c = 1 − 2α/(π + 2)
i_c = 0.5·[1 + √(1 − R_h/(A′·c_u))]

α must be entered in the formula in radians.

Notation

q — the total overburden (geostatic) pressure acting beside the foundation at the founding level
c_u — undrained shear strength
A′ = B′·L′ — the effective foundation area

3. Effective foundation area and treatment of groundwater

3.1 Effective foundation area — eccentricity

Under eccentric loading the load does not act at the centroid of the cross-section. For the calculation we use the effective foundation area, whose centroid coincides with the point of application of the resultant force:

B′ = B − 2·e_B L′ = L − 2·e_L

The eccentricity follows from the moment equation written about the foundation base. If a horizontal force H_k and an eccentric load Q_k act:

e_B,k = | (H_k·h) − (Q_k·e_q) | / V_k

Watch the signs! The moments of the horizontal force and of the eccentric vertical load can act in opposite directions — in that case the eccentricity decreases.

3.2 Effect of groundwater — the effective γ′

The value of γ′ in the width term is influenced by the depth of the groundwater below the foundation base (t_wd). The failure zone extends to a depth of B′; γ′ varies linearly with the ratio t_wd/B′:

t_wd ≤ 0.5·B′ → γ′ = γ_sat − γ_w   (water within the failure zone)
0.5·B′ < t_wd < 1.5·B′ → γ′ = (γ_sat − γ_w) + (γ_m − (γ_sat − γ_w))·(t_wd/B′ − 0.5)   (linear transition)
1.5·B′ ≤ t_wd → γ′ = γ_m   (water deep, no effect)

A negative t_wd means the groundwater is above the foundation base — then γ′ = γ_sat − γ_w.

4. Verification of adequacy

The bearing-resistance (GEO) limit state is satisfied if the design load does not exceed the design resistance:

V_d ≤ R_d

Design load — V_d

V_d = γ_G·G_k + γ_Q·Q_k

Partial factors: for permanent actions γ_G = 1.35, for variable (imposed) actions γ_Q = 1.5. G_k includes the self-weight of the superstructure, the foundation, and the overlying soil (fill).

Design resistance — R_d

R_d = R_k / γ_R

The resistance partial factor for ground failure of shallow foundations is γ_R = 1.4. The characteristic resistance R_k follows from the bearing-resistance formula.

If V_d > R_d: the foundation is inadequate. There are two remedies — (1) lower the founding level (q′ increases), or (2) widen the foundation (B increases). In practice, widening is usually the more practical option.

5. Sliding check (EQU)

If the force acting on the foundation base has a horizontal component (it is not perpendicular to the base), the safety against sliding must be checked. The condition for adequacy is:

H_d ≤ R_d + R_p;d

H_d — the design value of the horizontal load acting on the foundation base; it must also include the design value of the active earth pressure acting on the foundation.
R_d — the design value of the shearing (base) resistance mobilised at the foundation base.
R_p;d — the design value of the resistance arising from the (passive) earth pressure acting on the side face of the foundation.

Use R_p;d with caution! The soil in front of the face can be removed by erosion or human intervention, and in clay seasonal shrinkage can separate it from the face. The full passive earth resistance is mobilised only at large displacements, so in clay often only one-half to two-thirds of it may be relied upon.

Drained case — base friction

R_d = V′_k · tan δ_k / γ_R;h

V′_k — the characteristic value of the vertical load; only the variable load that is certainly concurrent with H_d may be included.
δ_k — the base–soil friction angle: for a cast-in-situ foundation δ_k = φ′_k, for a precast one δ_k = ⅔·φ′_k.
The cohesion c′ at the base is neglected on the safe side.
γ_R;h = 1.1 (DA-2*).

Undrained case — base adhesion

R_d = A_c · c_u;k / γ_R;h

A_c — the compressed base area (B·L, or the effective A′).
c_u;k — the characteristic undrained shear strength at the foundation base.
γ_R;h = 1.1.
Limit: because of a possible base–soil gap (water, air), the condition R_d ≤ 0.4·V_d must also be satisfied.

Worked examples

11 worked examples derived in detail from the practical material — drained and undrained strip and pad footings, concentric and eccentric, vertical and inclined loads, the effect and depth of groundwater, the effect of the direction of the horizontal force, the required foundation width, and the sliding check (drained and undrained) with passive earth resistance. Each example comes with a figure reconstructed from the practical material and a step-by-step solution.

Worked Example 1 — Checking a drained strip footing

Task

Check the adequacy against ground failure, in the drained case, of the strip footing (L ≫ B) shown in the figure. Loads: G_v,k = 220 kN/m, Q_v,k = 70 kN/m (concentric, vertical). Geometry: b = 0.3 m, B = 1.1 m, h = 0.6 m, with the foundation base 1.1 m below the ground surface. γ_b = 25 kN/m³. The groundwater is deep. saGr fill: γ = 17 kN/m³, φ = 30°, c = 10 kPa. siSa natural soil: γ = 18 kN/m³, φ = 24°, c = 12 kPa.

WE1

Figure

Solution

1. Action side — the design load

Weight of fill: G_f,k = (B−b)·t·γ = (1.1−0.3)·0.5·17 = 6.8 kN/m

Weight of the foundation: G_b,k = B·h·γ_b = 1.1·0.6·25 = 16.5 kN/m

Permanent loads: G_k = 220 + 6.8 + 16.5 = 243.3 kN/m

V_d = γ_G·G_k + γ_Q·Q_k = 1.35·243.3 + 1.5·70 = 434 kN/m

2. Geometry and overburden pressure

Concentric load → B = B′, A′ = B′·L′ = 1.1 m²/m.

q′ = Σγ_i·t_i = 17·0.8 + 18·0.3 = 19 kPa

Groundwater deep: t_wd/B′ > 1.5 → γ′ = γ_m = 18 kN/m³

3. Bearing-resistance factors (φ = 24°)

N_q = e^π·tan24°·tan²(45°+12°) = 9.63; N_c = (N_q−1)·cot24° = 19.32; N_γ = 2·(N_q−1)·tan24° = 7.66

Strip footing, concentric load, horizontal base → s = i = b ≈ 1.

4. Characteristic and design resistance

R_k/A′ = 0.5·γ′·B′·N_γ + q′·N_q + c·N_c = 0.5·18·1.1·7.66 + 19·9.63 + 12·19.32

= 76 + 182 + 232 = 490 kPa

R_k = 490·1.1 = 539 kN/m → R_d = R_k/γ_R = 539/1.4 = 385 kN/m

Result (B = 1.1 m): R_d = 385 kN/m < V_d = 434 kN/m → NOT ADEQUATE.

5. Remedy: widen the foundation — B = 1.3 m

New weights: G_f,k = (1.3−0.3)·0.5·17 = 8.5; G_b,k = 1.3·0.6·25 = 19.5 → G_k = 248 → V_d = 1.35·248 + 1.5·70 = 440 kN/m

R_k/A′ = 0.5·18·1.3·7.66 + 19·9.63 + 12·19.32 = 90 + 182 + 232 = 504 kPa

R_k = 504·1.3 = 655 → R_d = 655/1.4 = 468 kN/m

Result (B = 1.3 m): R_d = 468 kN/m > V_d = 440 kN/m → ADEQUATE. ✓

Mini calculator

WE1 — checking a drained strip footing

GEO

B width (m)

q′ overburden (kPa)

G_v,k (kN/m)

γ′ below the foundation (kN/m³)

Q_v,k (kN/m)

φ (°)

c cohesion (kPa)

Constants (for self-weight): b = 0.3 · h = 0.6 · t = 0.5 m · γ_fill = 17 · γ_b = 25 kN/m³

Worked Example 2 — Checking an undrained strip footing

Task

Same geometry and loading as WE1, but the natural soil beneath the foundation is now cohesive (Cl): φ_u = 0°, c_u = 150 kPa.
Check the strip footing in the undrained case (B = 1.1 m).

WE2

Solution

1. Action side

Same as WE1: G_k = 243.3 kN/m → V_d = 1.35·243.3 + 1.5·70 = 434 kN/m

2. Overburden pressure (total stress!)

Undrained case → we work with total stresses; water buoyancy is not taken into account.

q = Σγ_i·t_i = 17·0.8 + 18·0.3 = 19 kPa

3. Undrained resistance

Strip footing → s_c = 1; horizontal base → b_c = 1; vertical load → i_c = 1.

σ_t = R_v/A′ = (2+π)·c_u·b_c·s_c·i_c + q = (2+π)·150·1·1·1 + 19 = 771 + 19 = 790 kPa

R_k = 790·1.1 = 869 kN/m → R_d = 869/1.4 = 620 kN/m

Result: R_d = 620 kN/m > V_d = 434 kN/m → ADEQUATE. ✓ In the undrained state the stiff cohesive soil can carry the load even with a width of 1.1 m.

Mini calculator

WE2 — checking an undrained strip footing

undrained

B width (m)

q overburden (kPa)

G_v,k (kN/m)

c_u undrained shear str. (kPa)

Q_v,k (kN/m)

Constants (for self-weight): b = 0.3 · h = 0.6 · t = 0.5 m · γ_fill = 17 · γ_b = 25 kN/m³ · total stress (φ = 0)

Worked Example 3 — Eccentric and inclined load, with groundwater

Task

Check the strip footing in the drained case. Loads: G_z,k = 250 kN/m (concentric), Q_z,k = 170 kN/m applied eccentrically (e_q = 0.3 m), G_x,k = 30 kN/m horizontal force. Geometry: b = 0.3 m, B = 1.3 m, h = 1.0 m, t = 0.4 m, t_fill = 1.1 m, γ_b = 25 kN/m³. The groundwater is 0.6 m above the foundation base.
Sa fill: γ = 18 kN/m³, γ_sat = 19 kN/m³, φ = 30°, c = 5 kPa. siSa natural soil: γ = 18 kN/m³, γ_sat = 20 kN/m³, φ = 26°, c = 30 kPa.

WE3

Figure

Solution

1. Action side

G_f,k = (1.3−0.3)·0.4·18 = 7.2; G_b,k = 1.3·1.0·25 = 32.5 → G_k = 250 + 7.2 + 32.5 = 289.7 kN/m

V_d = 1.35·289.7 + 1.5·170 = 646.1 kN/m

2. Eccentricity of the resultant force

Characteristic vertical load: V_k = G_z,k+G_f,k+G_b,k+Q_z,k = 250+7.2+32.5+170 = 459.7 kN/m

Horizontal load: H_k = G_x,k = 30 kN/m

Moment equation about the axis of the foundation base: e_B,k = |(H_k·h) − (Q_z,k·e_q)| / V_k = |(30·1.0) − (170·0.3)| / 459.7 = 21/459.7 = 0.046 m

Effective width: B′ = B − 2·e_B,k = 1.3 − 2·0.046 = 1.208 m; A′ = 1.208 m²/m

3. Overburden pressure and γ′ (water above the base)

q′ = Σγ′_i·t_i = 18·0.8 + (19−10)·0.3 + (20−10)·0.3 = 20.1 kPa

Water above the base → t_wd/B′ < 0.5 → γ′ = γ_sat − γ_w = 20 − 10 = 10 kN/m³

4. Factors (φ = 26°)

N_q = 11.85; N_c = 22.25; N_γ = 10.59. Strip footing → s ≈ 1; horizontal base → b ≈ 1.

f = H_k/(V_k + B′·L′·c·cotφ) = 30/(459.7 + 1.208·30·cot26°) = 30/534 = 0.0562

m_B = 2 (strip footing) → i_γ = (1−f)³ = 0.841; i_q = (1−f)² = 0.891; i_c = (i_q·N_q−1)/(N_q−1) = 0.881

5. Resistance

R_k/A′ = 0.5·γ′·B′·N_γ·i_γ + q′·N_q·i_q + c·N_c·i_c

= 0.5·10·1.208·10.59·0.841 + 20.1·11.85·0.891 + 30·22.25·0.881 ≈ 54 + 212 + 588 = 854 kPa

R_k = 854·1.208 = 1031 kN/m → R_d = 1031/1.4 = 736 kN/m

Result: R_d = 736 kN/m > V_d = 646.1 kN/m → ADEQUATE. ✓

Mini calculator

WE3 — eccentric + inclined load, with groundwater

GEO

B width (m)

q′ overburden (kPa)

G_z,k concentric (kN/m)

γ′ below the foundation (kN/m³)

Q_z,k eccentric (kN/m)

φ (°)

e_q eccentricity of Q (m)

c cohesion (kPa)

G_x,k horizontal force (kN/m)

Constants (for self-weight): b = 0.3 · h = 1.0 · t = 0.4 m · γ_fill = 18 · γ_b = 25 kN/m³ · strip footing, m_B = 2

Worked Example 4 — Effect of the design groundwater depth

Task

The same strip footing as in WE3 — all data unchanged: B = 1.3 m, b = 0.3 m, h = 1.0 m, t = 0.4 m, t_fill = 1.1 m, γ_b = 25 kN/m³. Loads: G_z,k = 250 kN/m (concentric), Q_z,k = 170 kN/m (e_q = 0.3 m), G_x,k = 30 kN/m. Soils: Sa fill γ = 18 kN/m³, γ_sat = 19 kN/m³, φ = 30°, c = 5 kPa; siSa natural γ = 18 kN/m³, γ_sat = 20 kN/m³, φ = 26°, c = 30 kPa. The only difference: the design groundwater is now 2 m deeper — 1.4 m below the foundation base (t_wd = 1.4 m). Question: how does R_d change?

WE4

Figure

Solution

1. Action side — unchanged (same as WE3)

G_f,k = (1.3−0.3)·0.4·18 = 7.2; G_b,k = 1.3·1.0·25 = 32.5 → G_k = 250 + 7.2 + 32.5 = 289.7 kN/m

V_d = 1.35·289.7 + 1.5·170 = 646.1 kN/m (the load does not depend on the groundwater)

2. Eccentricity — unchanged

e_B,k = |H_k·h − Q_z,k·e_q|/V_k = |30·1.0 − 170·0.3|/459.7 = 0.046 m → B′ = 1.3 − 2·0.046 = 1.208 m; A′ = 1.208 m²/m

3. q′ and γ′ with the deeper groundwater — this changes!

The groundwater is 1.4 m below the foundation base (t_wd = 1.4 m), so there is no water above the base → q′ with the moist unit weights:

q′ = 18·1.1 + 18·0.3 = 25.2 kPa (in WE3 it was 20.1 kPa)

γ′ for the siSa layer beneath the foundation: since 0.5·B′ < t_wd < 1.5·B′ → linear transition:

γ′ = (20−10) + (18−(20−10))·(1.4/1.208 − 0.5) = 10 + 8·0.659 = 15.3 kN/m³ (in WE3 it was 10 kN/m³)

4. Factors (φ = 26°) — unchanged

N_q = 11.85; N_c = 22.25; N_γ = 10.59. f = 0.056 → i_γ = 0.841; i_q = 0.891; i_c = 0.881. Strip footing → s ≈ 1.

5. Resistance

R_k/A′ = 0.5·15.3·1.208·10.59·0.841 + 25.2·11.85·0.891 + 30·22.25·0.881 ≈ 82 + 266 + 588 = 936 kPa

R_k = 936·1.208 = 1131 kN/m → R_d = 1131/1.4 = 808 kN/m

Result: R_d = 808 kN/m > V_d = 646 kN/m → ADEQUATE. ✓ Because of the deeper groundwater, R_d increased (WE3: 736 → WE4: 808 kN/m): less buoyancy → larger γ′ and q′ → higher bearing resistance.

Mini calculator

WE4 — strip footing, deeper groundwater

GEO

B width (m)

q′ overburden (kPa)

G_z,k concentric (kN/m)

γ′ below the foundation (kN/m³)

Q_z,k eccentric (kN/m)

φ (°)

e_q eccentricity of Q (m)

c cohesion (kPa)

G_x,k horizontal force (kN/m)

Constants: b = 0.3 · h = 1.0 · t = 0.4 m · γ_fill = 18 · γ_b = 25 kN/m³ · strip footing. Effect of the deeper groundwater: q′ = 25.2 kPa, γ′ = 15.3 kN/m³ (in WE3 they were 20.1 and 10). Reset to these to recover the WE3 value.

Worked Example 5 — Effect of the direction of the horizontal force

Task

The same strip footing as in WE3 — all data unchanged: B = 1.3 m, b = 0.3 m, h = 1.0 m, t = 0.4 m, t_fill = 1.1 m, γ_b = 25 kN/m³; G_z,k = 250 kN/m (concentric), Q_z,k = 170 kN/m (e_q = 0.3 m), G_x,k = 30 kN/m; groundwater 0.6 m above the foundation base; Sa fill (γ = 18 kN/m³, γ_sat = 19 kN/m³, φ = 30°, c = 5 kPa) and siSa natural soil (γ = 18 kN/m³, γ_sat = 20 kN/m³, φ = 26°, c = 30 kPa). The only difference: the direction of the horizontal force G_x,k is reversed. Question: how does R_d change?

WE5

Figure

Solution

1. Action side — unchanged

G_k = 250 + 7.2 + 32.5 = 289.7 kN/m; V_d = 1.35·289.7 + 1.5·170 = 646.1 kN/m (the magnitude of the load is unchanged)

2. Eccentricity — this changes!

Originally (WE3) the moments of G_x and of the eccentric Q acted in opposite directions, so they partly cancelled:

e_B (WE3) = |H_k·h − Q_z·e_q|/V_k = |30·1.0 − 170·0.3|/459.7 = 21/459.7 = 0.046 m

If the direction of G_x is reversed, the two moments act in the same direction, so they add up:

e_B (reversed) = |H_k·h + Q_z·e_q|/V_k = |30·1.0 + 170·0.3|/459.7 = 81/459.7 = 0.176 m

B′ = 1.3 − 2·0.176 = 0.948 m; A′ = 0.948 m²/m (in WE3 it was 1.208 m)

3. q′, γ′ and the factors — unchanged

q′ = 20.1 kPa; γ′ = 10 kN/m³ (groundwater above the base); φ = 26° → N_q=11.85, N_c=22.25, N_γ=10.59. The load magnitude is the same → i_γ≈0.84, i_q≈0.89, i_c≈0.88.

4. Resistance with the smaller B′

R_k/A′ = 0.5·10·0.948·10.59·0.84 + 20.1·11.85·0.89 + 30·22.25·0.88 ≈ 42 + 211 + 585 = 839 kPa

R_k = 839·0.948 = 795 kN/m → R_d = 795/1.4 = 568 kN/m

Result: R_d = 568 kN/m < V_d = 646 kN/m → NOT ADEQUATE! Reversing G_x significantly reduces R_d (WE3: 736 → 568 kN/m): the two moments add up, the eccentricity grows (0.046 → 0.176 m), so the effective width B′ (and with it the bearing resistance) decreases.

Mini calculator

WE5 — reversed G_x (eccentricities add up)

GEO

B width (m)

q′ overburden (kPa)

G_z,k concentric (kN/m)

γ′ below the foundation (kN/m³)

Q_z,k eccentric (kN/m)

φ (°)

e_q eccentricity of Q (m)

c cohesion (kPa)

G_x,k horizontal force (kN/m)

Because of the reversed G_x, the eccentricity follows from the sum of the two moments: e_B = |H_k·h + Q_z·e_q|/V_k. Constants: b=0.3 · h=1.0 · t=0.4 m · γ_fill=18 · γ_b=25 kN/m³ · strip footing.

Worked Example 6 — Effect of increasing the eccentricity

Task

The same strip footing as in WE3 — all data unchanged: B = 1.3 m, b = 0.3 m, h = 1.0 m, t = 0.4 m, t_fill = 1.1 m, γ_b = 25 kN/m³; G_z,k = 250 kN/m, Q_z,k = 170 kN/m, G_x,k = 30 kN/m; groundwater 0.6 m above the foundation base; Sa fill and siSa natural soil. The only difference: the eccentricity of the load Q_z,k increases from e_q = 0.3 m to 0.5 m (50 cm). Question: how does R_d change?

WE6

Figure

Solution

1. Action side — unchanged

G_k = 250 + 7.2 + 32.5 = 289.7 kN/m; V_d = 1.35·289.7 + 1.5·170 = 646.1 kN/m (the magnitude of the load is unchanged)

2. Eccentricity — this grows!

Because of the larger e_q, the moment of Q (170·0.5 = 85) is now larger than the moment of G_x (30·1.0 = 30):

e_B = |H_k·h − Q_z·e_q|/V_k = |30·1.0 − 170·0.5|/459.7 = 55/459.7 = 0.120 m (in WE3 it was 0.046 m)

B′ = 1.3 − 2·0.120 = 1.06 m; A′ = 1.06 m²/m (in WE3 it was 1.208 m)

3. q′, γ′ and the factors — unchanged

q′ = 20.1 kPa; γ′ = 10 kN/m³; φ = 26° → N_q=11.85, N_c=22.25, N_γ=10.59; i_γ≈0.84, i_q≈0.89, i_c≈0.88.

4. Resistance with the smaller B′

R_k/A′ = 0.5·10·1.06·10.59·0.84 + 20.1·11.85·0.89 + 30·22.25·0.88 ≈ 47 + 212 + 587 = 845 kPa

R_k = 845·1.06 = 896 kN/m → R_d = 896/1.4 = 640 kN/m

Result: R_d = 640 kN/m < V_d = 646 kN/m → only just NOT ADEQUATE! Because of the larger eccentricity, R_d decreased (WE3: 736 → WE6: 640 kN/m): larger e_q → larger e_B → smaller effective width (B′) → lower bearing resistance.

Mini calculator

WE6 — larger eccentricity e_q

GEO

B width (m)

q′ overburden (kPa)

G_z,k concentric (kN/m)

γ′ below the foundation (kN/m³)

Q_z,k eccentric (kN/m)

φ (°)

e_q eccentricity of Q (m)

c cohesion (kPa)

G_x,k horizontal force (kN/m)

Set the e_q field to 0.3 to see the WE3 case (R_d = 736). Constants: b=0.3 · h=1.0 · t=0.4 m · γ_fill=18 · γ_b=25 kN/m³ · strip footing.

Worked Example 7 — Effect of an inclined foundation base

Task

The same strip footing as in WE3 — all data unchanged: B = 1.3 m, b = 0.3 m, h = 1.0 m, t = 0.4 m, t_fill = 1.1 m, γ_b = 25 kN/m³; G_z,k = 250 kN/m, Q_z,k = 170 kN/m (e_q = 0.3 m), G_x,k = 30 kN/m; groundwater 0.6 m above the foundation base; Sa fill and siSa natural soil. The only difference: the foundation base is inclined at 5° (sloping base). Question: how does R_d change?

WE7

Figure

Solution

1–2. Loads and eccentricity — unchanged

V_d = 646.1 kN/m; e_B = 0.046 m → B′ = 1.208 m (as in WE3).

3. q′, γ′ and the i factors — unchanged

q′ = 20.1 kPa; γ′ = 10 kN/m³; φ = 26° → N_q=11.85, N_c=22.25, N_γ=10.59; i_γ≈0.84, i_q≈0.89, i_c≈0.88.

4. Base-inclination factors (b) — this is the new part!

α = 5° = 0.0873 rad. b_q = b_γ = (1 − α·tan φ)² = (1 − 0.0873·tan26°)² = (1 − 0.0426)² = 0.917

b_c = b_q − (1 − b_q)/(N_c·tan φ) = 0.917 − 0.083/10.86 = 0.909

5. Resistance with the b factors

R_k/A′ = 54·0.917 + 212·0.917 + 588·0.909 ≈ 49 + 195 + 535 = 779 kPa

R_k = 779·1.208 = 941 kN/m → R_d = 941/1.4 = 672 kN/m

Result: R_d = 672 kN/m > V_d = 646 kN/m → ADEQUATE. ✓ Because of the inclined base, R_d decreases (WE3: 736 → WE7: 672 kN/m): the b < 1 base-inclination factors reduce all three terms — but the margin is still sufficient, so the foundation is adequate.

Mini calculator

WE7 — inclined base (base-inclination factors)

GEO

B width (m)

q′ overburden (kPa)

G_z,k concentric (kN/m)

γ′ below the foundation (kN/m³)

Q_z,k eccentric (kN/m)

φ (°)

e_q eccentricity of Q (m)

c cohesion (kPa)

G_x,k horizontal force (kN/m)

α base inclination (°)

Set α to 0 to see the WE3 case (R_d = 736). The base-inclination factors: b_q = b_γ = (1 − α·tan φ)², b_c = b_q − (1 − b_q)/(N_c·tan φ).

Worked Example 8 — Pad footing, checked in both directions

Task

Check the pad footing (B = 2.0 m, L = 1.5 m) against ground failure in the drained case. Column: b = l = 0.4 m; h = 1.2 m; t = 0.4 m; t_fill = 1.5 m; γ_b = 25 kN/m³. Loads: G_z,k = 1000 kN (concentric) and Q_x,k = 100 kN horizontal force (parallel to side B). The groundwater is 0.6 m above the foundation base (h_GWL). Sa fill: γ = 18 kN/m³, γ_sat = 19 kN/m³, φ = 30°, c = 5 kPa; saGr natural soil: γ = 19 kN/m³, γ_sat = 20 kN/m³, φ = 32°, c = 0 kPa. Since the horizontal load 'weakens' the side it is parallel to, both directions (B and L) must be checked.

WE8

Figure — cross-section

Figure — plan view

Solution

1. Vertical load acting at the foundation base

Fill: G_f,k = (B·L − b·l)·t·γ = (2·1.5 − 0.4·0.4)·0.4·18 = 20.5 kN; Foundation: G_b,k = B·L·h·γ_b = 2·1.5·1.2·25 = 90 kN

G_k = 1000 + 20.5 + 90 = 1110.5 kN; V_d = 1.35·1110.5 + 1.5·0 = 1499 kN (buoyancy is neglected)

2. Eccentricity of the resultant force

V_k = G_k = 1110.5 kN; H_k = Q_x,k = 100 kN

e_B = (H_k·h)/V_k = (100·1.2)/1110.5 = 0.11 m; e_L = 0

B′ = 2.0 − 2·0.11 = 1.78 m; L′ = 1.5 m; A′ = 1.78·1.5 = 2.67 m²

3. q′ and γ′ (with groundwater)

q′ = 18·1.0 + (19−10)·0.5 + (20−10)·0.1 = 18 + 4.5 + 1.0 = 23.5 kPa

Water above the base → t_wd/B′ < 0.5 → γ′ = γ_sat − γ_w = 20 − 10 = 10 kN/m³

4. Bearing-resistance factors (saGr: φ = 32°, c = 0)

N_q = 23.17; N_c = 35.48; N_γ = 27.71. Horizontal base → b_γ = b_q = b_c ≈ 1.

5. Checking both directions

Side B (weakened by Q_x — governing): s_γ=0.64, s_q=1.65, s_c=1.68; f = H_k/V_k = 0.09 → i_γ=0.793, i_q=0.871, i_c=0.865.

R_k/A′ = 0.5·10·1.78·27.71·0.64·0.793 + 23.5·23.17·1.65·0.871 + 0 = 125 + 782 + 0 = 907.7 kPa

R_k,B = 907.7·2.67 = 2424 kN → R_d,B = 2424/1.4 = 1731 kN

Side L: s_γ=0.75, s_q=1.46, s_c=1.48; the horizontal load is parallel to B, so in the L direction i = 1 → R_d,L > R_d,B. Hence side B governs.

Result: R_d = R_d,B = 1731 kN > V_d = 1499 kN → ADEQUATE. ✓ Side B governs, because the horizontal Q_x is parallel to this direction (i < 1).

Mini calculator

WE8 — pad footing, side B (governing)

GEO

B width (m)

q′ overburden (kPa)

L length (m)

γ′ below the foundation (kN/m³)

G_k permanent load (kN)

φ (°)

Q_z,k variable (kN)

c cohesion (kPa)

H_k = Q_x,k horizontal (kN)

h foundation depth (m)

The horizontal force is parallel to side B; the shape (s) and inclination (i) factors are taken into account. Side B governs.

Worked Example 9 — Determining the required foundation width

Task

Determine the required width B of the strip footing. The load given is a design value that already includes the self-weight of the foundation and the overlying soil: V_d = 255 kN/m, e_B = 0.02 m. b = 0.3 m, t_fill = 1.65 m, γ_b = 25 kN/m³. The groundwater is at the founding level. Layer 1 (fill): γ = 19 kN/m³, γ_sat = 20 kN/m³, φ = 15°, c = 5 kPa; Layer 2 (bearing stratum, below the base): γ = 19 kN/m³, γ_sat = 20 kN/m³, φ = 30°, c = 0 kPa. We design for the governing (just-adequate) condition: V_d = R_d.

WE9

Figure

Solution

1. The design condition

In the governing condition V_d = R_d = R_k/γ_R, so the required characteristic resistance is: R_k = V_d·γ_R = 255·1.4 = 357 kN/m

2. q′ and γ′ (groundwater at the base)

q′ = γ·t_fill = 19·1.65 = 31.35 kPa (water at the base → moist unit weight above it)

The bearing Layer 2: φ = 30°, c = 0; t_wd/B′ < 0.5 → γ′ = γ_sat − γ_w = 20 − 10 = 10 kN/m³

3. Bearing-resistance factors (φ = 30°)

N_q = e^{π·tan φ}·tan²(45°+φ/2) = 18.40; N_γ = 2·(N_q−1)·tan φ = 20.09. Strip footing → s ≈ 1; concentric (no inclined load) → i ≈ 1; c = 0 → the cohesion term vanishes.

4. The bearing-resistance formula — now B′ is the unknown

R_k/B′ = 0.5·B′·γ′·N_γ + q′·N_q → 0 = 0.5·γ′·N_γ·(B′)² + q′·N_q·B′ − R_k

With the numerical values: 0 = 100.45·(B′)² + 576.84·B′ − 357 (quadratic equation)

5. Solving the quadratic equation

B′_1,2 = [−576.84 ± √(576.84² + 4·100.45·357)] / (2·100.45) → B′₁ = −6.31 (meaningless), B′₂ = 0.56 m

B = B′ + 2·e_B = 0.56 + 2·0.02 = 0.60 m

Result: the required foundation width is B ≈ 0.60 m. In practice we round up to the chosen module (e.g. 0.6–0.7 m).

Mini calculator

WE9 — required foundation width

design

V_d design load (kN/m)

q′ overburden (kPa)

e_B eccentricity (m)

γ′ below the foundation (kN/m³)

φ (°)

c cohesion (kPa)

We design for the governing (just-adequate) condition: V_d = R_d. The solution follows from a quadratic equation (B = B′ + 2·e_B).

Worked Example 10 — Sliding check (EQU, drained)

Task

Check the adequacy of the pad footing (B = 2.0 m, L = 1.5 m) shown in the figure against horizontal sliding (EQU limit state, drained case). Column: b = l = 0.4 m; h = 1.2 m; t = 0.4 m; t_fill = 1.5 m; γ_b = 25 kN/m³. Loads (characteristic): G_z,k = 320 kN vertical (including the self-weight of the foundation and the soil) and Q_x,k = 100 kN horizontal column load. The groundwater is 0.6 m above the foundation base (h_GWL). Sa fill: γ = 18 kN/m³, γ_sat = 19 kN/m³, φ = 30°, c = 0 kPa; siSa natural soil: γ = 19 kN/m³, γ_sat = 20 kN/m³, φ = 28°, c = 0 kPa. The condition for adequacy: H_d ≤ R_d.

WE10

Figure — cross-section

Figure — plan view

Solution

1. Governing horizontal load at the foundation base

Horizontal column load: Q_x,k = 100 kN. Earth-pressure coefficients of the active earth pressure acting on the foundation:

Sa fill: K_a1 = tan²(45°−φ/2) = tan²(45°−15°) = 0.333; siSa natural: K_a2 = tan²(45°−14°) = 0.361

The resultant active earth pressure over the height of the foundation: E_a,k = 10.08 kN

The design value of the horizontal action: H_d = γ_G·E_a,k + γ_Q·Q_x,k = 1.35·10.08 + 1.5·100 = 163.6 kN

2. Sliding resistance — base friction

R_d = V′_k·tan δ_k / γ_R;h, where V′_k = G_z,k = 320 kN (includes the self-weight of the foundation and the soil).

Cast-in-situ reinforced-concrete foundation → δ_k = φ = 28° → tan δ_k = 0.532; γ_R;h = 1.1

R_d = (320·0.532)/1.1 = 155 kN → R_d = 155 < H_d = 163.6, so on its own it is not enough.

3. Adding earth resistance (at-rest earth pressure)

R_pd from at-rest or passive earth pressure? On the safe side we use the at-rest earth pressure: K₀ = 1 − sin φ.

The resultant at-rest earth pressure E_0,k = 15.1 kN → R_pd = E_0,k/γ_R;h = 15.1/1.1 = 13.7 kN

4. Adequacy

R_d + R_pd = 155 + 13.7 = 168.7 kN > H_d = 163.6 kN

Result: R_d + R_pd = 168.7 kN > H_d = 163.6 kN → ADEQUATE. ✓ Note: if the 'passive'/at-rest soil can disappear or heave, its resistance must not be taken into account.

Mini calculator

WE10 — sliding check (drained)

EQU

V′_k vertical load (kN)

Q_x,k horizontal column load (kN)

δ base friction angle (°)

E_a,k active earth pressure (kN)

γ_R;h resistance factor

E_0,k at-rest earth pressure (kN)

Cast-in-situ reinforced-concrete foundation → δ = φ. Condition: H_d ≤ R_d + R_pd. The at-rest earth pressure (R_pd) may be relied upon only if the soil is certain not to heave and cannot be removed.

Undrained case

The same pad footing and loading, but the natural soil beneath the base is now cohesive clay (Ci): φ_u = 0°, c_u = 65 kPa (the fill has c = 5 kPa). In the undrained case the sliding resistance comes not from base friction but from the adhesion mobilised over the whole base area (A_c·c_u).

1. The governing horizontal load is unchanged

The horizontal load does not depend on the drainage condition of the soil beneath the base: H_d = γ_G·E_a,k + γ_Q·Q_x,k = 1.35·10.08 + 1.5·100 = 163.6 kN

2. Sliding resistance — base adhesion

R_d = A_c·c_u,k / γ_R;h, where A_c = B·L = 2.0·1.5 = 3.0 m², c_u,k = 65 kPa, γ_R;h = 1.1.

R_d = (2.0·1.5·65)/1.1 = 177 kN (the whole base area is mobilised; no earth resistance is needed)

Result: R_d = 177 kN > H_d = 163.6 kN → ADEQUATE. ✓ The undrained base adhesion is sufficient on its own (in the drained case the earth resistance was also needed).

Mini calculator — undrained

WE10b — sliding check (undrained)

EQU

B width (m)

Q_x,k horizontal column load (kN)

L length (m)

E_a,k active earth pressure (kN)

c_u,k base adhesion (kPa)

γ_R;h resistance factor

Undrained sliding resistance over the whole base area: R_d = A_c·c_u,k/γ_R;h. Condition: H_d ≤ R_d.

Worked Example 11 — Sliding with passive earth resistance (EQU, drained)

Task

Check the adequacy of the pad footing (B = 0.95 m, L = 1.45 m) shown in the figure against horizontal sliding (EQU, drained). h = 0.95 m; t = 2.05 m; t_fill = 3.0 m; γ_b = 25 kN/m³. Loads: G_z,k = 490 kN (permanent) and Q_z,k = 610 kN (variable) vertical, plus Q_x,k = 250 kN horizontal; e_b = 0.04 m. The vertical loads include the weight of the foundation and of the soil lying above it. The groundwater is 0.95 m above the foundation base (h_GWL). Sa fill: γ = 19.5 kN/m³, γ_sat = 20.5 kN/m³, φ = 32°, c = 0 kPa; Si natural soil: γ = 19.5 kN/m³, γ_sat = 20.5 kN/m³, φ = 16°, c = 40 kPa. Between the foundation and the soil, δ = ⅔·φ is to be used.

WE11

Figure — cross-section

Figure — plan view

Solution

1. Governing horizontal load — active earth pressure + column load

Horizontal column load: Q_x,k = 250 kN. Coefficient of the active earth pressure acting on the foundation (Sa fill): K_a = tan²(45°−φ/2) = tan²(45°−16°) = 0.307

The vertical geostatic stress over the height of the foundation (groundwater at the top of the foundation): σ′_v,top = 19.5·2.05 = 39.98 kPa; σ′_v,bot = 39.98 + (20.5−10)·0.95 = 49.95 kPa

Horizontal active stress: σ′_h = K_a·σ′_v → 12.27 and 15.34 kPa. Resultant: E_a,k = (12.27+15.34)/2·h·L = 13.80·0.95·1.45 = 19 kN

H_d = γ_G·E_a,k + γ_Q·Q_x,k = 1.35·19 + 1.5·250 = 400.65 kN

2. Base friction resistance

Only the permanent vertical load contributes: V′_k = G_z,k = 490 kN (on the safe side, the variable Q_z,k and the cohesion at the base are neglected).

Precast foundation → δ_k = ⅔·φ = ⅔·16° = 10.67° → tan δ_k = 0.189; γ_R;h = 1.1

R_d = V′_k·tan δ_k / γ_R;h = (490·0.189)/1.1 = 84.2 kN

3. Passive earth resistance on the back face of the foundation

K_p = tan²(45°+φ/2) = tan²(45°+16°) = 3.255. The passive stress: σ′_h,p = K_p·σ′_v → 130.12 and 162.59 kPa.

E_p,k = (130.12+162.59)/2·h·L = 146.35·0.95·1.45 = 201.6 kN → R_pd = E_p,k/γ_R;h = 201.6/1.1 = 183.27 kN

4. Sliding check

R_d + R_pd = 84.2 + 183.27 = 267.47 kN < H_d = 400.65 kN

Result: R_d + R_pd = 267.47 kN < H_d = 400.65 kN → NOT ADEQUATE. ✗ Because of the large horizontal load, the base friction and the full passive earth resistance together are still not enough; a larger foundation width or deeper embedment is needed.

Mini calculator

WE11 — sliding + passive earth resistance

EQU

G_z,k permanent vertical (kN)

Q_x,k horizontal column load (kN)

φ below base (°)

E_a,k active earth pressure (kN)

E_p,k passive earth resistance (kN)

γ_R;h resistance factor

Precast foundation → δ = ⅔·φ. The base friction uses only the permanent vertical load. Condition: H_d ≤ R_d + R_pd.

Calculators

Three interactive calculators for checking a shallow foundation. Adjust the geometry, the loads and the soil parameters — the figure updates to scale, and the program computes the design resistance and compares it with the design load. Use the sub-menu at the top to switch between the three calculators.

Drained case — bearing-resistance check

Uses all three terms of the drained formula (width, surcharge, cohesion), accounting for the load inclination (i factors) and the eccentricity (B′). For a strip footing the shape factors ≈ 1.

Undrained case — bearing-resistance check

Fine-grained soil, rapid loading: the shear strength is described by the undrained value c_u (φ = 0), with total stresses. σ_t = (2+π)·c_u·b_c·s_c·i_c + q.

Sliding check

Checking the foundation against horizontal sliding. The condition for adequacy is: H_d ≤ R_d. In the drained case the resistance comes from base friction (V_d·tan δ), in the undrained case from the base adhesion c_u (A·c_u); optionally the passive earth pressure (E_p) can also be added.

Bearing resistance of shallow foundations

How to use this material

1. The calculation procedure

The usual design workflow

2. Bearing-resistance calculation

2.1 Drained case

Width (self-weight) term

Surcharge (depth) term

Cohesion term

Input parameters

Bearing-resistance (bearing capacity) factors

Correction factors

Shape factors (s)

Load-inclination factors (i)

Base-inclination factors (b)

The load-inclination exponent (m)

2.2 Undrained case

Factors (undrained)

Notation

3. Effective foundation area and treatment of groundwater

3.1 Effective foundation area — eccentricity

3.2 Effect of groundwater — the effective γ′

4. Verification of adequacy

Design load — Vd

Design resistance — Rd

5. Sliding check (EQU)

Drained case — base friction

Undrained case — base adhesion

Worked examples

Worked Example 1 — Checking a drained strip footing

Task

Figure

Solution

1. Action side — the design load

2. Geometry and overburden pressure

3. Bearing-resistance factors (φ = 24°)

4. Characteristic and design resistance

5. Remedy: widen the foundation — B = 1.3 m

Mini calculator

WE1 — checking a drained strip footing

Worked Example 2 — Checking an undrained strip footing

Task

Solution

1. Action side

2. Overburden pressure (total stress!)

3. Undrained resistance

Mini calculator

WE2 — checking an undrained strip footing

Worked Example 3 — Eccentric and inclined load, with groundwater

Task

Figure

Solution

1. Action side

2. Eccentricity of the resultant force

3. Overburden pressure and γ′ (water above the base)

4. Factors (φ = 26°)

5. Resistance

Mini calculator

WE3 — eccentric + inclined load, with groundwater

Worked Example 4 — Effect of the design groundwater depth

Task

Figure

Solution

1. Action side — unchanged (same as WE3)

2. Eccentricity — unchanged

3. q′ and γ′ with the deeper groundwater — this changes!

4. Factors (φ = 26°) — unchanged

5. Resistance

Mini calculator

WE4 — strip footing, deeper groundwater

Worked Example 5 — Effect of the direction of the horizontal force

Task

Figure

Solution

1. Action side — unchanged

2. Eccentricity — this changes!

3. q′, γ′ and the factors — unchanged

4. Resistance with the smaller B′

Mini calculator

WE5 — reversed Gx (eccentricities add up)

Design load — V_d

Design resistance — R_d

WE5 — reversed G_x (eccentricities add up)

WE6 — larger eccentricity e_q