Settlement Analysis on the Geotechnical Information Website (2024)

For comprehensive examples and calculations of settlement, stress analysis and pressure that is provided below, see:

• Settlement, stress and pressure calculations that are provided below

See below for settlement estimates that includes:

• immediate settlement of non-cohesive soils, NAVFAC method
• immediate settlement of non-cohesive soils, Alpan approximation
• immediate settlement of cohesive soils, Janbu approximation
• settlement of single pile, Vesic

See below for pressure distribution and stress analysis that includes:

• Boussinesq theory
• Total stress
• Effective stress
• Pore water pressure

Settlement of Coarse-Grained Granular Soils for Static Loads

Settlement analysis of non-cohesive, coarse-grained soils is usually limited to the immediate settlement analysis. Settlement of these soil types primarily occur from the re-arrangementof soil particles due to the immediate compression from the applied load. Typically, engineers justify ignoring the dissipation ofpore water pressure based on the assumption that these soil types have a large permeabilityrate, thus releasing the excess pore water as the load(s) is applied. Earthquakes and other ground movement such as machinery or blasting, may also cause settlements in non-cohesivesoils, which may be immediate settlements in the future. To learn more about settlements due toearthquakes and vibrations, review "Settlement Analysis" by the USACE in the settlement sectionof the geotechnical publications page.

Immediate Settlement of Non-Cohesive Soils, NAVFAC Method

DH_i = 4qB² for isolated shallow footings and B < 20 ft
K_vi(B + 1)²

DH_i = 2qB² for isolated shallow footings and B > 40 ft
K_vi(B + 1)²

DH_i = 2qB² for deep isolated foundations and B < 20 ft
K_vi(B + 1)²

Where:

DH_i = immediate settlement of footing (ft)
q = footing unit load, bearing pressure (tsf)
B = footing width (ft)
K_vi = Modulus of vertical subgrade reaction (tons/ft³)

Notes:
1. For continuous footings, multiply the computed settlement by 2.
2. Non-cohesive soils include gravels, sands and non-plastic silts.
3. Multiply K_vi by 0.5 if groundwater is at base of footing or above, multiplyK_vi by 1.0 if groundwater is at least 1.5B below base of footing, and interpolatemultiplication factor for groundwater between base of footing and 1.5B below the footing base.

Immediate Settlement of Non-Cohesive Soils, Alpan Approximation

p = m'[2B/(1 + B)]²(a/12)q

Where:

p = immediate settlement (ft)
m' = shape factor
= (L/B)^0.39
L = footing length (ft)
B = footing width (ft)
a = Alpan parameter
q = average bearing pressure applied by footing on soil (tsf)

Settlement of Cohesive Soils for Static Loads

Total settlement for cohesive soils are generally estimated by the sum of immediate settlement, primary consolidation and secondary compression, whereimmediate settlement usually constitutes a significant portion of the total settlement.

Immediate Settlement of Cohesive Soils, Janbu Approximation

p = (u₀)(u₁) qB
E_s

p = immediate settlement (ft)
u₀ = Influence factor for depth foundation
u₁ = Influence factor for shape of foundation
q = footing unit load, bearing capacity (tsf)
B = footing width (ft)
E_s = Young's modulus of soil (tsf)

Primary Consolidation

Review free and downloadable settlement publications in our Settlement Analysis Publications

Secondary Compression

Review free and downloadable settlement publications in our Settlement Analysis Publications

Settlement of Single Pile (Vesic)

The following calculations were first derived by Vesic, and can be found in the USACE manual. See USACE EM 1110-2-2906 - Design of PileFoundations.

w = w_s + w_f + w_p

where,

w = vertical settlement of a single pile at the top of pile, m (ft)
w_s = (Q_p +a_sQ_f) L m (ft)
AE
= amount of settlement due to the axial deformation of the pile shaft
w_f = Cs(Qs) m (ft)
Dq_p
= amount of settlement at the pile tip caused by load transmitted along the pile shaft
w_p = Cp(Qp)m (ft)
Bq_p
= amount of settlement at the pile tip due to the load transferred at the tip

and,

Q_p = Theoretical bearing capacity for tip of foundation, kN (lb)
see deep foundations inbearing capacity technical guidance
Q_f = Theoretical bearing capacity due to shaft friction, kN (lb)
see deep foundations inbearing capacity technical guidance
q_p = Theoretical unit tip bearing capacity, kN/m² (lb/ft²)
= see deep foundations inbearing capacity technical guidance
L = Length of pile, m (ft)
A = Cross-sectional area of pile, m² (ft²)
= p(B/2)² for closed end piles
E = Modulus of elasticity of the pile material, kN/m² (lb/ft²)
a_s = Alpha approximation. See referenced manual for long piles in dense soils or flexible shafts.
D = Embedment depth of the pile, m (ft)
B = Diameter of pile, m (ft)
C_s = C_p[0.93 + 0.16(D/B)^0.5]
C_p = empirical coefficient: See referenced manual for different soils within 10B of the pile tip.
soil type Driven PilesBored Piles
sand (dense to loose) 0.02 to 0.04 0.09 to 0.18
silt (dense to loose) 0.03 to 0.05 0.09 to 0.12
clay (stiff to soft) 0.02 to 0.03 0.03 to 0.06

Pressure Analysis

Inducing pressure on a soil from structural loads is sometimes important to calculate, especially for settlement analyses. The pressure, or stress, on the soil where the structure is in contact with the soil is simply the structural load. The methods provided below will allow us to determine the additional pressure on a soil at a certain depth below the contact point of the structure. This includes a pressure bulb, and the Boussinesq theory.

Pressure Bulb

Knowing the amount of applied building loads and the designed footing width, we can use charts to determine the additional stress on a soil at specified depths beneath the footing and points beyond the foundation footprint. See the following source from the NAVFAC manual:

• Pressure Bulb

Boussinesq Theory

The change in soil pressure due to an applied load may be calculated from the following methods:

Circular Foundation

P_v = 3q [1/(1+(r/z)²]^-2.5
2pz²

Where:

P_v = change in vertical stress at point z below the center of a circularly loaded area, and point r horizontally from the center of the circularly loaded area, lb/ft²
q = applied stress from structural load, lb/ft²
p = 3.1412...
z = depth below center of circularly loaded area in which a change in vertical stress is desired, ft
r = horizontal distance from the center of a circularly loaded area in which a change in vertical stress is desired, ft

Rectangular Foundation

Ds_v = SP_v

Where:

Ds_v = total change in vertical stress due to an applied stress, kN/m² (lb/ft²)
SP_v = summation of all stress components (i.e. P_v1 + P_v2 + .... + P_vn)
P_v = qI_s = change in vertical stress at point z below the corner of a rectangular loaded area, kN/m² (lb/ft²)
If the vertical stress is desired below the middle of the foundation, where the stress is maximum, the rectangular footing must be divided into 4 sections, or quadrants, so that the corner of each section is located in the middle of the foundation. Thus we calculate the change in vertical stress from 4 different sections. The total change in vertical stress is the summation of the 4 different sections.
q = applied stress from structural load, kN/m² (lb/ft²)
I_s = Influence value from Boussinesq chart. I_s is determined from chart using m and n values.
m = x
z
n = y
z
z = depth below corner of rectangular loaded area in which a change in vertical stress is desired, m (ft)
x = length of foundation, m (ft)
y = width of foundation, m (ft)

Stress Analysis of Soil

The following equations will aid in determining pore water pressure, total stress and effective stress. See examples below for sample calculations.

s' =s -m

Where:

s' =s -m, kN/m² (lb/ft²) effective stress
s =gD, kN/m² (lb/ft²) total stress
m =g_wd, kN/m² (lb/ft²) pore water pressure (see note below)

and,

g =unit weight of soil, kN/m² (lb/ft²)
D = depth of overburden, m (ft), or vertical distance between surface of soil to a subsurface depth at which total stress is determined.
g_w = 9.81 kN/m³ (62.4 lb/ft³)= unit weight of water, constant
d = depth of water, m (ft), or vertical distance between surface of water table to a subsurface depth at which pore water pressure is determined.

Notes: Capillary rise will increase the effective stress by creating a negative pore water pressure within the capillary zone. Capillary rise above the water table is sometimes calculated as:

h_c = 0.15
D₁₀

where D₁₀ = soil grain size diameter at the 10% finer of the particle size distribution

Examples for calculating settlement, stress & pressures

Example #1: The soil profile indicates a soil unit weight of 17.28 kN/m³ (110 lb/ft³) to a depth of 6.1 m (20 ft) below the ground surface. The water table is at 6.1 m 20 ft below the ground surface. A saturated soil unit weight of 20.42 kN/m³ (130 lb/ft³) extends to a depth of 7.6 m (25 ft) below the water table. A capillary rise of 1.5 m (5 ft) was determined to exist above the water table.

a) Calculate the effective stress at a depth of 3.0 m (10 ft) below the ground surface.
b) Calculate the effective stress at a depth of 5.5 m (18 ft) below the ground surface.
c) Calculate the effective stress at a depth of 6.1 m (20 ft) below the ground surface.
d) Calculate the effective stress at a depth of 13.7 m (45 ft) below the ground surface.

Given

• upper soil profile is 6.1 m (20 ft) deep
• upper soil profile has a unit weight, g, of 17.28 kN/m³ (110 lbs/ft³)
• lower soil profile is 7.6 m (25 ft) thick
• lower soil saturated unit weight, g, is 20.42 kN/m³ (130 lbs/ft³)
• water table is at 6.1 m (20 ft) below the ground surface
• capillary rise is 1.5 m (5 ft) above the water table

Solution

a)

s =gD
= (17.28 kN/m³)(3.0 m) = 51.8 kN/m²metric
= (110 lb/ft³)(10 ft) = 1100 lb/ft² standard

m =g_wd
= 0 because depth is above influence of the water table

s' =s -m
= 51.8 kN/m² - 0 = 51.8 kN/m²metric
= 1100 lb/ft² - 0 = 1100 lb/ft² standard

b)

s =gD
= (17.28 kN/m³)(5.5 m) = 95.0 kN/m²metric
= (110 lb/ft³)(18 ft) = 1980 lb/ft² standard

m =g_wd
= (9.81 kN/m³)(- 0.71 m) = - 6.97 kN/m²metric
= (62.4 lb/ft³)(- 2 ft) = - 124.8 lb/ft² standard

s' =s -m
= 95.0 kN/m² - (- 6.97 kN/m²) = 102.0 kN/m²metric
= 1980 lb/ft² - (- 124.8 lb/ft²) = 2105 lb/ft² standard

c)

s =gD
= (17.28 kN/m³)(6.1 m) = 105.4 kN/m²metric
= (110 lb/ft³)(20 ft) = 2200 lb/ft² standard

m =g_wd
= (9.81 kN/m³)(0 m) = 0metric
= (62.4 lb/ft³)(0 ft) = 0 standard

s' =s -m
= 105.4 kN/m² - 0 = 105.4 kN/m² metric
= 2200 lb/ft² - 0 = 2200 lb/ft² standard

d)

s =gD
= (17.3 kN/m³)(6.1 m) + (20.4 kN/m³)(7.6 m) = 260.6 kN/m² metric
= (110 lb/ft³)(20 ft) + (130 lb/ft³)(25 ft) = 5450 lb/ft² standard

m =g_wd
= (9.81 kN/m³)(7.6 m) = 74.6 kN/m²metric
= (62.4 lb/ft³)(25 ft) = 1560 lb/ft² standard

s' =s -m
= 260.6 kN/m² - 74.6 kN/m² = 186.0 kN/m²metric
= 5450 lb/ft² - 1560 lb/ft² = 3890 lb/ft² standard

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Example #2: Using the Boussinesq theory, calculate the change in vertical stress at 0.61 m (2 ft) below the middle of a 1.2 m x 1.8 m (4 ft x 6 ft) rectangular foundation. The applied building load on this foundation is 167.6 kN/m² (3500 lb/ft²).

Given

• z = 0.61 m (2 ft) *See the theory, equations and definitions provided above
• q = 167.6 kN/m² (3500 lb/ft²)
• 1.2 m x 1.8 m (4 ft x 6 ft) rectangular footing

Solution

Ds_v = SP_v

SP_v = summation of all stress components (i.e. P_v1 + P_v2 + .... + P_vn). In this case, we analyze the foundation in 4 equal but separate quadrants. Instead of a single 1.2 m x 1.8 m (4 ft x 6 ft) foundation, we have 4 separate 0.61 m x 0.91 m (2 ft x 3 ft) quadrants. This is done so that one corner of each quadrant is located in the center of the footing.

4P_v = 4qI_s Since the quadrants have equal dimensions with the same applied load, we simply multiply the equation by 4 (4 quadrants). If footing had varying applied loads or unequal quadrant shapes, then each stress summation must be done separately.

m = x = 0.91 m = 1.5 metric
z 0.61 m
m = x = 3.0 ft = 1.5standard
z 2.0 ft

n = y = 0.61 m = 1.0 metric
z 0.61 m
n = y = 2.0 ft = 1.0standard
z 2.0 ft

I_s = 0.2 Influence value from Boussinesq chart, where m = 1.5 and n= 1.0.

Ds_v = SP_v = 4P_v = 4qI_s

= 4(167.6 kN/m²)(0.2) = 134.1 kN/m² metric
= 4(3500 lb/ft²)(0.2) = 2800 lb/ft² standard

Conclusion

Additional pressure on the soil at a distance of 0.61 m (2 ft) below the center of the footing due to an applied building load of 167.6 kN/m² (3500 lb/ft²) is 134.1 kN/m² (2800 lb/ft²).

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