For comprehensive examples
and calculations of settlement, stress analysis and pressure that is
provided below, see:
See below for settlement estimates that includes:
 immediate settlement of noncohesive soils, NAVFAC method
 immediate settlement of noncohesive 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 CoarseGrained Granular Soils for Static Loads
Settlement analysis of noncohesive, coarsegrained soils is usually limited to the immediate
settlement analysis. Settlement of these soil types primarily occur from the rearrangement
of soil particles due to the immediate compression from the applied load. Typically, engineers justify ignoring the dissipation of
pore water pressure based on the assumption that these soil types have a large permeability
rate, 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 noncohesive
soils, which may be immediate settlements in the future. To learn more about settlements due to
earthquakes and vibrations, review "Settlement Analysis" by the USACE in the settlement section
of the
geotechnical publications page.
Immediate Settlement of NonCohesive Soils, NAVFAC Method
DH_{i} = 4qB^{2}
for isolated shallow footings and B < 20 ft
K_{vi}(B + 1)^{2}
DH_{i} = 2qB^{2}
for isolated shallow footings and B > 40 ft
K_{vi}(B + 1)^{2}
DH_{i} = 2qB^{2}
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/ft^{3})
Notes:
1. For continuous footings, multiply the computed settlement by 2.
2. Noncohesive soils include gravels, sands and nonplastic silts.
3. Multiply K_{vi} by 0.5 if groundwater is at base of footing or above, multiply
K_{vi} by 1.0 if groundwater is at least 1.5B below base of footing, and interpolate
multiplication factor for groundwater between base of footing and 1.5B below the footing base.
Immediate Settlement of NonCohesive 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, where
immediate settlement usually constitutes a significant portion of the total settlement.
Immediate Settlement of Cohesive Soils, Janbu Approximation
p = (u_{0})(u_{1}) qB
E_{s}
Where:p = immediate settlement (ft)
u_{0} =
Influence factor for
depth foundation
u_{1} =
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 111022906  Design of Pile
Foundations.
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_{s}Q_{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 in
bearing capacity technical
guidance
Q_{f} = Theoretical bearing capacity due to shaft friction, kN (lb)
see deep foundations in
bearing capacity technical
guidance
q_{p} = Theoretical unit tip bearing capacity, kN/m^{2 }(lb/ft^{2})
= see deep foundations in
bearing capacity technical
guidance
L = Length of pile, m (ft)
A = Crosssectional area of pile,
m^{2 }(ft^{2})
=
p(B/2)^{2} for closed end piles
E = Modulus of elasticity of the pile material, kN/m^{2 }(lb/ft^{2})
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 Piles Bored 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:
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}
2pz^{2}
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/ft^{2}
q = applied stress from structural load, lb/ft^{2}
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/m^{2} (lb/ft^{2})
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/m^{2} (lb/ft^{2})
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/m^{2} (lb/ft^{2})
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/m^{2}
(lb/ft^{2}) effective
stress
s =
gD, kN/m^{2}
(lb/ft^{2})
total stress
m =
g_{w}d, kN/m^{2}
(lb/ft^{2})
pore water pressure (see note below)
and,
g =
unit weight of soil, kN/m^{2} (lb/ft^{2})
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/m^{3}
(62.4 lb/ft^{3})
= 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
D_{10}
where D_{10 }= soil grain size diameter at the 10% finer of the
particle size distribution
Example #1: The soil profile indicates a soil
unit weight of 17.28 kN/m^{3} (110 lb/ft^{3}) 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/m^{3} (130 lb/ft^{3})
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/m^{3 }(110 lbs/ft^{3})

lower soil profile is 7.6 m (25 ft) thick

lower soil saturated unit weight,
g, is 20.42 kN/m^{3} (130 lbs/ft^{3})

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/m^{3})(3.0 m) = 51.8 kN/m^{2}
metric
=
(110 lb/ft^{3})(10 ft) = 1100 lb/ft^{2 }
standard
m =
g_{w}d
= 0
because depth is above influence of the water table
s' =
s 
m
= 51.8 kN/m^{2}  0 = 51.8 kN/m^{2}
metric
= 1100 lb/ft^{2}  0 = 1100 lb/ft^{2} standard
b)
s =
gD
= (17.28 kN/m^{3})(5.5 m) = 95.0 kN/m^{2}
metric
=
(110 lb/ft^{3})(18 ft) = 1980 lb/ft^{2 }
standard
m =
g_{w}d
= (9.81 kN/m^{3})( 0.71 m) =  6.97 kN/m^{2}
metric
= (62.4 lb/ft^{3})( 2 ft) =  124.8 lb/ft^{2}
standard
s' =
s 
m
= 95.0 kN/m^{2}  ( 6.97 kN/m^{2}) =
102.0 kN/m^{2} metric
= 1980 lb/ft^{2}  ( 124.8 lb/ft^{2}) =
2105 lb/ft^{2} standard
c)
s =
gD
= (17.28 kN/m^{3})(6.1 m) = 105.4 kN/m^{2}
metric
=
(110 lb/ft^{3})(20 ft) = 2200 lb/ft^{2 }
standard
m =
g_{w}d
= (9.81 kN/m^{3})(0 m) = 0
metric
= (62.4 lb/ft^{3})(0 ft) = 0^{
}
standard
s' =
s 
m
= 105.4 kN/m^{2}  0 = 105.4 kN/m^{2} metric
= 2200 lb/ft^{2}  0 = 2200 lb/ft^{2} standard
d)
s =
gD
= (17.3 kN/m^{3})(6.1 m) + (20.4 kN/m^{3})(7.6 m) =
260.6 kN/m^{2} metric
=
(110 lb/ft^{3})(20 ft) +
(130 lb/ft^{3})(25 ft) = 5450 lb/ft^{2 }
standard
m =
g_{w}d
= (9.81 kN/m^{3})(7.6 m) = 74.6 kN/m^{2}
metric
= (62.4 lb/ft^{3})(25 ft) = 1560 lb/ft^{2}
standard
s' =
s 
m
= 260.6 kN/m^{2}  74.6 kN/m^{2} =
186.0 kN/m^{2} metric
= 5450 lb/ft^{2}  1560 lb/ft^{2} =
3890 lb/ft^{2}
standard
******************************
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/m^{2} (3500 lb/ft^{2}).
Given

z = 0.61 m (2 ft)
*See the theory, equations and definitions provided above

q = 167.6 kN/m^{2 }(3500 lb/ft^{2})

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
= 1.5 standard
z 2.0 ft
n = y =
0.61 m = 1.0
metric
z 0.61 m
n = y =
2.0 ft
= 1.0 standard
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/m^{2})(0.2) = 134.1 kN/m^{2}
metric
= 4(3500 lb/ft^{2})(0.2) = 2800 lb/ft^{2}
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/m^{2}
(3500 lb/ft^{2}) is 134.1 kN/m^{2} (2800 lb/ft^{2}).
*************************
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