Settlement of shallow foundations

Learn the methods for calculating the stress beneath strip and pad footings and the resulting settlement, 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 four steps of a settlement calculation: stress distribution, calculation of strains, determination of the limit depth, and summation of the strains. Here you can learn the principle of the three stress-calculation methods — Kögler (straight-line spreading), Jáky (triangular stress diagram) and Kany (tabulated, beneath the characteristic point) — with interactive figures and a live demo.
Worked examples — seven practical examples derived in detail: WE1 Kögler stress calculation; WE2–WE4 Jáky settlement with the oedometer modulus; WE5–WE6 Jáky settlement with the compression curve; WE7 the Kany lamella method. Each comes with a figure reconstructed from the practical material, a step-by-step solution and its own calculator.
Calculators — a standalone, detailed calculator for each of the three methods (Kögler, Jáky, Kany). The foundation type and geometry are adjustable, as is the two-layer soil profile (layer thickness, unit weight, oedometer modulus); the figure updates to scale in real time, and the program also computes the limit depth and the oedometric settlement.

The sliders and input fields update the figures and results in real time — it is worth experimenting with the parameters. The limit depth is given by the Jáky formula for the Jáky method, and by the n = 5 rule for the Kögler and Kany methods.

The calculation procedure

A settlement calculation consists of four steps, regardless of which stress method is used. The four steps — each accessible separately from the sub-menu above — are discussed in detail below, each with explanatory figures.

Step 1 — Determining the stress distribution

Settlement is caused by the additional stress transmitted to the soil by the foundation — this must first be determined at various depths below the foundation base. We present three procedures: the elasticity-based, exact Kany method, then two approximate procedures based on an assumed stress distribution — the Kögler and the Jáky method.

Starting point — contact pressure, geostatic stress, additional stress

In an infinite half-space, beneath a uniformly distributed load the stress along a given vertical does not change with depth. Beneath foundations of finite extent, however, the stress also spreads laterally from the edges of the foundation, so beneath the foundation it gradually decreases with depth and after a while becomes practically zero.

σ_base = G_z,k / (B · L)  — average contact pressure (the load includes the self-weight of the foundation and of the overlying soil)
σ_g = Σ γ_i · h_i  — geostatic (overburden) stress at the foundation base
σ_z,0 = p₀ = σ_base − σ_g  — the additional stress at the foundation base that causes settlement

For settlement, only the additional stress (σ_z,0, p₀ in many sources) matters — i.e. what the foundation transmits to the soil in addition to the natural overburden stress. If a foundation pit or basement is excavated, the removed soil unloads the ground, so this too must be subtracted: σ_z,0 = σ_base − t₀·γ, where t₀ is the depth of excavation.

A) Kany method — based on the theory of elasticity (exact calculation)

The stress distribution computed beneath the characteristic point of a rectangular foundation is called the Kany method. The complex derivation is awkward to use directly, so the results are usually given and used in the form of a graph or a table.

Why the characteristic point?

Under uniform loading, the settlement beneath a rectangular foundation is greatest at the centre, smaller at the edge midpoints, and smallest at the corners. This curvature of the surface would be followed only by an infinitely flexible foundation; a rigid foundation keeps its plane, and its average settlement is a weighted average of these settlements. There exists a point — the characteristic point (Grasshof) — beneath which the computed settlement equals both the flexible and the rigid foundation's average settlement. Therefore, in settlement calculations we always use the (average) stresses arising in the vertical through the characteristic point.

Practical use of the Kany stress-calculation method

The table below gives the ratio σ_z/p₀ (the decay of the additional stress) as a function of two ratios:

z/B — the depth of the point below the foundation base divided by the smaller foundation dimension (B = the smaller foundation width)
B/L — the ratio of the two foundation dimensions: B/L = 0 for a strip footing, B/L = 1 for a square pad footing

We compute the ratios z/B and B/L, look up the corresponding σ_z/p₀ factor in the table — interpolating linearly between the tabulated values — and finally the required stress is: σ_z = (σ_z/p₀) · p₀.

Stress beneath the characteristic point (σ_z/p₀)

z/B	B/L=0 (strip)	0.20	0.40	0.60	0.80	1.00 (square)
0.00	1.000	1.000	1.000	1.000	1.000	1.000
0.05	0.990	0.990	0.989	0.988	0.985	0.981
0.10	0.945	0.944	0.941	0.932	0.918	0.898
0.20	0.826	0.824	0.804	0.770	0.731	0.694
0.30	0.739	0.730	0.689	0.637	0.593	0.557
0.40	0.677	0.660	0.601	0.544	0.502	0.470
0.50	0.630	0.603	0.532	0.477	0.438	0.409
0.60	0.590	0.553	0.477	0.425	0.389	0.362
0.80	0.524	0.469	0.392	0.348	0.316	0.289
1.00	0.467	0.399	0.290	0.290	0.260	0.234
1.50	0.360	0.278	0.226	0.193	0.166	0.144
2.00	0.288	0.206	0.163	0.134	0.111	0.094
3.00	0.203	0.128	0.095	0.072	0.057	0.047
4.00	0.155	0.088	0.060	0.044	0.034	0.028
5.00	0.125	0.065	0.041	0.029	0.023	0.018
6.00	0.113	0.056	0.035	0.024	0.020	0.015
7.00	0.100	0.047	0.029	0.020	0.016	0.013
8.00	0.088	0.039	0.023	0.016	0.013	0.010
10.00	0.063	0.021	0.011	0.008	0.006	0.005
15.00	0.050	0.015	0.007	0.005	0.004	0.003
20.00	0.032	0.006	0.003	0.002	0.001	0.001

The table gives individual points of the family of curves; we interpolate linearly between the listed z/B and B/L values. For a strip footing we use the B/L = 0 column.

Practical application — division into lamellae (Kany lamella method):

Immediately beneath the foundation, 20 cm lamellae — down to a depth of 2B for a strip footing, and to about 1–1.5B for a pad footing (the stress gradient is largest here)
Deeper down, 40 cm lamellae — all the way to the limit depth
A lamella boundary is mandatory at a change of layer and at the groundwater level
For each lamella we look up the Kany factor using the mid-height z, from it σ_z, then the average stress and the compression of the lamella (see Step 2)

B) Kögler method — a closed region bounded by straight lines

An approximate, quick method, mainly for strip footings. The bounding lines starting from the edges of the foundation (the z = 0 lines) make an angle α with the vertical, with no limit in depth. The method assumes that, between the bounding lines, the stress is uniform at any depth z. The stress thus spreads out with depth at the angle α.

The stress spreads out with depth at the angle α and is uniform within the region. From vertical force equilibrium (the force on the base of width B = the force on the plane of width B+2z·tan α):

Strip footing: σ_z = σ_z,0 · B / (B + 2·z·tan α)
Pad footing (two-way stress spreading): σ_z = σ_z,0 · B·L / [(B + 2·z·tan α) · (L + 2·z·tan α)]

α = φ (tan α = 0.5) is the classical assumption; in Hungarian practice α = 45° is more common. α = 30° is also used.

Limitations: simple and fast, but a crude approximation — it assumes a uniform stress within the region instead of the real parabolic distribution. The pad-footing formula is in many cases subject to large error, so it is not used in practice. On its own the method does not give a limit depth.

C) Jáky's procedure

A method that approximates the real stress distribution much better and is widespread in Hungarian practice. The Jáky procedure determines a limit depth (m₀), below which the stress from the foundation is already zero; down to this depth the stress decreases linearly. Since the stress is the same across the full width of the foundation, in the vertical sense it has a triangular distribution — so the settlement can be calculated simply, even in one's head.

From the foundation base to the limit depth the stress decreases by similar triangles, reaching zero at the limit depth. The Jáky theory gives the limit depth from the foundation dimensions:

Limit depth: m₀ = 2·B·(1 − B/(2·L))
Stress at depth z: σ_z = σ_z,0 · (m₀ − z) / m₀
Lateral stress spreading: ξ = B·z / (m₀ − z)

Case	m₀	Note
Strip footing (L → ∞)	2·B	B/L → 0
Square pad footing (L = B)	B	B/L = 1
Rectangular pad footing (B < L)	2·B·(1 − B/(2L))	general form

The calculation step by step

Limit depth (m₀) — from the foundation dimensions: m₀ = 2·B·(1 − B/(2·L)). For a strip footing (L → ∞) m₀ = 2B; for a square pad footing (L = B) m₀ = B.
Stress at a given depth z — the stress diagram runs linearly to zero from the foundation base to the limit depth, so by similar triangles:
σ_z / σ_z,0 = (m₀ − z) / m₀ ⟹ σ_z = σ_z,0 · (m₀ − z) / m₀
Derivation of the lateral stress spreading (ξ) — from vertical force equilibrium. The force transmitted over the width B at the foundation base equals the area of the stress diagram formed at depth z. This diagram is a plateau of width B (σ_z) and, on each side, a triangle of width ξ (½·σ_z·ξ):
B·σ_z,0 = B·σ_z + 2·(½·σ_z·ξ) = σ_z·(B + ξ)
substituting σ_z: B·σ_z,0 = σ_z,0·(m₀ − z)/m₀ · (B + ξ)
simplifying by σ_z,0: B = (m₀ − z)/m₀ · (B + ξ)
rearranging: B + ξ = B·m₀ / (m₀ − z)
hence ξ: ξ = B·m₀/(m₀ − z) − B = B·z / (m₀ − z)
Drawing the stress diagram — with the obtained σ_z and ξ values, the region between the verticals at the edges of the foundation can be drawn at any depth z. In the vertical through the characteristic point the stress is the same throughout (across the full width of the foundation), so in the vertical sense it has a triangular distribution — therefore the settlement can be calculated simply, even in one's head (see Steps 2 and 4).

With the Jáky method the limit depth follows automatically, and thanks to the triangular distribution the settlement is simple to compute. This is why it became popular in Hungarian practice — especially combined with the compression curve or the constrained (oedometer) modulus (see Steps 2 and 4).

🎯 Mini-demo — Kögler vs. Jáky stress distribution

Adjust the parameters: how does the stress (σ_z) change with depth according to the two methods beneath a strip footing?

B (foundation width): 2.0 m

σ_z,0 (kPa): 180 kPa

α (Kögler angle, deg): 45 °

Step 2 — Calculation of strains

Once the stress distribution beneath the foundation is known, the next step is to determine the vertical strains — the compressions — arising from the stresses. We do this layer by layer, and within each layer lamella by lamella: at the mid-height of each layer/lamella we determine the average stress increment, and from it the strain.

The stress–strain relationship can be linear or nonlinear. In the linear case the soil is treated as linearly elastic and the compression curve is replaced by a straight line over a given stress range — this mainly holds for dense granular soils (gravel, sand). We use two basic procedures: the constrained (oedometer) modulus, or the compression curve directly.

A) Calculation with the constrained (oedometer) modulus

Δε_z = Δσ_z,avg / E_oed

We divide the layer's (or lamella's) average stress increment by the layer's constrained (oedometer) modulus, E_oed. The modulus is approximated by a single constant value over a given stress range — this is a good approximation especially for granular soils. The method is simple and fast, so it is the most common in practice.

Example: if a layer's average stress increment is Δσ_z,avg = 92 kPa, and E_oed = 15 MPa = 15 000 kPa, then Δε_z = 92 / 15 000 = 0.00613.

B) Calculation with the compression curve

A more accurate procedure, because it accounts for the fact that the soil's stiffness changes with stress — the modulus is not constant. Each soil layer has a compression curve giving the strain (ε) as a function of the vertical stress (σ_z). The curve is steep at first (large compression at low stress), then gradually flattens.

The calculation for a layer's mid-height:

On the horizontal axis of the layer's compression curve we mark the initial (overburden) stress acting at the layer's mid-height, σ_z0,avg — this accounts for the fact that the layer was not unloaded before construction either.
Continuing from this, we mark the average stress increment caused by the structure, Δσ_z,avg; the layer's new stress is thus σ_z0,avg + Δσ_z,avg.
On the vertical axis we read the strain corresponding to the initial and the new stress — their difference is the layer's load-induced strain:

Δε_z = ε(σ_z0,avg + Δσ_z,avg) − ε(σ_z0,avg)

For nonlinear behaviour the compression curve is approximated semi-logarithmically, using the compression index (C_c):

Δε_z = C_c / (1 + e₀) · lg[(σ_z0 + Δσ_z) / σ_z0]

where e₀ is the layer's initial void ratio. Justified for heavily loaded soft fat clays and organic soils.

Important: when using the compression curve, one must always start from the layer's initial stress — we do not consider only the stress increment. The strain from the curve: the value read at the new stress minus the value read at the initial stress.

🎯 Mini-demo — computing Δε from Δσ and E_oed

How much strain develops in a lamella for a given stress increment?

Δσ_avg: 100 kPa

E_oed: 10 MPa

lamella thick.: 0.4 m

Step 3 — Determining the limit depth

Generally accepted

m₀: σ_z = σ_z,0/n, n = 5 or 10

EC7 Hungarian National Annex: n = 5.

Jáky's recommendation

m₀ = 2·B·(1 − B/(2L))

Strip footing: 2B; square pad footing: B.

From practical considerations

m₀ = at the top of a stiff layer (if reached before the limit depth).

For a raft (B > 10 m)

m₀ = 2/3·B … 1/2·B is justified.

Why is a limit depth needed? The σ_z(z) functions reach zero only at z = ∞, so the full integral would be infinite. Experience shows the settlement is finite — below the limit depth the stress causes no further grain movement (because of the friction threshold).

🎯 Mini-demo — Limit depth

Where does the σ_z curve intersect the σ_g/n line? Change the parameters and watch how m₀ changes for a strip footing (n = 5)!

B (m): 1.5 m

σ_z,0 (kPa): 250 kPa

γ (kN/m³): 18 kN/m³

Step 4 — Summation of the strains

s = ∫₀^{m₀} ε_z(z) dz ≈ Σ (layer thickness) · Δε_z,avg

In practice we usually integrate with the trapezoidal rule: the thickness of each lamella × the lamella's average strain. The settlement comes out in [m]; it is customary to give the result in mm.

Two-layer example (Jáky case)

Given a two-layer soil profile, down to the limit depth below the foundation base:

Δσ_base = σ_z,0 at the foundation base
Δσ_m0 = 0 at the limit depth
Δσ_LB = Δσ_base · (m₀ − Δz_LB) / m₀ at the layer boundary (linear interpolation)
Layer averages: Δσ_avg,1 = (Δσ_base + Δσ_LB)/2; Δσ_avg,2 = (Δσ_LB + 0)/2
Strain: Δε_i = Δσ_avg,i/E_oed,i
Settlement: s = (t₁−h)·Δε₁ + (m₀−(t₁−h))·Δε₂

3. Determining the limit depth

The limit depth (m₀) is the depth below which the additional stress no longer causes measurable strain — the stress increment does not exceed the friction threshold between the grains.

Why is a limit depth necessary?

The stress functions σ_z(z) reach 0 only as z → ∞, so the settlement integral would be infinite. Experience does not show this — below a certain depth the stress no longer causes grain movement.

Three methods for m₀

Generally accepted (n-rule)

m₀: Δσ_z(m₀) = σ_g/n

n = 5 (EC7 Hungarian National Annex)
n = 10 (other national annexes)

Jáky's recommendation

m₀ = 2·B·(1 − B/(2L))

Strip footing (L → ∞): m₀ = 2B; square pad footing (L = B): m₀ = B

Practical consideration

m₀ = at the top of a stiff layer, if reached before the limit depth.

For a raft (B > 10 m): m₀ = 2/3·B … 1/2·B is justified.

Examples

Foundation	B (m)	L (m)	m₀ per Jáky
Strip footing	1.5	→ ∞	2·1.5 = 3.00 m
Rectangular pad footing	2.0	3.0	2·2·(1−2/6) = 2.67 m
Square pad footing	1.2	1.2	2·1.2·(1−1.2/2.4) = 1.20 m
Rectangular pad footing	1.2	3.0	2·1.2·(1−1.2/6) = 1.92 m

4. Calculation of strains

Calculating the vertical strain (Δε) per lamella or per layer using the stress increment and the soil's elastic parameter.

Three methods for Δε

1. With the oedometer modulus (the most common)

Δε_z = Δσ_z,avg / E_oed

We divide the layer's average stress increment by the layer's oedometer modulus. E_oed is a constant characteristic of the soil (order of MPa).

E.g. Layer 1: Δσ_avg = 92 kPa, E_oed = 15 000 kPa → Δε = 92/15 000 = 0.00613

2. With the compression curve (more accurate)

Δε_z = ε(σ_init + Δσ_avg) − ε(σ_init)

From the curve we read the strain corresponding to the new and the initial stress — the difference is the layer's strain due to the load. This accounts for the modulus changing with stress (not constant).

3. With Hooke's law (theory of elasticity)

Δε_z = (1/E)·[Δσ_z − μ·(Δσ_x + Δσ_y)]

To be used when the horizontal stress increment is also known (e.g. from Boussinesq). The values E and μ (Poisson) are needed.

Relationship: E_oed = E·(1−μ)/(1−μ−2μ²)

Summation into settlement

s = ∫₀^{m₀} ε_z(z) dz ≈ Σ_i h_i · Δε_i

In practice we integrate with the trapezoidal rule: each lamella's thickness × the lamella's average strain. The result comes out in mm.

Tip: on the Worked examples tab (WE2–WE7) you can see how these methods apply in various situations (1 layer, 2 layers, compression curve, Kany lamella) with concrete numbers.

Worked examples

Seven worked examples derived in detail from the practical material — stress calculation by Kögler, settlement calculation by Jáky with the oedometer modulus and with the compression curve, and the Kany lamella method. Each example comes with a figure, a step-by-step solution and its own calculator.

Worked Example 1 — Stress calculation by the Kögler method

Task

At the base of a foundation, an additional stress of σ_z,0 = 180 kPa is transmitted to the soil. The foundation width is B = 2.5 m, and the spreading angle is α = 45°. Compute the vertical additional stress (σ_z) at depths z = 1, 2 and 4 m below the foundation base:

(a) for a strip footing
(b) for a pad footing of length L = 3.5 m

WE1

Figure

Solution

Strip footing: σ_z = σ_z,0 · B / (B + 2·z·tan α)
Pad footing: σ_z = σ_z,0 · B·L / [(B + 2·z·tan α) · (L + 2·z·tan α)]

α = 45° → tan α = 1, so the width (B + 2z) results with depth.

(a) Strip footing, B = 2.5 m:

z = 1 m: σ_z = 180 · 2.5 / (2.5 + 2·1) = 180 · 2.5/4.5 = 100.00 kPa
z = 2 m: σ_z = 180 · 2.5 / (2.5 + 4) = 180 · 2.5/6.5 = 69.23 kPa
z = 4 m: σ_z = 180 · 2.5 / (2.5 + 8) = 180 · 2.5/10.5 = 42.86 kPa

(b) Pad footing, B = 2.5 m, L = 3.5 m:

z = 1 m: σ_z = 180 · 2.5 · 3.5 / (4.5 · 5.5) = 1575/24.75 = 63.64 kPa
z = 2 m: σ_z = 180 · 8.75 / (6.5 · 7.5) = 1575/48.75 = 32.31 kPa
z = 4 m: σ_z = 180 · 8.75 / (10.5 · 11.5) = 1575/120.75 = 13.04 kPa

Observation: for the same foundation width, a pad footing (two-way spreading) 'sheds' the stress much faster than a strip footing (one-way). At z = 4 m depth, for instance, the pad footing's σ_z is only ~13 kPa, while the strip footing's is still ~43 kPa.

Calculator

Kögler stress — WE1

Kögler

Inputs

σ_z,0kPa

Lfor strip footing ≥ 100m

α°

Depths (m)

Comma-separated:

Mode

Worked Example 2 — Settlement by the Jáky method, with the oedometer modulus

Task

Pad footing, B = 2 m, L = 3 m, foundation height h = 1 m. The load is G_z,k = 820 kN (it also includes the self-weight of the foundation and of the soil above it). The ground surface is at 0.00 m (rel.), the foundation base at −1.00 m (rel.), and the layer boundary at −2.20 m (rel.) (1.20 m below the base). The groundwater is deep.

Layer 1: γ = 18 kN/m³, E_oed,1 = 15 MPa
Layer 2: γ = 18 kN/m³, E_oed,2 = 5 MPa

Determine the limit depth per Jáky, and compute the settlement of the foundation using the oedometer moduli.

WE2

Figure

Solution step by step

1. Contact pressure and additional stress

σ_base = G_z,k/(B·L) = 820/(2·3) = 136.67 kPa
σ_g = γ·h = 18·1 = 18 kPa
σ_z,0 = Δσ_base = σ_base − σ_g = 118.67 kPa

2. Limit depth (Jáky)

m₀ = 2·B·(1 − B/(2L)) = 2·2·(1 − 2/6) = 2.67 m

3. Δσ at the layer boundary (Δz_LB = 1.20 m below the base)

Δσ_LB = Δσ_base · (m₀ − Δz_LB)/m₀ = 118.67 · (2.67 − 1.20)/2.67 = 65.33 kPa

4. Layer averages and strains

Layer 1: Δσ_avg,1 = (118.67 + 65.33)/2 = 92.00 kPa; Δε₁ = 92.00/(15·1000) = 0.00613

Layer 2: Δσ_avg,2 = (65.33 + 0)/2 = 32.67 kPa; Δε₂ = 32.67/(5·1000) = 0.00653

5. Settlement by layer

Δs₁ = 1.20·0.00613 = 7.4 mm

Δs₂ = (2.67 − 1.20)·0.00653 = 9.6 mm

Result: s = Δs₁ + Δs₂ = 17.0 mm ✓

Calculator

Jáky + E_oed — WE2

Jáky

Geometry

Δz_LBbelow the basem

G_z,kkN

Soils

γ₁kN/m³

E_oed,1MPa

E_oed,2MPa

Worked Example 3 — What does doubling the load change?

Task

Same pad footing and soil profile as WE2, but the vertical load is doubled: G_z,k = 1640 kN. What will the settlement be?

WE3

Observation: since the foundation geometry is the same, the limit depth m₀ = 2.67 m is unchanged. The contact pressure, however, doubles, and the additional stress (σ_z,0) increases by a similar amount.

Figure

Solution

1. New stresses

σ_base = 1640/(2·3) = 273.33 kPa; σ_z,0 = 273.33 − 18 = 255.33 kPa

Δσ_LB = 255.33 · (2.67−1.20)/2.67 = 140.56 kPa

2. Layer averages and strains

Δε₁ = (255.33+140.56)/2 / 15 000 = 197.95/15 000 = 0.0132

Δε₂ = 140.56/2 / 5 000 = 70.28/5 000 = 0.01413

3. Settlement

Δs₁ = 1.20·0.0132 = 15.8 mm

Δs₂ = 1.47·0.01413 = 20.8 mm

Result: s = 36.6 mm ✓ — doubling the load increased the settlement ~2.15× (the geostatic stress is constant, so σ_z,0 increases by more than 2×).

Calculator

Jáky + E_oed — WE3

Jáky

Geometry

Δz_LBm

G_z,kkN

Soils

γ₁kN/m³

E_oed,1MPa

E_oed,2MPa

Worked Example 4 — What does a narrower foundation change (B = 1.2 m)?

Task

Same load as WE3 (G_z,k = 1640 kN), but the width decreases to B = 1.2 m (L = 3 m). How does the settlement turn out?

WE4

Two opposing effects:

Smaller area → larger contact pressure, i.e. σ_z,0
Smaller B → smaller limit depth m₀, i.e. the integration range shrinks

The two effects almost cancel each other — the total settlement barely changes.

Figure

Solution

1. New geometry and stress

m₀ = 2·1.2·(1 − 1.2/6) = 1.92 m

σ_base = 1640/(1.2·3) = 455.56 kPa; σ_z,0 = 455.56 − 18 = 437.56 kPa

Δσ_LB = 437.56·(1.92−1.2)/1.92 = 164.09 kPa

2. Strains

Δε₁ = (437.56+164.09)/2 / 15 000 = 0.0200

Δε₂ = 164.09/2 / 5 000 = 0.0164

3. Settlement

Δs₁ = 1.20·0.0200 = 24.0 mm

Δs₂ = (1.92−1.20)·0.0164 = 11.8 mm

Result: s = 35.8 mm ✓ — barely changes compared to WE3 (36.6 mm)!

Calculator

Jáky + E_oed — WE4

Jáky

Geometry

Δz_LBm

G_z,kkN

Soils

γ₁kN/m³

E_oed,1MPa

E_oed,2MPa

Worked Example 5 — Settlement by the Jáky method, with the compression curve

Task

Strip footing, B = 1.5 m, founding depth h = 1.2 m (foundation base at −1.2 m (rel.)). The load is G_z,k = 1100 kN/m. The layer boundary is at −2 m (rel.) (0.8 m below the foundation base). Both layers have a unit weight γ = 18 kN/m³; the strain is described by the given compression curve. The groundwater is deep.

WE5

Figure

Compression curves given for the problem

Solution

1. Contact pressure and additional stress

σ_base = 1100/(1.5·1) = 733.3 kPa; σ_g = 18·1.2 = 21.6 kPa; σ_z,0 = Δσ_base = 711.7 kPa

2. Limit depth (strip footing → m₀ = 2·B)

m₀ = 2·1.5 = 3.0 m

Δσ_LB = 711.7·(3.0−0.8)/3.0 = 521.9 kPa

3. Initial average stress in each layer

Layer 1 mid-height (0.4 m below the base): σ_z0,avg,1 = γ·(h + (t₁−h)/2) = 18·(1.2+0.4) = 28.8 kPa

Layer 2 mid-height ((0.8+3)/2 = 1.9 m below the base): σ_z0,avg,2 = 18·3.1 = 55.8 kPa

4. Increment averages

Δσ_avg,1 = (711.7+521.9)/2 = 616.8 kPa

Δσ_avg,2 = (521.9+0)/2 = 261.0 kPa

5. Strain from the compression curve

Layer 1: ε(28.8) → ε(28.8+616.8) = ε(645.6); read from the curve, e.g. Δε₁ ≈ 0.020

Layer 2: ε(55.8) → ε(55.8+261.0) = ε(316.8); read off, Δε₂ ≈ 0.008

6. Settlement

Δs₁ = 0.8·0.020 = 16 mm

Δs₂ = 2.2·0.008 = 18 mm

Result: s ≈ 34 mm ✓

Calculator

Jáky + compression curve — WE5

compr.

Geometry

Δz_LBm

G_z,kkN

γ₁kN/m³

γ₂kN/m³

Compression curve (σ:ε, comma-separated)

Layer 1:

Layer 2:

Worked Example 6 — Strip footing with a different soil profile, with the compression curve

Task

Strip footing, B = 1.3 m, founding depth h = 0.7 m. The load is G_z,k = 900 kN/m. The layer boundary is at −1.2 m (rel.) (0.5 m below the foundation base). Both layers have γ = 20 kN/m³. Each of the two layers has its own oedometric strain table. The groundwater is deep.

WE6

Figure

Oedometric strain table given for the problem

σ (kPa)	0	20	50	100	200	300	400	600	800
ε₁ (%) — Layer 1	0.00	0.20	0.50	0.80	1.00	1.10	1.20	1.30	1.35
ε₂ (%) — Layer 2	0.00	0.56	1.40	2.00	2.40	2.70	2.90	3.20	3.40

Solution (average stresses and reading off the compression curve)

1. Contact pressure and additional stress

σ_base = 900/(1.3·1) = 692.3 kPa; σ_g = 20·0.7 = 14 kPa; σ_z,0 = 678.3 kPa

2. Limit depth (strip footing)

m₀ = 2·B = 2.6 m

Δσ_LB = 678.3·(2.6−0.5)/2.6 = 547.9 kPa

3. Initial and final average stresses

Layer 1 mid-height 0.25 m below the base → σ_z0,avg,1 = 20·(0.7+0.25) = 19 kPa

Layer 2 mid-height: σ_z0,avg,2 = 20·(0.7+0.5+1.05) = 45 kPa

Δσ_avg,1 = (678.3+547.9)/2 = 613.1 kPa

Δσ_avg,2 = 547.9/2 = 274.0 kPa

4. Strain from the compression curve (example values)

Layer 1: ε(19) ≈ 0.002, ε(632) ≈ 0.012 → Δε₁ ≈ 0.010

Layer 2: ε(45) ≈ 0.012, ε(319) ≈ 0.027 → Δε₂ ≈ 0.015

5. Settlement

Δs₁ = 0.5·0.010 = 5 mm

Δs₂ = 2.1·0.015 = 32 mm

Result: s ≈ 37 mm (depending on the precise reading of the curves)

Calculator

Jáky + compr. curve — WE6

compr.

Geometry

Δz_LBm

G_z,kkN

γ₁kN/m³

γ₂kN/m³

Compression curve (σ:ε)

Layer 1:

Layer 2:

Worked Example 7 — Settlement by the Kany lamella method

Task

Square pad footing, B = L = 1.2 m, foundation base at −1.00 m (rel.) (h = 1.0 m). The load is G_z,k = 450 kN. The subsoil can be described as two layers:

Layer 1 (gravelly sand, 0–2.4 m below the base): γ = 20 kN/m³, E_oed,1 = 15 MPa
Layer 2 (silty sand, from 2.4 m below the base): γ = 20 kN/m³, E_oed,2 = 5 MPa

Determine the settlement with the Kany method beneath the characteristic point, dividing into lamellae.

WE7

Figure

Solution

1. Contact pressure and additional stress

σ_base = 450/(1.2·1.2) = 312.5 kPa; σ_g (h = 1.0 m, γ = 20) = 20 kPa

p₀ = σ_base − σ_g = 292.5 kPa

2. Defining the lamellae

For a pad footing, 0.2 m thick lamellae in the upper 1–1.5·B band, and 0.4 m deeper. A lamella boundary is also placed at the soil layer boundary (2.4 m below the base).

3. Kany factors and lamella stresses

At the bottom of each lamella: Δσ_z(z) = (Kany factor) · p₀. The factors are read from the B/L = 1.0 column.

E.g. z = 0.2 m → z/B = 0.167 → factor by interpolation ≈ 0.762 → Δσ_z = 222.9 kPa

z = 0.4 m → z/B = 0.333 → factor ≈ 0.528 → Δσ_z = 154.4 kPa

…and so on.

4. Determining the limit depth

m₀ is where Δσ_z = σ_g/n (n = 5). Here ≈ 3.30 m (already past the 2.4 m layer boundary, in the gravel layer).

5. Lamella strains

Δε_lamella = Δσ_z,avg / E_oed · 1000 (average of the stresses at the bottom and top lamella planes)

Δs_lamella = lamella thickness × Δε. Sum the settlements of all lamellae.

Result: s ≈ 19.1 mm (the calculator prints the full lamella table)

Calculator

Kany lamella — WE7

Kany

Geometry and loading

h_found.m

G_z,kkN

n—

h_gwwater below ground surfacem

Layers (depth_to_bottom_m, γ, γ_sat, E_oed)

Detailed calculators

A standalone, interactive calculator for each of the three stress-calculation methods. In each, the foundation type and geometry are adjustable, as are the thickness, unit weight and oedometer modulus of two soil layers. As you move the sliders, the figure is redrawn to scale, in real time; the program also computes the limit depth and the oedometric settlement.

Kögler method

A closed region bounded by straight lines, spreading at angle α. The limit depth is given by the n = 5 rule: where the additional stress falls to one fifth of the geostatic stress.

Jáky method

Triangular (linear) stress diagram: the stress decreases to 0 from the foundation base to the limit depth. The limit depth is given by the Jáky formula: m₀ = 2·B·(1 − B/(2·L)).

Kany method

The stress beneath the characteristic point from the Kany table (as a function of z/B and B/L). The limit depth is given by the n = 5 rule.

Settlement of shallow foundations

How to use this material

The calculation procedure

Step 1 — Determining the stress distribution

Starting point — contact pressure, geostatic stress, additional stress

A) Kany method — based on the theory of elasticity (exact calculation)

Why the characteristic point?

Practical use of the Kany stress-calculation method

Stress beneath the characteristic point (σz/p0)

B) Kögler method — a closed region bounded by straight lines

C) Jáky's procedure

The calculation step by step

🎯 Mini-demo — Kögler vs. Jáky stress distribution

Step 2 — Calculation of strains

A) Calculation with the constrained (oedometer) modulus

B) Calculation with the compression curve

🎯 Mini-demo — computing Δε from Δσ and E_oed

Step 3 — Determining the limit depth

Generally accepted

Jáky's recommendation

From practical considerations

For a raft (B > 10 m)

🎯 Mini-demo — Limit depth

Step 4 — Summation of the strains

Two-layer example (Jáky case)

3. Determining the limit depth

Why is a limit depth necessary?

Three methods for m0

Generally accepted (n-rule)

Jáky's recommendation

Practical consideration

Examples

4. Calculation of strains

Three methods for Δε

1. With the oedometer modulus (the most common)

2. With the compression curve (more accurate)

3. With Hooke's law (theory of elasticity)

Summation into settlement

Worked examples

Worked Example 1 — Stress calculation by the Kögler method

Task

Figure

Solution

Calculator

Kögler stress — WE1

Inputs

Depths (m)

Worked Example 2 — Settlement by the Jáky method, with the oedometer modulus

Task

Figure

Solution step by step

1. Contact pressure and additional stress

2. Limit depth (Jáky)

3. Δσ at the layer boundary (ΔzLB = 1.20 m below the base)

4. Layer averages and strains

5. Settlement by layer

Calculator

Jáky + Eoed — WE2

Geometry

Soils

Worked Example 3 — What does doubling the load change?

Task

Figure

Solution

1. New stresses

2. Layer averages and strains

3. Settlement

Calculator

Jáky + Eoed — WE3

Geometry

Soils

Worked Example 4 — What does a narrower foundation change (B = 1.2 m)?

Task

Figure

Solution

1. New geometry and stress

2. Strains

3. Settlement

Calculator

Jáky + Eoed — WE4

Stress beneath the characteristic point (σ_z/p₀)

Three methods for m₀

3. Δσ at the layer boundary (Δz_LB = 1.20 m below the base)

Jáky + E_oed — WE2

Jáky + E_oed — WE3

Jáky + E_oed — WE4

2. Limit depth (strip footing → m₀ = 2·B)

Layers (depth_to_bottom_m, γ, γ_sat, E_oed)