Basics of Foundation Engineering هندسة الأساسات & Eng. Ahmed S. Al-Agha

Basics of Foundation Engineering
with Solved Problems
Prepared By:
Ahmed S. Al-Agha
September -2015
Based on “Principles of Foundation Engineering, 7th
Edition”

Being rich is not about how
much you have, but is about how
much you can give

Chapter (2)
Subsoil Exploration

Page (1)
Foundation Engineering Subsoil Exploration
Ahmed S. Al-Agha
Introduction:
The soil mechanics course reviewed the fundamental properties of soils and
their behavior under stress and strain in idealized conditions. In practice,
natural soil deposits are not homogeneous, elastic, or isotropic. In some
places, the stratification of soil deposits even may change greatly within a
horizontal distance of 15 to 30 m. For foundation design and construction
work, one must know the actual soil stratification at a given site, the
laboratory test results of the soil samples obtained from various depths, and
the observations made during the construction of other structures built under
similar conditions. For most major structures, adequate subsoil exploration
at the construction site must be conducted.
Definition:
The process of determining the layers of natural soil deposits that will
underlie a proposed structure and their physical properties is generally
referred to as subsurface exploration.
Purpose of Subsurface Exploration:
The purpose of subsurface exploration is to obtain information that will aid
the geotechnical engineer in:
1. Determining the nature of soil at the site and its stratification.
2. Selecting the type and depth of foundation suitable for a given structure.
3. Evaluating the load-bearing capacity of the foundation.
4. Estimating the probable settlement of a structure.
5. Determining potential foundation problems (e.g., expansive soil,
collapsible soil, sanitary landfill, etc...).
6. Determining the location of water table.
7. Determining the depth and nature of bedrock, if and when encountered.
8. Performing some in situ field tests, such as permeability tests, van shear
test, and standard penetration test.
9. Predicting the lateral earth pressure for structures such as retaining
walls, sheet pile, and braced cuts.

Page (2)
Ahmed S. Al-Agha
Subsurface Exploration Program:
A soil exploration program for a given structure can be divided broadly into
three phases:
1. Collection of Preliminary Information:
This step includes obtaining information regarding the type of structure to
be built and its general use. The following are examples explain the needed
information about different types of structures:
 For the construction of building:
 The approximate column loads and their spacing.
 Local building-codes.
 Basement requirement.
 For the construction of bridge:
 The length of their spans.
 The loading on piers and abutments.
2. Reconnaissance:
The engineer should always make a visual inspection (field trip) of the site
to obtain information about:
 The general topography of the site, the possible existence of drainage
ditches, and other materials present at the site.
 Evidence of creep of slopes and deep, wide shrinkage cracks at regularly
spaced intervals may be indicative of expansive soil.
 Soil stratification from deep cuts, such as those made for the construction
of nearby highways and railroads.
 The type of vegetation at the site, which may indicate the nature of the
soil.
 Groundwater levels, which can be determined by checking nearby wells.
 The type of construction nearby and the existence of any cracks in walls
(indication for settlement) or other problems.
 The nature of the stratification and physical properties of the soil nearby
also can be obtained from any available soil-exploration reports on
existing structures.

Page (3)
Ahmed S. Al-Agha
3. Site Investigation:
This phase consists of:
 Planning (adopting steps for site investigation, and future vision for the
site)
 Making test boreholes.
 Collecting soil samples at desired intervals for visual observation and
laboratory tests.
Determining the number of boring:
There is no hard-and-fast rule exists for determining the number of borings
are to be advanced. For most buildings, at least one boring at each corner
and one at the center should provide a start. Spacing can be increased or
decreased, depending on the condition of the subsoil. If various soil strata
are more or less uniform and predictable, fewer boreholes are needed than in
nonhomogeneous soil strata.
The following table gives some guidelines for borehole spacing between for
different types of structures:
Approximate Spacing of Boreholes
Type of project Spacing (m)
Multistory building 10–30
One-story industrial plants 20–60
Highways 250–500
Residential subdivision 250–500
Dams and dikes 40–80
Determining the depth of boring:
The approximate required minimum depth of the borings should be
predetermined. The estimated depths can be changed during the drilling
operation, depending on the subsoil encountered (e.g., Rock).
To determine the approximate required minimum depth of boring, engineers
may use the rules established by the American Society of Civil Engineers
(ASCE 1972):
1. Determine the net increase in effective stress (∆σ′
) under a foundation
with depth as shown in the Figure below.

Page (4)
Ahmed S. Al-Agha
2. Estimate the variation of the vertical effective stress (σo
′
) with depth.
3. Determine the depth (D = D1) at which the effective stress increase
(∆𝛔′
) is equal to (
𝟏
𝟏𝟎
) q (q = estimated net stress on the foundation).
4. Determine the depth (D = D2) at which (∆𝛔′
/𝛔 𝐨
′
) = 𝟎. 𝟎𝟓.
5. Determine the depth (D = D3) which is the distance from the lower face
of the foundation to bedrock (if encountered).
6. Choose the smaller of the three depths, (D1, D2, and D3), just determined
is the approximate required minimum depth of boring.
After determining the value of (D) as explained above the final depth of
boring (from the ground surface to the calculated depth) is:
Dboring = Df + D
Because the Drilling will starts from the ground surface.
Determining the value of vertical effective stress (𝛔 𝐨
′
):
The value of (σo
′
) always calculated from the ground surface to the
required depth, as we previously discussed in Ch.9 (Soil Mechanics).
σo
′∆σ′

Page (5)
Ahmed S. Al-Agha
Determining the increase in vertical effective stress(∆𝛔′
):
The value of (∆σ′
) always calculated from the lower face of the foundation
as we discussed previously in soil mechanics course (Ch.10).
An alternative approximate method can be used rather than (Ch.10) in soil
mechanics course, this method is easier and faster than methods in (Ch.10).
This method called (2:1 Method). The value of (∆σ′
) can be determined
using (2:1 method) as following:
According to this method, the value of (∆σ′
) at depth (D) is:
∆σD
′
=
P
A
=
P
(B + D) × (L + D)
P = the load applied on the foundation (KN).
A = the area of the stress distribution at 𝐝𝐞𝐩𝐭𝐡 (𝐃).

Page (6)
Ahmed S. Al-Agha
Note that the above equation is based on the assumption that the stress from
the foundation spreads out with a vertical-to-horizontal slope of 2:1.
Now, the values of (D1 and D2) can be calculated easily as will be seen later.
Note: if the foundation is circular the value of (∆σ′
) at depth (D) can be
determined as following:
∆σD
′
=
P
Area at depth (D)
=
P
π
4
× (B + D)2
B = diameter of the foundation(m).

Page (7)
Ahmed S. Al-Agha
In practice: The number of boreholes and the depth of each borehole will be
identified according to the type of project and the subsoil on site, the
following is example for a 5 story residential building with dimensions of
(40 x 70) m:
 The required number of boreholes = 5 boreholes (one at each corner and
one at the center) as mentioned previously.
 The depth of each borehole for this project is (8-10) m up to a depth of
water table.
The following figure shows the distribution of boreholes on the land:

Page (8)
Ahmed S. Al-Agha
Procedures for Sampling Soil
There are two types of samples:
 Disturbed Samples: These types of samples are disturbed but
representative, and may be used for the following types of laboratory soil
tests:
 Grain size analysis.
 Determination of liquid and plastic limits.
 Specific gravity of soil solids.
 Determination of organic content.
 Classification of soil.
 But disturbed soil samples cannot be used for consolidation, hydraulic
conductivity, or shear tests, because these tests must be performed on
the same soil of the field without any disturbance (to be
representative)
The major equipment used to obtain disturbed sample is (Split Spoon)
which is a steel tube has inner diameter of 34.93 mm and outer diameter
of 50.8mm.
 Undisturbed Samples: These types of samples are used for the
following types of laboratory soil tests:
 Consolidation test.
 Hydraulic Conductivity test.
 Shear Strength tests.
These samples are more complex and expensive, and it’s suitable for
clay, however in sand is very difficult to obtain undisturbed samples.
The major equipment used to obtain undisturbed sample is (Thin-Walled
Tube).
Degree of Disturbance
If we want to obtain a soil sample from any site, the degree of disturbance
for a soil sample is usually expressed as:
AR(%) =
Do
2
− Di
2
Di
2 × 100
AR = area ratio (ratio of disturbed area to total area of soil)

Page (9)
Ahmed S. Al-Agha
Do = outside diameter of the sampling tube.
Di = inside diameter of the sampling tube.
If (AR) ≤ 10% → the sample is 𝐮𝐧𝐝𝐢𝐬𝐭𝐮𝐫𝐛𝐞𝐝.
If (AR) > 10% → the sample is 𝐝𝐢𝐬𝐭𝐮𝐫𝐛𝐞𝐝.
For a standard split-spoon sampler (which sampler for disturbed samples):
AR(%) =
(50.8)2
− (34.93)2
(34.93)2
× 100 = 111.5% > 10% → disturbed.
Standard Penetration Test (SPT)
This test is one of the most important soil tests for geotechnical engineers
because it’s widely used in calculating different factors as will explained
later. This test is performed according the following procedures:
1. Determining the required number and depth of boreholes in the site.
2. The sampler used in SPT test is (Standard Split Spoon) which has an
inside diameter of 34.39 mm and an outside diameter of 50.8 mm.
3. Using drilling machine, 1.5m are drilled.
4. The drilling machine is removed and the sampler will lowered to the
bottom of the hole.
5. The sampler is driven into the soil by hammer blows to the top of the
drill rod, the standard weight of the hammer is 622.72 N (63.48 Kg), and
for each blow, the hammer drops a distance of 76.2 cm.
6. The number of blows required for a spoon penetration of three 15 cm
intervals are recorded.
7. The first 15 cm drive is considered as seating load and is ignored.
8. The number of blows required for the last two intervals are added to
give the Standard Penetration Number (N) at that depth.
9. The sampler is then withdrawn and the soil sample recovered from the
tube is placed in a glass bottle and transported to laboratory.
10. Using the drilling machine to drill another 1.5m and then repeat the
above steps for each 1.5 m till reaching the specified depth of borehole.
11.Take the average for (N) value from each 1.5 m to obtain the final
Standard Penetration Number.
12.Split Spoon samples are taken at intervals (1.5m) because theses samples
are highly disturbed.

Page (10)
Ahmed S. Al-Agha
Drilling Machine

Page (11)
Ahmed S. Al-Agha
Correction to N value
There are several factors contribute to the variation of the standard
penetration number (N) at a given depth for similar profiles. Among these
factors are the SPT hammer efficiency, borehole diameter, sampling method,
and rod length.
In the field, the magnitude of hammer efficiency can vary from 30 to 90%,
the standard practice now is to express the N-value to an average energy
ratio of 60% (N60) (but we assume it 100%), so correcting for field
procedures is required as following:
N60 =
NηHηBηSηR
60
N= measured penetration number.
N60 = standard penetration number, corrected for the field conditions.
ηH = hammer efficiency (%).
ηB = correction for borehole diameter.
ηS = sampler correction.
ηR = correction for rod lenght.
Variations of ηH, ηB, ηS, and ηR are summarized in table 2.5 (page 84).
Note: take ηH = 0.6 (US safety hammer).
Correlations for N60:
N60 can be used for calculating some important parameters such as:
 Undrained shear strength (Cu) (page 84 in text book).
 Overconsolidation ratio (OCR) (page 85).
 Angle of internal friction (ϕ) (page 88 ).
 Relative Density (Dr)(page 87 ).
 Allowable bearing capacity (qall,net) and Settlement (Se)(Ch. 5 page 263).
Soil Report
Different soil reports will be discussed on the lecture.

Page (12)
Ahmed S. Al-Agha
Problems:
1.
Site investigation is to be made for a structure of 100m length and 70m
width. The soil profile is shown below, if the structure is subjected to 200
KN/m2
what is the approximate depth of borehole (Assume
𝛄 𝐰 =10KN/m3
).
Solution
Givens:
q = 200KN/m2
, structure dimensions = (70 × 100)m
→ P = 200 × (100 × 70) = 1.4 × 106
KN.
Df = 0.0 (Structure exist on the ground surface) , γsat = 18KN/m3
.
D3 = 130m (distance from the lower face of structure to the bedrock).
1. Calculating the depth (D1) at which ∆𝛔 𝐃 𝟏
′
= (
𝟏
𝟏𝟎
) × 𝐪 :
(
1
10
) × q = (
1
10
) × 200 = 20KN/m2
.
The following figure showing the distribution of stress under the structure at
depth (D1):
γsat = 18KN/m3

Page (13)
Ahmed S. Al-Agha
The increase in vertical stress (∆σ′
) at depth (D1) is calculated as follows:
∆σD1
′
=
P
A
=
1.4 × 106
(100 + D1) × (70 + D1)
@ D1 → ∆σ′
= (
1
10
) × q →
1.4×106
(100+D1)×(70+D1)
= 20 → D1 = 180 m.
2. Calculating the depth (D2) at which (
∆𝛔′
𝛔 𝐨
′ ) = 𝟎. 𝟎𝟓 ∶
The effective stress(σo
′
) at depth D2 is calculated as following:
σo,D2
′
= (γsat − γw) × D2
→ σo,D2
′
= (18 − 10) × D2 → σo,D2
′
= 8D2.
∆σD2
′
=
P
A
=
1.4 × 106
(100 + D2) × (70 + D2)
@ D2 → (
∆σ′
σo
′ ) = 0.05 →
1.4×106
(100+D2)×(70+D2)
= 0.05 × (8D2) → D2 = 101.4 m
So, the value of (D) is the smallest value of D1, D2, and D3 → D = D2 = 101.4 m.
→ Dboring = Df + D → Dboring = 0.0 + 101.4 = 101.4 m ✓.

Page (14)
Ahmed S. Al-Agha
2. (Mid 2005)
Site investigation is to be made for a structure of 100m length and 70m
width. The soil profile is shown below. Knowing that the structure exerts a
uniform pressure of 200 KN/m2
on the surface of the soil, and the load
transports in the soil by 2V:1H slope.
What is the approximate depth of borehole? (Assume 𝛄 𝐰 =10KN/m3
).
Solution
Givens:
q = 200KN/m2
, structure dimensions = (70 × 100)m
→ P = 200 × (100 × 70) = 1.4 × 106
KN.
Df = 0.0 (Structure exist on the ground surface).
D3 = 130m (distance from the lower face of structure to the bedrock).
1. Check if (D1<30m or D1>30m):
@ depth D=30 m if ∆σ′
< (
1
10
) × q → D1 < 30m, elseD1 > 30m →→
Because the magnitude of (∆σ′
) decreased with depth.
γsat = 19KN/m3
γsat = 17KN/m3

Page (15)
Ahmed S. Al-Agha
(
1
10
) × q = (
1
10
) × 200 = 20KN/m2
.
depth (30m):
) at depth (30m) is calculated as follows:
∆σ30m
′
=
P
A
=
1.4 × 106
(100 + 30) × (70 + 30)
= 107.7 KN/m2
.
→ ∆σ30m
′
> (
1
10
) × q → D1 > 30m.
′
= (
𝟏
𝟏𝟎
) × 𝐪 :
(
1
10
) × q = (
1
10
) × 200 = 20KN/m2
.
∆σD1
′
=
P
A
=
1.4 × 106
(100 + D1) × (70 + D1)
@ D1 → ∆σ′
= (
1
10
) × q →
1.4×106
(100+D1)×(70+D1)
= 20 → D1 = 180 m.

Page (16)
Ahmed S. Al-Agha
3. Check if (D2<30m or D2>30m):
@ depth D=30 m if (
∆σ′
σo
′ ) < 0.05 → D2 < 30m, elseD2 > 30m →→
Because the magnitude of (
∆σ′
σo
′ ) decreased with depth.
∆σ30m
′
= 107.7 KN/m2
(as calculated above).
The effective stress at depth (30m) is calculated as follows:
σo,30m
′
= (γsat − γw) × 30
→ σo,30m
′
= (17 − 10) × 30 → σo,30m
′
= 210KN/m2
→ (
∆σ′
σo
′
) = (
107.7
210
) = 0.51 > 0.05 → D2 > 30m.
∆𝛔′
𝛔 𝐨
′ ) = 𝟎. 𝟎𝟓 ∶
Let D2 = 30 + X (X: distance from layer (2)to reach(D2).
′
σo,D2
′
= (17 − 10) × 30 + (19 − 10) × X
→ σo,D2
′
= 210 + 9X.
∆σD2
′
=
P
A
=
1.4 × 106
(100 + D2) × (70 + D2)
, but D2 = 30 + X →→
∆σD2
′
=
1.4 × 106
(130 + X) × (100 + X)
@ D2 → (
∆σ′
σo
′ ) = 0.05 →
1.4×106
(130+X)×(100+X)
= 0.05 × (210 + 9X)
→ X = 69 m → D2 = 69 + 30 = 99 m
So, the value of (D) is the smallest value of D1, D2, and D3 → D = D2 = 99 m.
→ Dboring = Df + D → Dboring = 0.0 + 99 = 99 m ✓.

Page (17)
Ahmed S. Al-Agha
3. (Mid 2013)
For the soil profile shown below, if D1=10m and D2=2D1.
A- Determine the dimensions of the foundation to achieve the required depth
of borehole.
B- Calculate the load of column which should be applied on the foundation
to meet the required depth of boring.
Solution
Givens:
D1 = 10m , D2 = 2D1 → D2 = 2 × 10 = 20m , Df = 2m
D3 = 40m (distance from the lower face of foundation to the bedrock)
A. (B=??)
@D1 → ∆σD1
′
= (
1
10
) × q
γdry = 18 KN/m3
γsat = 22 KN/m3

Page (18)
Ahmed S. Al-Agha
depth (D1=10m):
) at depth (D1 = 10m) is calculated as
follows:
∆σD1
′
=
P
A
=
P
(B+10)×(B+10)
→ ∆σD1
′
=
P
(B+10)2
q =
P
A
=
P
(B×B)
→ (
1
10
) q =
P
10(B×B)
By equal 1&2→
P
(B+10)2 =
P
10B2 → B =4.62m✓.
B. (P=??)
D2 = 2D1 → D2 = 2 × 10 = 20m , B = 4.62m
@D2 → ∆σD2
′
= 0.05 × σo,D2
′
∆σD2
′
=
P
A
=
P
(B+20)×(B+20)
→ ∆σD2
′
=
P
(4.62+20)2
Eq.1
Eq.2
Eq.1

Page (19)
Ahmed S. Al-Agha
′
) at depth (D2=20m) is calculated as following:
σo
′
is calculated from the 𝐠𝐫𝐨𝐮𝐧𝐝 𝐬𝐮𝐫𝐟𝐚𝐜𝐞
σo,D2
′
= 18 × 2 + 18 × 10 + (22 − 10) × 10 = 336KN/m2
@D2 → ∆σD2
′
= 0.05 × σo,D2
′
→
P
(4.62 + 20)2
= 0.05 × 336
→ P = 10,183.2 KN✓.
4.
Site investigation is to be made for 2500 KN load carried on (3.0 m x 2.0 m)
footing. The foundation will be built on layered soil as shown in the figure
below, estimate the depth of bore hole. (Assume 𝛄 𝐰= 10KN/m3
).
Solution
Givens:
P = 2500 KN , foundation dimensions = (3 × 2)m
q =
P
A
=
2500
3 × 2
= 416.67 KN/m2
, Df = 1.5m
D3 = 100 − 1.5 = 98.5m
Sand γdry = 17 KN/m3
Sand γsat = 18.5 KN/m3
Clay γsat = 16.9 KN/m3

Page (20)
Ahmed S. Al-Agha
Without check, it’s certainly the values of D1 & D2 > 3.5m, but if you don’t
sure you should do the check at every change in soil profile (like problem 2).
′
= (
𝟏
𝟏𝟎
) × 𝐪 :
(
1
10
) × q = (
1
10
) × 416.67 = 41.67KN/m2
.
The following figure showing the distribution of stress under the foundation
at depth (D1):
∆σD1
′
=
P
A
=
2500
(3 + D1) × (2 + D1)
@ D1 → ∆σ′
= (
1
10
) × q →
2500
(3+D1)×(2+D1)
= 41.67 → D1 = 5.26 m.
∆𝛔′
𝛔 𝐨
′ ) = 𝟎. 𝟎𝟓 ∶
Let D2 = 3.5 + X (X: distance from the clay layer to reach(D2).
′
σo,D2
′
= 17 × 1.5 + 17 × 2 + (18.5 − 10) × 1.5 + (16.9 − 10) × X

Page (21)
Ahmed S. Al-Agha
→ σo,D2
′
= 72.25 + 6.9X → σo,D2
′
= 72.25 + 6.9 × (D2 − 3.5)
→ σo,D2
′
= 48.1 + 6.9D2
∆σD2
′
=
P
A
=
2500
(3 + D2) × (2 + D2)
@ D2 → (
∆σ′
σo
′ ) = 0.05 →
2500
(3+D2)×(2+D2)
= 0.05 × (48.1 + 6.9D2)
→ D2 = 15.47 m
So, the value of (D) is the smallest value of D1, D2, and D3 → D = D1 = 5.26 m.
→ Dboring = Df + D → Dboring = 1.5 + 15.26 = 6.76 m ✓.

Chapter (3)
Ultimate Bearing
Capacity of Shallow
Foundations

Page (23) Ahmed S. Al-Agha
Foundation Engineering Ultimate Bearing Capacity of Shallow Foundations
Introduction
To perform satisfactorily, shallow foundations must have two main
characteristics:
1. They have to be safe against overall shear failure in the soil that
supports them.
2. They cannot undergo excessive displacement, or excessive settlement.
Note: The term excessive settlement is relative, because the degree of
settlement allowed for a structure depends on several considerations.
Types of Shear Failure
Shear Failure: Also called “Bearing capacity failure” and it’s occur when
the shear stresses in the soil exceed the shear strength of the soil.
There are three types of shear failure in the soil:
1. General Shear Failure

The following are some characteristics of general shear failure:
 Occurs over dense sand or stiff cohesive soil.
 Involves total rupture of the underlying soil.
 There is a continuous shear failure of the soil from below the footing to
the ground surface (solid lines on the figure above).
 When the (load / unit area) plotted versus settlement of the footing, there
is a distinct load at which the foundation fails (Qu)
 The value of (Qu) divided by the area of the footing is considered to be
the ultimate bearing capacity of the footing(qu).
 For general shear failure, the ultimate bearing capacity has been defined
as the bearing stress that causes a sudden catastrophic failure of the
foundation.
 As shown in the above figure, a general shear failure ruptures occur and
pushed up the soil on both sides of the footing (In laboratory).
 However, for actual failures on the field, the soil is often pushed up on
only one side of the footing with subsequent tilting of the structure as
shown in figure below:

2. Local Shear Failure:
The following are some characteristics of local shear failure:
 Occurs over sand or clayey soil of medium compaction.
 Involves rupture of the soil only immediately below the footing.
 There is soil bulging )‫بروز‬ ‫او‬ ‫(انتفاخ‬ on both sides of the footing, but the
bulging is not as significant as in general shear. That’s because the
underlying soil compacted less than the soil in general shear.
 The failure surface of the soil will gradually (not sudden) extend outward
from the foundation (not the ground surface) as shown by solid lines in the
above figure.

 So, local shear failure can be considered as a transitional phase between
general shear and punching shear.
 Because of the transitional nature of local shear failure, the ultimate
bearing capacity could be defined as the firs failure load (qu,1) which occur
at the point which have the first measure nonlinearity in the load/unit area-
settlement curve (open circle), or at the point where the settlement starts
rabidly increase (qu) (closed circle).
 This value of (qu) is the required (load/unit area) to extends the failure
surface to the ground surface (dashed lines in the above figure).
 In this type of failure, the value of (qu) it’s not the peak value so, this
failure called (Local Shear Failure).
 The actual local shear failure in field is proceed as shown in the following
figure:
3. Punching Shear Failure:

The following are some characteristics of punching shear failure:
 Occurs over fairly loose soil.
 Punching shear failure does not develop the distinct shear surfaces
associated with a general shear failure.
 The soil outside the loaded area remains relatively uninvolved and there is
a minimal movement of soil on both sides of the footing.
 The process of deformation of the footing involves compression of the
soil directly below the footing as well as the vertical shearing of soil around
the footing perimeter.
 As shown in figure above, the (q)-settlement curve does not have a
dramatic break )‫مفاجئ‬ ‫,)تغير‬ and the bearing capacity is often defined as the
first measure nonlinearity in the (q)-settlement curve(qu,1).
 Beyond the ultimate failure (load/unit area) (qu,1), the (load/unit area)-
settlement curve will be steep and practically linear.
 The actual punching shear failure in field is proceed as shown in the
following figure:

Ultimate Bearing Capacity (𝐪 𝐮)
It’s the minimum load per unit area of the foundation that causes shear
failure in the underlying soil.
Or, it’s the maximum load per unit area of the foundation can be resisted
by the underlying soil without occurs of shear failure (if this load is
exceeded, the shear failure will occur in the underlying soil).
Allowable Bearing Capacity (𝐪 𝐚𝐥𝐥)
It’s the load per unit area of the foundation can be resisted by the underlying
soil without any unsafe movement occurs (shear failure) and if this load is
exceeded, the shear failure will not occur in the underlying soil till reaching
the ultimate load.
Terzaghi’s Bearing Capacity Theory
Terzaghi was the first to present a comprehensive theory for evaluation of
the ultimate bearing capacity of rough shallow foundation. This theory is
based on the following assumptions:
1. The foundation is considered to be sallow if (Df ≤ B).
2. The foundation is considered to be strip or continuous if (
B
L
→ 0.0).
(Width to length ratio is very small and goes to zero), and the derivation of
the equation is to a strip footing.
3. The effect of soil above the bottom of the foundation may be assumed to
be replaced by an equivalent surcharge (q = γ × Df). So, the shearing
resistance of this soil along the failure surfaces is neglected (Lines ab and cd
in the below figure)
4. The failure surface of the soil is similar to general shear failure (i.e.
equation is derived for general shear failure) as shown in figure below.
Note:
1. In recent studies, investigators have suggested that, foundations are
considered to be shallow if [ Df ≤ (3 → 4)B], otherwise, the foundation is
deep.
2. Always the value of (q) is the effective stress at the bottom of the
foundation.

Terzaghi’s Bearing Capacity Equations
As mentioned previously, the equation was derived for a strip footing and
general shear failure, this equation is:
qu = cNc + qNq + 0.5BγNγ (for continuous or strip footing)
Where
qu = Ultimate bearing capacity of the 𝐮𝐧𝐝𝐞𝐫𝐥𝐲𝐢𝐧𝐠 soil (KN/m2
)
c = Cohesion of 𝐮𝐧𝐝𝐞𝐥𝐲𝐢𝐧𝐠 soil (KN/m2
)
q = 𝐄𝐟𝐞𝐞𝐜𝐭𝐢𝐯𝐞 stress at the bottom of the foundation (KN/m2
)
Nc, Nq, Nγ = Bearing capacity factors (nondimensional)and are
functions 𝐨𝐧𝐥𝐲 of the 𝐮𝐧𝐝𝐞𝐫𝐥𝐲𝐢𝐧𝐠 soil friction angle, ϕ, →→
The variations of bearing capacity factors and underlying soil friction angle
are given in (Table 3.1, P.139) for general shear failure.
The above equation (for strip footing) was modified to be useful for both
square and circular footings as following:
For square footing:
qu = 1.3cNc + qNq + 0.4BγNγ
B = The dimension of each side of the foundation .
For circular footing:
B = The diameter of the foundation .
Note:
These two equations are also for general shear failure, and all factors in the
two equations (except, B,) are the same as explained for strip footing.

Now for local shear failure the above three equations were modified to be
useful for local shear failure as following:
qu =
2
3
cNc
′
+ qNq
′
+ 0.5BγNγ
′
(for continuous or strip footing)
qu = 0.867cNc
′
+ qNq
′
+ 0.4BγNγ
′
(for square footing)
qu = 0.867cNc
′
+ qNq
′
+ 0.3BγNγ
′
(for circular footing)
Nc
′
, Nq
′
, Nγ
′
= Modified bearing capacity factors and could be determined
by the following two methods:
1. (Table 3.2 P.140) variations of modified bearing capacity factors and
underlying soil friction angle.
2. [(Table 3.1 P.139)(if you don’t have Table 3.2)], variation of bearing
capacity factors and underlying soil friction angle, but you must do the
following modification for the underlying soil friction angle:
tanϕ (General Shear) =
2
3
× tanϕ (Local Shear) →→
ϕmodified,general = tan−1
(
2
3
tanϕlocal)
For example: Assume we have local shear failure and the value of ϕ = 30°
1. By using (Table 3.2) Nc
′
, Nq
′
, Nγ
′
= 18.99, 8.31, and 4.9 respectively
2. By using (Table 3.1) → ϕmodified,general = tan−1
(
2
3
tan30°
) = 21.05°
→
(Nc, Nq, Nγ)21.05,table 3.1 ≅ (Nc
′
, Nq
′
, Nγ
′
)30,table 3.2 = 18.92, 8.26, and 4.31 respectively
General Bearing Capacity Equation (Meyerhof Equation)
Terzagi’s equations shortcomings:
 They don’t deal with rectangular foundations (0 <
B
L
< 1).
 The equations do not take into account the shearing resistance along the
failure surface in soil above the bottom of the foundation (as mentioned
previously).
 The inclination of the load on the foundation is not considered (if exist).

To account for all these shortcomings, Meyerhof suggested the following
form of the general bearing capacity equation:
qu = cNcFcsFcdFci + qNqFqsFqdFqi + 0.5BγNγFγsFγdFγi
Where
c = Cohesion of the underlying soil
q = Effective stress at the level of the bottom of the foundation.
γ = unit weight of the underlying soil
B = Width of footing (= diameter for a circular foundation).
Nc, Nq, Nγ = Bearing capacity factors(will be discussed later).
Fcs, Fqs, Fγs = Shape factors (will be discussed later).
Fcd, Fqd, Fγd = Depth factors (will be discussed later).
Fci, Fqi, Fγi = Inclination factors (will be discussed later).
Notes:
1. This equation is valid for both general and local shear failure.
2. This equation is similar to original equation for ultimate bearing capacity
(Terzaghi’s equation) which derived for continuous foundation, but the
shape, depth, and load inclination factors are added to this equation
(Terzaghi’s equation) to be suitable for any case may exist.
Bearing Capacity Factors:
The angle α = ϕ (according Terzaghi theory in the last figure “above”) was
replaced by α = 45 +
ϕ
2
. So, the bearing capacity factor will be change.
The variations of bearing capacity factors (Nc, Nq, Nγ) and underlying soil
friction angle (ϕ) are given in (Table 3.3, P.144).
Shape Factors:
Fcs = 1 + (
B
L
) (
Nq
Nc
)
Fqs = 1 + (
B
L
) tanϕ
Fγs = 1 − 0.4 (
B
L
)

Notes:
1. If the foundation is continuous or strip →
B
L
= 0.0
2. If the foundation is circular→ B = L = diameter →
B
L
= 1
Depth Factors:
 𝐅𝐨𝐫
𝐃 𝐟
𝐁
≤ 𝟏
1. 𝐅𝐨𝐫 𝛟 = 𝟎. 𝟎
Fcd = 1 + 0.4 (
Df
B
)
Fqd = 1
Fγd = 1
2. 𝐅𝐨𝐫 𝛟 > 𝟎. 𝟎
Fcd = Fqd −
1 − Fqd
Nctanϕ
Fqd = 1 + 2 tanϕ (1 − sinϕ)2
(
Df
B
)
Fγd = 1
 𝐅𝐨𝐫
𝐃 𝐟
𝐁
> 𝟏
1. 𝐅𝐨𝐫 𝛟 = 𝟎. 𝟎
Fcd = 1 + 0.4 tan−1
(
Df
B
)
⏟
𝐫𝐚𝐝𝐢𝐚𝐧𝐬
Fqd = 1
Fγd = 1
2. 𝐅𝐨𝐫 𝛟 > 𝟎. 𝟎
Fcd = Fqd −
1 − Fqd
Nctanϕ

Fqd = 1 + 2 tanϕ (1 − sinϕ)2
tan−1
(
Df
B
)
⏟
Fγd = 1
Important Notes:
1. If the value of (B) or (Df)is required, you should do the following:
 Assume (
Df
B
≤ 1) and calculate depth factors in term of (B) or (Df).
 Substitute in the general equation, then calculate (B) or (Df).
 After calculated the required value, you must check your
assumption→ (
Df
B
≤ 1).
 If the assumption is true, the calculated value is the final required
value.
 If the assumption is wrong, you must calculate depth factors in case
of (
Df
B
> 1) and then calculate (B) or (Df) to get the true value.
2. For both cases (
Df
B
≤ 1) and (
Df
B
> 1) if ϕ > 0 → calculate Fqd firstly,
because Fcd depends on Fqd.
Inclination Factors:
Fci = Fqi = (1 −
β°
90
)
2
Fγi = (1 −
β°
ϕ°
)
β°
= Inclination of the load on the foundation with respect to the 𝐯𝐞𝐫𝐭𝐢𝐜𝐚𝐥
Note:
If β°
= ϕ → Fγi = 0.0, so you 𝐝𝐨𝐧′
𝐭 𝐧𝐞𝐞𝐝 to calculate Fγs and Fγd,
because the last term in Meyerhof equation will be zero.

Factor of Safety
From previous two equations (Terzaghi and Meyerhof), we calculate the
value of ultimate bearing capacity (qu) which the maximum value the soil
can bear it (i.e. if the bearing stress from foundation exceeds the ultimate
bearing capacity of the soil, shear failure in soil will be occur), so we must
design a foundation for a bearing capacity less than the ultimate bearing
capacity to prevent shear failure in the soil. This bearing capacity is
“Allowable Bearing Capacity” and we design for it (i.e. the applied stress
from foundation must not exceeds the allowable bearing capacity of soil).
qall,gross =
qu,gross
FS
→→ Applied stress ≤ qall,gross =
qu,gross
FS
qall,gross = Gross allowable bearing capacity
qu,gross = Gross ultimate bearing capacity (Terzaghi or Meyerhof equations)
FS = Factor of safety for bearing capacity ≥ 3
However, practicing engineers prefer to use the “net allowable bearing
capacity” such that:
qall,net =
qu,net
FS
qu,net = Net ultimate bearing capacity, and it’s the difference between the
gross ultimate bearing capacity (upward as soil reaction) and the weight of
the soil and foundation at the foundation level (downward), to get the net
pressure from the soil that support the foundation.
qu,net = qu,gross − γchc − γshs
Since the unit weight of concrete and soil are convergent, then
qu,net = qu,gross − q → qall,net =
qu,gross − q
FS
q = Effective stress at the level of foundation level.
If we deal with loads (Q)
qu,gross =
Qu,gross
Area
÷FS
→ qall,gross =
Qall,gross
Area

Modification of Bearing Capacity Equations for Water
Table
Terzaghi and Meyerhof equations give the ultimate bearing capacity based
on the assumption that the water table is located well below the foundation.
However, if the water table is close to the foundation, the bearing capacity
will decreases due to the effect of water table, so, some modification of the
bearing capacity equations (Terzaghi and Meyerhof) will be necessary.
The values which will be modified are:
1. (q for soil above the foundation) in the second term of equations.
2. (γ for the underlying soil) in the third (last) term of equations .
There are three cases according to location of water table:
Case I. The water table is located so that 0 ≤ D1 ≤ Df as shown in the
following figure:
 The factor ,q, (second term) in the bearing capacity equations will takes
the following form: (For the soil above the foundation)
q = effective stress at the level of the bottom of the foundation
→ q = D1 × γ + D2 × (γsat − γw)
 The factor , γ, (third term) in the bearing capacity equations will takes the
following form: (For the soil under the foundation)
γ = effective unit weight for soil below the foundation → γ′
= γsat − γw

Case II. The water table is located so that 0 ≤ d ≤ B as shown in the
following figure:
 The factor ,q, (second term) in the bearing capacity equations will takes
the following form: (For the soil above the foundation)
q = effective stress at the level of the bottom of the foundation
→ q = Df × γ
 The factor , γ, (third term) in the bearing capacity equations will takes the
following form: (For the soil under the foundation)
γ = effective unit weight for soil below the foundation 𝐚𝐭 𝐝𝐞𝐩𝐭𝐡 𝐝 = 𝐁
I.e. calculate the effective stress for the soil below the foundation from
(d = 0 to d = B), and then divide this value by depth (d = B) to get the
representative effective unit weight (γ̅) for this depth.
σ0→B
′
= d × γ + (B − d) × ( γsat − γw) → σ0→B
′
= d × γ + (B − d) × γ′
÷B
→
σ0→B
′
B
=
d × γ + B × γ′
− d × γ′
B
→ γ = γ′
+
d × (γ − γ′
)
B
Case III. The water table is located so that d ≥ B, in this case the water
table is assumed have no effect on the ultimate bearing capacity.

Eccentrically Loaded Foundation
If the load applied on the foundation is in the center of the foundation
without eccentricity, the bearing capacity of the soil will be uniform at any
point under the foundation (as shown in figure below) because there is no
any moments on the foundation, and the general equation for stress under the
foundation is:
Stress =
Q
A
±
Mx y
Ix
±
My X
Iy
In this case, the load is in the center of the foundation and there are no
moments so,
Stress =
Q
A
(uniform at any point below the foundation)
However, in several cases, as with the base of a retaining wall or neighbor
footing, the loads does not exist in the center, so foundations are subjected to
moments in addition to the vertical load (as shown in the below figure). In
such cases, the distribution of pressure by the foundation on the soil is not
uniform because there is a moment applied on the foundation and the stress

under the foundation will be calculated from the general relation:
Stress =
Q
A
±
Mx y
Ix
±
My X
Iy
(in case of two way eccentricity)
But, in this section we deal with (one way eccentricity), the equation will be:
Stress =
Q
A
±
M c
I
Since the pressure under the foundation is not uniform, there are maximum
and minimum pressures (under the two edges of the foundation) and we
concerned about calculating these two pressures.
General equation for calculating maximum and minimum pressure:
Assume the eccentricity is in direction of (B)
Stress = q =
Q
A
±
M c
I
A = B × L
M = Q × e
c =
B
2
(maximum distance from the center)
I =
B3
× L
12
(I is about the axis that resists the moment)
Substitute in the equation, the equation will
be:

q =
Q
B × L
±
Q × e × B
2 B3 × L
12
→ q =
Q
B × L
±
6eQ
B2L
→ q =
Q
B × L
(1 ±
6e
B
)
q =
Q
B × L
(1 ±
6e
B
) General Equation
Now, there are three cases for calculating maximum and minimum pressures
according to the values of (e and
B
6
) to maintain minimum pressure
always≥ 0
Case I. (For 𝐞 <
𝐁
𝟔
):
qmax =
Q
B × L
(1 +
6e
B
)
qmin =
Q
B × L
(1 −
6e
B
)
Note that when e <
B
6
the value of qmin
Will be positive (i.e. compression).
If eccentricity in (L) direction:
(For e <
L
6
):
qmax =
Q
B × L
(1 +
6e
L
)
qmin =
Q
B × L
(1 −
6e
L
)
Case II. (For 𝐞 =
𝐁
𝟔
):
qmax =
Q
B × L
(1 +
6e
B
)
qmin =
Q
B × L
(1 − 1) = 0.0

Note that when e =
B
6
the value of qmin will be zero (i.e. no compression and
no tension) and this case is the critical case and it is accepted.
(For e =
L
6
):
qmax =
Q
B × L
(1 +
6e
L
)
qmin =
Q
B × L
(1 − 1) = 0.0
Case III. (For 𝐞 >
𝐁
𝟔
):
As shown in the above figure (1) the value of (qmin) is negative (i.e. tension
in soil), but we know that soil can’t resist any tension, thus, negative
pressure must be prevented by making (qmin = 0) at distance (x) from point
(A) as shown in the above figure (2), and determine the new value of
(qmax)by static equilibrium as following:
R = area of triangle × L =
1
2
× qmax,new × X × L →→ (1)
∑ Fy = 0.0 → R = Q →→ (2)
1 2

∑ M@A = 0.0
→ Q × (
B
2
− e) = R ×
X
3
(but from Eq. 2 → R = Q) → X = 3 (
B
2
− e)
Substitute by X in Eq. (1) →
R = Q =
1
2
× qmax,new × 3 (
B
2
− e) × L → qmax,new =
4Q
3L(B − 2e)
(For e >
L
6
):
qmax,new =
4Q
3B(L − 2e)
Note:
All the above equations are derived for rectangular or square footing, but if
the foundation is circular you should use the original equation for calculating
the stress:
q =
Q
A
±
M c
I
Where
A =
π
4
D2
(D is the diameter of the circular foundation)
c =
D
2
I =
π
64
D4
And then calculate qmax and qmin

Ultimate Bearing Capacity under Eccentric
Loading‫ــــ‬One-Way Eccentricity
Effective Area Method:
As we discussed previously, if the load does not exist in the center of the
foundation, or if the foundation located to moment in addition to the vertical
loads, the stress distribution under the foundation is not uniform. So, to
calculate the ultimate (uniform) bearing capacity under the foundation, new
area should be determined to make the applied load in the center of this area
and to develop uniform pressure under this new area. This new area is called
Effective area. The following is how to calculate𝐪 𝐮 for this case:
1. Determine the effective dimensions of
the foundation:
Effective width = B′
= B − 2e
Effective Length = L′
= L
Bused
′
= min(B′
, L′
)
Lused
′
= max(B′
, L′
)
If the eccentricity were in the direction of
(L) of the foundation:
Effective width = B′
= B
Effective Length = L′
= L − 2e
Bused
′
= min(B′
, L′
)
Lused
′
= max(B′
, L′
)
2. If we want to use terzaghi’s equation
for example, for square footing:
The value of B (in last term) will be Bused
′
because the pressure is uniform for this
value of width and the pressure does not
uniform for width B. Other factors in the
equation will not change.

3. If we want to use Meyerhof Equation:
The value of B (in last term) will be Bused
′
to get uniform pressure on this
width.
In calculating of shape factors (Fcs, Fqs, Fγs) use Bused
′
and Lused
′
because
we concerned about the shape of the footing that make the pressure uniform.
In calculating of depth factors (Fcd, Fqd, Fγd) use the original value (B) and
don’t replace it by Bused
′
due to the following two reasons:
 Depth factors are used to consider the depth of the foundation and thereby
the depth of soil applied on the original dimensions of the foundation.
 In equations of depth factors, as the value of (B) decrease the depth
factors will increase and then the value of (qu) will increase, so for more
safety we use the larger value of width (B) to decreases depth factors and
thereby decrease (qu) which less than (qu) if we use Bused
′
(i.e. more safe).
4. If there is a water table (Case II), we need the following equation to
calculate (γ) in the last term of equations (Terzaghi and Meyerhof):
γ = γ′
+
d×(γ−γ′)
B
The value of B used in this equation should be the original value (B) because
we calculate the effective unit weight (γ) for depth (B) below the
foundation.
Safety Consideration
Calculate the gross ultimate load:
Qu = qu × (Lused
′
× Bused
′
)⏟
A′
(A′
= effective area)
The factor of safety against bearing capacity is: FS =
Qu
Qall
≥ 3
Maximum Applied Load ≤ Qall =
Qu
F.S
The factor of safety against qmax is: FS =
qu
qmax
≥ 3
The value of qall should be equal or more than qmax: qall ≥ qmax
The value of qmin should be equal or more than zero: qmin ≥ 0.0

Important Notes (before solving any problem)
1. The soil above the bottom of the foundation are used only to calculate the
term (q) in the second term of bearing capacity equations (Terzaghi and
Meyerhof) and all other factors are calculated for the underlying soil.
2. Always the value of (q) is the effective stress at the level of the bottom of
the foundation.
3. For the underlying soil, if the value of (c=cohesion=0.0) you don’t have to
calculate factors in the first term in equations (Nc in terzaghi’s equations)
and (Nc, Fcs, Fcd, Fci in Meyerhof equation).
4. For the underlying soil, if the value of (ϕ = 0.0) you don’t have to
calculate factors in the last term in equations (Nγ in terzaghi’s equations)
and (Nγ, Fγs, Fγd, Fγi in Meyerhof equation).
5. If the load applied on the foundation is inclined with an angle (β = ϕ) →
The value of (Fγi)will be zero, so you don’t have to calculate factors in the
last term of Meyerhof equation (Nγ, Fγs, Fγd).
6. Always if we want to calculate the eccentricity, it’s calculated as
following:
e =
Overall Moment
Vertical Loads
7. If the foundation is square, strip or circular, you may calculate (qu) from
terzaghi or Meyerhof equations (should be specified in the problem).
8. But, if the foundation is rectangular, you must calculate (qu) from
Meyerhof general equation.
9. If the foundation width (B) is required, and there exist water table below
the foundation at distance (d), you should assume d ≤ B, and calculate B,
then make a check for your assumption.

Problems
1.
The square footing shown below must be designed to carry a 2400 KN load.
Use Terzaghi’s bearing capacity formula and factor of safety = 3.
Determine the foundation dimension B in the following two cases:
1. The water table is at 1m below the foundation (as shown).
2. The water table rises to the ground surface.
Solution
1.
qu = qall × FS (qall =
Qall
Area
, FS = 3)
Applied load ≤ Qall → Qall = 2400kN
qall =
Qall
Area
=
2400
B2
, FS = 3 →→ qu =
3 × 2400
B2
c = 50 kN/m2
q(effective stress) = γ × Df = 17.25 × 2 = 34.5 kN/m2
Since the width of the foundation is not known, assume d ≤ B
2400 kN
ϕ = 32°
C = 50 kN/m2
γd = 17.25 kN/m3
γs = 19.5 kN/m3

γ = γ̅ = γ′
+
d × (γ − γ′
)
B
γ′
= γsat − γw = 19.5 − 10 = 9.5kN/m3
, d = 3 − 2 = 1m
→ γ̅ = 9.5 +
1 × (17.25 − 9.5)
B
→ γ̅ = 9.5 +
7.75
B
Assume general shear failure
Note:
Always we design for general shear failure (soil have a high compaction
ratio) except if we can’t reach high compaction, we design for local shear
(medium compaction).
For ϕ = 32°
→ Nc = 44.04, Nq = 28.52, Nγ = 26.87 (𝐓𝐚𝐛𝐥𝐞 𝟑. 𝟏)
Now substitute from all above factors on terzaghi equation:
7200
B2
= 1.3 × 50 × 44.04 + 34.5 × 28.52 + 0.4 × B × (9.5 +
7.75
B
) × 26.87
7200
B2
= 3923.837 + 102.106 B
Multiply both sides by (B2
) → 102.106 B3
+ 3923.837B2
− 7200 = 0.0
→ B = 1.33m✓.
2.
All factors remain unchanged except q and γ:
q(effective stress) = (19.5 − 10) × 2 = 19 kN/m2
γ = γ′
= 19.5 − 10 = 9.5 kN/m3
Substitute in terzaghi equation:
7200
B2
= 1.3 × 50 × 44.04 + 19 × 28.52 + 0.4 × B × 9.5 × 26.87
7200
B2
= 3404.48 + 102.106B
Multiply both sides by (B2
) → 102.106 B3
+ 3404.48 B2
− 7200 = 0.0
→ B = 1.42m✓.
Note that as the water table elevation increase the required width (B) will
also increase to maintain the factor of safety (3).

2.
Determine the size of square footing to carry net allowable load of 295 KN.
FS=3. Use Terzaghi equation assuming general shear failure.
Solution
Qall,net = 295 kN and we know qall,net =
Qall,net
Area
→ qall,net =
295
B2
Also, qall,net =
qu − q
FS
q(effective stress) = γ × Df = 18.15 × 1 = 18.15 kN/m2
, FS = 3
→
295
B2
=
qu − 18.15
3
→ qu =
885
B2
+ 18.15 →→ (1)
c = 50 kN/m2
q(effective stress) = 18.15 kN/m2
γ = 20 kN/m3
(for underlying soil)
For ϕ = 25°
→ Nc = 25.13, Nq = 12.72, Nγ = 8.34 (𝐓𝐚𝐛𝐥𝐞 𝟑. 𝟏)
Substitute from all above factor in Terzaghi equation:
ϕ = 35°
C = 0.0
γd = 18.15 kN/m3
γd = 20 kN/m3
ϕ = 25°
C = 50 kN/m2

qu = 1.3 × 50 × 25.13 + 18.15 × 12.72 + 0.4 × B × 20 × 8.34
→ qu = 1864.318 + 66.72B
Substitute from Eq. (1):
885
B2
+ 18.15 = 1864.318 + 66.72B
Multiply both side by B2
:
66.72 B3
+ 1846.168B2
− 885 = 0.0
→ B = 0.68 m✓.

3.
For the square footing (2.5m x 2.5m) shown in the figure below, determine
the allowable resisting moment (M), if the allowable load P = 800 KN and
F.S = 3. (Using Meyerhof Equation).
Solution
M = Q × e = 800e
The first term in the equation will be zero because(c = 0), so the equation
will be:
qu = qNqFqsFqdFqi + 0.5BγNγFγsFγdFγi
q(effective stress) = γ × Df = 16.8 × 1.5 = 25.2 kN/m2
Calculating the new area that maintains 𝐪 𝐮 uniform:
B′
= B − 2e → B′
= 2.5 − 2e , L′
= 2.5
Bused
′
= min(B′
, L′) = 2.5 − 2e , Lused
′
= 2.5 m
qu = qall × FS (qall =
Qall
A′
→ A′
= Bused
′
× Lused
′
, FS = 3)
Applied load ≤ Qall → Qall = 800kN
qall =
800
(2.5 − 2e) × 2.5
=
320
2.5 − 2e
, →→ qu = 3 ×
320
2.5 − 2e
=
960
2.5 − 2e
ϕ = 35°
C = 0
γd = 16.8 kN/m3
γs = 20 kN/m3

d = 1m ≤ B = 2.5m → water table will effect on qu →→
γ = γ̅ = γ′
+
d × (γ − γ′
)
B
(Use B not Bused
′
as we explained previously)
γ′
= γsat − γw = 20 − 10 = 10 kN/m3
, d = 1m , γ = 16.8 kN/m3
→
γ̅ = 10 +
1 × (16.8 − 10)
2.5
= 12.72 kN/m3
For ϕ = 35°
→ Nc = 46.12, Nq = 33.3, Nγ = 48.03 (𝐓𝐚𝐛𝐥𝐞 𝟑. 𝟑)
Shape Factors:
As we explained previously, use Bused
′
and Lused
′
Fcs = 1 + (
Bused
′
Lused
′ ) (
Nq
Nc
) does not required (because c = 0.0)
Fqs = 1 + (
Bused
′
Lused
′ ) tanϕ = 1 + (
2.5 − 2e
2.5
) × tan35 = 1.7 − 0.56 e
Fγs = 1 − 0.4 (
Bused
′
Lused
′ ) = 1 − 0.4 × (
2.5 − 2e
2.5
) = 0.6 + 0.32 e
Depth Factors:
As we explained previously, use B not Bused
′
Df
B
=
1.5
2.5
= 0.6 < 1 and ϕ = 35 > 0.0 →→→
Fcd = Fqd −
1 − Fqd
Nctanϕ
does not required (because c = 0.0)
Fqd = 1 + 2 tanϕ (1 − sinϕ)2
(
Df
B
)
= 1 + 2 tan35 × (1 − sin35)2
× 0.6 = 1.152
Fγd = 1
The load on the foundation is not inclined, so all inclination factors are (1).
Now substitute from all above factors in Meyerhof equation:
960
2.5 − 2e
= 25.2 × 33.3 × (1.7 − 0.56 e) × 1.152
+0.5 × (2.5 − 2e) × 12.72 × 48.03 × (0.6 + 0.32 e)

960
2.5 − 2e
= 2101.6 − 663.54 e − 195.5e2
Multiply both sides by (2.5 − 2e) →→
960 = 5254 − 4203.2e − 1658.85e + 1327.08e2
− 488.75e2
+ 391e3
→ 391e3
+ 838.33e2
− 5862.05e + 4294 = 0.0
Solve for e→ e = −5.33 or e = 2.29 or e = 0.89
Now, the value of (e) must be less than
B
2
and must be positive value
B
2
=
2.5
2
= 1.25 < 2.29 → reject the value of e = 2.29 and negative value
→ e = 0.89 m
Before calculate the value of moment, we check for qmax:
qall =
320
2.5 − 2e
=
320
2.5 − 2 × 0.89
= 444.44 kN/m2
To calculate qmax we firstly should check the value of (e = 0.89m)
B
6
=
2.5
6
= 0.416m → e = 0.89 >
B
6
= 0.416 →→
qmax = qmax,new =
4Q
3L(B − 2e)
qmax,new =
4 × 800
3 × 2.5 × (2.5 − 2 × 0.89)
= 592.6 kN/m2
> qall = 444.44
Now, we calculate the adequate value of “e” (that makes qall = qmax)
444.44 =
4 × 800
3 × 2.5 × (2.5 − 2 × e)
→ e = 0.77m
M = Q × e = 800 × 0.77 = 616 kN. m ✓.
Note that the only variable in this problem is e, so we calculate the value of e
that insure that the maximum pressure qmax does not exceed the allowable
pressure qall.

4.
For the soil profile is given below, determine the allowable bearing capacity
of the isolated rectangular footing (2m x 2.3m) that subjected to a vertical
load (425 kN) and moment (85 kN.m), FS=3.
Solution
qu = qall × FS → qall =
qu
FS
→ qall =
qu
3
Note that the value of (c) for the soil under the foundation equal zero, so the
first term in the equation will be terminated (because we calculate the
bearing capacity for soil below the foundation) and the equation will be:
q(effective stress) = γ × Df = 16 × 1.5 = 24 kN/m2
Note that the eccentricity in the direction of (L=2.3)
e =
M
Q
=
85
425
= 0.2m
ϕ = 20°
C = 35 kN/m2
γd = 16 kN/m3
γs = 19kN/m3
ϕ = 25°
C = 0

B′
= B = 2 m → , L′
= L − 2e → L′
= 2.3 − 2 × 0.2 = 1.9m
Bused
′
= min(B′
, L′) = 1.9 m , Lused
′
= 2 m
Effective Area (A′) = 1.9 × 2 = 3.8 m2
Water table is at the bottom of the foundation → γ = γ′
= γs − γw
→ γ = γ′
= 19 − 10 = 9 kN/m3
For ϕ = 25°
→ Nc = 20.72, Nq = 10.66, Nγ = 10.88 (𝐓𝐚𝐛𝐥𝐞 𝟑. 𝟑)
Shape Factors:
′
and Lused
′
Fcs = 1 + (
Bused
′
Lused
′ ) (
Nq
Nc
Fqs = 1 + (
Bused
′
Lused
′ ) tanϕ = 1 + (
1.9
2
) × tan25 = 1.443
Fγs = 1 − 0.4 (
Bused
′
Lused
′ ) = 1 − 0.4 × (
1.9
2
) = 0.62
Depth Factors:
′
Df
B
=
1.5
2
= 0.75 < 1 and ϕ = 25 > 0.0 →→→
Fcd = Fqd −
1 − Fqd
Nctanϕ
Fqd = 1 + 2 tanϕ (1 − sinϕ)2
(
Df
B
)
= 1 + 2 tan25 × (1 − sin25)2
× 0.75 = 1.233
Fγd = 1
qu = 24 × 10.66 × 1.443 × 1.233 + 0.5 × 1.9 × 9 × 10.88 × 0.62 × 1
→ qu = 512.87 kN/m2

qall =
qu
3
=
512.87
3
= 170.95 kN/m2
✓.
Now, we check for qmax → qmax ≤ qall
Now, to calculate qmax we firstly should check the value of (e = 0.2m)
L
6
=
2.3
6
= 0.38m → e = 0.2 <
B
6
= 0.38 →→
qmax =
Q
B × L
(1 +
6e
L
)
qmax =
425
2 × 2.3
(1 +
6 × 0.2
2.3
) = 140.6kN/m2
< qall = 170.95 kN/m2
So, the allowable bearing capacity of the foundation is 170.95 kN/m2
✓.
Important Note:
If the previous check is not ok, we say (without calculations): the allowable
bearing capacity of 170.95 is not adequate for qmax, so the footing
dimensions (B or L) must be enlarged to be adequate, the dimension (B or L)
is the dimension in the direction of eccentricity (L in this problem).
But, if you are asked to calculate the new dimension of the footing:
Put: qmax = qall and then substitute in equation of qmax to calculate
the new dimension

5.
An eccentrically loaded rectangular foundation (6ft x 8ft) shown below. Use
factor of safety of 3 and if e = 0.5ft, determine the allowable load that the
foundation could carry. (The factor of safety is based on the maximum
stress along the base of the footing).
Solution
Note that the factor of safety is for qmax → FS =
qu
qmax
≥ 3 (As required)
c = 800 Ib/ft2
q(effective stress) = 110 × 3 + (122.4 − 62.4) × 4 = 570 Ib/ft2
Note that the eccentricity in the direction of (B=6)
e = 0.5ft
B′
= B − 2e = 6 − 2 × 0.5 = 5ft , L′
= L = 8ft
Bused
′
= min(B′
, L′) = 5ft , Lused
′
= 8ft
Effective Area (A′) = 5 × 8 = 40 ft2
γd = 110pcf
γs = 122.4pcf
C = 800psf
ϕ = 15°

Water table is above the bottom of the foundation → γ = γ′
= γs − γw
→ γ = γ′
= 122.4 − 62.4 = 60 Ib/ft3
For ϕ = 15°
→ Nc = 10.98, Nq = 3.94, Nγ = 2.65 (𝐓𝐚𝐛𝐥𝐞 𝟑. 𝟑)
Shape Factors:
′
and Lused
′
Fcs = 1 + (
Bused
′
Lused
′ ) (
Nq
Nc
) = 1 + (
5
8
) (
3.94
10.98
) = 1.224
Fqs = 1 + (
Bused
′
Lused
′ ) tanϕ = 1 + (
5
8
) × tan15 = 1.167
Fγs = 1 − 0.4 (
Bused
′
Lused
′ ) = 1 − 0.4 × (
5
8
) = 0.75
Depth Factors:
′
Df
B
=
7
6
= 1.16 > 1 and ϕ = 15 > 0.0 →→→
Fqd = 1 + 2 tanϕ (1 − sinϕ)2
tan−1
(
Df
B
)
⏟
tan−1
(
Df
B
)
⏟
= tan−1
(
7
6
) = 0.859
→ Fqd = 1 + 2 tan(15) × (1 − sin15)2
× 0.859 = 1.252
Fcd = Fqd −
1 − Fqd
Nctanϕ
= 1.252 −
1 − 1.252
10.98 × tan(15)
= 1.337
Fγd = 1
qu = 800 × 10.98 × 1.224 × 1.337 + 570 × 3.94 × 1.167 × 1.252
+0.5 × 5 × 60 × 2.65 × 0.75 × 1
→ qu = 17954.34 Ib/ft2

Now, to calculate qmax we firstly should check the value of (e = 0.5ft)
B
6
=
6
6
= 1ft → e = 0.5 <
B
6
= 1 →→
qmax =
Q
B × L
(1 +
6e
B
)
qmax =
Q
6 × 8
(1 +
6 × 0.5
6
) = 0.03125Q
FS =
qu
qmax
= 3 → qu = 3qmax
→ 17954.34 = 3 × 0.03125Q → Q = 191512.96 Ib = 191.5 Kips ✓.
6.
For the rectangular foundation (2m x 3m) shown below:
a) Compute the net allowable bearing capacity (FS=3).
b)If the water table is lowered by 2m. What effect on bearing capacity
would occur due to the water lowering?
γd = 18 kN/m3
ϕ = 25°
C = 0.0
γs = 21 kN/m3

Solution
Important Note:
The load on the foundation is considered inclined when this load is applied
directly on the foundation, however if the load does not applied directly on
the foundation (like this problem), this load is not considered inclined.
The analysis of the inclined load (700 KN) on the column will be as shown
in figure below:
The inclined load on the column will be divided into two components
(vertical and horizontal):
Vertical component = 700 × sin60 = 606.2 KN
Horizontal component = 700 × cos60 = 350 kN
The horizontal component will exerts moment on the foundation in the
direction shown in figure above:
M = 350 × 1.5 = 525 kN. m
e =
Overall moment
Vertical Load
=
525
606.2
= 0.866 m
a)
qall,net =
qu − q
FS
Note that the value of (c) for the soil under the foundation equal zero, so the
first term in the equation will be terminated (because we calculate the
bearing capacity for soil below the foundation) and the equation will be:
q(effective stress) = 18 × 0.5 + (21 − 10) × 1 = 20 kN/m2

Note that the eccentricity in the direction of (L=3)
e = 0.866m
B′
= B = 2 m → , L′
= L − 2e → L′
= 3 − 2 × 0.866 = 1.268m
Bused
′
= min(B′
, L′) = 1.268 m , Lused
′
= 2 m
Effective Area (A′) = 1.268 × 2 = 2.536 m2
Water table is above the bottom of the foundation → γ = γ′
= γs − γw
→ γ = γ′
= 21 − 10 = 11 kN/m3
For ϕ = 25°
→ Nc = 20.72, Nq = 10.66, Nγ = 10.88 (𝐓𝐚𝐛𝐥𝐞 𝟑. 𝟑)
Shape Factors:
′
and Lused
′
Fcs = 1 + (
Bused
′
Lused
′ ) (
Nq
Nc
Fqs = 1 + (
Bused
′
Lused
′ ) tanϕ = 1 + (
1.268
2
) × tan25 = 1.296
Fγs = 1 − 0.4 (
Bused
′
Lused
′ ) = 1 − 0.4 × (
1.268
2
) = 0.746
Depth Factors:
′
Df
B
=
1.5
2
= 0.75 < 1 and ϕ = 25 > 0.0 →→→
Fcd = Fqd −
1 − Fqd
Nctanϕ
Fqd = 1 + 2 tanϕ (1 − sinϕ)2
(
Df
B
)
= 1 + 2 tan25 × (1 − sin25)2
× 0.75 = 1.233
Fγd = 1

qu = 20 × 10.66 × 1.296 × 1.233 + 0.5 × 1.268 × 11 × 10.88 × 0.746
→ qu = 397.29 kN/m2
qall,net =
qu − q
FS
=
397.29 − 20
3
= 125.76 kN/m2
✓.
qall =
qu
FS
=
397.3
3
= 132.4 kN/m2
To calculate qmax we firstly should check the value of (e = 0.866m)
L
6
=
3
6
= 0.5m → e = 0.866 >
B
6
= 0.5 →→
qmax = qmax,new =
4Q
3B(L − 2e)
qmax,new =
4 × 606.2
3 × 2 × (3 − 2 × 0.866)
= 318.7kN/m2
> qall = 132.4
So, the allowable bearing capacity of the foundation is 132.4 kN/m2
is not
adequate for qmax and the dimensions of the footing must be enlarged.
b)
This case is shown in the below figure:

All factors remain unchanged except q and γ:
q(effective stress) = γ × Df = 18 × 1.5 = 27kN/m2
d = 1m ≤ B = 2m → water table will effect on qu →→
γ = γ̅ = γ′
+
d × (γ − γ′
)
B
(Use B not Bused
′
γ′
= γsat − γw = 21 − 10 = 11 kN/m3
, d = 1m , γ = 18 kN/m3
→
γ̅ = 11 +
1 × (18 − 11)
2
= 14.5 kN/m3
Substitute in Meyerhof equation:
qu = 27 × 10.66 × 1.296 × 1.233 + 0.5 × 1.268 × 14.5 × 10.88 × 0.746
→ qu = 534.54 kN/m2
The effect of water lowering is increase qu by 534.5 − 397.3 = 137.2kN/m2
✓.
7.
For the rectangular footing (2.5m x 3m) shown below, if e = 0.35m
and qmax = 410 kN/m2
. Calculate the factor of safety against bearing
capacity, and determine whether the design is good or not.
Solution
γd = 15 kN/m3
ϕ = 30°
C = 30 kN/m2
γs = 21 kN/m3

Note that the inclined load is applied directly on the foundation, so it is an
inclined load with angle (β = 90 − 60 = 30°
with vertical).
FS =
Qu
Qall
, Qu = qu × A′
, Qall =? ?
Since β = ϕ = 30°
, the inclination factor Fγi will equal zero, so the last term
in equation will be terminated and the equation will be:
qu = cNcFcsFcdFci + qNqFqsFqdFqi
c = 30 kN/m2
q(effective stress) = 15 × 0.5 + (21 − 10) × 1 = 18.5 kN/m2
Note that the eccentricity in the direction of (L=3)
e = 0.35m
B′
= B = 2.5 m → , L′
= L − 2e → L′
= 3 − 2 × 0.35 = 2.3m
Bused
′
= min(B′
, L′) = 2.3 m , Lused
′
= 2.5 m
Effective Area (A′) = 2.3 × 2.5 = 5.75 m2
For ϕ = 30°
→ Nc = 30.14, Nq = 18.4, Nγ = 22.4 (𝐓𝐚𝐛𝐥𝐞 𝟑. 𝟑)
Shape Factors:
′
and Lused
′
Fcs = 1 + (
Bused
′
Lused
′ ) (
Nq
Nc
) = 1 + (
2.3
2.5
) (
18.4
30.14
) = 1.56
Fqs = 1 + (
Bused
′
Lused
′ ) tanϕ = 1 + (
2.3
2.5
) × tan30 = 1.53
Fγs = 1 − 0.4 (
Bused
′
Lused
′ ) does not required (because β = ϕ = 30°
)

Depth Factors:
′
Df
B
=
1.5
2.5
= 0.6 < 1 and ϕ = 30 > 0.0 →→→
Fqd = 1 + 2 tanϕ (1 − sinϕ)2
(
Df
B
)
Fqd = 1 + 2 tan(30) × (1 − sin30)2
× 0.6 = 1.173
Fcd = Fqd −
1−Fqd
Nctanϕ
= 1.173 −
1−1.173
30.14×tan(30)
= 1.183
Fγd = 1
Fci = Fqi = (1 −
β°
90
)
2
= (1 −
30
90
)
2
= 0.444
Fγi = 0.0
qu = 30 × 30.14 × 1.56 × 1.183 × 0.444
+18.5 × 18.4 × 1.53 × 1.173 × 0.444
→ qu = 1012.14 kN/m2
Qu = qu × A′
= 1012.14 × 5.75 = 5819.8 KN
e = 0.35 m ,
L
6
=
3
6
= 0.5m → e = 0.35 <
L
6
= 0.5 →→→
We used term (L) because eccentricity in L direction
qmax = 410 =
Qall
B × L
(1 +
6e
L
) → 410 =
Qall
2.5 × 3
× (1 +
6 × 0.35
3
) →→
Qall = 1808.3KN
FS =
Qu
Qall
=
5819.8
1808.3
= 3.22✓.
Since the factor of safety is larger than 3, the design is good✓.

8.
A square footing 2.5m x 2.5m is shown in the figure below. If the maximum
pressure on the foundation should not exceed the allowable bearing capacity.
Using factor of safety (FS=3), find the maximum horizontal force that the
foundation can carry if the water table is 1m below the foundation.
(Use Terzaghi equation)
Solution
The following figure explains the analysis of the given loads:
e =
Overall moment
Vertical Load
=
165 + 1.5H
300
= 0.55 + 0.005H →→ (1)
γd = 17 kN/m3
ϕ = 30°
C = 50 kN/m2
γsat = 19.5 kN/m3

qmax ≤ qall(given) →→ qall = qmax (To get maximum value of H)
→ FS =
qu
qall
→ 3 =
qu
qall
→ qu = 3qall so, qu = 3qmax
c = 50 kN/m2
q(effective stress) = 17 × 1.5 = 25.5 kN/m2
B′
= B − 2e = 2.5 − 2e → , L′
= B = 2.5
Bused
′
= min(B′
, L′) = 2.5 − 2e , Lused
′
= 2.5m
Effective Area (A′) = (2.5 − 2e) × 2.5 = 6.25 − 5e
d = 1m ≤ B = 2.5m → water table will effect on qu →→
γ = γ̅ = γ′
+
d × (γ − γ′
)
B
(Use B not Bused
′
γ′
= γsat − γw = 19.5 − 10 = 9.5 kN/m3
, d = 1m , γ = 17 kN/m3
→
γ̅ = 9.5 +
1 × (17 − 9.5)
2.5
= 12.5 kN/m3
For ϕ = 30°
→ Nc = 37.16, Nq = 22.46, Nγ = 19.13 (𝐓𝐚𝐛𝐥𝐞 𝟑. 𝟏)
Substitute from all above factors in Terzaghi equation:
qu = 1.3 × 50 × 37.16 + 25.5 × 22.46 + 0.4 × (2.5 − 2e) × 12.5 × 19.13
qu = 3227.25 − 191.3e
Calculating of 𝐪 𝐦𝐚𝐱:
B
6
=
2.5
6
= 0.416 , e = 0.55 + 0.005H
(Note that the first term of e = 0.55 >
B
6
= 0.416 → e >
B
6
→→
Use the modified equation for qmax:
qmax,modified =
4Q
3L(B − 2e)
=
4 × 300
3 × 2.5 × (2.5 − 2e)
=
160
2.5 − 2e
qu = 3qmax → 3227.25 − 191.3e = 3 ×
160
2.5 − 2e

Multiply both side by (2.5 − 2e):
382.6e2
− 6932.75e + 7588.125 = 0.0 → e = 1.17 m
Substitute in Eq.(1):
1.17 = 0.55 + 0.005H → H = 124 kN✓.
9.
For the soil profile given below, determine the net allowable bearing
capacity of the isolated rectangular footing (2.5m x 3m) that subjected to a
given load as shown. Use FS=3.
For ϕ = 20°
→ Nc = 14.83, Nq = 6.4, Nγ = 5.39
For ϕ = 32°
→ Nc = 35.49, Nq = 23.18, Nγ = 30.22
Solution
The analysis of the inclined load (800 KN) on the column will be as shown
in figure below:

e =
Overall moment
Vertical Load
=
202.87
692.8
= 0.29m
qall,net =
qu − q
FS
But c = 0.0 for the soil under the foundation →
→ qu = qNqFqsFqdFqi + 0.5BγNγFγsFγdFγi
q(effective stress) = 16 × 1.2 = 19.2 kN/m2
Eccentricity in the direction of (L=3)
e = 0.29m
B′
= B = 2.5 m → , L′
= L − 2e → L′
= 3 − 2 × 0.29 = 2.42m
Bused
′
= min(B′
, L′) = 2.42 m , Lused
′
= 2.5 m
Water table is at distance (2.7m) below the foundation base
→ B = 2.5m < 2.7 → No effect of water table → use γ = 18kN/m3
For ϕ = 32°
→ Nc = 35.49, Nq = 23.18, Nγ = 30.22 (𝐆𝐢𝐯𝐞𝐧𝐬)

Shape Factors:
Fcs = 1 + (
Bused
′
Lused
′ ) (
Nq
Nc
Fqs = 1 + (
Bused
′
Lused
′ ) tanϕ = 1 + (
2.42
2.5
) × tan32 = 1.6
Fγs = 1 − 0.4 (
Bused
′
Lused
′ ) = 1 − 0.4 × (
2.42
2.5
) = 0.61
Depth Factors:
Df
B
=
1.2
2.5
= 0.48 < 1 and ϕ = 32 > 0.0 →→→
Fcd = Fqd −
1 − Fqd
Nctanϕ
Fqd = 1 + 2 tanϕ (1 − sinϕ)2
(
Df
B
)
= 1 + 2 tan32 × (1 − sin32)2
× 0.48 = 1.13
Fγd = 1
qu = 19.2 × 23.18 × 1.6 × 1.13 + 0.5 × 2.42 × 18 × 30.22 × 0.61 × 1
= 1206.16 kN/m2
qall,net =
qu − q
FS
=
1206.16 − 19.2
3
= 395.65 kN/m2
✓.

qall =
qu
FS
=
1206.16
3
= 402 kN/m2
Now, to calculate qmax we firstly should check the value of (e = 0.29m)
L
6
=
3
6
= 0.5m → e = 0.29 <
B
6
= 0.5 →→
qmax =
Q
B × L
(1 +
6e
L
)
qmax =
692.8
2.5 × 3
(1 +
6 × 0.29
3
) = 145.95 kN/m2
< qall = 402 kN/m2
So, the allowable bearing capacity of the foundation is 402 kN/m2
✓.

Chapter (4)
Ultimate Bearing
Capacity of Shallow
Foundations
(Special Cases)

Foundation Engineering Ultimate B.C. of Shallow Foundations (Special Cases)
Introduction
The ultimate bearing capacity theories discussed in Chapter 3 assumed that
the soil supporting the foundation is homogeneous (i.e. one layer) and
extends to a great depth below the bottom of the foundation. They also
assume that the ground surface is horizontal. However, that is not true in all
cases: It is possible to encounter a soil may be layered and have different
shear strength parameters, and in some cases it may be necessary to
construct foundations on or near a slope.
All of above cases are special cases from Chapter 3, and will be discussed in
this Chapter.
Bearing Capacity of Layered Soils: Stronger soil
Underlain by Weaker Soil
The bearing capacity equations presented in Chapter 3 involved cases in
which the soil supporting the foundation is homogeneous and extend to a
great depth (i.e. the cohesion, angle of friction, and unit weight of soil were
assumed to remain constant for the bearing capacity analysis). However, in
practice, layered soil profiles are often encountered (more than one layer). In
such instances, the failure surface at ultimate load may extend in two or
more soil layers. This section features the procedures for estimating the
bearing capacity for layered soils (stronger soil layer, underlain by a weaker
soil layer that extends to a great depth).
Notes:
1. Always the factors of top soil are termed by (1) and factors of bottom soil
are termed by (2) as shown in the following table:
2. The equation will be derived for continuous or strip footing and then will
be modified to be valid for rectangular, square, and circular footings.
Soil Properties
Layer
Unit
weight
Friction
angle
Cohesion
Top γ1 ϕ1 c1
Bottom γ2 ϕ2 c2

Let the depth ,H, is the distance from the bottom of the foundation to the top
of weaker soil (bottom soil layer) and ,B, is the width of continuous or strip
footing (i.e. equation will be derived for continuous footing), the failure
surface in layered soil below the foundation may have two cases:
Case I: If the depth H is relatively small compared with the foundation
width B (upper layer can’t resist overall failure due to its small thickness), a
punching shear failure will occur in the top soil layer, followed by a general
shear failure in the bottom soil layer (due to its large extend downward), so
the ultimate bearing capacity in this case will equal the ultimate bearing
capacity of bottom layer (because general shear failure occur on it) in
addition to punching shear resistance from top layer.
qu = qb + Punching shear resistance from top layer (qpunching)
(qpunching)can be calculated as following (see the above figure):
qpunching =
(2Ca + 2PP sin δ)
B × 1⏟
Upward
− γ1 × H⏟
Downward
Ca = adhesive force (between concrete and soil) → Ca = ca × H
ca = adhesion between concrete and soil along the thickness H

δ = inclination of the passive, PP, force with the horizontal
PP = passive force per unit length along the thickness H applied from
soil to the foundation and can be calculated as following:
PP =
1
2
H × vertical effective stress × K
1
2
H × vertical effective stress = area of the vertical pressure diagram
vertical effective stress = γ1 × H
KPH = Kcosδ → K =
KPH
cosδ
K = Coefficient used to transform vertical pressure to the direction
of passive force
KPH = Horizontal component of passive earth pressure coefficient(K)
Now the equation of PP will be:
PP =
1
2
H × (γ1 × H) × K =
1
2
× γ1 × H2
×
KPH
cosδ
Now substitute in equation ofqpunching:
qpunching =
2ca × H
B
+
2 × (
1
2
× γ1 × H2
×
KPH
cosδ
) × sin δ
B
− γ1 × H
qpunching =
2ca × H
B
+ γ1H2
×
KPH tanδ
B
− γ1 × H
Now correction for depth factors (according Terzaghi assumption) should be
established. This modification will be in punching shear term as following:
qpunching =
2ca × H
B
+ γ1H2
(1 +
2Df
H
) ×
KPH tanδ
B
− γ1 × H
From several experiments, investigators found that KPH tanδ = Ks tanϕ1
Ks = Punching shear coefficient
→ qpunching =
2ca × H
B
+ γ1H2
(1 +
2Df
H
) ×
Ks tanϕ1
B
− γ1 × H
Now substitute in equation of (qu):
qu = qb +
2ca × H
B
+ γ1H2
(1 +
2Df
H
) ×
Ks tanϕ1
B
− γ1 × H

Case II: If the depth ,H, is relatively large (thickness off top layer is large),
then the failure surface will be completely located in the top soil layer and
the ultimate bearing capacity for this case will be the ultimate bearing
capacity for top layer alone (qt).
qu = qt = c1Nc(1) + q Nq(1) + 0.5Bγ1Nγ(1)
Nc(1), Nq(1), Nγ(1) = Meyerhof bearing capacity factors (for ϕ1)(𝐓𝐚𝐛𝐥𝐞𝟑. 𝟑)
All depth factors will equal (1) because their considered in punching term.
All shape factors will equal (1) because strip or continuous footing.
Assume no inclination so, all inclination factors equal (1).
Combination of two cases:
As mentioned above, the value of qt is the maximum value of qu can be
reached, so it should be an upper limit for equation of qu:
qu = qb +
2ca × H
B
+ γ1H2
(1 +
2Df
H
) ×
Ks tanϕ1
B
− γ1 × H ≤ qt
The above equation is the derived equation for strip or continuous footing,
but if the foundation is square, circular and rectangular the equation will be
modified to be general equation for all shapes of footings:

qu = qb + (1 +
B
L
) ×
2ca×H
B
+γ1H2
× (1 +
B
L
) (1 +
2Df
H
) ×
Ks tanϕ1
B
− γ1 × H ≤ qt
qt = c1Nc(1)Fcs(1) + q Nq(1)Fqs(1) + 0.5Bγ1Nγ(1)Fγs(1)
q = effective stress at the top of layer(1) = γ1 × Df
qb = c2Nc(2)Fcs(2) + q Nq(2)Fqs(2) + 0.5Bγ2Nγ(2)Fγs(2)
q = effective stress at the top of layer(2) = γ1 × (Df + H)
All depth factors will equal (1) because their considered in punching term.
Assume no inclination so, all inclination factors equal (1).
Note:
All factors and equations mentioned above are based on Meyerhof theory
discussed in Chapter 3.
All of above factors are known except Ks andca
Ks = f (
q2
q1
, ϕ1) and
ca
c1
= f (
q2
q1
) → to find Ks andca: (
q2
q1
) must be known.
Calculating of q1 andq2 is based on the following three main assumptions:
1. The foundation is always strip foundation even if it’s not strip
2. The foundation exists on the ground surface (Df = 0.0) and the second
term on equation will be terminated.
3. In calculating q1 we assume the top layer only exists below the foundation
to a great depth, and the same in calculating of q2.
q1 = c1Nc(1) + 0.5Bγ1Nγ(1)
q2 = c2Nc(2) + 0.5Bγ2Nγ(2)
→ (
q2
q1
) = ✓
Calculating of Ks:
Ks = f (
q2
q1
, ϕ1)

Ks can be calculated easily from (Figure 4.9) according the values of
(
q2
q1
andϕ1)
Calculating of ca:
ca
c1
= f (
q2
q1
)
ca
c1
can be calculated easily from (Figure 4.10) according the value of (
q2
q1
)
ca
c1
= ✓ and c1 = ✓ →→ ca = ✓
Important Notes:
1. If there is a water table near the foundation (above or below foundation),
the three cases discussed in Chapter 3 should be considered (i.e. the factor q
for top and bottom layers may be modified and γ1and γ2 for top and bottom
layers may also be modified according to the existing case of water table.
2. If the strong layer and the weak layer are not clear (cohesion and friction
angle for each layer are convergent), to know the strong and the weak layer
do the following:
 Calculate q1 and q2 and then calculate (
q2
q1
)
If (
q2
q1
) < 1 → The top layer is the stronger layer and the
bottom is the weaker layer
If (
q2
q1
) > 1 → The top layer is the weaker layer and the
bottom is the stronger layer.
3. Any special cases can be derived from the general equation above.

Bearing Capacity of Layered Soils: Weaker soil
Underlain by Stronger Soil
Let the depth ,H, is the distance from the bottom of the foundation to the top
of stronger soil (bottom soil layer) and ,B, is the width of the foundation and
,D, is the depth of failure beneath the foundation.
As shown on the above figure, there are two cases:
Case I: For (H < D →
H
D
< 1) →The failure surface in soil at ultimate load
will pass through both soil layers (i.e. the ultimate bearing capacity of soil
will be greater than the ultimate bearing capacity for bottom layer alone).
Case II: (H > D →
H
D
> 1) →The failure surface on soil will be fully located
on top ,weaker soil layer, (i.e. the ultimate bearing capacity in this case is
equal the ultimate bearing capacity for top layer alone).
Case I
Case II
D

For these two cases, the ultimate bearing capacity can be given as following:
For (H ≤ D →
H
D
≤ 1)
qu = qt + (qb − qt) (1 −
H
D
)
2
Note that if
H
D
= 1 →the value of qu will equal qt, and this is logical,
because in this special case the failure surface will be exist on whole depth
of top (weaker layer).
For (H > D →
H
D
> 1)
qu = qt
Because failure surface is fully located on top (weaker soil).
qt,weak = c1Nc(1)Fcs(1) + q Nq(1)Fqs(1) + 0.5Bγ1Nγ(1)Fγs(1)
q = effective stress at the top of layer(1) = γ1 × Df
qb,strong = c2Nc(2)Fcs(2) + q Nq(2)Fqs(2) + 0.5Bγ1Nγ(2)Fγs(2)
q = effective stress at the top of layer(1) by assuming the foundation
is located directly above stronger soil layer at depth of Df
→ q = γ2 × Df
Important Note:
D = B (for 𝐥𝐨𝐨𝐬𝐞 𝐬𝐚𝐧𝐝 and 𝐜𝐥𝐚𝐲)
D = 2B (for 𝐝𝐞𝐧𝐬𝐞 𝐬𝐚𝐧𝐝 )

Bearing Capacity of Foundations on Top of a Slope
In some instances, foundations need to be constructed on top of a slope, thus
calculating of bearing capacity of soil under such conditions will differ from
Chapter 3. This section explains how we can calculate the bearing capacity
of soil under these conditions.
H = height of slope , β = angle between the slope and horizontal
b = distance from the edge of the foundation to the top of the slope
The ultimate bearing capacity for continuous or strip footing can be
calculated by the following theoretical relation:
qu = cNcq + 0.5BγNγq
For purely granular soil (c = 0.0):
qu = 0.5BγNγq
For purely cohesive soil (ϕ = 0.0):
qu = cNcq
Calculating of 𝐍 𝛄𝐪:
The value of Nγq can be calculated from (Figure 4.15 P.204) according the
following steps:
1. Calculate the value of (
Df
B
).
2. If (
Df
B
) = 0.0 → use 𝐬𝐨𝐥𝐢𝐝 𝐥𝐢𝐧𝐞𝐬 on the figure.

3. If (
Df
B
) = 1 → use 𝐝𝐚𝐬𝐡𝐞𝐝 𝐥𝐢𝐧𝐞𝐬 on the figure.
b
B
) which the horizontal axis aof the figure.
5. According the values of (ϕ, β and factors mentioned above ) we can
calculate the value of Nγq on vertical axis of the figure.
Note:
If the value of
Df
B
is in the following range: (0 <
Df
B
< 1) do the following:
 Calculate Nγq at (
Df
B
) = 1.
 Calculate Nγq at (
Df
B
) = 0.
 Do interpolation between the above two values of Nγq to get the required
value of Nγq.
Calculating of 𝐍 𝐜𝐪:
The value of Ncq can be calculated from (Figure 4.16 P.205) according the
following steps:
Df
B
).
2. If (
Df
B
) = 0.0 → use 𝐬𝐨𝐥𝐢𝐝 𝐥𝐢𝐧𝐞𝐬 on the figure.
3. If (
Df
B
) = 1 → use 𝐝𝐚𝐬𝐡𝐞𝐝 𝐥𝐢𝐧𝐞𝐬 on the figure.
4. Determining the horizontal axis of the figure:
 If B < H → the horizontal axis of the figure is (
b
B
)
 If B ≥ H → the horizontal axis of the figure is (
b
H
)
5. Calculating the value of stability number for clay (Ns):
 If B < H → use Ns = 0.0 in the figure
 If B ≥ H → calculate Ns from this relation Ns =
γH
c
to be used in
the figure.
6. According the values of (ϕ, β and factors mentioned above) we can
calculate the value of Ncq on vertical axis of the figure.
Note:
If the value of
Df
B
is in the following range: (0 <
Df
B
< 1) →
Do interpolation as mentiond above.

Problems
1.
The figure below shows a continuous foundation.
a) If H=1.5 m, determine the ultimate bearing capacity,qu
b) At what minimum depth ,H, will the clay layer not have any effect on the
ultimate bearing capacity of the foundation?
Solution
The first step in all problems like this one is determining whether the two
soils are stronger soil and weaker soil as following:
q1 = c1Nc(1) + 0.5Bγ1Nγ(1) (c1 = 0.0) → q1 = 0.5Bγ1Nγ(1)
B = 2m , γ1 = 17.5 kN/m3
For ϕ1 = 40°
→ Nγ(1) = 109.41 (𝐓𝐚𝐛𝐥𝐞𝟑. 𝟑)
→ q1 = 0.5 × 2 × 17.5 × 109.41 = 1914.675 kN/m2
q2 = c2Nc(2) + 0.5Bγ2Nγ(2) (ϕ2 = 0.0) → q2 = c2Nc(2)
c2 = 30 kN/m2
, For ϕ2 = 0°
→ Nc(2) = 5.14 (𝐓𝐚𝐛𝐥𝐞𝟑. 𝟑)
q2 = 30 × 5.14 = 154.2 kN/m2
Sand
γ1 = 17.5 kN/m3
ϕ1 = 40°
C1 = 0
Clay
γ2 = 16.5 kN/m3
ϕ2 = 0.0
C2 = 30 kN/m2

q2
q1
=
154.2
1914.675
= 0.08 < 1 → The top layer is stronger soil and bottom
layer is weaker soil.
1. For strip footing:
qu = qb +
2ca × H
B
+ γ1H2
(1 +
2Df
H
) ×
Ks tanϕ1
B
− γ1 × H ≤ qt
qt = c1Nc(1) + q Nq(1) + 0.5Bγ1Nγ(1)
c1 = 0.0 , q = γ1 × Df = 17.5 × 1.2 = 21kN/m2
, B = 2m
For ϕ1 = 40°
→ Nc(1) = 75.31, Nq(1) = 64.2 , Nγ(1) = 109.41 (𝐓𝐚𝐛𝐥𝐞𝟑. 𝟑)
qt = 0 + 21 × 64.2 + 0.5 × 2 × 17.5 × 109.41 = 3262.875 kN/m2
qb = c2Nc(2) + q Nq(2) + 0.5Bγ2Nγ(2)
c2 = 30 , q = γ2 × (Df + H) = 17.5 × (1.2 + 1.5) = 47.25 kN/m2
For ϕ2 = 0°
→ Nc(2) = 5.14, Nq(2) = 1 , Nγ(2) = 0 (𝐓𝐚𝐛𝐥𝐞𝟑. 𝟑)
qb = 30 × 5.14 + 47.25 × 1 + 0 = 201.45 kN/m2
Calculating of 𝐜 𝐚:
q2
q1
= 0.08
From figure (4.10) →
ca
c1
= 0.7 → ca = 0.7 × 0 = 0
Calculating of 𝐊 𝐬:
q2
q1
= 0.08
From figure (4.9) → Ks = 2.4
qu = 201.45 + 0 + 17.5 × 1.52
(1 +
2 × 1.2
1.5
) ×
2.4 tan40
2
− 17.5 × 1.5
qu = 278 kN/m2
✓.
2. The minimum depth that make the clay layer have no effect on qu is
occur when qu = qt and qb = 0.0
qu = qt = 0 +
2ca × H
B
+ γ1H2
(1 +
2Df
H
) ×
Ks tanϕ1
B
− γ1 × H
3262.875 = 0 + 0 + 17.5 × H2
(1 +
2 × 1.2
H
) ×
2.4 tan40
2
− 17.5 × H
→ H = 12.92 m✓.

2.
A rectangular footing of size 6m x 8m is founded at a depth of 3 m in a clay
stratum of very stiff consistency overlying a softer clay stratum at a depth of
5 m from the ground surface. The soil parameters of the two layers of soil
are as shown in the Figure below. If the top layer has been removed and
kN/m3
ultimate bearing capacity of the footing.
For ϕ = 35°
→ Nc = 46.12, Nq = 33.3, Nγ = 48.03
For ϕ = 0 → Nc = 5.14, Nq = 1, Nγ = 0
Solution
The top layer is dense sand with ϕ1 = 35°
and γ1 = 19 kN/m3
Determine which layer is strong:
q1 = c1Nc(1) + 0.5Bγ1Nγ(1) butc1 = 0.0 → q1 = 0.5Bγ1Nγ(1)
at ϕ1 = 35°
→ Nγ(1) = 48.03
→ q1 = 0.5 × 6 × 19 × 48.03 = 2737.71kN/m2
q2 = c2Nc(2) + 0.5Bγ2Nγ(2) but ϕ2 = 0.0 → q2 = c2Nc(2)
at ϕ2 = 0 → Nc(2) = 5.14 → q2 = 100 × 5.14 = 514 kN/m2
C = 100 kN/m2

q2
q1
=
514
2737.7
= 0.187 < 1
→ The top layer is strong and the bottom is weak
General equation fir rectangular foundation:
qu = qb + (1 +
B
L
) ×
2ca × H
B
+ γ1H2
× (1 +
B
L
) (1 +
2Df
H
) ×
Ks tanϕ1
B
−γ1 × H ≤ qt
But c1 = 0.0 → ca = 0.0 → so the equation will be:
qu = qb + γ1H2
× (1 +
B
L
) (1 +
2Df
H
) ×
Ks tanϕ1
B
− γ1 × H ≤ qt
Calculation of qt:
qt = c1Nc(1)Fcs(1) + q Nq(1)Fqs(1) + 0.5Bγ1Nγ(1)Fγs(1) butc1 = 0.0
→ qt = q Nq(1)Fqs(1) + 0.5Bγ1Nγ(1)Fγs(1)
q = 3 × 19 = 57 kN/m2
at ϕ1 = 35°
→ Nq(1) = 33.3 , Nγ(1) = 48.03
Fqs(1) = 1 + (
B
L
) tanϕ1 = 1 + (
6
8
) tan35 = 1.525
Fγs(1) = 1 − 0.4 (
B
L
) = 1 − 0.4 (
6
8
) = 0.7
→ qt = 57 × 33.3 × 1.525 + 0.5 × 6 × 19 × 48.03 × 0.7
= 4811 kN/m2
Calculation of qb:
qb = c2Nc(2)Fcs(2) + q Nq(2)Fqs(2) + 0.5Bγ2Nγ(2)Fγs(2) but ϕ2 = 0.0
→ qb = c2Nc(2)Fcs(2) + q Nq(2)Fqs(2)
q = (3 + 2) × 19 = 95 kN/m2

at ϕ2 = 0 → Nc(2) = 5.14 , Nq(2) = 1
Fcs(2) = 1 + (
B
L
) (
Nq(2)
Nc(2)
) = 1 +
6
8
×
1
5.14
= 1.146
Fqs(2) = 1 + (
B
L
) tanϕ2 = 1 + (
6
8
) tan0 = 1
→ qb = 100 × 5.14 × 1.146 + 95 × 1 × 1 = 684 kN/m2
Determination of Ks
q2
q1
= 0.187 and ϕ1 = 35°
→ Ks = 2.5 (from the given chart).
Now apply in the equation an calculate qu
qu = 684 + 19 × 22
× (1 +
6
8
) (1 +
2 × 3
2
) ×
2.5 × tan35
6
− 19 × 2
= 801.21 kN/m2
< qt = 4811 kN/m2
→ qu = 801.21 kN/m2
✓.

3.
Solve examples 4.4 and 4.5 in your text book.
4.
Solve example 4.6 in your text book, but use this equation for
calculating(qu):
qu = qt + (qb − qt) (1 −
H
D
)
2
Because the equation in text book for this case doesn’t true.
5.
For the soil profile shown below, determine the ultimate bearing capacity of
the continuous footing.
Solution
From the figure: B = 2.5m , b = 1.25 m, H = 5m , Df = 2.5m , β = 45°
qu = cNcq + 0.5BγNγq butϕ = 0.0 → qu = cNcq
c = 40 kN/m2
Calculating of 𝐍 𝐜𝐪 (Figure 4.16):
Df
B
=
2.5
2.5
= 1 → use dashed lines on figure
B = 2.5 < H = 5 → the horizontal axis of the figure is (
b
B
) =
1.25
2.5
= 0.5
B = 2.5 < H = 5 → use Ns = 0.0 in the figure
From the figure, the value of Ncq ≅ 5.7 → qu = 40 × 5.7 = 228 kN/m2
✓.
γ = 17.5 kN/m3
C = 40 kN/m2
ϕ = 0.0

6.
For the soil profile shown below, determine the ultimate bearing capacity of
the continuous footing.
Solution
From the figure: B = 3m , b = 1.25 m, H = 2.5m , Df = 0.0m , β = 45°
qu = cNcq + 0.5BγNγq butϕ = 0.0 → qu = cNcq
c = 25 kN/m2
Calculating of 𝐍 𝐜𝐪 (Figure 4.16):
Df
B
=
0
2.5
= 0 → use solid lines on figure
B = 3 > H = 2.5 → the horizontal axis of the figure is (
b
H
) =
1.25
2.5
= 0.5
B = 3 > H = 2.5 → use Ns =
γ × H
c
=
20 × 2.5
25
= 2 in the figure
From the figure, the value of Ncq ≅ 3 → qu = 25 × 3 = 75 kN/m2
✓.
γ = 20 kN/m3
C = 25 kN/m2
ϕ = 0.0

Chapter (5)
Allowable Bearing
Capacity and
Settlement

Foundation Engineering Allowable Bearing Capacity and Settlement
Introduction
As we discussed previously in Chapter 3, foundations should be designed for
both shear failure and allowable settlement. So the allowable settlement of
shallow foundations may control the allowable bearing capacity. The
allowable settlement itself may be controlled by local building codes.
For example; the maximum allowable settlement for mat foundation is 50
mm, and 25 mm for isolated footing. These foundations should be designed
for these limiting values of settlement (by calculating the allowable bearing
capacity from the allowable settlement). Thus, the allowable bearing
capacity is the smaller of the following two conditions:
qall = smallest of {
qu
FS
(to control shear failure Ch.3)
qall,settlement(to control settlement)
In this Chapter, we will learn how to calculate the allowable bearing
capacity for settlement (qall,settlement), but firstly we want to calculate the
total settlement of the foundation.
The settlement of a foundation can be divided into two major categories:
a) Immediate or elastic settlement (𝐒 𝐞):
Elastic or immediate settlement occurs during or immediately after the
application of the load (construction of structure) without change in the
moisture content of the soil.
b) Consolidation Settlement (𝐒 𝐜):
Consolidation settlement occur over time, such that pore water is extruded
from the void spaces of saturated clayey soil submerged in water.
Consolidation settlement comprises two phases: Primary and secondary.
To calculate foundation settlement (both elastic and consolidation), it is
required to estimate the vertical stress increase in the soil mass due to the
net load applied on the foundation (exactly as discussed previously in soil
mechanics course “Ch.10” ). Hence, this chapter is divided into the
following three parts:
1. Calculation of vertical stress increase (Ch.10 in soil mechanics course).

2. Elastic Settlement Calculation (Main topic of this chapter).
3. Consolidation settlement calculation (Ch.11 in soil mechanics course).
Vertical Stress Increase in a Soil Mass Caused by
Foundation Load
Stress Due to a Concentrated (Point) Load:
We calculate the vertical stress increase at any point at any depth due to the
applied point load as following:
Consider we want to calculate the vertical stress increase at point A in figure
below:
∆σz,A =
3. P. Z3
2π(r2 + Z2)
5
2
, and the same at any point.
r = √X2 + Y2
X , Y and Z are measured from the point of applied load as shown in figure
above.
Note: If there are more than one point load applied on the soil profile at
different positions , you should calculate ∆σz for each load and then :
∆σz,t = ∆σz,1 + ∆σz,2 + ∆σz,3 + ⋯ + ∆σz,n

Stress Due to a Circularly Loaded Area:
If we want to calculate the vertical stress below the center of the circular
foundation or at any point at distance (r) from the center of the foundation,
do the following:
Let R =
B
2
= Radius of circular area
r = distance from the center of the foundation to the required point
z = depth of point at which the vertical stress increas is required
We can calculate the value of (
∆σz
qo
) from (Table 5.1 P. 226) which gives the
variation of (
∆σz
qo
) with (
r
B/2
) and (
z
B/2
) {For 0 ≤ (
r
B/2
) ≤ 1}
qo = the stress at the base of the foundation (
column load
foundation area
)
Note that, if the value of (
r
B/2
) = 0.0 →The point is below the center of the
foundation

Vertical Stress Caused by a Rectangularly Loaded
Area
Consider we want to calculate the vertical stress increase at point A in figure
below:
We calculate the vertical stress increase at point below the corner of
rectangular loaded area as following:
∆σz = qI
I = Influence factor = f(m, n) (From Table 5.2 P. 228 and 229)
m =
B
Z
, n =
L
Z
B: Smaller dimension , L: Larger dimension

If we want to calculate ∆𝛔 𝐳 below the center of rectangular
area there are two methods:
1. Divide this area into 4 areas to make point “A” under the corner of each
area:
We note that, point “A” is under the corner
of each rectangular area, so:
∆σz,total = q(I1 + I2 + I3 + I4)
Because the total area is rectangular and
divided into 4 areas it is clear that the four
areas are equal so:
I1 = I2 = I3 = I4 = I
∆σz,total = q(4I)
2.
∆σz,total = qIc
Ic = f(m1, n1) (From Table 5.3 P. 230)
m1 =
L
B
, n1 =
Z
b
=
2Z
B

Approximate Method (2:1 Method)
An alternative approximate method can be used rather than (Ch.10) in soil
mechanics course, this method is easier and faster than methods in (Ch.10).
This method called (2:1 Method). The value of (∆σ′)at depth D can be
determined using (2:1 method) as following:
According to this method, the value of (∆σ′
) at depth (D) under the center
of the foundation is:
∆σD
′
=
P
A
=
P
(B + D) × (L + D)
A = the area of the stress distribution at 𝐝𝐞𝐩𝐭𝐡 (𝐃).
Note that the above equation is based on the assumption that the stress from
the foundation spreads out with a vertical-to-horizontal slope of 2:1.

Note: if the foundation is circular the value of (∆σ′
) at depth (D) under the
center of the foundation can be determined as following:
∆σD
′
=
P
Area at depth (D)
=
P
π
4
× (B + D)2
B = diameter of the foundation(m).

Average Vertical Stress Increase Due to a
Rectangularly Loaded Area
In many cases, the average stress increase (∆σav
′
) below the corner of the
rectangular foundation is required (to calculate the consolidation settlement
below the corner of rectangular foundation), this can be calculated by the
following method:
The average stress increase for layer between z = H1 and z = H2 can be
calculated using the following equation:
∆σav(H2/H1)
′
= qo [
H2Ia(H2) − H1Ia(H1)
H2 − H1
]
qo = Stress at the base of the foundation
Ia(H2) = Ia for z = 0 to z = H2 = f (m2 =
B
H2
, n2 =
L
H2
)
Ia(H1) = Ia for z = 0 to z = H1 = f (m2 =
B
H1
, n2 =
L
H1
)
Values of Ia(H2) and Ia(H1) can be calculated from (𝐅𝐢𝐠𝐮𝐫𝐞 𝟓. 𝟕 𝐏. 𝟐𝟑𝟒)

Basics of Foundation Engineering هندسة الأساسات & Eng. Ahmed S. Al-Agha

Basics of Foundation Engineering هندسة الأساسات & Eng. Ahmed S. Al-Agha

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Basics of Foundation Engineering هندسة الأساسات & Eng. Ahmed S. Al-Agha