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
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)

If we want to calculate ∆𝛔 𝐚𝐯(𝐇 𝟐/𝐇 𝟏)
′
below the center of
rectangular area:
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:
Because the total area is rectangular and
divided into 4 areas it is clear that the four
areas are equal so:
∆σav(H2/H1),center
′
= 4 × ∆σav(H2/H1),corner
′
Solve Example (5.2)
There is another method used to calculate ∆σav
′
:
∆σav
′
=
∆σt
′
+ 4∆σm
′
+ ∆σb
′
6
(under the center or under the cornere)
∆σt
′
= Increase in effective stress at the top of soil layer.
∆σm
′
= Increase in effective stress at the middle of soil layer.
∆σb
′
= Increase in effective stress at the bottom of soil layer.
∆σt
′
, ∆σm
′
and ∆σb
′
can be caculated under the 𝐜𝐞𝐧𝐭𝐞𝐫 or 𝐜𝐨𝐫𝐧𝐞𝐫𝐞 of the
foundation (as required).
If ∆σav
′
is required under the center of the foundation, we can use (2:1)
method to calculate ∆σt
′
, ∆σm
′
and ∆σb
′
.

Stress Increase under Embankment Loading
below:
We can calculate the vertical stress increase at any point due to the
embankment load as following:
∆σz = qI′
q = Embankment Load = γH
The value of I′
can be taken from (Figure 5.11 P.237) according to the
values of
B1
Z
and
B2
Z
B1 → Always for rectangle
B2 → Always for triangle

Elastic Settlement
Elastic Settlement of Foundations on Saturated Clay
This method used for calculating elastic settlement only for saturated clay
from the following equation:
Se = A1A2
qoB
Es
A1 = f (
H
B
&
L
B
) = factor depends on the shape of the loaded area
and can be estimated from (𝐅𝐢𝐠𝐮𝐫𝐞 𝟓. 𝟏𝟒 𝐏. 𝟐𝟒𝟒)
A2 = f (
Df
B
) = factor depends on the depth of the footing
and can be estimated from (𝐅𝐢𝐠𝐮𝐫𝐞 𝟓. 𝟏𝟒 𝐏. 𝟐𝟒𝟒)
H = Thickness of clay layer under the bottom of the foundation
B = Foundation width , L = Foundation length
Es = Modulus of elsticity of clay layer = βcu
cu = undrained shear strength for clay
β = f(PI, OCR)
PI = Plasticity Index = LL − PL
OCR = Overconsolidated Ratio =
Preconsolidation Pressure
Present Effective Pressure
=
σc
′
σo
′
Now, the value of β can be taken from (Table 5.7 P.244) and then calculate
the value of Es.

Settlement Based on the Theory of Elasticity
Elastic or immediate settlement (Se) occurs directly after the application of
the load without change in the moisture content of the soil.
The magnitude of the elastic settlement will depend on the flexibility of the
foundation (flexible or rigid), and on the type of material (soil) that the
foundation will rest on it (i.e. this method is valid for both sand and clay).
Elastic Settlement under a flexible foundation can be calculated from the
following equation:
Se = ∆σ(αB′
)
1 − μs
2
Es
IsIf
L = Larger dimension of the foundation
B = Smaller dimension of the foundation
∆σ = Applied Net Pressure = Load/Area
Es = Modulus of Elsticity of the soil through the depth H.
∆σ

If there are more than one soil layer through the depth H we can find the
modulus of elasticity for all layers by weighted average method:
Es =
∑ Es(i) × ∆Z
Z
Es(i) = modulus of elasticity of each layer.
∆Z = depth of each layer.
Z = H or 5B whichever is 𝐬𝐦𝐚𝐥𝐥𝐞𝐫
H = Distance from the face of the footing to the nearest rock layer (as
shown in figure above)
Now, as shown in figure above, the elastic settlement under a flexible
foundation at the center is larger than at the corner, thus there are some
differences in calculating (Se) under the center and under the corner of the
footing.
These differences can be considered in the values of (B′
and α) .
For calculating 𝐒 𝐞 under the center of the foundation:
B′
=
B
2
and α = 4
For calculating 𝐒 𝐞 under the corner of the foundation:
B′
= B and α = 1
Where, α = factor depends on the location on the foundation where
settlement is being calculated.
Is = shape factor.
Is = F1 +
1−2μs
1−μs
F2
F1(can be calculated from 𝐓𝐚𝐛𝐥𝐞 𝟓. 𝟖 𝐏. 𝟐𝟒𝟖)
F2(can be calculated from 𝐓𝐚𝐛𝐥𝐞 𝟓. 𝟗 𝐏. 𝟐𝟓𝟎)
To get the values of F1and F2 from tables we need the values of m′
and n′
m′
=
L
B
, n′
=
H
B′
If = factor depens on depth of excavation, footing dimensions and soil type
The value of (If) can be calculated from (Table 5.10 P.252).
Note: If Df = 0.0 →→ If = 1

Elastic Settlement under a rigid foundation:
From the figure above (page 93) it is noted that the elastic settlement under a
rigid foundation is constant and less than Se for flexible foundation (at
center).
So, the value of Se under a rigid foundation can be estimated as following:
Se(rigid) = 0.93Se(flexible,center)
See (example 5.5 P.252) in your textbook.
Settlement of Sandy Soil: Using Strain Influence Factor
This method is one of the most famous methods used to calculate elastic
settlement for sandy soil. According to this method, the settlement is:
Se = C1C2(q̅ − q) ∑
Iz
Es
∆z
z2
0
C1 = correction factor for depth of foundation embedment
C1 = 1 − 0.5 ×
q
q̅ − q
C2 = correction factor to account for creep in soil
(i. e. at what time after construction, you want to calculate Se? )
C2 = 1 + 0.2 log (
Time in years
0.1
)
q̅ = gross stress at the level of the foundation =
Applied gross Load
Foundation Area
=
P
A
q = 𝐞𝐟𝐟𝐞𝐜𝐭𝐢𝐯𝐞 𝐬𝐭𝐫𝐞𝐬𝐬 at the base of the foundation
Es = modulus of elasticity of soil
∆z = thickness of each soil layer
Iz = influence line factor
Note:
q̅ − q = 𝐧𝐞𝐭 𝐬𝐭𝐫𝐞𝐬𝐬 applied at the base of the foundation, so if you are
given the net load → q̅ − q =
Net Load
Foundation Area

Procedures for calculating elastic settlement using strain influence
factor:
1. Plot the variation of 𝐈 𝐳 with depth (under the base of the foundation):
To do this, three values of Iz at three depths should be calculated:
A- Calculate the value of Iz,initial at depth z = 0.0 (i.e.at the base of the
foundation) by the following equation:
Iz,initial = 0.1 + 0.0111 (
L
B
− 1) ≤ 0.2
Special cases:
 For square or circular footings →
L
B
= 1 → Iz,initial = 0.1
 For strip or continuous footings→
→
L
B
≥ 10 → Iz,initial = 0.2 (max value at
L
B
= 10)
 For rectangular footing→ apply on the equation by
L
B
to get Iz,initial
B- Calculate the value of Iz,max at depth z = z1from the base of the
foundation using the following equation:
Iz,max = 0.5 + 0.1√
q̅ − q
qz(1)
′
qz(1)
′
= effective stress from the ground surface to depth z1befor
construction the foundation
So to calculate qz(1)
′
and then calculate Iz,max, we fiestly must calculate
(z1) by the following equation:
z1
B
= 0.5 + 0.0555 (
L
B
− 1) ≤ 1
Special cases:
L
B
= 1 → z1 = 0.5 B
→
L
B
≥ 10 → z1 = B (max value at
L
B
= 10)
L
B
to get z1
Now, you can easily calculate qz(1)
′
an then calculate Iz,max

C- The value of Iz,final = 0.0 at depth z = z2from the base of the
foundation, so we need to calculate the value of (z2) by the following
equation:
z2
B
= 2 + 0.222 (
L
B
− 1) ≤ 4
Special cases:
L
B
= 1 → z2 = 2 B
→
L
B
≥ 10 → z2 = 4B (max value at
L
B
= 10)
L
B
to get z2
Now, you can draw the variation of Iz with depth as following:
Iz,initial

2. Plot the variation of 𝐄 𝐬 with depth (under the base of the foundation):
Firstly we want to calculate Es for each layer under the base of the
foundations as following:
 Using correlation from cone penetration test:
Es = 2.5 qc (For square or circular foundation)
Es = 3.5 qc (For strip or continuous foundation)
qc = cone penetration resistance
For rectangular foundation:
Es(rectangle) = (1 + 0.4log
L
B
) Es(square)
 Using correlation from standard penetration test:
Es = 500 N60 (KN/m2
) for all types of foundations, or you may given
different rule according the type of sand.
Now, calculate the value of Es for each soil layer and then draw the variation
of Es with depth on the same graph of Iz with depth as following:

3. Divide the soil layer from 𝐳 = 𝟎 𝐭𝐨 𝐳 = 𝐳 𝟐into a number of layers:
Each layer must satisfy the following two conditions:
 The value of Es must be constant along the layer thickness.
 The slope of Iz line must be constant along the layer thickness, if not you
have to divide this layer into two layers although the value of Es is constant.
After divided the layers, match between Es and Iz charts by horizontal line
as shown below:
4. Calculate the value of 𝐈 𝐳 at the middle of each layer by interpolation:

5. Prepare table (such the following table) to obtain ∑
𝐈 𝐳
𝐄 𝐬
∆𝐳
𝐳 𝟐
𝟎 :
Layer No. ∆z Es
Iz at the middle
of each layer
Iz
Es
∆z
1 ∆z1 Es(1) Iz(1)
IIz(1)
Es(1)
∆z1
2 ∆z2 Es(2) Iz(2)
IIz(2)
Es(2)
∆z2
3 ∆z3 Es(3) Iz(3)
IIz(3)
Es(3)
∆z3
4 ∆z4 Es(4) Iz(4)
IIz(4)
Es(4)
∆z4
n ∆zn Es(n) Iz(n)
IIz(n)
Es(n)
∆zn
∑
Iz
Es
∆z
z2
0
6. Finally calculate the value of Se:
Se = C1C2(q̅ − q) ∑
Iz
Es
∆z
z2
0

Settlement of Foundation on Sand Based on Standard
Penetration Resistance
Bowles (1977) proposed a correlation for a net allowable bearing capacity
for foundations:
qall,net(kN/m2
) =
N60
0.05
× Fd (
Se
25
) For B ≤ 1.22 m
qall,net(kN/m2
) =
N60
0.08
(
B + 0.3
B
)
2
Fd (
Se
25
) For B > 1.22 m
Fd = 1 + 0.33
Df
B
≤ 1.33
Se = elastic settlement in (mm)
N60 = corrected standard penetration resistance (number).
Se(mm) =
1.25×qall,net
N60×Fd
(
B
B+0.3
)
2
For B ≤ 1.22m
Se(mm) =
2×qall,net
N60×Fd
(
B
B+0.3
)
2
For B > 1.22m
Consolidation Settlement
Primary Consolidation Settlement
As we discussed previously in soil mechanics course (Ch.11), the
consolidation settlement is the process of water expulsion from the voids of
saturated clay with time. But here, we want to calculate total primary
consolidation settlement (i.e. @ t = ∞ ) using the following three equations:
1. For normally consolidated clay:
For Normally consolidated clay → OCR =
σc
′
σo
′
= 1
Sc =
Cc × H
1 + eo
× log (
σo
′
+ ∆σav
′
σo
′
)
Sc = primary consolidation settlement
Cs = swell index , H = thickness of clay layer , eo = initial void ratio
σo
′
= present effective overburden pressure (at the middle of clay layer)
∆σav
′
= average effective stress increase for clay layer

2. For Overconsolidated clay
For Overconsolidated clay → OCR =
σc
′
σo
′
> 1
There are two cases:
Case One (σc
′
≥ σo
′
+ ∆σav
′
):
Sc =
Cs×H
1+eo
× log (
σo
′
+∆σav
′
σo
′
)
Cs = swell index
Case Two (σc
′
< σo
′
+ ∆σav
′
):
Sc =
Cs × H
1 + eo
× log (
σc
′
σo
′
) +
Cc × H
1 + eo
× log (
σo
′
+ ∆σav
′
σc
′
)
If the stress at the base of the foundation is decreased with depth, we can
calculate the value of ∆σav
′
by Simpson’s rule as following:
∆σav
′
=
∆σt
′
+ 4∆σm
′
+ ∆σb
′
6
∆σt
′
= Increase in effective stress at the top of clay layer.
∆σm
′
= Increase in effective stress at the middle of clay layer.
∆σb
′
= Increase in effective stress at the bottom of clay layer.
Secondary Consolidation Settlement
At the end of primary consolidation (after complete dissipation of pore water
pressure), some settlement is observed that is due to the plastic adjustment of
soil fabrics. This stage of consolidation is called secondary consolidation.
Ss =
H × Cα
1 + ep
× log (
t2
t1
)
ep = void ratio at the end of primary consolidation stage.
t1 = time at the end of primary consolidation settlement (start secondary
consolidation settlement).
t2 =any time after beginning secondary consolidation settlement.
the value of Cα is depend on type of clay (N.C.Clay or O.C.Clay or Organic
clay) and there is a typical values for each type.

Plate Load Test
The ultimate bearing capacity of foundations, as well as the allowable
bearing capacity based on tolerable settlement considerations, can be
effectively determined from the field load test, generally referred as plate
load test.
Plate Properties:
The plate used in this test is made of steel and have the following
dimensions:
 If the plate is circular, the diameter will be (150mm to 762mm) with
25mm thickness.
 If the plate is square, the dimensions are (305mm x 305mm) with 25mm
thickness.
Test Mechanism:
 To conduct a test, a hole is excavated with a minimum diameter of 4B (B
is the diameter of the test plate) to depth Df (depth of proposed foundation).
 The plate is at the center of the hole, and the load is applied on the plate
and increased gradually.
 As the load increase, the settlement of the plate is observed on dial gauge.
 The test should be conducted until failure, or the settlement of the plate
became 25mm.
 The value of load at which the test is finished is the ultimate load can be
resisted by the plate.
 Divide the ultimate load on the plate area to get ultimate bearing capacity
of the plate qu(P).
For test in clay:
qu(F) = qu(P)
qu(F) = ultimate bearing capacity of the proposed foundation
For test in sandy soil:
qu(F) = qu(P)
BF
BP
BF = width of the foundation , BP = width of the test plate

Problems
1.
A 3m×3m square footing is shown in the figure below, if the net load on the
foundation is 2000 kN.
A. Use strain influence factor method to calculate elastic settlement of the
footing after 6 years of construction. Given to you values of tip resistance
from cone penetration test:
Depth (m) from
the ground surface
qc(kN/m2
)
0-2 8000
2-4 10000
4-6 9000
6-8 8500
B. If the corrected standard penetration number (N60) = 10 and the
allowble settlement is 25.4mm. Calculate the net allowable bearing
capacity using Bowles equations.
Solution
A. Se =? ? (using influence line factor method).
Se = C1C2(q̅ − q) ∑
Iz
Es
∆z
z2
0
C1 = 1 − 0.5 ×
q
q̅ − q

q = γDf = 17 × 2 = 34 kN/m2
q̅ − q =
Net Load
Foundation Area
=
2000
3 × 3
= 222.22 kN/m2
→ C1 = 1 − 0.5 ×
34
222.22
= 0.923
C2 = 1 + 0.2 log (
Time in years
0.1
) time = 6years
→ C2 = 1 + 0.2 log (
6
0.1
) = 1.356
Calculation of ∑
𝐈 𝐳
𝐄 𝐬
∆𝐳
𝐳 𝟐
𝟎 :
1. Calculation of Iz with depth:
 At depth z = 0 (at the base of the foundation) → Iz = Iz,initial
Iz,initial = 0.1 + 0.0111 (
L
B
− 1) but for square footing → Iz,initial = 0.1
 At depth z = z1 → Iz = Iz(max)
Iz(max) = 0.5 + 0.1√
q̅ − q
qz(1)
′
Before calculating Iz(max) we need to calculate qz(1)
′
at depth z1
z1
B
= 0.5 + 0.0555 (
L
B
− 1) but for square footing → z1 = 0.5 B
→ z1 = 0.5 B = 0.5 × 3 = 1.5m
qz(1)
′
= 17 × (2 + 1.5) = 59.5 kN/m2
→ Iz(max) = 0.5 + 0.1√
222.22
59.5
= 0.69
 At depth z = z2 → Iz = 0.0
z2
B
= 2 + 0.222 (
L
B
− 1) but for square footing → z2 = 2 B
→ z2 = 2B = 2 × 3 = 6m
I.e. we concerned about soil to depth of 6m below the foundation or (6+2) m
from the ground surface.

2. Calculation of Eswith depth:
The value of Es will vary with depth according the values of qc and for
square footing → Es = 2.5 qc →→
Depth (m) from
the ground surface
qc(kN/m2
)
Es = 2.5 qc
(kN/m2
)
0-2 8000 20000
2-4 10000 25000
4-6 9000 22500
6-8 8500 21250
3. Now we can draw the variation of Iz and Es with depth and then divide
the soil for layers according the values of Es and the slope of Iz:
Note that the layer (2) has the same value of Es for layer (1), but because the
slope is differing, we divided them into two layers.

Now we want to calculate the value of Iz at the middle of each layer by
interpolation:
By interpolation (using calculator):
Iz(1) = 0.395 , Iz(2) = 0.651 , Iz(3) = 0.46 , Iz(4) = 0.153
4. Now we can calculate the value of ∑
Iz
Es
∆z
z2
0 :
Layer No. ∆z(m) Es(kN/m2
)
Iz at the middle
of each layer
Iz
Es
∆z
1 1.5 25000 0.395 2.37 × 10−5
2 0.5 25000 0.651 1.30 × 10−5
3 2 22500 0.46 4.08 × 10−5
4 2 21250 0.153 1.44 × 10−5
∑
Iz
Es
∆z
z2
0
= 9.192 × 10−5
→ Se = 0.923 × 1.356 × 222.222 × 9.192 × 10−5
= 0.02557m = 25.57 mm✓.
B. qall,net =? ? (if Se = 25.4mm and N60 = 10)
qall,net(kN/m2
) =
N60
0.08
(
B + 0.3
B
)
2
Fd (
Se
25
) For B > 1.22 m
Fd = 1 + 0.33
Df
B
≤ 1.33 → Fd = 1 + 0.33 ×
2
3
= 1.22
qall,net(kN/m2
) =
10
0.08
(
3 + 0.3
3
)
2
× 1.22 (
25.4
25
) = 184.52 kN/m2
✓.

2.
A continuous foundation resting on a deposit of sand layer is shown in the
figure below, along with the variation of the modulus of elasticity of soil
(Es). Assuming γ = 115 Ib/ft3
and the time = 10 years. Calculate the elastic
settlement using strain influence factor.
Solution
Se = C1C2(q̅ − q) ∑
Iz
Es
∆z
z2
0
C1 = 1 − 0.5 ×
q
q̅ − q
q = γDf = 115 × 5 = 575b/ft2
, q̅ = 4000 Ib/ft2
q̅ − q = 4000 − 575 = 3425 b/ft2
→ C1 = 1 − 0.5 ×
575
3425
= 0.916
C2 = 1 + 0.2 log (
Time in years
0.1
) time = 10years
q̅ = 4000 Ib/ft2

→ C2 = 1 + 0.2 log (
10
0.1
) = 1.4
Calculation of ∑
𝐈 𝐳
𝐄 𝐬
∆𝐳
𝐳 𝟐
𝟎 :
1. Calculation of Iz with depth:
 At depth z = 0 (at the base of the foundation) → Iz = Iz,initial
Iz,i = 0.1 + 0.0111 (
L
B
− 1) but for continuous footing → Iz,i = 0.2
 At depth z = z1 → Iz = Iz(max)
Iz(max) = 0.5 + 0.1√
q̅ − q
qz(1)
′
Before calculating Iz(max) we need to calculate qz(1)
′
at depth z1
z1
B
= 0.5 + 0.0555 (
L
B
− 1) but for square footing → z1 = B
→ z1 = B = 8 ft.
qz(1)
′
= 115 × (5 + 8) = 1495 Ib/ft2
→ Iz(max) = 0.5 + 0.1√
3425
1495
= 0.651
 At depth z = z2 → Iz = 0.0
z2
B
= 2 + 0.222 (
L
B
− 1) but for continuous footing → z2 = 4 B
→ z2 = 4B = 4 × 8 = 32 ft
2. Calculation of Eswith depth:
The variations of Es with depth is given.
Now we can draw the variation of Iz and Es with depth:

Now we want to calculate the value of Iz at the middle of each layer by
interpolation:
By interpolation (using calculator):
Iz(1) = 0.369 , Iz(2) = 0.595 , Iz(3) = 0.488 , Iz(4) = 0.163
Now, don’t forget to transform the unit of Es from (Ib/in2
to Ib/ft2
) →
→ 875Ib/in2
= 875 × 144 = 126000 Ib/ft2
and so on, ,,

3. Now we can calculate the value of ∑
Iz
Es
∆z
z2
0 :
Layer No. ∆z(ft) Es(Ib/ft2
)
Iz at the middle
of each layer
Iz
Es
∆z
1 6 126000 0.369 1.757 × 10−5
2 2 250560 0.595 0.449 × 10−5
3 12 250560 0.488 2.337 × 10−5
4 12 208800 0.163 0.937 × 10−5
∑
Iz
Es
∆z
z2
0
= 5.53 × 10−5
→ Se = 0.916 × 1.4 × 3425 × 5.53 × 10−5
= 0.242ft = 0.074m = 74mm✓.
3.
Solve example 5.7 in your textbook

Chapter (6)
Geometric Design of
Shallow Foundations

Foundation Engineering Geometric Design of Shallow Foundations
Introduction
As we stated in Chapter 3, foundations are considered to be shallow if
if [Df ≤ (3 → 4)B]. Shallow foundations have several advantages: minimum
cost of materials and construction, easy in construction “labor don’t need
high experience to construct shallow foundations”. On the other hand, the
main disadvantage of shallow foundations that if the bearing capacity of the
soil supporting the foundation is small, the amount of settlement will be
large.
Types of Sallow Foundations
1. Isolated Footings (spread footings).
2. Combined Footings.
3. Strap Footings.
4. Mat “Raft” Foundations.
Geometric Design of Isolated Footings
The most economical type of foundations, and usually used when the loads
on the columns are relatively small and the bearing capacity of the soil
supporting the foundations is large.
In practice, we usually use isolated square footing because is the most
economical type if the following condition is satisfied:
The distance between each footing should be more than 30 cm from all
direction, if not; we use isolated rectangular footing (if possible) to make the
distance more than 30 cm. The following figure explains this condition:

Note that for the middle foundation the distance from two sides is less than
30, so we use rectangular isolated footing as shown in figure (right side).
If using of rectangular footing is not possible, we use combined footing as
we will explain later.
Design Procedures:
1. Calculate the net allowable bearing capacity:
The first step for geometric design of foundations is to calculate the
allowable bearing capacity of the foundations as we discussed in previous
chapters.
qall,net =
qu,net
FS
qu,net = qu,gross − γchc − γshs
2. Calculate the required area of the footing:
Areq =
Qservice
qall,net
= B × L
Assume B or L then find the other dimension.
If the footing is square:
Areq = B2
→ B = √Areq
Qservice = PD + PL
Why 𝐰𝐞 𝐮𝐬𝐞 𝐐 𝐬𝐞𝐫𝐯𝐢𝐜𝐞:
Areq =
Qservice
qall,net
= Areq =
Qservice
qu,net
FS
=
FS × Qservice
qu,net
In the above equation the service load "Qservice" is multiplied by FS and
when we multiply it by load factor, this is not economical and doesn’t make
sense.
Note:
The equation of calculating the required area (Areq =
Qservice
qall,net
) is valid only
if the pressure under the base of the foundation is uniform.

Geometric Design of Combined Footings
Types:
1. Rectangular Combined Footing (two columns).
2. Trapezoidal Combined Footing (two columns).
3. Strip Footing (more than two columns and may be rectangular or
trapezoidal).
Usage:
1. Used when the loads on the columns are heavy and the distance between
these columns is relatively small (i.e. when the distance between isolated
footings is less than 30 cm).
2. Used as an alternative to neighbor footing which is an eccentrically loaded
footing and it’s danger if used when the load on the column is heavy.
Design of Rectangular Combined Footings:
There are three cases:
1. Extension is permitted from both side of the footing:
The resultant force R is more closed to the column which have largest load.

To keep the pressure under the foundation uniform, the resultant force of all
columns loads (R) must be at the center of the footing, and since the footing
is rectangular, R must be at the middle of the footing (at distance L/2) from
each edge to keep uniform pressure.
Areq =
∑ Qservice
qall,net
=
Q1+Q2
qall,net
= B × L
How we can find L:
L2 = distance between centers of the two columns and it′
s always known
Xr = distance between the resultant force and column (1)
OR column (2) as u like  .
L1and L3 = extensions from left and right " usually un knowns"
Now take summation moments at C1 equals zero to find Xr:
∑ MC1
= 0.0 → Q2L2 + (Wfooting + Wsoil) × Xr = R × Xr → Xr = ✓.
(Wfooting + Wsoil) are located at the center of the footing
If we are not given any information about (Wfooting + Wsoil) →
Q2L2 = R × Xr → Xr = ✓.
Now, to keep uniform pressure under the foundation:
Xr + L1 =
L
2
(Two unknowns "L1" and "L")
The value of L1can be assumed according the permitted extension in site.
→ L = ✓ →→ B =
Areq
L
= ✓.
2. Extension is permitted from one side and prevented from other side:

The only difference between this case and the previous case that the
extension exists from one side and when we find Xr we can easily find L:
To keep the pressure uniform → Xr +
column width
2
=
L
2
→ L = ✓.
3. Extension is not permitted from both sides of the footing:
In this case the resultant force R doesn’t in
the center of rectangular footing because Q1
and Q2 are not equals and no extensions from
both sides. So the pressure under the
foundation is not uniform and we design the
footing in this case as following:
L = L1 + W1 + W2 = ✓.
How we can find e:
∑ Mfoundation center = 0.0
Q1 × (
L
2
−
W1
2
) − Q2 × (
L
2
−
W2
2
) = R × e
→ e = ✓.
Note: the moment of Wf + Ws is zero
because they located at the center of footing.
The eccentricity in the direction of L:
Usually e <
L
6
(because L is large)
qmax =
R
B × L
(1 +
6e
L
)
qall,gross ≥ qmax → qall,gross = qmax (critical case)
qall,gross =
R
B × L
(1 +
6e
L
) → B = ✓.
Check for B:
qmin =
R
B × L
(1 −
6e
L
) must be ≥ 0.0
If this condition doesn’t satisfied, use the modified equation for qmax to find
B:
qmax,modified =
4R
3B(L − 2e)
→ B = ✓.

Design of Trapezoidal Combined Footings:
Advantages:
1. More economical than rectangular combined footing if the extension is
not permitted from both sides especially if there is a large difference
between columns loads.
2. We can keep uniform contact pressure in case of “extension is not
permitted from both sides” if we use trapezoidal footing because the
resultant force “R” can be located at the centroid of trapezoidal footing.
Design:
Q1 > Q2 → B1at Q1 and B2at Q2
L = L1 + W1 + W2 = ✓.
Areq =
∑ Qservice
qall,net
=
Q1 + Q2
qall,net
Q1 + Q2
qall,net
=
L
2
(B1 + B2) →→ Eq. (1)
Now take summation moments at C1 equals
zero to find Xr:
∑ MC1
= 0.0 → Q2L1 +
(Wf + Ws) × Xr = R × Xr → Xr = ✓.
Xr +
W1
2
= X̅ = ✓.
X̅ =
L
3
(
B1 + 2B2
B1 + B2
) →→ Eq. (2)
Solve Eq. (1)and Eq. (2) →→
B1 = ✓ and B2 = ✓.
X̅

Geometric Design of Strap Footing (Cantilever Footing)
Usage:
1. Used when there is a property line which prevents the footing to be
extended beyond the face of the edge column. In addition to that the edge
column is relatively far from the interior column so that the rectangular and
trapezoidal combined footings will be too narrow and long which increases
the cost. And may be used to connect between two interior foundations one
of them have a large load require a large area but this area not available, and
the other foundation have a small load and there is available area to enlarge
this footing, so we use strap beam to connect between these two foundations
to transfer the load from largest to the smallest foundation.
2. There is a “strap beam” which connects two separated footings. The edge
Footing is usually eccentrically loaded and the interior footing is centrically
loaded. The purpose of the beam is to prevent overturning of the
eccentrically loaded footing and to keep uniform pressure under this
foundation as shown in figure below.
Note that the strap beam doesn’t touch the ground (i.e. there is no contact
between the strap beam and soil, so no bearing pressure applied on it).
This footing also called “cantilever footing” because the overall moment on
the strap beam is negative moment.

Design:
R = Q1 + Q2 = R1 + R2 but, Q1 ≠ R1 and Q2 ≠ R2
Q1 and Q2 are knowns but R1 and R2 are unknowns
Finding 𝐗 𝐫:
∑ MQ1
= 0.0 (before use of strap beam) → R × Xr = Q2 × d → Xr = ✓.
a = Xr +
w1
2
−
L1
2
(L1 should be assumed "if not given")
b = d − Xr
Finding 𝐑 𝟏:
∑ MR2
= 0.0(after use of strap beam) → R1 × (a + b) = R × b → R1 = ✓.
Finding 𝐑 𝟐:
R2 = R − R1
Design:
A1 =
R1
qall,net
, A2 =
R2
qall,net
Mat Foundation
Usage:
We use mate foundation in the following cases:
1. If the area of isolated and combined footing > 50% of the structure area,
because this means the loads are very large and the bearing capacity of the
soil is relatively small.
2. If the bearing capacity of the soil is small (usually < 15 t/m2
).
3. If the soil supporting the structure classified as (bad soils) such as:
 Expansive Soil: Expansive soils are characterized by clayey material
that shrinks and swells as it dries or becomes wet respectively. It is
recognized from high values of Plasticity Index, Plastic Limit and
Shrinkage Limit.

 Compressible soil: It contains a high content of organic material and
not exposed to great pressure during its geological history, so it will be
exposed to a significant settlement, so mat foundation is used to avoid
differential settlement.
 Collapsible soil: Collapsible soils are those that appear to be strong and
stable in their natural (dry) state, but which rapidly consolidate under
wetting, generating large and often unexpected settlements. This can yield
disastrous consequences for structures unwittingly built on such deposits.
Types:
 Flat Plate (uniform thickness).
 Flat plate thickened under columns.
 Beams and slabs.
 Slabs with basement walls as a part of the mat.
(To see these types open Page 295 in your textbook)
Compensated Footing
The net allowable pressure applied on the mat foundation may be
expressed as:
q =
Q
A
− γDf
Q = Service Loads of columns on the mat + own weight of mat
A = Area of the mat "raft" foundation.
In all cases q should be less than qall,net of the soil.
The value of q can be reduced by increasing the depth Df of the mat, this
called compensated footing design (i.e. replacing (substituting) the weight
of the soil by the weight of the building) and is extremely useful when the
when structures are to be built on very soft clay.
At the point which q = 0.0 → the overall weight of the soil above this point
will replaced (substituted) by the weight of the structure and this case is
called fully compensated footing (fully safe).
The relationship for fully compensated depth Df can be determined as
following:
0.0 =
Q
A
− γDf → Df =
Q
Aγ
(fully compensated depth )

Geometric Design of Mat Foundation (Working Loads)
Procedures:
1. Determine the horizontal and vertical axes (usually at the center line of the
horizontal and vertical edge columns) as shown.
2. Calculate the centroid of the mat [ point C (X̅, Y̅)]with respect to X and Y
axes:
X̅ =
∑ Xi × Ai
∑ Ai
Y̅ =
∑ Yi × Ai
∑ Ai
Ai = shapes areas.
Xi = distance between y − axis and the center of the shape.
Yi = distance between y − axis and the center of the shape.
If the mat is rectangular:
X̅ =
L
2
−
wvertical edge columns
2
Y̅ =
B
2
−
w horizontal edge columns
2
Y̅
X̅

3. Calculate the resultant force R:
R = ∑ Qi
4. Calculate the location of resultant force R (XR, YR) with respect to X and
Y axes:
To find XR take summation moments about Y-axis:
XR =
∑ Qi × Xri
∑ Qi
To find YR take summation moments about X-axis:
YR =
∑ Qi × Yri
∑ Qi
Qi = load on column
Xri = distance between columns center and Y − axis
Yri = distance between columns center and X − axis
5. Calculate the eccentricities:
ex = |XR − X̅| ey = |YR − Y̅|
6. Calculate moments in X and Y directions:
Mx = ey × ∑ Qi My = ex × ∑ Qi
7. Calculate the stress under each corner of the mat:
q =
∑ Qi
A
±
My
Iy
X ±
Mx
Ix
Y
How we can know the sign (+ or −):
If compression (+) If tension (−)
Compression if the arrow of moment is at the required point
Tension if the arrow of moment is far away from the required point
X and Y are distances from the 𝐜𝐞𝐧𝐭𝐫𝐨𝐢𝐝 𝐭𝐨 𝐭𝐡𝐞 𝐫𝐞𝐪𝐮𝐢𝐫𝐞𝐝 𝐩𝐨𝐢𝐧𝐭
X and Y (take them always positive)
Ix = IX̅ = moment of inertia about centoid of mat (in x − direction)
Iy = IY̅ = moment of inertia about centoid of mat (in y − direction)
Ix =
B3
L
12
IY =
L3
B
12
(in case of rectangular foundation

8. Check the adequacy of the dimensions of mat foundation:
Calculate qmax (maximum stress among all corners of the mat)
Calculate qmin (minimum stress among all corners of the mat)
qmax ≤ qall,net
qmin ≥ 0.0
If one of the two conditions doesn’t satisfied, increase the dimensions of the
footing.
Structural Design of Mat Foundation (Ultimate Loads)
In structural design we want to draw shear force and bending moment
diagrams, to do this we have to subdivide mat foundations into a strips in
both directions, each strip must contains a line of columns, such that the
width of strip is related to the loads of the columns including in this stip.
For the previous mat, let we take a strip of width B1 for the columns 5,6,7
and 8 as shown in figure below:
Procedures for structural design (drawing shear and moments
diagrams):
We will take the strip as shown in the figure above, and the following
procedures are the same for any other strip:

1. Locate the points E and F at the middle of strip edges.
2. Calculate the factored resultant force (Ru):
Ru = ∑ Qui
3. The eccentricities in X and Y directions remains unchanged because the
location of the resultant force will not change since we factored all columns
by the same factor.
4. Calculate the factored moment in X and Y directions:
Mu,x = ey × ∑ Qui Mu,y = ex × ∑ Qui
5. Calculate the stresses at points E and F (using factored loads and
moments):
q =
∑ Qui
A
±
Mu,y
Iy
X ±
Mu,x
Ix
Y
6. Since the stress at points E and F is not equal, the pressure under the strip
in not uniform, so we find the average stress under the strip:
qu,avg =
qE + qF
2
7. Now, we check the stability of the strip:
(∑ Qui)
strip
= qu,avg × Astrip
If this check is ok, draw the SFD and BMD and then design the strip.
If not :>>>> Go to step 8.
8. We have to modified the loads to make the strip stable by the following
steps:
 Calculate the average load on the strip:
Average Load =
(∑ Qui)strip + qu,avg × Astrip
2
 Calculate the modified columns loads:
(Qui)mod = Qui ×
Average Load
(∑ Qui)strip

 Calculate the modified soil pressure:
(qu,avg)mod = qu,avg ×
Average Load
qu,avg×Astrip
Now, (∑ Qui)
strip
= qu,avg × Astrip → Draw SFD and BMD
Important Note:
If the SFD and BMD and you are given only service loads, do the above
steps for service loads and draw SFD and BMD for service loads.

Problems
1.
Design the foundation shown below to support the following two columns
with uniform contact pressure:
Column (1): PD=500 kN , PL=250 kN , cross section (50cm × 50cm).
Column (2): PD=700 kN , PL=350 kN , cross section (50cm × 50cm).
Assume the net allowable soil pressure is 200 kN/m2
Solution
To keep uniform contact pressure
under the base, the resultant force R
must be at the center of the
foundation.
Since the extension is permitted from
right side, we can use rectangular
combined footing.
Q1 = 500 + 250 = 750 KN
Q2 = 700 + 350 = 1050 kN
R = Q1 + Q2 = 750 + 1050
R = 1800 kN
The weight of the foundation and the
soil is not given, so we neglect it.

∑ Mc1
= 0.0 → 1050 × 5 = 1800 × Xr → Xr = 2.92 m
To keep uniform pressure: Xr +
0.5
2
=
L
2
L
2
= 2.92 +
0.5
2
= 3.17m
L
2
= 3.17m → L = 3.17 × 2 = 6.34 m ✓.
Calculation of B:
Areq =
Qservice
qall,net
= B × L →
1800
200
= B × 6.34 → B = 1.42 m✓.
Check for B:
Available value for B:
The permitted extension for the width B is 0.5 m, so the available width is
0.5+ (column width)+0.5 = 0.5+0.5+0.5 =1.5 m
B = 1.42 m < 1.5m Ok

2.
For the combined footing shown below.
 Find distance X so that the contact pressure is uniform.
 If qall,net=140 kN/m2
, find B.
 Draw S.F. and B.M diagrams.
Solution
To keep uniform contact pressure under the base, the resultant force R must
be at the center of the foundation.
Q1 = 1000 KN
Q2 = 660 kN
R = Q1 + Q2 = 1000 + 660
R = 1660 kN
The weight of the foundation and the soil is not given, so we neglect it.

∑ Mc1
= 0.0 → 660 × 5 = 1660 × Xr → Xr = 1.98 m
To keep uniform pressure: Xr + X =
L
2
L
2
= (5 − 1.98) + 1 = 4.02 m
L
2
= 4.02m → L = 4.02 × 2 = 8.04 m
Xr + X =
L
2
→ 1.98 + X = 4.02 → X = 2.04 m✓.
Calculation of B:
Areq =
Qservice
qall,net
= B × L →
1660
140
= B × 8.04 → B = 1.47 m✓.
Drawing SFD and BMD:
To draw SFD and BMD we use factored loads (if givens), but in this
problem we given the service loads directly, so we use service loads.
The free body diagram for the footing, SFD and BMD are shown in figure
below:

Distance 2.8 ≅
580
206
(
Shear
W
)
Note that the moment and shear for the footing is the opposite for beams
such that the positive moments is at the supports and the minimum moments
at the middle of spans, so when reinforced the footing, the bottom
reinforcement mustn’t cutoff at supports but we can cut it at the middle of
the span, also the top reinforcement mustn’t cutoff at the middle of the span
and we can cut it at the supports (Exactly).

3.
A. For the rectangular combined footing shown below, to make the soil
reaction uniform, determine the live load for the column C1, knowing that
dead load is one and half live load for this column.
Column DL (tons) LL (tons)
C1 DL1 LL1
C2 80 46
C3 55 35
B. Design a combined footing for the same figure above if C1 has a dead
load of 90 tons and live load of 54 tons, knowing that the extension is not
permitted and the soil reaction is uniform. (qall,net=20.8 t/m2
).
Solution
A.
For rectangular footing, to keep uniform pressure, the resultant force R must
be in the center of the foundation.
Q1 = DL + LL
DL = 1.5 LL → Q1 = 1.5 LL + LL = 2.5 LL
Q2 = 80 + 46 = 126 ton
Q3 = 55 + 35 = 90 ton
R = Q1 + Q2 + Q3 = Q1 + 126 + 90
R = Q1 + 216
The weight of the foundation and the soil is not given, so we neglect it.
L = 0.4 + 4.8 + 3.8 + 0.4 = 9.4m → 0.5L = 4.7m

Xr = 4.7 − 0.2 = 4.5m
∑ Mc1
= 0.0 → 126 × 5 + 90 × 9 = (Q1 + 216) × 4.5 → Q1 = 104 ton
Q1 = 104 = 2.5 LL → LL = 4.6 ton ✓.
B.
If Q1 = 90 + 54 = 144 ton,
Design the footing to keep uniform contact pressure .
Note that if we want to use rectangular footing, the pressure will be uniform
only when Q1 = 104 𝑡𝑜𝑛 otherwise if we want to use rectangular footing the
pressure will not be uniform, so to maintain uniform pressure under the
given loading we use trapezoidal combined footing (The largest width at
largest load and smallest width at smallest column load) as shown:

Q1 = 144 ton
Q2 = 126 ton
Q3 = 90 ton
R = Q1 + Q2 + Q3 = 144 + 126 + 90 = 360 ton
L = 0.4 + 4.8 + 3.8 + 0.4 = 9.4m
Areq =
Qservice
qall,net
=
L
2
(B1 + B2) →
360
20.8
=
9.4
2
(B1 + B2)
→ (B1 + B2) = 3.68 → B1 = 3.68 − B2 →→ Eq. (1)
∑ Mc1
= 0.0 → 126 × 5 + 90 × 9 = 360 × Xr → Xr = 4 m
Xr + 0.2 = X̅ → 4.2 =
L
3
(
B1 + 2B2
B1 + B2
) → 1.34 = (
B1 + 2B2
B1 + B2
) →→ Eq. (2)
Substitute from Eq.(1) in Eq.(2):
1.34 = (
3.68 − B2 + 2B2
3.68 − B2 + B2
) → B2 = 1.25 m ✓.
B1 = 3.68 − 1.25 = 2.43 m ✓.
X̅

4.
Determine B1 and B2 of a trapezoidal footing for a uniform soil pressure of
300 kN/m2
. (Consider the weight of the footing "γconcrete = 24 KN/m3
" ).
Solution
Note that we are given the thickness of the footing and the unit weight of
concrete (i.e. the weight of the footing) that must be considered.
The largest Width B1 will be at the largest load (4000) as shown below:
R = ∑ Q = 9000 kN
L = 2 + 6 + 6 + 2 = 16m
Footing Weight = Footing Volume × γc
Footing Volume =
L
2
(B1 + B2) × 1 = 8(B1 + B2)
Footing Weight = 8(B1 + B2) × 24 = 192(B1 + B2)
Note:
Areq =
Qservice,net
qall,net
=
Qservice,gross
qall,gross

Qservice,net = applied column load on the foundation
Qservice,gross = Applied column load + footing weight + soil weight
In this problem the footing weight is given and the soil weight is not given,
so we neglect it.
The given bearing capacity is allowable gross bearing capacity.
Areq =
Qservice,gross
qall,gross
= 8(B1 + B2) →
9000 + 192(B1 + B2)
300
B1 + B2 = 4.07 → B1 = 4.07 − B2 →→ Eq. (1)
Now, take summation moments about the center of the footing (to eliminate
the moments of the resultant force and the weight of the footing).
∑ Mcenter = 0.0 → 3000 × (14 − X̅) + 600 + 2000(8 − X̅) + 800
−4000 × (X̅ − 2) − 1200 = 0.0 →→ X̅ = 7.36 m
X̅ = 7.36 =
L
3
(
B1 + 2B2
B1 + B2
) → 7.36 =
16
3
(
B1 + 2B2
B1 + B2
)
1.38 = (
B1 + 2B2
B1 + B2
) →→ Eq. (2) substitute from Eq. (1)
1.38 = (
4.07 − B2 + 2B2
4.07 − B2 + B2
) → B2 = 1.55 m✓ → B1 = 2.52 m ✓.
X̅

5.
A combined footing consists of four columns as shown in figure, determine
the width of the rectangular combined footing B.
Allowable gross soil pressure = 255 kN/m2
.
γc = 24 KN/m3
and γsoile = 20 KN/m3
Solution
Note that the foundation is rectangular (as given) but the pressure in this
case will not be uniform, so we will design the footing for eccentric
loadings.
Wf + Ws = 28 × B × (1 × 20 + 1.5 × 24) = 1568 B
∑ Q = Rservice,gross = 11500 + 1568 B
Now calculate the moment at the centroid of the footing:
∑ M
@center
= 1500 × (14 − 2.5) + 2500 + 3500 × 4 − 4000 × 4
−2500(14 − 2.5) − 2500 = −13500 kN. m
e =
∑ M@center
∑ Q
=
13500
11500 + 1568 B
The eccentricity in L direction and since L is so large e <
L
6
→
q =
Q
B × L
(1 ±
6e
L
)
qmax =
Q
B × L
(1 +
6e
L
) by equating qmax and qall,gross →

255 =
11500 + 1568 B
B × 28
(1 +
6 (
13500
11500 + 1568 B
)
28
)
→ B = 2.58 m ✓.
Check For B:
e =
13500
11500 + 1568 × 2.58
= 0.868 m
∑ Q = 11500 + 1568 × 2.58 = 15545.4
qmin =
Q
B × L
(1 −
6e
L
) =
15545.4
2.58 × 28
(1 −
6 × 0.868
28
) = 175.2kN/m2
> 0 Ok
6.
For the strap footing shown below, if qall,net =250 kN/m2
, determine Q1 and Q2
Solution
R1 = A1 × qall,net = (2 × 3) × 250 = 1500 KN
R2 = A2 × qall,net = (4 × 4) × 250 = 4000 KN
R = R1 + R2 = 1500 + 4000 = 5500 KN = Q1 + Q2
a + b = 10 + 0.15 − 1 = 9.15m
∑ MR2
= 0.0(after use of strap ) → 1500 × 9.15 = 5500 × b → b = 2.5 m
→ a = 9.15 − 2.5 = 6.65 m → Xr = 6.65 + 1 − 0.15 = 7.5m

∑ M@ Q1
= 0.0 → 10Q2 = 5500 × 7.5 → Q2 = 4125 kN ✓.
→ Q1 = R − Q2 = 5500 − 4125 = 1375 kN ✓.
7.
For the shown mat foundation, If qall,net=150 kN/m2
.
1. Check the adequacy of the foundation dimensions.
2. Draw SFD and BMD for the strip ABCD which is 2m width.
Interior Columns Edge Columns Corner Columns
Columns Dimensions 60cm x 60cm 60 cm x 40 cm 40 cm x 40 cm
Service Loads 1800 kN 1200 kN 600 kN
Factored Loads 2700 kN 1800 kN 900 kN

Solution
Firstly we take the horizontal and vertical axes as shown in figure above.
Calculate the centroid of the footing with respect to these axes:
B = 5 + 8 + 2 × 0.2 = 13.4m (horizontal dimension)
L = 5 + 8 + 4 + 2 × 0.2 = 17.4m (certical dimension)
X̅ =
13.4
2
−
0.4
2
= 6.5m (from y − axis)
Y̅ =
17.4
2
−
0.4
2
= 8.5 m (from x − axis)
Calculate the resultant force R:
R = ∑ Qi (service loads) = 2 × 1800 + 6 × 1200 + 4 × 600 = 13200 kN
Calculate the location of resultant force R (𝐗 𝐑, 𝐘 𝐑) with respect to X and
Y axes:
XR =
∑ Qi × Xri
∑ Qi
=
2 × 1800 × 5 + 2 × 1200 × 5 + 2 × 1200 × 13 + 2 × 600 × 13
13200
XR = 5.82 m (from y − axis)
Note that the moments of the first vertical line of columns will equal zero
because y-axis is at the centerline of these columns.
YR =
∑ Qi × Yri
∑ Qi
=
2 × 1200 × 4 + 1800 × 4 + 2 × 1200 × 12 + 1800 × 12 + 2 × 600 × 17 + 1200 × 17
13200
YR = 8.2 m (from x − axis)
Note that the moments of the first horizontal line of columns will equal zero
because x-axis is at the centerline of these columns.
Calculate the eccentricities:
ex = |XR − X̅| = |5.82 − 6.5| = 0.68
ey = |YR − Y̅| = |8.2 − 8.5| = 0.3m

Calculate moments in X and Y directions:
Mx = ey × ∑ Qi = 0.3 × 13200 = 3960 kN. m
My = ex × ∑ Qi = 0.68 × 13200 = 8976 kN. m
Calculate moment of inertia in X and Y directions:
Ix =
B L3
12
=
13.4 × 17.43
12
= 5882.6 m4
Iy =
L B3
12
=
17.4 × 13.43
12
= 3488.85 m4
Not that the value which perpendicular to the required axis will be tripled
because it’s the value that resist the moment.
Calculate the stresses:
q =
∑ Qi
A
±
My
Iy
X ±
Mx
Ix
Y → q =
13200
13.4 × 17.4
±
8976
3488.85
X ±
3960
5882.6
Y
q = 56.61 ± 2.57 X ± 0.67 Y
If compression, use (+) sign
If Tension, use (‫)ـــ‬ sign.
But we want to calculate qmax and qmin
qmax = 56.61 + 2.57 X + 0.67 Y
qmin = 56.61 − 2.57 X − 0.67 Y
X = maximum horizontal distance from centroid to the edge of mat
=
13.4
2
= 6.7 m
Y = maximum vertical distance from centroid to the edge of mat
=
17.4
2
= 8.7 m
→ qmax = 56.61 + 2.57 × 6.7 + 0.67 × 8.7 = 79.66 kN/m2
→ qmin = 56.61 − 2.57 × 6.7 − 0.67 × 8.7 = 33.56 kN/m2
qmax = 79.66 < qall,net = 150 → Ok
qmin = 33.56 > 0.0 → Ok
The foundation Dimensions are adequate✓.

Now we want to Draw SFD and BMD for strip ABCD (using factored
loads)
Locate point E at the middle of the upper edge of strip (between A and B)
and point F at the middle of the lower edge of strip (between C and D).
Ru = ∑ Qui = 2 × 2700 + 6 × 1800 + 4 × 900 = 19800 kN
The eccentricities will not change since we factored all loads by the same factor
Mu,x = ey × ∑ Qui = 0.3 × 19800 = 5940 kN. m
Mu,y = ex × ∑ Qui = 0.68 × 19800 = 13464 kN. m
qu =
∑ Qui
A
±
Mu,y
Iy
X ±
Mu,x
Ix
Y → qu =
19800
13.4 × 17.4
±
13464
3488.85
X ±
5940
5882.6
Y
→ qu = 84.92 ± 3.86 X ± 1.01 Y
X = horizontal distance from centroid to the points E and F at strip
=
13.4
2
−
2
2
= 5.7 m
Y = vertical distance from centroid to points E and F at strip
=
17.4
2
= 8.7 m
The shown figure explains the
moments signs (compression or
tension) and the strip ABCD with
points E and F:
At point E:
My = compression (+)
Mx = tension (−)
At point F:
My = compression (+)
Mx = compression (+)

qu,E = 84.92 + 3.86 × 5.7 − 1.01 × 8.7 = 98.13 kN/m2
qu,F = 84.92 + 3.86 × 5.7 + 1.01 × 8.7 = 115.7 kN/m2
qu,avg =
qu,E + qu,F
2
=
98.13 + 115.7
2
= 106.9 kN/m2
Now, we check the stability of the strip:
(∑ Qui)
strip
= 2 × 1800 + 2 × 900 = 5400 kN
qu,avg × Astrip = 106.9 × (17.4 × 2) = 3720
(∑ Qui)
strip
≠ qu,avg × Astrip → Not stable → Loads must modified
Average Load =
(∑ Qui)strip + qu,avg × Astrip
2
=
5400 + 3720
2
= 4560 kN
(Qui)mod = Qui ×
Average Load
(∑ Qui)strip
= Qui ×
4560
5400
= 0.84Qui
(Qu,corner columns)mod = 0.84 × 900 = 756 Kn
(Qu,edge columns)mod = 0.84 × 1800 = 1512 kN
(qu,avg)mod = qu,avg ×
Average Load
qu,avg × Astrip
= 106.9 ×
4560
3720
= 131kN/m2
Now, we check the stability of the strip (after modification):
(∑ Qui)
strip
= 2 × 1512 + 2 × 756 = 4536 kN
qu,avg × Astrip = 131 × (17.4 × 2) = 4558.8 kN
(∑ Qui)
strip
≅ qu,avg × Astrip → can be considered stable
The final loads on the strip are shown below:

Now you can draw SFD and BMD but you will observe that the SFD and
BMD will not enclosed to zero (at the end) and this because the vertical
loads not exactly the same 4536 ≅ 4558.8.
I try to say “ don’t confused if SFD and BMD” are not enclosed to zero (i.e.
trust yourself ). (see problem No.2).

8.
For the mat foundation shown in the figure below:
A. Calculate the depth of embedment for fully compensated foundation
(ignore the mar self weight).
B. Find the pressure below points A, B, C and D.
C.Draw the SFD fir the strip ABCD.
D. For the soil profile shown below, calculate the primary consolidation
settlement for clay layer using Df =1.5m.
Hint: Use 2:1 method and all columns are 80 cm x 80 cm.
Solution
A-
The net applied pressure from the structure is:
q =
Q
A
− γDf but for fully compensated footing → q = 0.0 →
Q
A
= γDf
→ Df =
Q
A × γ
γ = 17 kN/m3
Sand
γ = 18 kN/m3
c = 50 kN/m2
Cc = 0.32, Cs = 0.06
eo = 0.8 , OCR = 0.2
Sand

Q = ∑ Qi = 1000 + 1000 + 1300 + 900 + 1300 + 1400 + 1100
+1200 + 1300 + 1200 = 10400 KN
A = area of mat foundation = B × L
B = 5 + 5 + 2 × 0.8 + 2 ×
0.8
2
= 12.4 m
L = 6 + 6 + 2 × 0.8 + 2 ×
0.8
2
= 14.4 m
A = 12.4 × 14.4 = 178.56 m2
γ = unit weight of the soil that will be excavated = 17 kN/m3
→ Df =
10400
178.56 × 17
= 3.426 m ✓.
B-
Firstly we take the horizontal and vertical axes as shown in figure above.
Calculate the centroid of the footing with respect to these axes:
B = 12.4m (vertical dimension)
L = 14.4m (horizontal dimension)
X̅ =
14.4
2
− 0.8 −
0.8
2
= 6 m (from y − axis)
Y̅ =
12.4
2
− 0.8 −
0.8
2
= 5 m (from x − axis)
Calculate the resultant force R:
R = ∑ Qi = 10400 kN (as calculated above)
Calculate the location of resultant force R (𝐗 𝐑, 𝐘 𝐑) with respect to X and
Y axes:
XR =
∑ Qi × Xri
∑ Qi
=
(1300 + 1400 + 1100) × 6 + (2 × 1200 + 1300) × 12
10400
XR = 6.461 m (from y − axis)
Note that the moments of the first vertical line of columns will equal zero
because y-axis is at the centerline of these columns.

YR =
∑ Qi × Yri
∑ Qi
=
(1000 + 1400 + 1300) × 5 + (1000 + 1300 + 1200) × 10
10400
YR = 5.14 m (from x − axis)
Note that the moments of the first horizontal line of columns will equal zero
because x-axis is at the centerline of these columns.
Calculate the eccentricities:
ex = |XR − X̅| = |6.461 − 6| = 0.461m
ey = |YR − Y̅| = |5.144 − 5| = 0.144 m
Calculate moments in X and Y directions:
Mx = ey × ∑ Qi = 0.144 × 10400 = 1497.6 kN. m
My = ex × ∑ Qi = 0.461 × 10400 = 4794.4 kN. m
Calculate moment of inertia in X and Y directions:
Ix =
L B3
12
=
14.4 × 12.43
12
= 2287.95 m4
Iy =
B L3
12
=
12.4 × 14.43
12
= 3085.5 m4
Not that the value which perpendicular to the required axis will be tripled
because it’s the value that resist the moment.
Calculate the stresses:
q =
∑ Qi
A
±
My
Iy
X ±
Mx
Ix
Y → q =
10400
12.4 × 14.4
±
4794.4
3085.5
X ±
1497.6
2287.95
Y
q = 58.24 ± 1.55X ± 0.6545 Y
At point A:
My = tension (−)
Mx = tension (−)
XA =
L
2
=
14.4
2
= 7.2 m YA =
B
2
=
12.4
2
= 6.2 m

qA = 58.24 − 1.55 × 7.2 − 0.6545 × 6.2 = 43.02 kN/m2
✓.
At point B:
My = tension (−)
Mx = tension (−)
XB =
L
2
− 3 =
14.4
2
− 3 = 4.2 m YB =
B
2
=
12.4
2
= 6.2 m
qB = 58.24 − 1.55 × 4.2 − 0.6545 × 6.2 = 47.67 kN/m2
✓.
At point C:
My = tension (−)
Mx = tension (+)
XC =
L
2
− 3 =
14.4
2
− 3 = 4.2 m YC =
B
2
=
12.4
2
= 6.2 m
qC = 58.24 − 1.55 × 4.2 + 0.6545 × 6.2 = 55.78 kN/m2
✓.
At point D:
My = tension (−)
Mx = tension (+)
XD =
L
2
=
14.4
2
= 7.2 m YD =
B
2
=
12.4
2
= 6.2 m
qD = 58.24 − 1.55 × 7.2 + 0.6545 × 6.2 = 51.14 kN/m2
✓.
C-
Locate point E at the middle of the upper edge of strip (between C and D)
and point F at the middle of the lower edge of strip (between A and B).
Note that we are not given factored load, so we use the given loads to draw
SFD and BMD.
The relation for calculating the stress (q) will remains unchanged because
the loads on columns doesn’t changed.
q = 58.24 ± 1.55X ± 0.6545 Y
At point E:
My = tension (−)
Mx = tension (+)
XE =
L
2
−
3
2
=
14.4
2
−
3
2
= 5.7 m YE =
B
2
=
12.4
2
= 6.2 m
qE = 58.24 − 1.55 × 5.7 + 0.6545 × 6.2 = 53.46 kN/m2

At point F:
My = tension (−)
Mx = tension (−)
XF =
L
2
−
3
2
=
14.4
2
−
3
2
= 5.7 m YF =
B
2
=
12.4
2
= 6.2 m
qF = 58.24 − 1.55 × 5.7 − 0.6545 × 6.2 = 45.35 kN/m2
✓.
qavg =
qE + qF
2
=
53.46 + 45.35
2
= 49.4 kN/m2
Now, we check the stability of the strip:
(∑ Qi)
strip
= 2 × 1000 + 900 = 2900 kN
qu,avg × Astrip = 49.4 × (12.4 × 3) = 1837.68 kN
(∑ Qi)
strip
≠ qavg × Astrip → Not stable → Loads must modified
Average Load =
(∑ Qi)strip + qavg × Astrip
2
=
2900 + 1837.68
2
= 2368.8 kN
(Qi)mod = Qi ×
Average Load
(∑ Qi)strip
= Qi ×
2368.8
2900
= 0.816Qi
(Q 1)mod = 0.816 × 1000 = 816 KN
(Q2)mod = 0.816 × 900 = 734.4 kN
(qavg)mod = qavg ×
Average Load
qavg × Astrip
= 49.4 ×
2368.8
1837.68
= 63.67kN/m2
Now, we check the stability of the strip (after modification):
(∑ Qi)
strip
= 2 × 816 + 734.4 = 2366.4 kN
qu,avg × Astrip = 63.67 × (12.4 × 3) = 2368.52 kN
(∑ Qui)
strip
≅ qu,avg × Astrip → can be considered stable
The final loads on the strip are shown below:

The SFD and BMD are shown below:
Note that the BMD not enclosed to zero, because we assume that the
pressure under the strip is uniform and this assumption is not accurate.

D-
To calculate consolidation settlement, the following parameters should be
calculated firstly:
Calculation of present effective stress at the
middle of clay layer (σ 𝑜
′ ):
Df = 1.5 m (as given)
σo
′
= 17 × 4.5 + 18 ×
5
2
= 121.5 kN/m2
Calculation of preconsolidation pressure (σ 𝑐
′
):
OCR =
σc
′
σo
′
→ 2 =
σc
′
121.5
→ σc
′
= 243 kN/m2
Calculation of average stress increase on the
clay layer (∆σav
′ ):
∆σav
′
=
∆σt
′
+ 4∆σm
′
+ ∆σb
′
6
We calculate the stress increase at the top,
middle and bottom of clay layer by using
(2:1) method (as given):
∆σz
′
=
P
(B + z) × (L + z)
, P = ∑ Qi = 10400 kN, B = 12.4m, L = 14.4m
At the top of the clay layer (∆σt
′
):
z = 3m (distance from mat foundation to the required point)
∆σt
′
=
10400
(12.4 + 3) × (14.4 + 3)
= 38.8 kN/m2
At the middle of the clay layer (∆σm
′
):
z = 5.5
∆σm
′
=
10400
(12.4 + 5.5) × (14.4 + 5.5)
= 29.2 kN/m2
At the bottom of the clay layer (∆σb
′
):
z = 5.5
∆σb
′
=
10400
(12.4 + 8) × (14.4 + 8)
= 22.75 kN/m2

→ ∆σav
′
=
38.8 + 4 × 29.2 + 22.75
6
= 29.72 kN/m2
σo
′
+ ∆σav
′
= 121.5 + 29.72 = 151.22 < σc
′
= 243 →→
Use the following equation for calculating consolidation settlement:
For (σc
′
≥ σo
′
+ ∆σav
′
):
Sc =
Cs × H
1 + eo
× log (
σo
′
+ ∆σav
′
σo
′
)
Cs = 0.06 , eo = 0.8 , H = 5m →
Sc =
0.06 × 5
1 + 0.8
× log (
151.22
121.5
) = 0.01583 m = 15.83 mm ✓.

9.
For the shown mat foundation, If qall,net=150 kN/m2
, check the adequacy of
the foundation dimensions.
Solution
What is shear walls??
It’s a structural element that used when the number of stories of the building
became more than 7 stories to resist the horizontal forces (earthquake and
wind).
Theses horizontal forces will exert a moments on shear walls, and thereby
shear walls will resist these moment about its strongest axis only.
Also shear walls can resist vertical loads as a column.
Interior
Columns
Edge
Columns
Corner
Columns
SW1 SW2
Dimensions 60cm x 60cm 60cm x 40cm 40cm x 40cm 40cm x 5m 40cm x 4m
Loads 1800 kN 1200 kN 600 kN 1500 kN 1300 kN
Moment 0.0 0.0 0.0 800 KN.m 600 KN.m

The following figure explains how shear walls resist the applied moment
(about its strongest axis):
Now, return to problem above, the procedures of solution will not differ
from the other problem discussed above except in calculating the center of
resultant force (XR , YR) and for this problem will be calculated as
following:
R = ∑ Q = 4 × 600 + 2 × 1800 + 2 × 1200 + 2 × 1500 + 2 × 1300
= 12802 kN
To calculate XR we take summation moments about y-axis:
Note that the moment of SW1 is about y-axis so will be considered here, but
the moment of SW2 is about x-axis and will not be considered here.
The loads of all shear walls will exert moment exactly as columns.
Always take the moments on shear walls in the same direction of the
moments from columns loads because this will exerts the worst case
(maximum eccentricity).
∑ M@y−axis = 0.0 → 2 × 800 + 2 × 1500 × (5 − 0.2)
+2 × 1800 × (5 − 0.2) + 2 × 600 × (10 − 0.2 − 0.2)
+1300 × (10 − 0.2 − 0.2) + 1200 × (10 − 0.2 − 0.2) = 12802XR
→ XR = 5.374 m

To calculate YR we take summation moments about X-axis:
Note that the moment of SW2 is about X-axis so will be considered here, but
the moment of SW1 is about Y-axis and will not be considered here.
The loads of all shear walls will exert moment exactly as columns.
∑ M@X−axis = 0.0 → 2 × 1200 × (3.5 − 0.2)+1800 × (3.5 − 0.2)
+2 × 600 + 2 × 1300 × (8.5 − 0.2) + 1800 × (8.5 − 0.2)
+2 × 600 × (13 − 0.2 − 0.2) + 1500 × (13 − 0.2 − 0.2) = 12802YR
→ YR = 6.686 m
Note:
The loads on the shear wall may be given as distributed load (kN/m) in this
case multiply this load by the length of shear wall to be a concentrated load.
Now you can complete the problem exactly with the same procedures
discussed in previous problems.
10.
Calculate the base pressure at the points indicated below the mat foundation
shown. Total vertical load acting on the foundation is 26000 kN and it’s
location as shown in the figure.

Solution
Any shape you given it the first step is to calculate the centroid of this
shape, but because the shape of the mat shown above is symmetry, the
centroid will be at the middle of horizontal and vertical distance.
The idea in this problem is how to calculate moments of inertia in x and y
directions as following:
Iyy =
103
× 5
12
+
63
× 5
12
+
103
× 5
12
= 923.33 m4
Ixx = Ibox − Itwo rectangles
=
153
× 10
12
−
53
× 2
12
−
53
× 2
12
= 2770.83 m4
Also we can calculate Ixx by using parallel axis theorem:
Ixx =
53
× 6
12
+ 2 × (
53
× 10
12
+ Ad2
)
2: because the upper and the lower rectangles are the same.
A = area of upper and lower rectangles = 10 × 5 = 50
d = distance from the center of the rectangle to x − axis = 5m
→ Ixx =
53
× 6
12
+ 2 × (
53
× 10
12
+ 50 × 52
) = 2770.83
q =
∑ Qi
A
±
My
Iy
X ±
Mx
Ix
Y
But, as shown in the given figure, no eccentricity in y-direction, so the
moment about x-axis equal zero and the equation will be:
q =
∑ Qi
A
±
My
Iy
X
∑ Qi = 26000 Kn
A = mat area = 15 × 10 − 2 × (5 × 2) = 130 m2
My = ∑ Qi × ex
ex = 5.75 − 5 = 0.75 m → My = 26000 × 0.75 = 19500 kN. m
Now, compute the stress at any point you want with sign convention
discussed previously.

Chapter (7)
Lateral Earth
Pressure

Foundation Engineering Lateral Earth Pressure
Introduction
Vertical or near vertical slopes of soil are supported by retaining walls,
cantilever sheet-pile walls, sheet-pile bulkheads, braced cuts, and other
similar structures. The proper design of those structures required estimation
of lateral earth pressure, which is a function of several factors, such as (a)
type and amount of wall movement, (b) shear strength parameters of the
soil, (c) unit weight of the soil, and (d) drainage conditions in the backfill.
The following figures shows a retaining wall of height H. For similar types
of backfill.

As shown in figure above, there are three types of Lateral Earth Pressure
(LEP):
1. At Rest Lateral Earth Pressure:
The wall may be restrained from moving, for example; basement wall is
restrained to move due to slab of the basement and the lateral earth force in
this case can be termed as" Po".
2. Active Lateral Earth Pressure:
In case of the wall is free from its upper edge (retaining wall), the wall may
move away from the soil that is retained with distance " + ∆H " (i.e. the soil
pushes the wall away) this means the soil is active and the force of this
pushing is called active force and termed by " Pa".
3. Passive Lateral Earth Pressure:
For the wall shown above (retaining wall) in the left side there exist a soil
with height less than the soil in the right and as mentioned above the right
soil will pushes the wall away, so the wall will be pushed into the left soil
(i.e. soil compresses the left soil) this means the soil has a passive effect and
the force in this case is called passive force and termed by " PP".
Now, we want to calculate the lateral pressure from water firstly and from
earth (3 cases mentioned above) secondly.
Lateral Pressure from Water
As we learned previously in fluid mechanics course, the pressure of static
fluid at a specific point is the same in all directions “Pascal’s Law”. So if
there exist water in the soil (saturated soil) we must calculate the vertical
stress for soil alone (effective stress) and calculate the vertical pressure for
water alone (because the horizontal pressure of water is the same as vertical
pressure), but for soil, each one –according soil parameters- having different
transformation factor from vertical to horizontal pressure as we will discuss
later. The following figure showing that the horizontal pressure of water
against a wall is the same as vertical pressure:

At Rest Lateral Earth Pressure:
As stated above, the soil in this case is static and can’t pushes the wall with
any movement, the transformation factor of vertical pressure to horizontal
pressure in this case is "Ko" and the lateral earth force is termed by "Po"
Calculation of at rest lateral earth force "𝐏𝐨" for different cases:
Firstly the value of Ko can be calculated as following:
Ko = 1 − sinϕ
Always, (at rest) lateral earth pressure at any depth (z) may be calculated as
following:
σh,o = vertical effective pressure (σv
′ ) × Ko + Pore water pressure (u)
σv
′
= q + γ′
× z , u = γw × z
→ σh,o = σv
′
× Ko + u
Note that the value of (u) doesn’t multiplied by any factor since the
horizontal pressure of water is the same as vertical pressure.

As shown there is no water table.
Firstly we calculate the vertical stress at each depth (each change):
At depth z = 0.0:
σv
′
= q
At depth z = H:
σv
′
= q + γ × H
So, lateral at rest pressure at each depth now can be calculated:
At depth z = 0.0:
σh,o = qKo
At depth z = H:
σh,o = (q + γ × H) × Ko
Now calculate the lateral forces P1 and P2 and then calculate Po:
P1 = Area of rectangle (1) = qKo × H (per unit lenght)
P2 = Area of triangle (2) =
1
2
× (γ × H × Ko) × H =
1
2
× γ × H2
× Ko
Po = P1 + P2
To find the location of Po, take summation moments at point A (for ex.):
P1 ×
H
2
+ P2 ×
H
3
= Po × z̅ →→ z̅ = ✓.
qKo
(q + γH)Ko
Po
P1
P2
z̅
γ, ϕ

In case of water table (if exist):
Firstly we calculate the vertical stress at each depth (each change):
At depth z = 0.0:
σv
′
= q
At depth z = H1:
σv
′
= q + γ × H1 u = 0.0
At depth z = H2:
σv
′
= q + γ × H1 + γ′
H2 (γ′
= γsat − γw)
u = γ 𝑤H2
Now calculate the lateral forces P1, P2, P3, P4 and P5 then calculate Po:
P1 = area of recangle(1) = qKoH1
P2 = area of triangle(2) =
1
2
γH1
2
Ko
P3 = area of recangle(3) = (q + γH1)Ko × H2
1
2
γ′
H2
2
Ko
1
2
γwH2
2
(factor for water = 1)
Po = P1 + P2 + P3 + P4 + P5
To find the location of Po, take summation moments at point A (for ex.):
(𝐍𝐨𝐭𝐞, P5 must be included in moment equation)
(See example 7.1 Page 327)
qKo
(q + γH1)Ko
(q + γH1 + γ′
H2)Ko γ 𝑤H2
γ, ϕ
γsat

Lateral Earth PressureTheories
Two theories are used to calculate lateral earth pressure (active and passive):
Rankine Earth Pressure theory and Coulomb’s Earth Pressure theory.
Firstly we will learn rankine earth pressure theory (the most important) and
then coulomb earth pressure theory.
Rankine Active Lateral Earth Pressure
This theory is based mainly on the assumption of neglecting friction between
the soil and the wall, so no shear forces are developed on soil particles.
As previously introduced, the soil in this case pushes the wall far away.
The transformation factor of vertical pressure to horizontal pressure in this
case is "Ka" and the lateral earth force is termed by "Pa"
Firstly the value of Ka can be calculated as following:
Ka = tan2
(45 −
ϕ
2
)
There are different cases:
In case of granular soil (pure sand):
Exactly as the case of at rest LEP but here the transformation factor is Ka
Pa = P1 + P2
If the soil is 𝐂 − 𝛟 𝐬𝐨𝐢𝐥:
The clay exerts a lateral earth pressure with value of 2c√K (in general).
In case of active earth pressure the value of K is Ka, and when the wall
moves away from soil, the soil particles will disturbed and the cohesion of
soil will decreased, so in case of active earth pressure we subtract the lateral
earth pressure of clay because the cohesion of clay decreased.
qKa
qKa
γHKa (q + γH)Ka
γ , ϕ

The value of 2c√K is constant along the layer, and differ when the value of
C or ϕ change (i.e. constant for each layer)
As we see, at depth z=0.0, the lateral earth pressure is −2c√Ka this negative
value (i.e. the soil will exerts a tensile stress on the wall) and this tensile
stress will causes cracking on the wall from depth z=0.0 to depth = zc.
Calculation of 𝐳 𝐜:
The lateral pressure will be zero at depth 𝐳 𝐜:
γ zcKa − 2c√Ka = 0.0 → zc =
2c√Ka
γKa
=
2c√Ka
γ √Ka × √Ka
=
2c
γ √Ka
But, we know that soil can’t develop any tension, so in design (in practice)
we modify negative pressure to be zero and design for it (more safe because
we enlarge lateral pressure) as shown:
2c√Ka
γHKa
γHKa − 2c√Ka
−2c√Ka
γHKa − 2c√Ka
γ
C, ϕ

If there exist surcharge:
So the final equation for active lateral earth pressure at and depth z can be
calculated as following:
σh,a = (q + γH)Ka − 2c√Ka
Note:
If there exist a water table, calculate the lateral force from water alone and
then added it to the lateral force from soil to get total active force.
Generalized Case for Rankine Active Pressure:
In previous section, we calculate rankine active lateral pressure for the
vertical wall and horizontal backfill, but here we will calculate the lateral
earth pressure for general case (inclined wall and inclined backfill):
This General case for Granular soil only (pure sand)
γ
C, ϕ
qKa γHKa
2c√Ka
𝛼
β
θ θ
𝛼
β
90 − θ
90 − θ
θ

α = inclination of backfill with horizontal
θ = inclination of wall with vertical
β = inclination of Pa with the normal to the wall
From trigonometry, the angle between the normal to the wall and horizontal
is θ.
Calculation of 𝐏𝐚:
Pa = Vertical force × Ka
Vertical force = area of vertical pressure digram =
1
2
γH2
The value of Ka in this case is calculated from the following equation:
Ka =
cos(α − θ) √1 + sin2ϕ − 2 sinϕ cos ψa
cos2θ(cosα + √sin2ϕ − sin2α)
ψa = sin−1
(
sinα
sinϕ
) − α + 2θ
The angle β is:
β = tan−1
(
sinϕ sinψa
1 − sinϕ cosψa
)
Pa =
1
2
γH2
× Ka
The location of Pa is
H
3
from base as shown above
If we need the horizontal and vertical components of Pa:
Pa,h = Pa cos(β + θ)
Pa,v = Pa sin(β + θ)

Case of vertical wall and inclined backfill:
Two cases:
1. For pure sand
2. For 𝐂 − 𝛟 𝐬𝐨𝐢𝐥
For Pure sand:
Here Pa is inclined with angle 𝛼 with horizontal.
Effective vertical pressure = γH
Pa =
1
2
γH2
Ka
Ka in this case is calculated from the following equation:
Ka = cos α
cos α − √cos2α − cos2ϕ
cos α + √cos2α − cos2ϕ
Or, by using (Table 7.1Page 337) easier than the equation above.
𝛼
𝛼

For 𝐂 − 𝛟 𝐬𝐨𝐢𝐥
Here the force Pa is inclined with angle 𝛼 with horizontal, but the
calculation of Pa will differ because there exist clay and there is a tensile
stress from clay at depth Zc as shown:
The value of Zc in this case is calculated as following:
Zc =
2c
γ
√
1 + sinϕ
1 − sinϕ
Calculation of Pa:
Vertical pressure= γH
Horizontal pressure = γHKa
Ka = Ka
′
cosα → Horizontal pressure = γHKa
′
cosα
Pa =
1
2
× (γHKa
′
cosα) × (H − Zc)
Ka
′
can be calculated from (𝐓𝐚𝐛𝐥𝐞 𝟕. 𝟐 𝐏𝐚𝐠𝐞 𝟑𝟑𝟖)
Note
The calculated value of Pa is inclined by angle α with horizontal, so:
𝛼
𝛼

Pa,h = Pa cos(α)
Pa,v = Pa sin(α)
Rankine Passive Lateral Earth Pressure
As previously introduced, the wall in this case pushed into the soil.
The transformation factor of vertical pressure to horizontal pressure in this
case is "KP" and the lateral earth force is termed by "PP"
Firstly the value of KP can be calculated as following:
KP = tan2
(45 +
ϕ
2
)
The only difference between passive and active is in the formula of
calculating K.
If the soil is 𝐂 − 𝛟 𝐬𝐨𝐢𝐥:
In case of passive earth pressure the value of K is KP, and when the wall
moves into the soil, the soil particles will converges and the cohesion of soil
will increased, so in case of passive earth pressure we add the lateral earth
pressure of clay because the cohesion of clay increased.
The value of 2c√K is constant along the layer, and differ when the value of
C or ϕ change (i.e. constant for each layer)
If there exist surcharge, will be added to vertical pressure and the final form
for passive rankine pressure will be:
σh,P = (q + γH)KP + 2c√KP
+2c√KP
2c√KP
γHKP γHKP + 2c√KP

Rankine Passive Pressure (Vertical wall and inclined backfill):
The same as rankine active pressure for this case, the only difference is in
the equation of calculating KP (negative sign in active transformed to
positive sign in passive). (Page 363).
Coulomb’s Lateral Earth Pressure Theory
The main assumption of this theory is considering the friction between the
wall and the soil, this friction angle between the soil and the wall is (δ).
So there exist shear stresses on the soil particles and the equations for
calculating passive lateral earth coefficient will differ from equations of
active lateral earth coefficient.
This theory deal only with granular soil (pure sand).
Coulomb’s Active Lateral Earth Pressure
General case (inclined wall and inclined backfill):
α = inclination of backfill with horizontal
θ = inclination of wall with vertical
β = inclination of wall with the horizontal
δ = friction angle between soil and wall
From trigonometry, the angle between the normal to the wall and horizontal
is θ.
𝛼
δ
θ θ
𝛼
δ
θ
ββ
β

Calculation of 𝐏𝐚:
Pa = Vertical force × Ka
Vertical force = area of vertical pressure digram =
1
2
γH2
The value of Ka in this case is calculated from equation 7.26 Page 342.
The value of δ is less than ϕ since δ is friction angle between wall and soil,
but ϕ is friction angle between soil itself.
In general, the value of δ = (
1
2
→
2
3
) ϕ so there are tables for calculating Ka
for δ =
2
3
ϕ and for δ =
1
2
ϕ
(Table 7.4 Page 343) for δ =
2
3
ϕ →
Kais obtained according the following angles:
α, β and ϕ
(Table 7.5 Page 344) for δ =
1
2
ϕ
Now, we can easily calculate the value of Pa:
Pa =
1
2
γH2
× Ka
Pa,h = Pa cos(δ + θ) Pa,v = Pa sin(δ + θ)
Special case (when the wall is vertical "β = 90" and the backfill is
horizontal "α = 0") .
Pa =
1
2
γH2
× Ka (Inclined by angle δ with horizontal)
Ka can be calculated from( 𝐓𝐚𝐛𝐥𝐞 𝟕. 𝟑 𝐏𝟑𝟒𝟐)according the values of
δ and ϕ, so for this special case:
Pa,h = Pa cos δ Pa,v = Pa sin δ
Coulomb’s Passive Lateral Earth Pressure
The same as coulomb active pressure for this case, the only difference is in
the equation of calculating KP (negative sign in active transformed to
positive sign in passive). (Page 365).

Problems:
For the shown figure below. Plot the pressure diagram and find the resultant
force F and its location under active conditions.
Solution
Note that the value of ϕ is differ for each layer, so the value of Ka will differ
for each layer, thus the first step is to calculate the value of Ka for each layer.
Ka = tan2
(45 −
ϕ
2
)
For ϕ = 32 → Ka = tan2
(45 −
32
2
) = 0.307
Calculate Ka for each layer by the same way.
The values of Ka are written on each layer on the figure above.
ϕ = 32 𝜊
γ = 110 pcf
ϕ = 30 𝜊 γ = 125 pcf
ϕ = 10 𝜊 γ = 126 pcf
c = 600 pcf
ϕ = 0 𝜊
γ = 120 pcf
c = 800 pcf
ϕ = 20 𝜊 γ = 120 pcf
c = 400 pcf
Ka = 0.307
Ka = 0.333
Ka = 0.704
Ka = 1
Ka = 0.49

Now, calculate the lateral earth pressure at each depth (each change) from
soil alone (i.e. vertical effective pressure x Ka), then the water will
considered alone.
Before calculating the pressure, at each change we calculate the lateral earth
pressure just before and just after the layer because the value of Ka is differ
before and after the layer.
The general formula for active lateral earth pressure at any depth is:
σh,a = (q + γH)Ka − 2c√Ka γw = 62.4pcf
q = 2ksf = 2000psf
@𝐳 = 𝟎. 𝟎:
σh,a = (2000 + 0) × 0.307 − 0 = 614 psf
@𝐳 = 𝟔𝐟𝐭:
Just before (Ka = 0.307, c = 0):
σh,a = (2000 + 110 × 6) × 0.307 − 0 = 816.62 psf
Just after (Ka = 0.333 , c = 0):
σh,a = (2000 + 110 × 6) × 0.333 − 0 = 885.8 psf
@𝐳 = 𝟖𝐟𝐭:
Just before (Ka = 0.333, c = 0):
σh,a = (2000 + 110 × 6 + (125 − 62.4) × 2) × 0.333 − 0 = 927.5 psf
Just after (Ka = 0.704 , c = 600):
σh,a = (2000 + 110 × 6 + (125 − 62.4) × 2) × 0.704
−2 × 600 × √0.704 = 953.9 psf
@𝐳 = 𝟏𝟕𝐟𝐭:
Just before (Ka = 0.704, c = 600):
σh,a = (2000 + 110 × 6 + (125 − 62.4) × 2 + (126 − 62.4) × 9) × 0.704
−2 × 600 × √0.704 = 1356.9 psf
Just after (Ka = 1 , c = 800):
σh,a = (2000 + 110 × 6 + (125 − 62.4) × 2 + (126 − 62.4) × 9) × 1
−2 × 800 × √1 = 1757.6 psf

@𝐳 = 𝟐𝟓𝐟𝐭:
Just before (Ka = 1, c = 800):
σh,a = (
2000 + 110 × 6 + (125 − 62.4) × 2 + (126 − 62.4) × 9
+(120 − 62.4) × 8
) × 1
−2 × 800 × √1 = 2218.4 psf
Just before (Ka = 0.49, c = 400):
σh,a = (
2000 + 110 × 6 + (125 − 62.4) × 2 + (126 − 62.4) × 9
+(120 − 62.4) × 8
) × 0.49
−2 × 400 × √0.49 = 1311 psf
@𝐳 = 𝟑𝟎𝐟𝐭:
σh,a = (
2000 + 110 × 6 + (125 − 62.4) × 2 + (126 − 62.4) × 9
+(120 − 62.4) × 8 + (120 − 62.4) × 5
) × 0.49
−2 × 400 × √0.49 = 1452 psf
Now we calculate the pore water pressure:
Water starts at depth 6ft to the end (i.e. depth of water is 24 ft)
u = 62.4 × 24 = 1497.6 psf
Now we draw the pressure diagram as following:
.

Now calculate the force for each shape (1 to 11) i.e. area of each shape and
then sum all of these forces to get total active lateral force Pa.
Pa ≅ 57214 Ib/ft′
✓.
To calculate the location of Pa, take summation moments at point A
(Include the moment from water force “don’t forget it”)
The location of Pa is ≅ 10.7 ft (above point A) ✓.

Foundation Engineering Retaining Walls
Introduction
Retaining walls must be designed for lateral earth pressure. The procedures
of calculating lateral earth pressure was discussed previously in Chapter7.
Different types of retaining walls are used to retain soil in different places.
Three main types of retaining walls:
1. Gravity retaining wall (depends on its weight for resisting lateral earth
force because it have a large weigh)
2. Semi-Gravity retaining wall (reduce the dimensions of the gravity
retaining wall by using some reinforcement).
3. Cantilever retaining wall (reinforced concrete wall with small dimensions
and it is the most economical type and the most common)
Note:
Structural design of cantilever retaining wall is depend on separating each
part of wall and design it as a cantilever, so it’s called cantilever R.W.
The following figure shows theses different types of retaining walls:
There are another type of retaining wall called “counterfort RW” and is a
special type of cantilever RW used when the height of RW became larger
than 6m, the moment applied on the wall will be large so we use spaced
counterforts every a specified distance to reduce the moment RW.

Where we use Retaining Walls
Retaining walls are used in many places, such as retaining a soil of high
elevation (if we want to construct a building in lowest elevation) or retaining
a soil to save a highways from soil collapse and for several applications.
The following figure explain the function of retaining walls:

Elements of Retaining Walls
Each retaining wall divided into three parts; stem, heel, and toe as shown for
the following cantilever footing (as example):
Application of Lateral Earth Pressure Theories to
Design
Rankine Theory:
Rankine theory discussed in Ch.7 was modified to be suitable for designing
a retaining walls.
This modification is drawing a vertical line from the lowest-right corner till
intersection with the line of backfill, and then considering the force of soil
acting on this vertical line.
The soil between the wall and vertical line is not considered in the value of
Pa, so we take this soil in consideration as a vertical weight applied on the
stem of the retaining wall as will explained later.

The following are all cases of rankine theory in designing a
retaining wall:
1. The wall is vertical and backfill is horizontal:
Here the active force Pa is horizontal and can be calculated as following:
Pa =
1
2
γH2
Ka , Ka = tan2
(45 −
ϕ
2
)
2. The wall is vertical and the backfill is inclined with horizontal by
angle (𝛂):
γ, ϕ
γ, ϕ
α
α

Here the active force Pa is inclined with angle (α) and can be calculated as
following:
Pa =
1
2
γH′2
Ka
Why H′
? → Because the pressure is applied on the vertical line (according
active theory) not on the wall, so we need the height of this vertical line H′
H′
= H + d →→ d = L tanα
Ka is calculating from (𝐓𝐚𝐛𝐥𝐞 𝟕. 𝟏 𝐏𝐚𝐠𝐞 𝟑𝟑𝟕)
Now the calculated value of Pa is inclined with an angle (α), so its analyzed
in horizontal and vertical axes and then we use the horizontal and vertical
components in design as will explained later.
Pa,h = Pa cos(α) , Pa,v = Pa sin(α)
3. The wall is inclined by angle (𝛉)with vertical and the backfill is
inclined with horizontal by angle (𝛂):
Note that the force Pa is inclined with angle (α) and not depend on the
inclination of the wall because the force applied on the vertical line
and can be calculated as following:
Pa =
1
2
γH′2
Ka
α
α
θ

What about Ka ? ? ?
Ka is depend on the inclination of the wall and inclination of the backfill
because it’s related to the soil itself and the angle of contact surface with
this soil, so Ka can be calculated from the following equation (Ch.7):
Ka =
cos(α − θ) √1 + sin2ϕ − 2 sinϕ cos ψa
cos2θ(cosα + √sin2ϕ − sin2α)
ψa = sin−1
(
sinα
sinϕ
) − α + 2θ
Pa,h = Pa cos(α) , Pa,v = Pa sin(α)
Coulomb’s Theory:
Coulomb’s theory discussed in Ch.7 will remains unchanged (without any
modifications) in this chapter. The force Pa is applied directly on the wall, so
whole soil retained by the wall will be considered in Pa and thereby the
weight of soil will not apply on the heel of the wall.
θ
β
θ
δ
α

Pa =
1
2
γH2
Ka
Why H??? Here the force Pa is applied directly on the wall, so the lateral
pressure of the soil is applied on the wall from start to end, so we only take
the height of the wall (in coulomb theory).
Ka is calculated from (Table 7.4 and 7.5 Page 343) according the following
angles:
ϕ, α, β and δ
As shown, the force Pa is inclined with angle (δ + θ) with horizontal, so:
Pa,h = Pa cos(δ + θ) , Pa,v = Pa sin(δ + θ)
What about Passive Force
You can always calculate passive force from rankine theory even if its
require to solve the problem based on coulomb’s theory, because we
concerned about rankine and coulomb’s theories in active lateral pressure.
Important Note:
Coulomb’s theory can’t be used in the following cases:
1. If the soil retained by the wall is C − ϕ soil, because coulomb’s theory
deal only with granular soil (pure sand).
2. If wall friction angle between retained soil and the wall is equal zero.
3. If we asked to solve the problem using rankine theory .
Stability of Retaining Wall
A retaining wall may be fail in any of the following:
1. It may overturn about its toe.
2. It may slide along its base.
3. It may fail due to the loss of bearing capacity of the soil supporting the
base.
4. It may undergo deep-seated shear failure.
5. It may go through excessive settlement.
We will discuss the stability of retaining wall for the first three types of
failure (overturning, sliding and bearing capacity failures).

We will use rankine theory to discusses the stability of these types of
failures. Coulomb’s theory will be the same with only difference mentioned
above (active force applied directly on the wall).
Stability for Overturning
The horizontal component of active force will causes overturning on
retaining wall about point O by moment called “overturning moment”
MOT = Pa,h ×
H
3
This overturning moment will resisted by all vertical forces applied on the
base of retaining wall:
1. Vertical component of active force Pa,v(if exist).
2. Weight of all soil above the heel of the retaining wall.
3. Weight of each element of retaining wall.
4. Passive force (we neglect it in this check for more safety).
Now, to calculate the moment from these all forces (resisting moment) we
prepare the following table:
Force = Volume × unit weight but, we take a strip of 1 mlength
→ Force = Area × unit weight

Section Area
Weight/unit length
of the wall
Moment arm
measured from O
Moment
about O
1 A1 W1=A2 × γ1 X1 M1
2 A2 W2 = A2 × γc X2 M2
3 A3 W3 = A2 × γc X3 M3
4 A4 W4 = A2 × γc X4 M4
Pa,v(if exist). B MV
∑ ∑ V ∑ M = MR
γ1 = unit weight of the soil above the heel of RW
FSOT =
MR
MOT
≥ 2
Note:
If you asked to consider passive force→ consider it in the resisting moment
and the factor of safety remains 2. (So we neglect it here for safety).
Stability for Sliding along the Base

Also, the horizontal component of active force may causes movement of the
wall in horizontal direction (i.e. causes sliding for the wall), this force is
called driving force Fd = Pa,h.
This driving force will be resisted by the following forces:
1. Adhesion between the soil (under the base) and the base of retaining wall:
ca = adhesion along the base of RW (KN/m)
Ca = ca × B = adhesion force under the base of RW (KN)
ca can be calculated from the following relation:
ca = K2c2 c2 = cohesion of soil under the base
So adhesion force is:
Ca = K2c2B
2. Friction force due to the friction between the soil and the base of RW:
Always friction force is calculated from the following relation:
Ffr = μsN
Here N is the sum of vertical forces calculated in the table of the first check
(overturning)
→ N = ∑ V (including the vertical component of active force)
μs = coefficient of friction (related to the friction between soil and base)
μs = tan(𝛿2) δ2 = K1ϕ2 →→ μs = tan(K1ϕ2)
ϕ2 = friction angle of the soil under the base.
→ Ffr = ∑ V × tan(K1ϕ2)
Note:
K1 = K2 = (
1
2
→
2
3
) if you are not given them → take K1 = K2 =
2
3
3. Passive force PP.(Calculated using rankine theory).
So the total resisting force FR can be calculated as following:
FR = ∑ V × tan(K1ϕ2) + K2c2B + PP
Factor of safety against sliding:
FSS =
FR
Fd
≥ 2 (if we consider PP in FR)
FSS =
FR
Fd
≥ 1.5 (if we dont consider PP in FR)

Check Stability for Bearing Capacity Failure
As we see, the resultant force (R) is not applied on the center of the base of
retaining wall, so there is an eccentricity between the location of resultant
force and the center of the base, this eccentricity may be calculated as
following:
From the figure above, take summation moments about point O:
MO = ∑ V × X̅
From the first check (overturning) we calculate the overturning moment and
resisting moment about point O, so the difference between these two
moments gives the net moment at O.
MO = MR − MOT
→ MR − MOT = ∑ V × X̅ →→ X̅ =
MR − MOT
∑ V
e =
B
2
− X̅ = ✓ (see the above figure).
Since there exist eccentricity, the pressure under the base of retaining wall is
not uniform (there exist maximum and minimum values for pressure).
𝐗̅

We calculate qmax and qmin as stated in chapter 3:
Eccentricity in B-direction and retaining wall can be considered strip footing
If 𝐞 <
𝐁
𝟔
qmax =
∑ V
B × 1
(1 +
6e
B
)
qmin =
∑ V
B × 1
(1 −
6e
B
)
If 𝐞 >
𝐁
𝟔
qmax,new =
4 ∑ V
3 × 1 × (B − 2e)
Now, we must check for qmax:
qmax ≤ qall → qmax = qall (at critical case)
FSB.C =
qu
qmax
≥ 3
Calculation of 𝐪 𝐮:
quis calculated using Meyerhof equation as following:

Where
c = Cohesion of soil under the base
q = Effective stress at the level of the base of retaining wall.
q = γ2 × Df
Df here is the depth of soil above the toe = D (above figure)
→ q = γ2 × D
γ = unit weight of the soil under the base of the RW.
Important Note:
May be a water table under the base or at the base or above the base (three
cases discussed in chapter 3) is the same here, so be careful don’t forget
Ch.3.
B = B′
= B − 2e
Nc, Nq, Nγ = Myerhof bearing capacity factors (Table3.3)according
the friction angle for the soil under the base
Fcs = Fqs = Fγs = 1 (since RW is considered a strip footing)
Depth factors: (We use B not B′
)
Here since the depth D is restively small to width of the base B, in most
cases
𝐃
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

Note that the resultant force applied on the base of the foundation is not
vertical, but it is inclined with angle β = Ψ (with vertical), this angle can
be calculated as following:
β = Ψ = tan−1
(
Pa,h
∑ V
)
Fci = Fqi = (1 −
β°
90
)
2
Fγi = (1 −
β°
ϕ°
)
Problems:
1.
The cross section of the cantilever retaining wall shown below. Calculate the
factor of safety with respect to overturning, sliding, and bearing capacity.
γ1 = 18 kN/m3
ϕ1 = 30°
C1 = 0.0
γ2 = 19 kN/m3
ϕ2 = 24°
C2 = 40kN/m2
γc = 24 kN/m3

Solution
Since it is not specified a method for solving the problem, directly we use
Rankine theory.
Now draw a vertical line starts from the right-down corner till reaching the
backfill line and then calculate active force (Pa):
tan 10 =
d
2.6
→ d = 2.6 × tan 10 = 0.458m
H′
= 6.7 + d = 6.7 + 0.458 = 7.158m
Now we calculate Pa:
Pa =
1
2
× γ1 × H′2
× Ka
Since the backfill is inclined and the wall is vertical, Ka is calculated from
Table 7.1 according the values of α = 10 and ϕ1 = 30:
Ka = 0.3495
→ Pa =
1
2
× 18 × 7.1582
× 0.3495 = 161.2 kN
Location of Pa:
Location=
H′
3
=
7.158
3
= 2.38
The force Pa is inclined with angle α = 10 with horizontal:
Pa,h = 161.2 cos(10) = 158.75 , Pa,v = 161.2 sin(10) = 28

Check for Overturning:
MOT = 158.75 × 2.38 = 337.8 KN. m
Now to calculate MR we divided the soil and the concrete into rectangles and
triangles to find the area easily (as shown above) and to find the arm from
the center of each area to point O as prepared in the following table:
Section Area
Weight/unit length
of the wall
Moment arm
measured from O
Moment about
O
1 0.595 0.595 × 18 = 10.71 4 −
2.6
3
= 3.13 33.52
2 15.6 15.6 × 18 = 280.8 1.4 + 1.3 = 2.7 758.16
3 3 3 × 24 = 72 1.4 − 0.25 = 1.15 82.8
4 0.6 0.6 × 24 = 14.4 0.9 −
0.2
3
= 0.833 12
5 2.8 2.8 × 24 = 67.2
4
2
= 2 134.4
Pa,v = 28 B=4 112
∑ ∑ V = 470.11 MR = 1132.88
Note that we neglect passive force because it is not obligatory.

FSOT =
MR
MOT
=
1132.88
377.8
= 2.99 > 2 → 𝐎𝐊 ✓.
Check for Sliding:
FSS =
FR
Fd
It is preferable to consider passive force in this check.
Applying rankine theory on the soil in the left (draw vertical line till
reaching the soil surface).
kP is calculated for the soil using rankine theory without considering
any iniclination of the wall, because it is calculated for the soil below the bas
kP = tan2
(45 +
ϕ2
2
) = tan2
(45 +
20
2
) = 2.04
P1 = (rectangle area) = (2 × 40 × √2.04) × 1.5 = 171.4 kN
P2 = (triangle area) =
1
2
× (19 × 1.5 × 2.04) × 1.5 = 43.6 kN
PP = P1 + P2 = 171.4 + 43.6 = 215 kN
Fd = Pa,h = 158.75 Kn
FR = ∑ V × tan(K1ϕ2) + K2c2B + PP
Take K2 = K2 = 2/3 ∑ V = 470.11 (from table of first check)
2C2√kP
γ2 × 1.5 × kP

FR = 470.11 × tan (
2
3
× 20) +
2
3
× 40 × 4 + 215 = 433.1 kN
→ FSS =
433.1
158.75
= 2.72 > 2 → 𝐎𝐊 ✓.
Check for Bearing Capacity Failure:
As stated previously, X̅ can be calculated as following:
X̅ =
MR − MOT
∑ V
=
1132.88 − 377.8
470.11
= 1.6 m
e =
B
2
− X̅ = 2 − 1.6 = 0.4m
B
6
=
4
6
= 0.667 → e = 0.4 <
B
6
→→→
qmax =
∑ V
B × 1
(1 +
6e
B
) =
470.11
4 × 1
(1 +
6 × 0.4
4
) = 188.04 kN/m2
qmin =
∑ V
B × 1
(1 −
6e
B
) =
470.11
4 × 1
(1 −
6 × 0.4
4
) = 47 kN/m2
𝐗̅

Calculation of 𝐪 𝐮 (𝐟𝐨𝐫 𝐭𝐡𝐞 𝐬𝐨𝐢𝐥 𝐛𝐞𝐥𝐨𝐰 𝐭𝐡𝐞 𝐛𝐚𝐬):
c = 40 , q = 1.5 × 19 = 28.5 , γ = 19
B = B′
= B − 2e = 4 − 2(0.4) = 3.2m
Shape factors = 1 (RW can be considered strip footing).
For ϕ = 20 → Nc = 14.83 , Nq = 6.4 , Nγ = 5.39 (from table 3.3)
)
D
B
=
1.5
4
= 0.375 < 1 and ϕ = 20 > 0.0 →→
Fqd = 1 + 2 tanϕ (1 − sinϕ)2
(
Df
B
)
= 1 + 2 tan20 (1 − sin20)2 (0.375) = 1.12
Fcd = Fqd −
1 − Fqd
Nctanϕ
= 1.12 −
1 − 1.12
14.83 × tan20
= 1.14
Fγd = 1
β = Ψ = tan−1
(
Pa,h
∑ V
) = tan−1
(
158.75
470.11
) = 18.6 (with vertical)
Fci = Fqi = (1 −
β°
90
)
2
(1 −
18.6
90
)
2
= 0.63
Fγi = (1 −
β°
ϕ°
) = (1 −
18.6
20
) = 0.07
→ qu = 40 × 14.83 × 1.14 × 0.63 + 28.5 × 6.4 × 1.12 × 0.63
+0.5 × 3.2 × 19 × 5.39 × 1 × 0.07
→ qu = 566.2 kN/m2
FSB.C =
qu
qmax
=
566.2
188.04
= 3.01 > 3 (slightly satisfied)𝐎𝐊 ✓.

2. (Example 8.2)
A gravity retaining wall is shown in figure below. Use δ =
2
3
ϕ1 and
coulomb's active earth pressure theory.
(See the solution from textbook) with the following notes:
γ1 = 18.5 kN/m3
ϕ1 = 32 𝑜
kN/m3
c1 = 0.0
γ2 = 18 kN/m3
ϕ2 = 24 𝑜
kN/m3
c2 = 30 kN/m2
δ

1. As seen the force Pa is applied directly on the wall.
2. The force Pa is inclined with angle δ with the normal to the wall and
inclined with angle (δ + 15) with horizontal.
3. The distance 2.83 (arm of vertical component of Pa from toe corner) is not
given and can be calculated as following:
The location of force Pa is
H
3
=
6.5
3
= 2.167
2.167 − 0.8 = 1.367
tan(75) =
1.367
X
→ X =
1.367
tan(75)
→ X = 0.366 m
2.83 = (3.5 − 0.3) − 0.366
4.
Note that when we want to calculate passive force (in overturning pressure)
we use rankine theory for the following two reasons:
 The soil below the base is C − ϕ soil, so we can’t use coulomb’s theory
because it deals only with granular soil.
 It is required to use coulomb’s theory in calculating of active force,
however in calculating passive force we can always use rankine theory.
5. In calculating summation of vertical forces, the weighs if soil above the
heel are not taken in consideration because the force is applied directly on
the wall.
6. Always when calculating Pa the height of the wall (H) is always taken
even if the backfill is inclined because the force applied directly on the wall.
7. (After these notes, solve the problem by yourself ).

3.
For the retaining wall shown below.
a. Find the lateral earth pressure distribution.
b. Compute Pa (Rankine).
c. Calculate Overturning stability.
d. Compute Sliding safety factor.
e. Locate the resultant on the base of the footing and determine the
eccentricity.
f. Calculate the factor of safety against bearing capacity failure.
Solution
We use rankine theory because there is no friction between the soil and the
wall.
a.
Calculation of active lateral earth pressure distribution:
σh,a = (q + γH)Ka − 2c√Ka Ka = tan2
(45 −
ϕ
2
)
γ = 115 pcf
C = 400 psf
ϕ = 28o
γ = 130 pcf
C = 750 psf
ϕ = 35o
𝛄 𝐜 = 𝟏𝟓𝟎 𝐩𝐜𝐟

Ka = tan2
(45 −
28
2
) = 0.361 (for the soil retained by the wall)
@z = 0.0 (right side)
σh,a = (500 + 0) × 0.361 − 2 × 400 × √0.361 = −300.2 psf
@z = H = 22ft (right side)
σh,a = (500 + 115 × 22) × 0.361 − 2 × 400 × √0.361 = 613.2 psf
Since the pressure at the top is negative, so there are some depth causing
cracking on the wall and may be calculated as following:
(500 + 115 × Zc) × 0.361 − 2 × 400 × √0.361 = 0.0 → Zc = 7.23 ft
Calculation of passive lateral earth pressure distribution:
σh,P = (q + γH)KP + 2c√KP KP = tan2
(45 +
ϕ
2
)
KP = tan2
(45 +
35
2
) = 3.69 (for the soil below the wall "left soil")
@z = 0.0 (left side)
σh,P = (0 + 0) × 3.69 + 2 × 750 × √3.69 = 2881.4 psf
@z = 4ft (left side)
σh,P = (0 + 130 × 4) × 3.69 + 2 × 750 × √3.69 = 4800.2 psf
Now the LEP distribution on the wall is as following:

Now, for design purposes, we modify this pressure to be zero at the top of
the wall (for more safety), so the LEP distribution after modification is:
b.
We calculate the active rankine force using the modified LEP because we
use it in designing process.
Pa = (area of modified triangle) =
1
2
× 613.2 × 22 = 6745.2 Ib.
Location of Pa is at
22
3
= 7.33 ft (from the base)
If we want to calculate passive force (for sliding check):
PP = (area of trapezoidal) =
1
2
× (2881.4 + 4800.2) × 4 = 15363.2 Ib.
c. Overturning Stability:

Note that there is no vertical component of active force.
MOT = 6745.2 × 7.33 = 49442.3 Ib. ft.
Section Area
Weight/unit length of
the wall
Moment arm
measured from O
Moment about
O
1 120 120 × 115 = 13800 12 − 3 = 9 124200
2 20 20 × 150 = 3000 6 − 0.5 = 5.5 16500
3 10 10 × 150 = 1500 5 −
1
3
= 4.67 7005
4 24 24 × 150 = 3600
12
2
= 6 21600
∑ ∑ V = 21900 MR = 169305
FSOT =
MR
MOT
=
169305
49442.3
= 3.42 > 2 → 𝐎𝐊 ✓.
The most important note here, the surcharge (q) is considered only when we
calculate the pressure and it is not develop any moment about O because it is
not applied force on the wall, and doesn’t considered in vertical forces
because we considered it in pressure calculation.
If the overturning check is not satisfied, what modification you shall do
to make the foundation stable against overturning?? (Important)
In this case, we want to increase the resisting moment (MR) by increasing
the vertical forces and these arms about O.
This can be satisfied by increasing the width of the footing (increase the
width of heel) to increase the weight of soil above the base and the weight of
the base itself as following:

MOT = 6745.2 × 7.33 = 49442.3 Ib. ft.
The table of calculating MRis:
Sectio
n
Area
Weight/unit length of the
wall
Moment arm
measured
from O
Moment about O
1 20B′ 20B′ × 115 = 2300B′ 0.5B′
+ 6 1150B′2
+ 13800B′
2 20 20 × 150 = 3000
6 − 0.5
= 5.5
16500
3 10 10 × 150 = 1500 5 −
1
3
= 4.67 7005
4 12 12 × 150 = 1800
6
2
= 3 5400
5 2B′ 2B′
× 150 = 300B′
0.5B′
+ 6 150B′2
+ 1800B′
∑ ∑ V = 2600B′
+ 6300 MR
MR = 1300B′2
+ 15600B′
+ 28905
Now, put FSOT = 2 (critical case) to calculate B′
2 =
MR
MOT
=
1300B′2
+ 15600B′
+ 28905
49442.3

1300B′2
+ 15600B′
− 69979.6 = 0.0
Now, in this problem if we calculate B′
here, it will be less than 6ft, because
calculated the FS is 3.42>2 (as calculated above) and here we put it 2.
So, if the FS is not satisfied (<2) do the above procedures and calculate the
new value of B′
and then:
The final footing width is: B = 6 + B′
(must be larger than original value of
B).
d. Check for sliding:
FSS =
FR
Fd
Fd = Pa,h = 6745.2 Ib
FR = ∑ V × tan(K1ϕ2) + K2c2B + PP
Take K2 = K2 = 2/3 ∑ V = 21900 (from table of first check)
FR = 21900 × tan (
2
3
× 35) +
2
3
× 750 × 12 + 15363.2 = 30809.94 Ib.
→ FSS =
30809.94
6745.2
= 4.57 > 2 → 𝐎𝐊 ✓.
If the Sliding stability not satisfied, what modifications you shall do:
Solution (1):
Increase the base width of the footing (width of the heel) to increase vertical
forces:
FR = ∑ V × tan(K1ϕ2) + K2c2B + PP
∑ V = 2600B′
+ 6300 (as calculted above in terms of B′
)
FSS =
FR
Fd
= 2 (at critical case)
Now the value of B′
can be calculated and then calculate the new width of
the footing.
Solution (2):
Use a base key (beam) of depth D under the base of the wall, this base key
increase the passive force as following:

As we see, this base key increase the passive force and thereby increase the
value of FR and factor of safety.
@z = 4 + D
σh,P = (0 + 130 × (4 + D)) × 3.69 + 2 × 750 × √3.69 = 4800.2 psf
PP =
1
2
× (2881.4 + 4800.2 + 479.7 D) × (4 + D)
You may ask the following tow question (in this case):
1. If a base key of depth 1.5m is constructed under the base of the
foundation, calculate the factor of safety against sliding.
PP =
1
2
× (2881.4 + 4800.2 + 479.7 × 1.5) × (4 + 1.5) = 23103.16 Ib.
FR = 21900 × tan (
2
3
× 35) +
2
3
× 750 × 12 + 23103.16 = 38549.94 Ib
→ FSS =
38549.94
6745.2
= 5.7
Note that the FS increase when we use base key.
2. If the sliding stability is not satisfied, find the depth of base key located
under the base key to make the wall sable against sliding.
Here, the passive force is a function of D (base key depth), so calculate the
value of FR in terms of D, and then:

FSS =
FR
6745.2
= 2 → FR = ✓ → D = ✓.
e.
X̅ =
MR − MOT
∑ V
=
169305 − 49442.3
21900
= 5.47ft
e =
B
2
− X̅ = 6 − 5.47 = 0.53 ft.
B
6
=
12
6
= 2 → e = 0.53 <
B
6
→→→
qmax =
∑ V
B × 1
(1 +
6e
B
) =
21900
12 × 1
(1 +
6 × 0.53
12
) = 2308.6 psf
qmin =
∑ V
B × 1
(1 −
6e
B
) =
21900
12 × 1
(1 −
6 × 0.53
12
) = 1341.4 psf
c = 750 , q = 4 × 130 = 520 , γ = 130
B = B′
= B − 2e = 12 − 2(0.53) = 10.94 ft
X̅

For ϕ = 35 → Nc = 46.12 , Nq = 33.3 , Nγ = 48.03 (from table 3.3)
)
D
B
=
4
12
= 0.333 < 1 and ϕ = 35 > 0.0 →→
Fqd = 1 + 2 tanϕ (1 − sinϕ)2
(
Df
B
)
= 1 + 2 tan35 (1 − sin35)2 (0.333) = 1.084
Fcd = Fqd −
1 − Fqd
Nctanϕ
= 1.084 −
1 − 1.084
46.12 × tan35
= 1.086
Fγd = 1
β = Ψ = tan−1
(
Pa,h
∑ V
) = tan−1
(
6745.2
21900
) = 17.12 (with vertical)
Fci = Fqi = (1 −
β°
90
)
2
(1 −
17.12
90
)
2
= 0.65
Fγi = (1 −
β°
ϕ°
) = (1 −
17.12
35
) = 0.51
→ qu = 750 × 46.12 × 1.086 × 0.65 + 520 × 33.3 × 1.084 × 0.65
+0.5 × 10.94 × 130 × 48.03 × 1 × 0.51
→ qu = 54036.54
FSB.C =
qu
qmax
=
54036.54
2308.6
= 23.4 > 3 𝐎𝐊 ✓.

4.
A gravity retaining wall shown in the figure below is required to retain 5 m
of soil. The backfill is a coarse grained soil with saturated unit weight =18
kN/m3
, and friction angle of ϕ = 30o
. The existing soil below the base has
the following properties; γsat = 20 kN/m3
, ϕ = 36o
. The wall is
embedded 1m into the existing soil, and a drainage system is provided as
shown. The ground water table is at 4.5m below the base of the wall.
Determine the stability of the wall for the following conditions (assume
K1=K2 = 2/3):
a- Wall friction angle is zero.
b-Wall friction angle is 20o
.
c- The drainage system becomes clogged during several days of rainstorm
and the ground water rises to the surface of backfill (use rankine).
γconcrete = 24 kN/m3
Solution
a- (wall friction angle = δ = 0.0)
Since δ = 0.0 (we use rankine theory).

(The unit weight of the soil (natural) is not given, so we consider the
saturated unit weight is the natural unit weight).
Ka = tan2
(45 −
ϕ
2
) = tan2
(45 −
30
2
) = 0.333 (for the retained soil)
KP = tan2
(45 +
ϕ
2
) = KP = tan2
(45 +
36
2
) = 3.85 (for soil below the base)
Calculation of active lateral earth pressure distribution:
@z = H = 5m (right side)
σh,a = (0 + 18 × 5) × 0.333 − 0 = 29.97 kN/m2
Calculation of passive lateral earth pressure distribution:
σh,P = (q + γH)KP + 2c√KP
@z = 1m(left side)
σh,P = (0 + 20 × 1) × 3.85 + 0 = 77 kN/m2
Calculation of active force:
Pa = (area of right triangle) =
1
2
× 29.97 × 5 = 74.9 kN
Calculation of passive force:
PP = (area of left triangle) =
1
2
× 77 × 1 = 38.5 kN

Overturning Stability:
MOT = 74.9 × 1.67 = 125.08 kN. m
Note that since there is no heel for the wall, the force is applied directly on
the wall.
Section Area
Weight/unit length of
the wall
Moment arm
measured from O
Moment about
O
1 3 3 × 24 = 72 3.9 280.8
2 9 9 × 24 = 216 2.4 518.4
∑ ∑ V = 288 MR = 799.2
Note that there is no vertical component of active force
FSOT =
MR
MOT
=
799.2
125.08
= 6.39 > 2 → 𝐎𝐊 ✓.

Sliding Stability:
FSS =
FR
Fd
Fd = Pa,h = 74.9 kN/m2
FR = ∑ V × tan(K1ϕ2) + K2c2B + PP
Take K2 = K2 = 2/3 ∑ V = 288 (from table of first check)
PP = 38.5 kN/m2
(as calculated above)
FR = 288 × tan (
2
3
× 36) +
2
3
× 0 × 4.2 + 38.4 = 166.62 kN.
→ FSS =
166.62
74.9
= 2.2 > 2 → 𝐎𝐊 ✓.
Bearing capacity check:
X̅ =
MR − MOT
∑ V
=
799.2 − 125.08
288
= 2.34 m
e =
B
2
− X̅ =
4.2
2
− 2.34 = −0.24 m (R is at right of base center)
X̅

B
6
=
4.2
6
= 0.7 → e = 0.24 <
B
6
→→→
qmax =
∑ V
B × 1
(1 +
6e
B
) =
288
4.2 × 1
(1 +
6 × 0.24
4.2
) = 92.08 kN/m2
qmin =
∑ V
B × 1
(1 −
6e
B
) =
288
4.2 × 1
(1 −
6 × 0.24
4.2
) = 45.06 kN/m2
c = 0.0 , q = 1 × 20 = 20
Water table is at distance 4.5m > B=4.2m >>> no effect of water table.
→ γ = 20
B = B′
= B − 2e = 4.2 − 2(0.24) = 3.72m
For ϕ = 36 → Nc = 50.59, Nq = 37.75 , Nγ = 56.31 (from table 3.3)
)
D
B
=
1
4.2
= 0.238 < 1 and ϕ = 36 > 0.0 →→
Fqd = 1 + 2 tanϕ (1 − sinϕ)2
(
Df
B
)
= 1 + 2 tan36 (1 − sin36)2 (0.238) = 1.058
Fcd = Fqd −
1 − Fqd
Nctanϕ
= 1.058 −
1 − 1.058
50.59 × tan36
= 1.06
Fγd = 1
β = Ψ = tan−1
(
Pa,h
∑ V
) = tan−1
(
74.9
288
) = 14.6 °
(with vertical)
Fci = Fqi = (1 −
β°
90
)
2
(1 −
14.6
90
)
2
= 0.7
Fγi = (1 −
β°
ϕ°
) = (1 −
17.12
36
) = 0.59
→ qu = 0.0 + 20 × 37.75 × 1.058 × 0.7
+0.5 × 3.72 × 20 × 56.31 × 1 × 0.59

→ qu = 1795 kN/m2
FSB.C =
qu
qmax
=
1795
92.08
= 19.5 > 3 𝐎𝐊 ✓.
b-(wall friction angle = δ = 20)
Since δ = 20 (we use coulomb′
s theory for active pressure ).
Here, the active force is not horizontal, but it is inclined by angle δ = 20
with horizontal:
Pa=
1
2
× γ × H2
× ka
δ = 20 , ϕ = 30 → δ =
2
3
ϕ →→
ka in this case is calculated from (Table 7.4 P.343)
β = 90 , α = 0 , ϕ = 30 →→ ka = 0.2973
Pa =
1
2
× 18 × 52
× 0.2973 = 66.9 kN
This force having horizontal and vertical components:
Pa,h = 66.9 cos(20) = 62.86 KN
Pa,v = 66.9 sin(20) = 22.88 kN
Calculation of passive force always done by rankine theory (i.e. passive
force doesn’t change from first required)
PP = 38.5 kN (as calculate in first required)
Now, you can complete the solution without any problems  .
δ = 20°

c- When the ground water rises to the surface, the retaining wall is shown
below:
What differ???
If we want to use rankine theory (force from soil is gorizontal):
1. Calculation of active force:
Don’t forget we calculate effective stress every change, and the we add
water alone.
P1 = (force due to effective soil) =
1
2
× 13.32 × 5 = 33.3 kN
P2 = (force due to water) =
1
2
× 50 × 5 = 125 KN

Pa,h = P1 + P2 = 33.3 + 125 = 158.33 KN
Loacation of Pa,h:
Take the moment at the bottom of the wall to get the location, but here the
two forces have the same location, so the resultant of the two forces will
have the same location (1.67 from base).
2. Calculation of passive force:
PP,h = P1 + P2
3. In calculation of vertical forces due to the soil weight always take the
effective unit weight and multiply it by the area to get the effective force
but this is not required in this problem because the force applied
directly on the wall.
4. In calculating of bearing capacity for the soil below the base, since the
water table is above the base (case 3) we take γ = γ′
in Meyerhof Eq.
Now you can complete the solution with the same procedures without any
problem .
Now, If the water table is at distance 2m below the surface, what’s new???

Calculation of Active force:
Here we calculate the effective stress every change, and then added the
water alone from its beginning:
Pa,h = P1 + P2 + P3+P4
To find the location of Pa,h take summation moment at the base of the wall.
Calculation of passive force will not change
The weight of soil above heel (when heel exist), we divide the soil above
the heel for two areas, soil above water table and soil below water table.
The area of soil above water table is multiplied by natural unit weight,
and the area of soil below water table is multiplied by effective unit
weight.
In calculating of bearing capacity, the water is still above the base, so we
use effective unit weight in Meyerhof Eq.

The last idea in this chapter:
If you are asked to solve this problem (in case of water table) using
Coulomb’s theory:
As stated above, the force here will be inclined by angle δ = 20 with
horizontal.
Calculation of Active force:
Calculation of P1:
P1 =
1
2
× γ × H2
× ka γ = γ′
= 18 − 10 = 8
δ = 20 , ϕ = 30 → δ =
2
3
ϕ →→

ka in this case is calculated from (Table 7.4 P.343)
β = 90 , α = 0 , ϕ = 30 →→ ka = 0.2973
Pa =
1
2
× 8 × 52
× 0.2973 = 29.73 kN
This force having horizontal and vertical components:
P1,h = 29.73 cos(20) = 27.93 KN
P1,v = 29.73 sin(20) = 10.17 kN
Calculation of P2:
P2 = (force due to water) =
1
2
× 50 × 5 = 125 KN
Calculation of 𝐚𝐜𝐭𝐢𝐯𝐞 𝐟𝐨𝐫𝐜𝐞:
Pa,h = P1,h + P2 = 27.93 + 125 = 152.93 kN
Pa,v = P1,v = 10.17 kN
Calculation of 𝐩𝐚𝐬𝐬𝐢𝐯𝐞 𝐟𝐨𝐫𝐜𝐞:
Will not changes because always we can use rankine theory in calculating of
passive force.
Other notes remains the same.

Foundation Engineering Sheet Pile Walls
Introduction
Sheet piles are a temporary structures used to retain a soil or water for a
specific period of time, to build a structure in the other side of this wall.
For example; if we want to build a structure with three basement floors
(underground) and this structure surrounded by other structures, when the
excavation process starts, if the soil under the surrounding structures doesn’t
retained by a sheet pile, this soil will fail and will moves to the excavation
site, and the structure above this soil may collapse suddenly, so before
establishment of excavation process, sheet pile must be constructed to retain
this soil and prevent it from fails and after completion of constructed the
structure , we can remove this sheet pile because it’s function was end.
Another example; if we wanna build a structure in the sea (waterfront
structures) we can use sheet piles to retain sea water from flowing to the
required area, and then withdraw the water confined between sheet piles and
thereby build the required structures, finally remove sheet piles because
there functions were end.
The following figures are some explanation of the applications of sheet piles
and the shape of sheet pile itself:
Sheet Pile in basement Sections of steel Sheet Piles

Notes:
1. Sheet piles may be made from steel, concrete or wood.
2. As seen in the above pictures, sheet piles must penetrates a specified
distance in earth (from both sides) to be stable against applied lateral loads,
this depth called depth of penetration, and the following figure explain the
main parts of sheet piles:
The line at which the sheet pile starts penetrating in soil from both sides is
known by dredge line, and the depth of penetration of sheet pile under this
line is D: depth of penetration.
Sheet Piles in Waterfront structure

Designing of sheet piles mainly is to calculate the depth of penetration D and
determining the section of sheet pile as will be discussed later.
Types of Sheet Piles
There are two main types of sheet piles:
1. Cantilever Sheet Piles.
2. Anchored Sheet piles.
Now, we will learn how to analyze and design each type.
Cantilever Sheet Piles
Cantilever sheet pile walls are usually recommended to use for walls of
moderate height (≤ 6m) measured above the dredge line. In such walls, the
sheet piles are act as a wide cantilever beam above the dredge line.
The main step in analyzing cantilever sheet pile is to knowing the deflection
of cantilever sheet pile with depth, and knowing (from deflection shape) the
type of LEP (active or passive).
The following figure clarifies the deflection of the cantilever sheet pile due
to lateral earth pressure:

As you see, due to the lateral earth pressure the wall will pushed out the soil
above the dredge line so the type of LEP above the dredge line is active
pressure and no passive pressure because there is no soil exist in the other
side (Zone A) in the above figure.
Below the dredge line there exists a soil in both sides of the wall the wall
and the wall still moves out (left side) till reaching point O (point of
rotation) after point O the wall will moves to right side as shown.
So, soil below dredge line can be divided into two zones; zone B between
dredge line and point O, in this zone the wall moves to the left, so the soil on
the right exerts active pressure and the soil on the left exerts passive
pressure. Zone C between point O to the end of sheet pile, in this zone the
wall moves to the right, so the soil on the right exerts passive pressure and
the soil on the left exerts active pressure as seen in the figure above.
There are three cases for cantilever sheet piles:
 Cantilever Sheet Pile penetrating in Sandy Soil.
 Cantilever Sheet Pile penetrating in Clayey Soil.
 Cantilever Sheet Pile penetrating in C − ϕ Soil.
Before discussing each type, the following notes are very important:
 The first step in designing the sheet pile is to draw the net LEP
distribution with depth along the sheet pile, the net LEP is the difference
between passive LEP and active LEP at every change in soil with depth
i. e. net pressure = Δσ = σpassive − σactive
 Designing a sheet pile consists of the following two steps:
1. Calculation the depth of penetration (D).
2. Determining the section Modulus (S) where: S =
Mmax
σall
Mmax = maximum moment along sheet pile
σall = maximum allwable flexural stress (for sheet pile material).
 Always in this chapter we will use Rankine LEP theory.
The most important one, is drawing the LEP distribution along sheet pile
(especially below the dredge line) correctly (If you did, completion of
designing process will be easy), so now we want to learn how to draw the
LEP distribution for all cases of cantilever sheet pile.

Cantilever Sheet Piles Penetrating Sandy Soil
Consider the following example:
The first step always is calculating 𝐊 𝐚 𝐚𝐧𝐝 𝐊 𝐩 for each layer, but here all
layers have the same friction angle, so:
Ka = tan2
(45 −
ϕ
2
) = tan2
(45 −
30
2
) = 0.333
Kp = tan2
(45 +
ϕ
2
) = tan2
(45 +
30
2
) = 3
Now we calculate the LEP at each depth:
σh,p = (q + γH)Kp + 2c√Kp
@ z = 0.0 (above dredge line → active pressure only)
σh,a = (0 + 16 × 0) × 0.333 − 0 = 0.0
∆σh = σh,p − σh,a = 0.0
@ z = 2m (above dredge line → active pressure only)
just before = just after because Ka is the same
σh,a = (0 + 16 × 2) × 0.333 − 0 = 10.65kN/m2
∆σh = σh,p − σh,a = 0 − 10.65 = −10.65 kN/m2
The negative sign means we draw this value at right side (side of active
pressure because is the largest pressure at this depth.
ϕ = 30°
γsat = 19 kN/m3
ϕ = 30°
ϕ = 30°
γsat = 19 kN/m3
Dredge Line

@ z = 5m (just before → active pressure only)
σh,a = (0 + 16 × 2 + (19 − 10) × 3) × 0.333 − 0 = 19.65 kN/m2
∆σh = σh,p − σh,a = 0 − 19.65 = −19.65 kN/m2
(Don′
t forget, effective pressure always)
@ z = 5m (just after → active pressure at right, passive pressure at left)
Since Ka is the same before and after, the pressure will be the same
σh,a = (0 + 16 × 2 + (19 − 10) × 3) × 0.333 − 0 = 19.65kN/m2
σh,p = (0 + (19 − 10) × 0)3 + 0 = 0.0
∆σh = σh,p − σh,a = 0 − 19.65 = −19.65 kN/m2
@ z = 5 + D (passive pressure at right, active pressure at left)
Note that at this depth the types of pressure changes as explained above
Active pressure at left side:
σh,a = (0 + (19 − 10) × D) × 0.333 − 0 = 3D
Passive pressure at right side:
σh,p = (0 + 16 × 2 + (19 − 10) × 3 + (19 − 10) × D)3 + 0 = 177 + 27D
∆σh = σh,p − σh,a = (177 + 27D) − 3D = 177 + 24D
Note that the value of (177 + 24D)is positive which means we draw this
value as a line in the passive zone at this depth (right side).
Now, we draw the LEP distribution along the wall:

Now, under dredge line the net lateral pressure ∆σhalways will increase by
the value of γz′
(Kp − Ka) where z′
: depth below dredge line at any point.
And since Kp is always larger than Ka (when ϕ > 0) the increase in
pressure will always in the direction of passive zones. So, at the specific
point (point O) the pressure will changes from active to passive (right side)
and thereby the LEP distribution will tend to moves in the direction of
passive zone as shown.
Cantilever Sheet Piles Penetrating Clay
The first step always is calculating 𝐊 𝐚 𝐚𝐧𝐝 𝐊 𝐩 for each layer:
Ka = tan2
(45 −
ϕ
2
)
Ka1 = tan2
(45 −
30
2
) = 0.333
Ka2 = tan2
(45 −
35
2
) = 0.27
Ka3 = tan2(45 − 0) = 1
γ = 16 kN/m3
ϕ = 35°
γ = 18 kN/m3
γ = 17 kN/m3
C = 40 kN/m2
ϕ = 30°
ϕ = 0.0
Dredge Line

Kp = tan2
(45 +
ϕ
2
)
Kp1 = tan2
(45 +
30
2
) = 3
Kp2 = tan2
(45 +
35
2
) = 3.69
Kp3 = tan2(45 + 0) = 1
σh,a = (0 + 16 × 0) × 0.333 − 0 = 0.0
∆σh = σh,p − σh,a = 0.0
just before
σh,a = (0 + 16 × 2) × 0.333 − 0 = 10.65 kN/m2
∆σh = σh,p − σh,a = 0 − 10.65 = −10.65 kN/m2
just after
σh,a = (0 + 16 × 2) × 0.27 − 0 = 8.64 kN/m2
∆σh = σh,p − σh,a = 0 − 8.64 = −8.64 kN/m2
σh,a = (0 + 16 × 2 + 18 × 3) × 0.27 − 0 = 23.2 kN/m2
∆σh = σh,p − σh,a = 0 − 23.2 = −23.2 kN/m2
And there is a value for cohesion( C = 40) below the drege line
σh,a = (0 + 16 × 2 + 18 × 3) × 1 − 2 × 40 × √1 = 6kN/m2
σh,p = (0 + 0) × 1 + 2 × 40 × √1 = 80
∆σh = σh,p − σh,a = 80 − 6 = 74 kN/m2
The positive sign means we draw this value at left side (side of passive
pressure because is the largest pressure at this depth.

σh,a = (0 + 17 × D) × 1 − 2 × 40 × √1 = 17D − 80
σh,p = (0 + 16 × 2 + 18 × 3 + 17 × D) × 1 + 2 × 40 × √1 = 166 + 17D
∆σh = σh,p − σh,a = (166 + 17D) − (17D − 80) = 246
Note that the value of (246)is positive which means we draw this value as
a line in the passive zone at this depth (right side).
the value of γz′
(Kp − Ka) . But here since (ϕ = 0) → Kp = Ka = 1.
So, the increase in pressure will be zero till reaching point O (i.e. the
pressure will be constant at this depth). At the specific point (point O) the
pressure will changes from active to passive (right side) and thereby the LEP
distribution will tend to moves in the direction of passive zone as shown.

Cantilever Sheet Piles Penetrating 𝐂 − 𝛟 Soil
The first step always is calculating 𝐊 𝐚 𝐚𝐧𝐝 𝐊 𝐩 for each layer:
Ka = tan2
(45 −
ϕ
2
)
Ka1 = tan2
(45 −
30
2
) = 0.333
Ka2 = tan2
(45 −
35
2
) = 0.27
Ka3 = tan2
(45 −
20
2
) = 0.49
Kp = tan2
(45 +
ϕ
2
)
Kp1 = tan2
(45 +
30
2
) = 3
Kp2 = tan2
(45 +
35
2
) = 3.69
Kp3 = tan2
(45 +
20
2
) = 2.04
γ = 16 kN/m3
ϕ = 35°
γ = 18 kN/m3
γ = 17 kN/m3
C = 40 kN/m2
ϕ = 30°
ϕ = 20°
Dredge Line

σh,a = (0 + 16 × 0) × 0.333 − 0 = 0.0
∆σh = σh,p − σh,a = 0.0
just before
σh,a = (0 + 16 × 2) × 0.333 − 0 = 10.65 kN/m2
∆σh = σh,p − σh,a = 0 − 10.65 = −10.65 kN/m2
just after
σh,a = (0 + 16 × 2) × 0.27 − 0 = 8.64 kN/m2
∆σh = σh,p − σh,a = 0 − 8.64 = −8.64 kN/m2
σh,a = (0 + 16 × 2 + 18 × 3) × 0.27 − 0 = 23.2 kN/m2
∆σh = σh,p − σh,a = 0 − 23.2 = −23.2 kN/m2
And there is a value for cohesion( C = 40) below the drege line
σh,a = (0 + 16 × 2 + 18 × 3) × 0.49 − 2 × 40 × √0.49 = −13.86 kN/m2
σh,p = (0 + 0) × 2.04 + 2 × 40 × √2.04 = 114.3 kN/m2
∆σh = σh,p − σh,a = 114.3 − (−13.86) = 128.12 kN/m2
σh,a = (0 + 17 × D) × 0.49 − 2 × 40 × √0.49 = 8.33D − 56
σh,p = (0 + 16 × 2 + 18 × 3 + 17 × D) × 2.04 + 2 × 40 × √2.04
= 200.26 + 34.68D
∆σh = σh,p − σh,a = (200.26 + 34.68D) − (8.33D − 56)
= 256.26 + 26.35D

Note that the value of (256.26 + 26.35D )is positive which means we
draw this value as a line in the passive zone at this depth (right side).
the value of γz′
(Kp − Ka) . And since ϕ = 20 → Kp > Ka →the increase in
pressure will be in the direction of passive zones. So, after the value of
128.12 the stress will be increased gradually in the direction of passive zone
till reaching (point O) the pressure will changes from active to passive (right
side) and thereby the LEP distribution will tend to moves in the direction of
passive zone as shown.
Now If we draw the net LEP distribution correctly, we can calculate the
depth of penetration D by applying equilibrium equations:
∑ Fx = 0.0 (Along the sheet pile)
∑ M = 0.0 (At the bottom of sheet pile)
Note:
If Mmax is required always take a section with distance (x) above point O
Because in cantilever sheet piles the maximum moment always above O.
See examples 9.1 and 9.3

Anchored Sheet Piles
When the height of the backfill soil behind a sheet pile exceeds 6m, the
deflection on the sheet pile will be great and thereby the depth of penetration
and the section of sheet pile will be large to meet this large deflection. To
reduce this deflection, sheet pile should be supported from its upper edge
(usually at distance 1m-2m from the top), this support is called anchor and
the sheet pile with anchor called Anchored Sheet Pile.
There are two ways for analysis of anchored sheet piles:
 Free Earth Support Method.
 Fixed Earth Support Method.
In our discussion we will mainly discuss free earth support method.
The following figure shows anchored sheet pile:
Anchored Sheet pile (Free Earth Support Method)
In this method, the soil is assumed as a simply support (pin support) at the
end of sheet pile, and also the wall is simply supported from its upper edge
by anchor. So the deflection of sheet pile will be similar to the deflection of
simply supported beam as shown in the following figure:

Important notes on the above figure:
1. As you see, the deflection of the sheet pile is similar to the deflection of
simply supported beam, so if we need Mmax we take a section above the
dredge line (at point of maximum deflection “zero shear”).
2. Note that the soil in right side at all depths pushes the wall to the left side,
so the soil in the right side will exerts active LEP at all depths of sheet pile
and no inflection point (as in cantilever sheet pile), also, under the dredge
line, the soil on the wall will pushed into the left side soil, thus the LEP of
the left soil is passive pressure to the end without any inflection (‫.)إنقالب‬
3. Depending on note(2), when we drawing the net pressure distribution
under the dredge line, the increase in pressure will be always in the left side
(passive side) to the end without any inflection.
Now, we will sketch the pressure distribution for anchored sheet pile using
free earth support method when the sheet pile penetrating in different types
of soil.

Anchored Sheet Piles Penetrating Sandy Soil
The net pressure distribution along the sheet pile will be as following:
𝐏𝐮𝐫𝐞 𝐒𝐚𝐧𝐝

Below the dredge line:
The net lateral pressure ∆σhalways will increased by the value of
γz′
(Kp − Ka) . And since the soil is 𝐩𝐮𝐫𝐞 𝐬𝐚𝐧𝐝 (ϕ > 0) → Kp > Ka →the
increase in pressure will be in the direction of passive zones. So, below
dredge line, the stress will be increased gradually in the direction of passive
zone till reaching the end of sheet pile because there is no inflection point
(no change in LEP types) below the dredge line.
Anchored Sheet Piles Penetrating Clay
𝐏𝐮𝐫𝐞 𝐂𝐥𝐚𝐲

The net lateral pressure ∆σhalways will increased by the value of
γz′
(Kp − Ka) . But since the soil is 𝐩𝐮𝐫𝐞 𝐜𝐥𝐚𝐲 (ϕ = 0) → Kp = Ka →the
increase in pressure will zero till reaching the end of sheet pile because
there is no changes in the type of LEP.
Anchored Sheet Piles Penetrating 𝐂 − 𝛟 Soil
𝐂 − 𝛟 𝐒𝐨𝐢𝐥

The net lateral pressure ∆σhalways will increase by the value of
γz′
(Kp − Ka) . And since the soil is 𝐂 − 𝛟 𝐬𝐨𝐢𝐥 (ϕ > 0) → Kp > Ka →the
increase in pressure will be in the direction of passive zones. So, below
dredge line, the stress will be increased gradually in the direction of passive
zone till reaching the end of sheet pile because there is no inflection point
(no change in LEP types) below the dredge line.
Important Note:
Read the fixed earth support method from textbook page 476 and read
example 9.9 pages (477 – 479).
Holding Capacity of Anchor Plates in Sand (Page 488)
Solve Example 9.11page (493-494) (Important)

Problems
1.
For the anchored sheet pile shown below, do the following:
1. Draw the lateral earth pressure distribution with depth.
2. Calculate the depth of penetration (D).
3. Calculate the anchor force per unit length of the sheet pile.
4. Calculate section modulus if σall=175 MPa.
Solution
The first is calculating 𝐊 𝐚 𝐚𝐧𝐝 𝐊 𝐩 for each layer:
Ka = tan2
(45 −
ϕ
2
)
Ka1 = tan2
(45 −
33
2
) = 0.29
Ka2 = tan2
(45 −
28
2
) = 0.36
q = 70 kN/m2

Ka3 = tan2
(45 −
22
2
) = 0.45
Kp = tan2
(45 +
ϕ
2
)
The required value of Kp is the value of the third layer (layer below the
dredge line)
Kp3 = tan2
(45 +
22
2
) = 2.2
Now we calculate the net LEP at each depth:
σh,a = (70 + 18 × 0) × 0.29 − 2 × 17 × √0.29 = 2 kN/m2
∆σh = σh,p − σh,a = 0 − 2 = −2 (2 in active direction)
just before (Ka = 0.29 , C = 17)
σh,a = (70 + 18 × 3) × 0.29 − 2 × 17 × √0.29 = 17.65 kN/m2
∆σh = σh,p − σh,a = 0 − 17.65 = −17.65 kN/m2
just after (Ka = 0.36 , C = 27)
σh,a = (70 + 18 × 3) × 0.36 − 2 × 27 × √0.36 = 12.24 kN/m2
∆σh = σh,p − σh,a = 0 − 12.24 = −12.24 kN/m2
@ z = 9m (just before → active pressure only)(Ka = 0.36 , C = 27)
σh,a = (70 + 18 × 3 + (19 − 10) × 6) × 0.36 − 2 × 27 × √0.36
= 31.68 kN/m2
∆σh = σh,p − σh,a = 0 − 31.68 = −31.68 kN/m2
And(Ka = 0.45 , Kp = 2.2 , C = 50) below the drege line
Active pressure (at right)
σh,a = (70 + 18 × 3 + (19 − 10) × 6) × 0.45 − 2 × 50 × √0.45
= 13 kN/m2

Passive pressure (at left)
σh,p = (0 + (19 − 10) × 0) × 2.2 + 2 × 50 × √2.2
= 148.3 kN/m2
∆σh = σh,p − σh,a = 148.3 − 13 = 135.3 kN/m2
@ z = 9 + D (Active pressure at right, Passive pressure at left "no inflection")
Active pressure at (at right):
σh,a = (70 + 18 × 3 + (19 − 10) × 6 + (19 − 10) × D) × 0.45
−2 × 50 × √0.45 = 13 + 4.05 D
Passive pressure (at left):
σh,p = (0 + (19 − 10) × D) × 2.2 + 2 × 50 × √2.2
= 148.3 + 19.8 D
∆σh = σh,p − σh,a = (148.3 + 19.8 D) − (13 + 4.05 D)
= 135.3 + 15.75 D
Note that the value of (135.3 + 15.75 D ) is positive which means we
draw this value as a line in the passive zone at this depth (left side).
Now we draw the net stress distribution along the sheet pile:

As you see, there are two unknowns (D and F).
The most suitable method to find D and F is:
To find D → take ∑ M@F = 0.0
To find F → take ∑ Fx = 0.0
Assume forces 1 ,2 ,3 and 4 are positive and forces F , 5 and 6 are negative.
We prepare the following table (to simplified the solution):
Force Magnitude Arm from F Moment about F
P1 2 × 3 = 6 (+) 0.0 0.0
P2
0.5 × (17.65 − 2) × 3
= 23.48 (+) 1.5 −
3
3
= 0.5 11.74 (+)
P3 12.24 × 6 = 73.44 (+) 3 + 1.5 = 4.5 330.48 (+)
P4
0.5 × (31.68 − 12.24) × 6
= 58.32 (+) 6 −
6
3
+ 1.5 = 5.5 320.76 (+)
P5 135.3 D (−) 7.5 + 0.5 D 1014.75D + 67.65D2
(−)
P6
0.5 × (15.75D) × D
= 7.88 D2
(−)
7.5 +
2
3
D 59.1D2
+ 5.25D3
(−)
F F (−) 0.0 0.0
∑ M@F = 0.0 → 11.74 + 330.48 + 320.76 = 1014.75D + 67.65D2
+59.1D2
+ 5.25D3
→ 5.25 D3
+ 126.75 D2
+ 1014.75 D − 663 = 0.0
By trial and error or by calculati → D = 0.606 m ✓.
∑ MFx = 0.0 → 6 + 23.48 + 73.44 + 58.32 = 135.3 × 0.606
+7.88 × 0.6062
+ F →→ F = 76.35 kN ✓.
Now to find section modulus:
S =
Mmax
σall
So we need to calculate Mmax, and we mentioned previously, the maximum
moment in anchored sheet pile will be above dredge line (at point of zero
shear), so we can make a section above the dredge line and calculate Mmax
as following:

Now, to calculate the areas 3 and 4 in terms of X we must calculate the
pressure at the distance X (σ 𝑋) by interpolation.
@X = 0.0 → σ = 12.24 , @X = 6 → σ = 31.68 @X = X → σ = σ 𝑋
31.68 − 12.24
6 − 0
=
σX − 12.24
X − 0
→ σX = 3.24 X + 12.24
Now we can calculate the force 3 and 4 in terms of X:
P3 = 12.24 X
P4 = 0.5 × (σX − 12.24)X
= 0.5 × (3.24 X + 12.24 − 12.24)X = 1.62X2
Now at distance X, summation forces must be zero (point of zero shear) to
get maximum moment:
F = P1 + P2 + P3 + P4 (F, P1 and P2 are taken from table above)
76.35 = 6 + 23.48 + 12.24 X + 1.62 X2
→ 1.62 X2
+ 12.24 X − 46.87 = 0.0
→ X = 2.8 m
→ P3 = 12.24 × 2.8 = 34.27 and P4 = 3.24 × 2.82
= 25.4
σX
E

Now to get maximum moment, take summation moment at point E (point of
zero shear):
Mmax = ∑ M@E
Mmax = 76.35 × (1.5 + 2.8) − 6 × (1.5 + 2.8) − 23.48 × (
3
3
+ 2.8)
−34.27 × (
2.8
2
) − 25.4 × (
2.8
3
)
→→ Mmax = 141.6 kN. m
σall = 175 MPa. = 175,000 kPa. (kN/m2
)
S =
Mmax
σall
=
141.6
175000
= 0.81 × 10−3
m3
/m ofwall ✓.
Important Note:
In the above problem, the water table is at both sides of sheet pile, so pore
water pressure will canceled from both sides, however if the water table is at
one side of the sheet pile (right side) as shown in the following figure:
When calculating the pressure at right side (active pressure), pore water
pressure must be added as following:
σh,a = (q + γH)Ka − 2c√Ka + γw × hw (at each depth below the GWT)

2.
For the anchored sheet pile shown below, do the following:
1. Draw the lateral earth pressure distribution with depth.
2. Calculate the depth of penetration (D).
3. Calculate the anchor force per unit length of the sheet pile.
4. Calculate Maximum moment.
Solution
Note that there is no water table in this problem.
Solve the problem by yourself  with the same procedures in the problem
above.
Final Answers:
D = 2.06 m F = 149 kN Mmax = 351 kN. m
See Examples 9.5 and 9.10 in your textbook
γ = 17 kN/m3
ϕ = 36°
γ = 17 kN/m3
ϕ = 32°
γ = 18 kN/m3
ϕ = 20°
C = 30 kN/m2

Foundation Engineering Pile Foundations
Introduction
Piles are structural members that are made of steel, concrete, or timber. They
are used to build pile foundations (classified as deep foundations) which cost
more than shallow foundations (discussed in Chapters 3, 4, 5, and 6).
Despite the cost, the use of piles often is necessary to ensure structural
safety. The most case in which pile foundations are required, is when the soil
supporting the structure is weak soil (expansive soil, or collapsible soil, etc...)
we use piles to transmit the foundation load to the nearest bed rock layer,
and if bed rock is not encountered, we use piles to transmit the load to the
nearest stronger soil layer to ensure the safety for the structure.
The following figure clarifies the function of pile foundation (which
mentioned above):

Capacity of Piles
The ultimate load capacity of the pile may be expressed as:
Qu = QP + Qs
When the pile penetrates weak soil to rest on strong soil or bed rock, the pile
will supported by the bed rock or the strong soil from at the pile end (end
point of pile), So:
QP = Load carried at the pile end point
In addition, when the pile penetrates the soil, the shearing resistance between
the soil and the pile should be considered in Qs where:
Qs = Load carried by the skin friction developed at the sides of the pile
(caused by shearing resistance between the soil and the pile)

Types of Pile
1.Point Bearing Piles:
If the soil supporting the structure is weak soil, pile foundation will used to
transmit the load to the strong soil layer or to the bed rock (if encountered),
here the pile will resist the entire load depending on its end point load QP
and the value of Qs (frictional resistance) is very small in this case, so:
Qu = QP + Qs Qs ≈ 0.0 → Qu = QP (Point Bearing Piles)
2.Friction Piles:
When no strong layer or rock is present at reasonable depth at a site, point
bearing piles becomes very long (to reach strong layer) and uneconomical.
In these type of soil profiles, piles are driven through the softer (weaker) soil
to specified depth, and here the point bearing load (QP) is very small and
can be considered zero, however the load on the pile will resisted mainly by
the frictional resistance between soil and pile (Qs) so:
Qu = QP + Qs QP ≈ 0.0 → Qu = Qs (Friction Piles)

In practice, we assume the pile resist the applied loads by its point bearing
load and its frictional resistance to estimate the ultimate load the pile can
carry.
Qu = QP + Qs
In the following sections, we will learn how to calculate the value of
QP and Qs and thereby Qu for sand and clay and C − ϕ soil

Calculation of Point Bearing Load(𝐐 𝐏)
We will use Meyerhof method to calculate the value of QP for sand and
clay.
Calculation of 𝐐 𝐏 for sand:
QP = AP × qP ≤ QL
AP =Cross-sectional area of the end point of the pile (bearing area
between pile and soil).
qP = q′
× Nq
∗
q′
= Effective vertical stress at the level of the end of the pile.
Nq
∗
= Load capacity Factor (depends only on ϕ − value)
Nq
∗
is calculated from (𝐅𝐢𝐠𝐮𝐫𝐞 𝟏𝟏. 𝟏𝟑 P. 557) or (𝐓𝐚𝐛𝐥𝐞 𝟏𝟏. 𝟏𝟓 P. 558)
QL = Limiting value for point resistance
QL = 0.5 × AP × pa × Nq
∗
× tanϕ
pa = atmospheric pressure = (100 kN/m2
or 2000 Ib/ft2
)
So, for sandy soil the value of 𝐐 𝐏 𝐢𝐬:
QP = AP × q′
× Nq
∗
≤ 0.5 × AP × pa × Nq
∗
× tanϕ
Important Note:
The soil profile may consists of several sand layers, the value of friction
angle (ϕ) which used to calculate QP as shown in the above equation is the
friction angle for the soil that supporting the pile end (for last soil layer).
Calculation of 𝐐 𝐏 for Clay:
QP = AP × cu × Nc
∗
cu = Cohesion for the soil supported the pile at its end.
Nc
∗
= Bearing capacity factor for clay = 9 (when ϕ = 0.0)
QP = 9 × AP × cu
Calculation of 𝐐 𝐏 for 𝐂 − 𝛟 𝐒𝐨𝐢𝐥𝐞:
If the supporting the pile from its end is C − ϕ soil:
QP = (AP × q′
× Nq
∗
≤ 0.5 × AP × pa × Nq
∗
× tanϕ) + AP × cu × Nc
∗
But here, the value of ϕ ≠ 0.0 → Nc
∗
≠ 9 (you will given it according ϕ)
But, if you are not given the value of Nc
∗
at the existing value of ϕ →
Assume Nc
∗
= 9 and complete the solution.

Calculation of Frictional Resistance(𝐐 𝐬)
Calculation of 𝐐 𝐬 for sand:
The general formula for calculating Qs is:
Qs = P × ∑ fi × Li
P = pile perimeter = π × D (if the pile is circular, D = Pile diameter)
= 4 × D (if the pile is square, D = square dimension)
fi = unit friction resistance at any depth
Li = depth of each soil layer
Now, how we calculate the value of 𝐟𝐢 (𝐟𝐨𝐫 𝐞𝐚𝐜𝐡 𝐬𝐨𝐢𝐥 𝐥𝐚𝐲𝐞𝐫):
f = μs × N
Here the value of (f) is vertical, so N must be perpendicular to f (i.e. N must
be horizontal) as shown in the following figure:
μs = friction coefficient between soil and pile = tan δ
δ = soil − pile friction angle = 0.8ϕ → μs = tan(0.8ϕ) (for each layer)
N = Horizontal stress from the soil to the pile
→ N = σv
′
× K (for each soil layer)
σv
′
= vertical effective stress for each layer
But, to calculate σv
′
for each soil layer, to be representative, we take the
average value for σv
′
for each layer.
K = Effective earth pressure coefficient

K = 1 − sinϕ or K = 0.5 + 0.008 Dr Dr = relative density (%)
If you are not given the relative density for each layer, use K = 1 − sinϕ or
you may given another formula to calculate K.
Now, N = σv,av
′
→→ f = tan(0.8ϕ) × σv,av
′
Now, how we calculate the value of σv,av
′
for each soil layer:
We draw the vertical effective stress along the pile, but the stress will
linearly increase to a depth of (15D), after this depth the stress will be
constant and will not increase. (this is true only if we deal with sandy soil).
If there is one soil layer before reaching 15D:
σv,av,1
′
=
0 + σv
′
2
= 0.5σv
′
σv,av,2
′
=
σv
′
+ σv
′
2
= σv
′
σv
′
σv,av,1
′
σv,av,2
′

If there are more than one soil layer before reaching 15D:
σv,av,1
′
=
0 + σv,1
′
2
= 0.5σv,1
′
σv,av,2
′
=
σv,1
′
+ σv,2
′
2
σv,av,3
′
=
σv,2
′
+ σv,3
′
2
σv,av,4
′
=
σv,3
′
+ σv,3
′
2
= σv,3
′
Finally we can calculate the value of Qs as following:
Qs = P × ∑ tan(0.8ϕi) × σv,av,i
′
× Ki × Li
i = each soil layer
Note:
We take soil layer every change in soil properties or every change in slope
of vertical stress.
σv,1
′
σv,2
′
σv,3
′
σv,av,1
′
σv,av,2
′
σv,av,3
′
σv,av,4
′

Calculation of 𝐐 𝐬 for clay:
There are three methods used to calculate Qs in clay:
1. 𝛌 𝐌𝐞𝐭𝐡𝐨𝐝
Qs = P × ∑ fi × Li
But here we take the entire length of the pile:
Qs = P × L × ∑ fi
∑ fi = fav = λ × (σv,av
′
+ 2 cu,av)
σv,av
′
= mean effective vertical stress for the entire embedment length
cu,av = mean undrained shear strength for the entire embedment length
λ = function of pile length (L) (calculated from 𝐓𝐚𝐛𝐥𝐞 𝟏𝟏. 𝟗 P. 576)
Calculation of 𝛔 𝐯,𝐚𝐯
′
𝐚𝐧𝐝 𝐜 𝐮,𝐚𝐯:
We prepare the following graph (assuming three soil layers):
Note that the soil is clay, and the stress is not constant after 15D, the stress is
constant (after 15D) in sand only.
cu,av =
L1 × cu,1 + L2 × cu,2 + L3 × cu,3
L
σv,av
′
=
A1 + A2 + A3
L

2. 𝛂 𝐌𝐞𝐭𝐡𝐨𝐝
Qs = P × ∑ fi × Li
fi = αi × cu,i
αi = function of (
cu,i
patm
) (calculated from 𝐓𝐚𝐛𝐥𝐞 𝟏𝟏. 𝟏𝟎 P. 577)
Qs = P × ∑ αi × cu,i × Li
3. 𝛃 𝐌𝐞𝐭𝐡𝐨𝐝
Qs = P × ∑ fi × Li
fi = βi × σv,av,i
′
σv,av,i
′
= average vertical effective stress for each clay layer
βi = Ki × tanϕR,i
ϕR = drained friction angle of remolded clay (given for each layer)
Ki = earth pressure coefficient for each clay layer
K = 1 − sinϕR (for normally consolidated clay)
K = (1 − sinϕR) × √OCR (for overconsolidated clay)
Important Note:
If the soil is (C − ϕ)soil, we calculate Qs for sand alone and for clay
alone and then sum the two values to get the total Qs
Now, from all above methods, we can calculate the ultimate load that the
pile could carry:
Qu = Qp + Qs
If we want to calculate the allowable load:
Qall =
Qu
FS
(FS ≥ 3)

Problems
1.
Determine the ultimate load capacity of the 800 mm diameter concrete bored
pile given in the figure below.
Solution
Calculation of 𝐐 𝐏:
Note that the soil supporting the pile at its end is clay, so:
QP = AP × cu × Nc
∗
Nc
∗
= 9 (pure clay ϕ = 0.0)
AP =
π
4
× 0.82
= 0.502 m2
cu = 100 kN/m2
(for the soil supporting the pile at its end)
QP = 0.502 × 100 × 9 = 452.4 KN
QL = 0.0 since tanϕ = 0.0 → QP = 452.4 KN
Calculation of 𝐐 𝐬:
Since there are one sand layer and two clay layer, we solve firstly for sand
and then for clay:
For sand:
The stress will increase till reaching 15D
15D = 15 × 0.8 = 12m
Now we draw the vertical effective stress with depth:
Cu = 60 kN/m2
γ = 20 kN/m3
𝐂𝐥𝐚𝐲
𝐒𝐚𝐧𝐝
γ = 20 kN/m3
ϕ = 30°
𝐂𝐥𝐚𝐲
Cu = 100 kN/m2
γ = 18 kN/m3

Qs = P × ∑(tan(0.8ϕi) × σv,av,i
′
× Ki) × Li
Note that the value of ϕ for layers 1, 3, and 4 is zero (clay), so we calculate
Qs only for the layer 2 (sand layer).
P = π × D = π × 0.8 = 2.51 m
σv,av,2
′
=
72 + 132
2
= 102 kN/m2
L2 = 6m
ϕ2 = 30°
K2 = 1 − sinϕ2 = 1 − sin30 = 0.5
→ Qs,sand = 2.51 × (tan(0.8 × 30) × 102 × 0.5) × 6 = 342 kN
For Clay:
If we want to use λ Method:
Qs = P × L × fav
fav = λ × (σv,av
′
+ 2 cu,av)
P = 2.51m L = 4 + 6 + 5 = 15m
λ = 0.2 (at L = 15m from Table 11.9)
cu,av =
L1 × cu,1 + L2 × cu,2 + L3 × cu,3
L
cu,av =
4 × 60 + 6 × 0 + 5 × 100
15
= 49.33 kN/m2

σv,av
′
=
A1 + A2 + A3 + ⋯ An
L
We draw the vertical effective pressure with depth:
A1 =
1
2
× 72 × 4 = 144
A2 =
1
2
× (72 + 132) × 6 = 612
A3 =
1
2
× (132 + 182) × 5 = 785
σv,av
′
=
144 + 612 + 785
15
= 102.73 kN/m2
fav = 0.2 × (102.73 + 2 × 49.33) = 40.28
→ Qs,clay = 2.51 × 15 × 40.28 = 1516.54 kN
If we want to use α Method:
For layer (1)
cu,1 = 60 →
cu,1
p 𝑎𝑡𝑚
=
60
100
= 0.6 → α1 = 0.62

For layer (2)
cu,2 = 0.0 →
cu,2
p 𝑎𝑡𝑚
=
0
100
= 0 → α2 = 0
For layer (3)
cu,3 = 100 →
cu,3
patm
=
100
100
= 1 → α3 = 0.48
Qs,clay = 2.51 × [(0.62 × 60 × 4) + 0 + (0.48 × 100 × 5)] = 975.88 kN
Qs,total = Qs,sand + Qs,clay
Qs,total = 342 + 1516.54 = 1858.54 kN (when using λ − method)
Qs,total = 342 + 975.88 = 1317.88 kN (when using α − method)
Qu = Qp + Qs,total
Qu = 452.4 + 1858.54 = 2310.94 kN (when using λ − method)✓.
Qu = 452.4 + 1317.88 = 1770.28 kN (when using α − method)✓.
2.
A pile is driven through a soft cohesive deposit overlying a stiff clay, the
average un-drained shear strength in the soft clay is 45 kPa. and in the lower
deposit the average un-drained shear strength is 160 kPa. The water table is
5 m below the ground and the stiff clay is at 8 m depth. The unit weights are
17.5 kN/m3
and 19 kN/m3
for the soft and the stiff clay respectively.
Estimate the length of 500 mm diameter pile to carry a load of 500 kN with
a safety factor of 4. Using (a). α − method (b). λ − method
Solution
There is no given graph in this problem, so you should understand the
problem and then draw the following graph by yourself:

Qall = 500 kN , FS = 4 → Qu = 500 × 4 = 2000 KN
Qu = QP + Qs
Calculation of 𝐐 𝐏:
Note that the soil supporting the pile from its end is clay, so:
QP = AP × cu × Nc
∗
Nc
∗
= 9 (pure clay ϕ = 0.0)
AP =
π
4
× 0.52
= 0.196 m2
cu = 160 kN/m2
(for the soil supporting the pile at its end)
QP = 0.196 × 160 × 9 = 282.24 KN
QL = 0.0 since tanϕ = 0.0 → QP = 282.24 KN
Calculation of 𝐐 𝐬:
Note that all layers are clay.
(a). α − method
P = π × D = π × 0.5 = 1.57m
γ = 17.5 kN/m3 C = 45 kN/m2
γ = 19 kN/m3
C = 160 kN/m2

For layer (1)
cu,1 = 45 →
cu,1
patm
=
45
100
= 0.45 → α1 = 0.71 (by interpolation from table)
For layer (2)
cu,2 = 160 →
cu,2
patm
=
160
100
= 1.6 → α2 = 0.38
Qs = 1.57 × [(0.71 × 45 × 8) + (0.38 × 160 × X)] = 401.3 + 95.45X Kn
Qu = QP + Qs = 282.24 + 401.3 + 95.45X = 683.54 + 95.45X
But Qu = 2000 → 2000 = 683.54 + 95.45X → X = 13.8m
→→ L = 8 + 13.8 = 21.8 ≅ 22m✓.
(b). λ − method
Qs = P × L × fav
fav = λ × (σv,av
′
+ 2 cu,av)
P = 1.57m L = 8 + X
We want to calculate λ from the table, but λ is a function of pile length
which is required, so in this types of problems when the solution is required
according λ − method you are strongly recommended to assume a
reasonable value of L.
Assume X = 7 m → L = 8 + 7 = 15
λ = 0.2 (at L = 15m from Table 11.9)
cu,av =
L1 × cu,1 + L2 × cu,2 + L3 × cu,3
L
cu,av =
8 × 45 + 7 × 160
15
= 98.67 kN/m2
σv,av
′
=
A1 + A2 + A3 + ⋯ An
L
For the upper layer (Soft clay) assume the saturated unit weight is the same
as the natural unit weight (17.5) because no enough information about them.

A1 =
1
2
× 87.5 × 5 = 218.75
A2 =
1
2
× (87.5 + 110) × 3 = 296.25
A3 =
1
2
× (110 + 173) × 7 = 990.5
σv,av
′
=
218.75 + 296.25 + 990.5
15
= 100.36 kN/m2
fav = 0.2 × (100.36 + 2 × 98.67 ) = 59.54
→ Qs = 1.57 × 15 × 59.54 = 1402.167 kN
Qu = Qp + Qs = 282.24 + 1402.167 = 1684.4 < 2000 →→→ We need
to increase L to be closed from 2000
Try X = 12 m → L = 8 + 12 = 20
λ = 0.173 (at L = 20m from Table 11.9)
cu,av =
L1 × cu,1 + L2 × cu,2 + L3 × cu,3
L
cu,av =
8 × 45 + 12 × 160
20
= 114 kN/m2

σv,av
′
=
A1 + A2 + A3 + ⋯ An
L
A1 =
1
2
× 87.5 × 5 = 218.75
A2 =
1
2
× (87.5 + 110) × 3 = 296.25
A3 =
1
2
× (110 + 218) × 12 = 1968
σv,av
′
=
218.75 + 296.25 + 1968
20
= 124.15 kN/m2
fav = 0.173 × (124.15 + 2 × 114 ) = 60.92
→ Qs = 1.57 × 20 × 60.92 = 1912.94 kN
Qu = Qp + Qs = 282.24 + 1912.94 = 2195.12 > 2000
Note that at X= 12 m the value of Qu is closed to 2000 but its need to
slightly decrease, so we can say X ≅ 10m → L ≅ 18m✓.
As you see, the solution is by trial and error, so when you assume the value
for L, be logic and be realistic to save time.   

3.
A concrete pile is 20 m length and 360 mm x 360 mm in cross section. The
pile is fully embedded in sand which unit weight is 16.8 kN/m3
and ϕ = 30°
You are given alsoNq
∗
= 56.7. Calculate:
a) The ultimate load (Qp), by using Meyerhof’s method.
b) Determine the frictional resistance (Qs), if k = 1.3 and δ = 0.8ϕ.
c) Estimate the allowable load carrying capacity of the pile (Use FS = 4).
Solution
a) 𝐐 𝐏 =? ?
QP = AP × q′
× Nq
∗
≤ QL
QP = AP × q′
× Nq
∗
AP = 0.36 × 0.36 = 0.1296 m2
q′
= 16.8 × 20 = 336 kN/m2
Nq
∗
= 56.7 (given)
→ QP = 0.1296 × 336 × 56.7 = 2469 KN
Now we check for Upper limit:
QL = 0.5 × AP × pa × Nq
∗
× tanϕ
pa = atmospheric pressure = 100 kPa , ϕ = 30°
(given)
QL = 0.5 × 0.1296 × 100 × 56.7 × tan30 = 212.13 kN
QL < QP → QP = QL = 212.13 kN ✓.

b) 𝐐 𝐬 =? ?
The soil is pure sand, so:
Qs = P × ∑ tan(0.8ϕi) × σv,av,i
′
× Ki × Li
But, since the soil is sand the stress will varies at depth of 15D then will be
constant on the remained pile length.
15D = 15 × 0.36 = 5.4m
The shown figure is the pressure distribution with depth:
P = 4 × 0.36 = 1.44 (Square cross section)
K1 = K2 = 1.3 (given)
0.8ϕi = 0.8 × 30 = 24°
Qs = 1.44 × 1.3 × tan(24) [σv,av,i
′
× Li]
Qs = 0.8334[45.36 × 5.4 + 90.72 × 14.6]
→ Qs = 1308 kN ✓.
c) 𝐐 𝐚𝐥𝐥 =? ?
Qall =
Qu
FS
Qu = QP + Qs = 212.13 + 1308 = 1520.13 kN , FS = 4 (given)
→ Qall =
1520.13
4
= 380 kN ✓.

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

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