1.
Boca Raton London NewYork Washington, D.C.
CRC Press
SHALLOW
FOUNDATIONS
Bearing Capacity and SettlementBearing Capacity and Settlement
Braja M. Das

2.
PREFACE
Shallow Foundations: Bearing Capacity and Settlement is intended for use as
a reference book by university faculty members and graduate students in the
area of geotechnical engineering as well as by consulting engineers.
The text is divided into eight chapters. Chapters 2, 3, and 4 present the
varioustheoriesdevelopedduringthepastfiftyyearsforestimatingtheultimate
bearing capacity of shallow foundations under various types of loading and
subsoil conditions.
Chapter 5 discusses the principles for estimating the settlement of
foundations—both elastic and consolidation. In order to calculate the founda-
tion settlement, it is desirable to know the principles for estimating the stress
increase in a soil mass supporting a foundation which carries the load trans-
mitted from the superstructure. These principles are also discussed in this
Chapter 5. Recent developments regarding the ultimate bearing capacity of
shallowfoundations due to earthquake loading are presented in Chapter 6. Also
included in Chapter 6 are some details regarding the permanent founda-tion
settlement due to cyclic and transient loading derived from experimental
observations obtained from laboratory and field tests.
During the past fifteen years, steady progress has been made to evaluate
the possibility of using reinforcement in soil to increase the ultimate and
allowable bearing capacities of shallow foundations and also to reduce their
settlement under various types of loading conditions. The reinforcement
materials include galvanized steel strips, geotextile, and geogrid. Chapter 7
presents the state-of-the-art on this subject.
Shallow foundations (such as transmission tower foundations) are, on
some occasions, subjected to uplifting forces. The theories relating to the esti-
mation of the ultimate uplift capacity of shallow foundations in granular and
clay soils are presented in Chapter 8.
Example problems to illustrate the theories are given in each chapter.
I am grateful to my wife, Janice, for typing the manuscript in camera-
ready form and preparing the necessary artwork. It will be satisfying to know
from the users of the text if it serves the intended purpose.
Braja M. Das
Sacramento, California
© 1999 by CRC Press LLC

3.
To Janice and Valerie
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4.
CONTENTS
ONE INTRODUCTION
1.1 Shallow Foundations—General
1.2 Types of Failure in Soil at Ultimate Load
1.3 Settlement at Ultimate Load
1.4 Ultimate and Allowable Bearing Capacities
References
TWO ULTIMATE BEARING CAPACITY THEORIES—
CENTRIC VERTICAL LOADING
2.1 Introduction
2.2 Terzaghi’s Bearing Capacity Theory
2.3 Terzaghi’s Bearing Capacity Theory for
Local Shear Failure
2.4 Meyerhof’s Bearing Capacity Theory
2.5 General Discussion on the Relationships
of Bearing Capacity Factors
2.6 Other Bearing Capacity Theories
2.7 Scale Effects on Bearing Capacity
2.8 Effect of Water Table
2.9 General Bearing Capacity Equation
2.10 Effect of Soil Compressibility
2.11 Bearing Capacity of Foundations on
Anisotropic Soil
2.12 Allowable Bearing Capacity With Respect
to Failure
2.13 Interference of Continuous Foundations in
Granular Soil
References
THREE ULTIMATE BEARING CAPACITY UNDER
INCLINED AND ECCENTRIC LOADS
3.1 Introduction
FOUNDATIONS SUBJECTED TO INCLINED LOAD
3.2 Meyerhof’s Theory (Continuous Foundation)
3.3 General Bearing Capacity Equation
3.4 Other Results For Foundations With Centric
Inclined Load
3.5 Continuous Foundation With Eccentric Load
3.6 Ultimate Load on Rectangular Foundations
References
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5.
FOUR SPECIAL CASES OF SHALLOW FOUNDATIONS
4.1 Introduction
4.2 Foundation Supported by a Soil With a Rigid
Rough Base at a Limited Depth
4.3 Foundation on Layered Saturated Anisotropic
Clay
4.4 Foundation on Layered c– Soil—Stronger Soil
Underlain by Weaker Soil
4.5 Foundation on Layered c– Soil—Weaker Soil
Underlain by a Stronger Soil
4.6 Continuous Foundation on Weak Clay With a
Granular Trench
4.7 Shallow Foundations Above a Void
4.8 Foundations on a Slope
4.9 Foundations on Top of a Slope
References
FIVE SETTLEMENT AND ALLOWABLE BEARING
CAPACITY
5.1 Introduction
5.2 Stress Increase in Soil Due to Applied Load
ELASTIC SETTLEMENT
5.3 Flexible and Rigid Foundations
5.4 Settlement Under a Circular Area
5.5 Settlement Under a Rectangular Area
5.6 Effect of a Rigid Base at a Limited Depth
5.7 Effect of Depth of Embedment
5.8 Elastic Parameters
5.9 Settlement of Foundations on Saturated Clay
5.10 Settlement of Foundations on Sand
5.11 Field Plate Load Tests
CONSOLIDATION SETTLEMENT
5.12 General Principles of Consolidation Settlement
5.13 Relationships for Primary Consolidation
Settlement Calculation
5.14 Three-Dimensional Effect on Primary
Consolidation Settlement
5.15 Secondary Consolidation Settlement
DIFFERENTIAL SETTLEMENT
5.16 General Concepts of Differential Settlement
5.17 Limiting Values of Differential Settlement
Parameters
References
SIX DYNAMIC BEARING CAPACITY
AND SETTLEMENT
6.1 Introduction
(! = 0)
!
!
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6.
6.2 Effect of Load Velocity on Ultimate
Bearing Capacity
6.3 Ultimate Bearing Capacity Under
Earthquake Loading
6.4 Settlement of Foundations on Granular Soil
Due to Earthquake Loading
6.5 Foundation Settlement Due to Cyclic Loading—
Granular Soil
6.6 Foundation Settlement Due to Cyclic Loading
in Saturated Clay
6.7 Settlement Due to Transient Load on Foundation
References
SEVEN SHALLOW FOUNDATIONS ON
REINFORCED SOIL
7.1 Introduction
FOUNDATIONS ON METALLIC STRIP-REINFORCED
GRANULAR SOIL
7.2 Failure Mode
7.3 Force in Reinforcement Ties
7.4 Factor of Safety Against Tie Breaking
and Tie Pullout
7.5 Design Procedure for a Continuous Foundation
FOUNDATIONS ON GEOTEXTILE-REINFORCED SOIL
7.6 Laboratory Model Test Results
7.7 Comments on Geotextile Reinforcement
FOUNDATIONS ON GEOGRID-REINFORCED SOIL
7.8 General Parameters
7.9 Relationships for Critical Nondimensional
Parameters for Foundations on Geogrid-
Reinforced Sand
7.10 Relationship Between BCRu and BCRs in Sand
7.11 Critical Nondimensional Parameters for
Foundations on Geogrid-Reinforced Clay
( = 0 condition)
7.12 Bearing Capacity Theory
7.13 Settlement of Foundations on Geogrid-
Reinforced Soil Due to Cyclic Loading
7.14 Settlement Due to Impact Loading
References
EIGHT UPLIFT CAPACITY OF SHALLOW
FOUNDATIONS
8.1 Introduction
FOUNDATIONS IN SAND
8.2 Balla’s Theory
8.3 Theory of Meyerhof and Adams
!
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7.
8.4 Theory of Vesic 5
8.5 Saeedy’s Theory
8.6 Discussion of Various Theories
8.7 Effect of Backfill on Uplift Capacity
FOUNDATIONS IN SATURATED CLAY
( = 0 CONDITION)
8.8 Ultimate Uplift Capacity—General
8.9 Vesic’s Theory
8.10 Meyerhof’s Theory
8.11 Modifications to Meyerhof’s Theory
8.12 Factor of Safety
References
APPENDIX
Conversion Factors
!
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8.
FIGURE 1.1 Individual footing
CHAPTER 1
INTRODUCTION
1.1 SHALLOW FOUNDATIONS—GENERAL
The lowest part of a structure which transmits its weight to the underlying soil
or rock is the foundation. Foundations can be classified into two major cate-
gories: that is, shallow foundations and deep foundations. Individual footings
(Fig. 1.1) square or rectangular in plan which support columns, and strip
footings which support walls and other similar structures are generally referred
to as shallow foundations. Mat foundations, also considered shallow founda-
tions, are reinforced concrete slabs of considerable structural rigidity which
support a number of columns and wall loads. When the soil located immediate-
ly below a given structure is weak, the load of the structure may be transmitted
to a greater depth by piles and drilled shafts, which are considered deep
foundations. This book is a compilation of the theoretical and experimental
evaluations presently available in literature as they relate to the load-bearing
capacity and settlement of shallow foundations.
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9.
FIGURE 1.2 General shear failure in soil
The shallow foundation shown in Fig. 1.1 has a width B and a length L.
The depth of embedment below the ground surface is equal to Df . Theoreti-
cally, when B/L is equal to zero (that is, L = !), a plane strain case will exist in
the soil mass supporting the foundation. For most practical cases when B/L "
1/5 to 1/6, the plane strain theories will yield fairly good results. Terzaghi [1]
defined a shallow foundation as one in which the depth, Df , is less than or
equal to the width B (Df /B " 1). However, research studies conducted since
then have shown that, for shallow foundations, Df /B can be as large as 3 to 4.
1.2 TYPES OF FAILURE IN SOIL AT ULTIMATE LOAD
Figure 1.2 shows a shallow foundation of width B located at a depth Df below
the ground surface and supported by a dense sand (or stiff clayey soil). If this
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10.
FIGURE 1.3 Local shear failure in soil
foundation is subjected to a load Q which is gradually increased, the load per
unit area, q = Q/A ( A = area of the foundation), will increase and the foun-
dation will undergo increased settlement. When q becomes equal to qu at
foundation settlement S = Su , the soil supporting the foundation undergoes
sudden shear failure. The failure surface in the soil is shown in Fig. 1.2a, and
the q versus S plot is shown in Fig. 1.2b. This type of failure is called general
shear failure, and qu is the ultimate bearing capacity. Note that, in this type of
failure, a peak value q = qu is clearly defined in the load-settlement curve.
If the foundation shown in Fig. 1.2a is supported by a medium dense sand
or clayey soil of medium consistency (Fig. 1.3a), the plot of q versus S will be
as shown in Fig. 1.3b. Note that the magnitude of q increases with settlement
up to q = q´u , which is usually referred to as the first failure load [2]. At this
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11.
time, the developed failure surface in the soil will be like that shown by the
solid lines in Fig. 1.3a. If the load on the foundation is further increased, the
load-settlementcurvebecomessteeperanderraticwiththegradualoutwardand
upward progress of the failure surface in the soil (shown by the broken line in
Fig. 1.3b) under the foundation. When q becomes equal to qu (ultimate bearing
capacity), the failure surface reaches the ground surface. Beyond that, the plot
of q versus S takes almost a linear shape, and a peak load is never observed.
This type of bearing capacity failure is called local shear failure.
Figure 1.4a shows the same foundation located on a loose sand or soft
clayey soil. For this case, the load-settlement curve will be like that shown in
Fig. 1.4b. A peak value of load per unit area, q, is never observed. The ultimate
bearing capacity, qu , is defined as the point where ∆S/∆q becomes the largest
FIGURE 1.4 Punching shear failure in soil
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12.
FIGURE 1.5 Nature of failure in soil with relative density of sand (Dr) and Df /R
and almost constant thereafter. This type of failure in soil is called punching
shear failure. In this case, the failure surface never extends up to the ground
surface.
The nature of failure in soil at ultimate load is a function of several factors
such as the strength and the relative compressibility of soil, the depth of the
foundation (Df ) in relation to the foundation width (B), and the width-to-length
ratio (B/L) of the foundation. This was clearly explained by Vesic [2] who
conducted extensive laboratory model tests in sand. The summary of Vesic’s
findings is shown in a slightly different form in Fig. 1.5. In this figure, Dr is the
relative density of sand, and the hydraulic radius, R, of the foundation is
defined as
R
A
P
= (1.1)
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13.
R
BL
B L
=
+2( )
R
B
=
4
where A = area of the foundation = BL
P = perimeter of the foundation = 2(B + L)
Thus
(1.2)
For a square foundation, B = L. So
(1.3)
From Fig. 1.5 it can be seen that, when Df /R # about 18, punching shear
failure occurs in all cases, irrespective of the relative density of compaction of
1.3 SETTLEMENT AT ULTIMATE LOAD
The settlement of the foundation at ultimate load, Su , is quite variable and
depends on several factors. A general sense can be derived from the laboratory
model test results in sand for surface foundations (Df /B = 0) provided byVesic
[3] and which are presented in Fig. 1.6. From this figure it can be seen that, for
any given foundation, a decrease in the relative density of sand results in an
increase in the settlement at ultimate load.
Basedonlaboratoryandfieldtestresults,the approximaterangesof values
of Su in various types of soil are given below.
Soil
D
B
f S
B
u
(%)
Sand
Sand
Clay
Clay
0
Large
0
Large
5 to 12
25 to 28
4 to 8
15 to 20
1.4 ULTIMATE AND ALLOWABLE BEARING CAPACITIES
For a given foundation to perform to its optimum capacity, one must ensure
that the load per unit area of the foundation does not exceed a limiting value,
thereby causing shear failure in soil. This limiting value is the ultimate bearing
capacity, q Considering the ultimate bearing capacity and the uncertainties
sand.
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14.
q
q
FS
u
all =
FIGURE 1.6 Variation of Su /B for surface foundation (Df /B) on sand (after Vesic [3])
involved in evaluating the shear strength parameters of the soil, the allowable
bearing capacity, qall , can be obtained as
(1.4)
A factor of safety of 3 to 4 is generally used. However, based on limiting
settlement conditions, there are other factors which must be taken into account
in deriving the allowable bearing capacity. The total settlement, St , of a
foundation will be the sum of the following:
1. Elastic or immediate settlement, S (described in Section 1.3), and
2. Primary and secondary consolidation settlement, Sc , of a clay layer
(located below the ground water level) if located at a reasonably small
depth below the foundation.
Most building codes provide an allowable settlement limit for a foundation
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15.
FIGURE 1.7 Settlement of a structure
which may be well below the settlement derived corresponding to qall given by
Eq. (1.4). Thus, the bearing capacity corresponding to the allowable settlement
must also be taken into consideration.
A given structure with several shallow foundations may undergo uniform
settlement (Fig. 1.7a). This occurs when a structure is built over a very rigid
structural mat. However, depending on the load on various foundation
components, a structure may experience differential settlement. A foundation
may undergo uniform tilt (Fig. 1.7b) or nonuniform settlement (Fig. 1.7c). In
these cases, the angular distortion, ∆, can be defined as
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16.
∆ =
−
′
S S
L
t t(max) (min)
(
1
for nonuniform settlement)
(1.5)
and
(1.6)
Limits for allowable differential settlement of various structures are also
available in building codes. Thus, the final decision on the allowable bearing
capacity of a foundation will depend on (a) the ultimate bearing capacity, (b)
the allowable settlement, and (c) the allowable differential settlement for the
structure.
REFERENCES
1. Terzaghi, K., Theoretical Soil Mechanics. Wiley, New York, 1943.
2. Vesic, A. S., Analysis of ultimate loads of shallow foundations. J. Soil Mech.
Found. Div., ASCE, 99(1), 45, 1973.
3. Vesic, A. S., Bearing capacity of deep foundations in sand. Highway Res. Rec.
39, National Research Council, Washington, D.C.,112, 1963.
∆ =
−
′
S S
L
t t(max) (min)
(for uniform tilt)
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17.
r r e= 0
θ φtan
CHAPTER TWO
ULTIMATE BEARING CAPACITY THEORIES –
CENTRIC VERTICAL LOADING
2.1 INTRODUCTION
During the last fifty years, several bearing capacity theories were proposed for
estimating the ultimate bearing capacity of shallow foundations. This chapter
summarizes some of the important works developed so far. The cases consid-
ered in this chapter assume that the soil supporting the foundation extends to
a great depth and also that the foundation is subjected to centric vertical
loading. The variation of the ultimate bearing capacity in anisotropic soils is
also considered.
2.2 TERZAGHI’S BEARING CAPACITY THEORY
In 1948, Terzaghi [1] proposed a well-conceived theory to determine the ultimate
bearing capacity of a shallow rough rigid continuous (strip) foundation
supported by a homogeneous soil layer extending to a great depth. Terzaghi
defined a shallow foundation as a foundation where the width,B, is equal to or
less than its depth, Df .The failure surface in soil at ultimate load (that is, qu , per
unit area of the foundation) assumed by Terzaghi is shown in Fig. 2.1. Referring
to Fig. 2.1, the failure area in the soil under the foundation can be divided into
three major zones. They are:
1. Zone abc. This is a triangular elastic zone located immediately below
the bottom of the foundation. The inclination of sidesac andbc of the
wedge with the horizontal is " = N (soil friction angle).
2. Zone bcf. This zone is the Prandtl’s radial shear zone.
3. Zone bfg. This zone is the Rankine passive zone. Theslip lines inthis
zone make angles of ±(45 − N/2) with the horizontal.
Note that a Prandtl’s radial shear zone and a Rankine passive zone are also
located to the left of the elastic triangular zone abc; however, they are not
shown in Fig. 2.1.
Line cf is an arc of a log spiral, defined by the equation
(2.1)
Lines bf and fg are straight lines. Line fg actually extends up to the ground
surface. Terzaghi assumed that the soil located above the bottom of the
foundation could be replaced by a surcharge q = (Df .
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18.
FIGURE 2.1 Failure surface in soil at ultimate load for a continuous rough rigid foundation as assumed by Terzaghi
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19.
s c= ′ +σ φtan





 φ
+==
2
45tan
2
)1( ddpp
qHHqKP
The shear strength, s, of the soil can be given as
(2.2)
where F´ = effective normal stress
c = cohesion
The ultimate bearing capacity, qu , of the foundation can be determined if
we consider faces ac andbc of the triangular wedgeabc and obtain the passive
force on each face required to cause failure. Note that the passive force Pp will
be a function of the surcharge q = (Df , cohesionc, unit weight (, and angle of
friction of the soil N. So, referring to Fig. 2.2, the passive force Pp on the face
bc per unit length of the foundation at right angles to the cross section is
Pp = Ppq + Ppc + Pp( (2.3)
where Ppq , Ppc , and Pp( = passive force contributions of q, c, and (,
respectively
It is important to note that the directions of Ppq , Ppc , and Pp( are vertical,
since the face bc makes an angle N with the horizontal, and Ppq , Ppc , and Pp(
must make an angle N to the normal drawn tobc. In order to obtainPpq , Ppc , and
Pp( , the method of superposition can be used; however, it will not be an exact
solution.
Relationship for Ppq (NN …… 0, (( = 0, q …… 0, c = 0)
Consider the free body diagram of the soil wedge bcfj shown in Fig. 2.2 (also
shown in Fig. 2.3). For this case the center of the log spiral, of whichcf is an arc,
will be at point b. The forces per unit length of the wedge bcfj due to the
surcharge q only are shown in Fig. 2.3a, and they are:
1. Ppq
2. Surcharge, q
3. The Rankine passive force, Pp(1)
4. The frictional resisting force along the arc cf, F
The Rankine passive force, Pp(1) , can be expressed as
(2.4)
where Hd = bbf j
Kp = Rankine passive earth pressure coefficient = tan2
(45 + N/2)
According to the property of a log spiral defined by the equation r =
r0
e
2tanN
, the radial line at any point makes an angle N with the normal. Hence,
the line of action of the frictional force F will pass through b, the center of the
log spiral (as shown in Fig. 2.3a). Taking the moment about point b
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20.
FIGURE 2.2 Passive force on the face bc of wedge abc shown in Fig. 2.1
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21.
FIGURE 2.3 Determination of Ppq (N… 0, (= 0, q …0, c' 0)
22
)(
4
)1(
d
ppq
H
P
bj
bjq
B
p +








=





φ





== sec
2
0
B
rbc
(2.5)
Let
(2.6)
From Eq. (2.1)
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22.





 φ
−=
2
45cos1
rbj





 φ
−=
2
45sin1
rHd
2
2
45tan
2
45sin
2
2
45cos
4
222
1
22
1 




 φ
+




 φ
−
+





 φ
−
=
qrqr
BPpq











 φ
−=
2
45cos
4 22
1
qr
B
Ppq





 φ
+
=











 φ
−








φ=
φ




 φ
−
π
φ




 φ
−
π
2
45cos4
2
45cossec
2
tan
24
3
2
2
tan
24
3
2
2
qBe
eqBPpq
(2.7)
So
(2.8)
and
(2.9)
Combining Eqs. (2.4), (2.5), (2.8), and (2.9)
or
(2.10)
Now, combining Eqs.(2.6), (2.7), and (2.10)
(2.11)
Considering the stability of the elastic wedge abc under the foundation as
shown in Fig.2.3b
φ




 φ
−
π
==
tan
24
3
01
errbf
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23.
q
qN
pq
q
qN
e
q
B
P
q =

















 φ
+
==
φ




 φ
−
π
444 3444 21
2
45cos2
2
2
tan
24
3
2





 φ
+==
2
45tan22)2( ddpp
cHHKcP
cppc
M
r
P
B
P +

















 φ
−
=





2
2
45sin
4
1
)2(
qq (B × 1) = 2Ppq
where qq = load per unit area on the foundation, or
(2.12)
Relationship for Ppc (NN …… 0, (( = 0, q = 0, c …… 0)
Figure 2.4 shows the free body diagram for the wedgebcfj (also refer to Fig. 2.2).
As in the case of Ppq , the center of the arc of the log spiral will be located at
point b. The forces on the wedge which are due to cohesion c are also shown
in Fig.2.4, and they are
1. Passive force, Ppc
2. Cohesive force, C c bc= ×( )1
3. Rankine passive force due to cohesion,
4. Cohesive force per unit area along arc cf, c.
Taking the moment of all the forces about point b
(2.13)
where Mc = moment due to cohesion c along arc cf
= (2.14)
c
r r
2
1
2
0
2
tan
( )
φ
−
So
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24.
FIGURE 2.4 Determination of Ppc (N… 0, ( = 0, q = 0, c … 0)
)(
tan2
2
2
45sin
2
45tan2
4
2
0
2
1
1
rr
c
r
cH
B
P dpc
−





φ
+

















 φ
−











 φ
+=





(2.15)
The relationships for Hd , r0 , and r1 in terms of B and N are given in Eqs.
(2.9), (2.6), and (2.7), respectively. Combining Eqs. (2.6), (2.7), (2.9), and (2.15),
and noting that sin2
(45 ! N/2) × tan(45 + N/2) = ½cosN
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25.








φ





φ
+





 φ








φ=
φ




 φ
−
π
φ




 φ
−
π
tan
24
3
2
2
tan
24
3
2
2
sec
tan2
2
cos
)(sec
e
Bc
eBcP pc
φ+
φ
φ
−
φ
φ
+φ=
φ




 φ
−
π
φ




 φ
−
π
tan
tan
sec
tan
sec
sec
2
tan
24
3
22tan
24
3
2
c
c
e
c
ecqc






φ−
φ
φ
−





φ
φ
+φ=
φ




 φ
−
π
tan
tan
sec
tan
sec
sec
22
tan
24
3
2
cceq c

















 φ
+
φ=






φ
φ+
φ=
φφ
+
φ
=
φ
φ
+φ
2
45cos2
1
cot
cos
sin1
cot
sincos
1
cos
1
tan
sec
sec
2
2
2
(2.16)
Considering the equilibrium of the soil wedge abc (Fig. 2.4b)
qc (B × 1) = 2C sinN + 2Ppc
or
qc B = cB secN sinN + 2Ppc (2.17)
where qc = load per unit area of the foundation
Combining Eqs. (2.16) and (2.17)
(2.18)
or
(2.19)
However
(2.20)
Also
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26.
φ=





φ
φ
φ=






φ
φ
−
φ
φ=
φ−φφ=φ−
φ
φ
cot
cos
cos
cot
cos
sin
cos
1
cot
)tan(seccottan
tan
sec
2
2
2
2
2
22
2
)1(cot1
2
45cos2
cot
2
tan
24
3
2
−φ==












−





 φ
+
φ=
φ




 φ
−
π
qc
cN
c
NccN
e
cq
4444 34444 21





 φ
+γ=
2
45tan
2
1 22
)3( dp
HP
P l Wl P lp p w p Rγ = + ( )3
(2.21)
Substituting Eqs. (2.20) and (2.21) into Eq. (2.19)
(2.22)
Relationship for Pp(( (NN …… 0, (( …… 0, q = 0, c = 0)
Figure 2.5a shows the free body diagram of wedge bcfj. Unlike the free body
diagrams shown in Figs. 2.3 and 2.4, the center of the log spiral of whichbfis an
arc is at a pointOalong line bf and not at b. This is because the minimumvalue
of Pp( has to be determined by several trials. PointOis only one trial center. The
forces per unit length of the wedge that need to be considered are:
1. Passive force, Pp(
2. The weight of wedge bcfj, W
3. The resultant of the frictional resisting force acting along arc cf, F
4. The Rankine passive force, Pp (3)
The Rankine passive force Pp (3) can be given by the relation
(2.23)
Also note that the line of action of force F will pass through O. Taking the
moment about O
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27.
FIGURE 2.5 Determination of Pp( (N… 0, (… 0, q = 0, c = 0)
P
l
Wl P lp
p
w p Rγ = +
1
3( )
or
(2.24)
If a number of trials of this type are made by changing the location at the center
of the log spiral O along line bf, then the minimum value of Pp( can be deter-
mined.
Considering the stability of wedge abc as shown in Fig. 2.5, we can write
that
q( B = 2Pp( !Ww (2.25)
© 1999 by CRC Press LLC
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28.






φγ−=
φγ=
γγ
tan
4
2
1
tan
4
2
2
B
P
B
q
B
W
p
w
So
φγ=




 φ
γ=γ= γγγγ
22
2
2
tan
8
1
2
tan
2
1
2
1
pppp
KBK
B
KhP
γ
γ
γ
γγ
γ=




 φ
−φγ=






φγ−φγ=
BNKB
B
KB
B
q
N
p
p
2
1
2
tan
tan
2
1
2
1
tan
4
tan
4
11
2
2
22
4444 34444 21
q cN qN BNu c q= + +
1
2
γ γ
where q( = force per unit area of the foundation
Ww = weight of wedge abc
However,
(2.26)
(2.27)
The passive force Pp( can be expressed in the form
(2.28)
where Kp( = passive earth pressure coefficient
Substituting Eq. (2.28) into Eq. (2.27)
(2.29)
Ultimate Bearing Capacity
The ultimate load per unit area of the foundation (that is, the ultimate bearing
capacity qu ) for a soil with cohesion, friction, and weight can now be given as
qu = qq + qc + q( (2.30)
Substituting the relationships for qq , qc , and q( given byEqs.(2.12),(2.22),and
(2.29) into Eq. (2.30) yields
(2.31)
where Nc , Nq , and N( = bearing capacity factors, and
© 1999 by CRC Press LLC
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29.





 φ
+
=
φ




 φ
−
π
2
45cos2 2
tan
24
3
2
e
Nq
N Kpγ γ φ
φ
= −
1
2 2
2
tan
tan
N
N
c
q
=
+
−
=
+
−
228 4 3
40
40 5
40
. φ
φ
φ
φ
(2.32)
Nc = cot!(Nq!1) (2.33)
(2.34)
Table 2.1 gives the variations of the bearing capacitiy factors with soil
friction angle ! given by Eqs. (2.32), (2.33), and (2.34). The values of N" were
obtained by Kumbhojkar [2].
TABLE 2.1 Terzaghi’s Bearing Capacity Factors—Eqs. (2.32), (2.33), and (2.34)
! Nc Nq N" ! Nc Nq N" ! Nc Nq N"
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
5.70
6.00
6.30
6.62
6.97
7.34
7.73
8.15
8.60
9.09
9.61
10.16
10.76
11.41
12.11
12.86
13.68
1.00
1.1
1.22
1.35
1.49
1.64
1.81
2.00
2.21
2.44
2.69
2.98
3.29
3.63
4.02
4.45
4.92
0.00
0.01
0.04
0.06
0.10
0.14
0.20
0.27
0.35
0.44
0.56
0.69
0.85
1.04
1.26
1.52
1.82
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
14.60
15.12
16.57
17.69
18.92
20.27
21.75
23.36
25.13
27.09
29.24
31.61
34.24
37.16
40.41
44.04
48.09
5.45
6.04
6.70
7.44
8.26
9.19
10.23
11.40
12.72
14.21
15.90
17.81
19.98
22.46
25.28
28.52
32.23
2.18
2.59
3.07
3.64
4.31
5.09
6.00
7.08
8.34
9.84
11.60
13.70
16.18
19.13
22.65
26.87
31.94
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
52.64
57.75
63.53
70.01
77.50
85.97
95.66
106.81
119.67
134.58
151.95
172.28
196.22
224.55
258.28
298.71
347.50
36.50
41.44
47.16
53.80
61.55
70.61
81.27
93.85
108.75
126.50
147.74
173.28
204.19
241.80
287.85
344.63
415.14
38.04
45.41
54.36
65.27
78.61
95.03
115.31
140.51
171.99
211.56
261.60
325.34
407.11
512.84
650.87
831.99
1072.80
Krizek [3] gave simple empirical relations for Terzaghi’s bearing capacity
factors Nc , Nq , and N" with a maximum deviation of 15%. They are as follows:
(2.35a)
(2.35b)
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30.
φ−
φ
=γ
40
6
N
q
c q B
u =
+ + + +
−
° °
( . ) ( )228 4 3 40 5 3
40
φ φ φγ
φ
φ(for = 0 to 35 )
(2.35c)
where N = soil friction angle, in degrees
Equations (2.35a), (2.35b), and (2.35c) are valid for N = 0 to 35E. Thus,
substituting Eqs. (2.35) into (2.31)
(2.36)
For foundations that are rectangular or circular in plan, a plane strain
condition in soil at ultimate load does not exist. Therefore, Terzaghi [1]
proposed the following relationships for square and circular foundations.
qu = 1.3cNc + qNq + 0.4(BN( (square foundation; plan B × B) (2.37)
and
qu = 1.3cNc + qNq + 0.3(BN( (circular foundation; plan B × B) (2.38)
SinceTerzaghi’sfoundingwork,numerousexperimentalstudiestoestimate
the ultimate bearing capacity of shallow foundations have been conducted.
Based on these studies, it appears that Terzaghi’s assumption of the failure
surface in soil at ultimate load is essentially correct. However, the angle " that
the sides ac and bc of the wedge (Fig. 2.1) make with the horizontal is closer to
45 + N/2, and not N as assumed by Terzaghi. In that case, the nature of the soil
failure surface would be as shown in Fig. 2.6.
The method of superposition was used to obtain the bearing capacity
factors, Nc , Nq , and N( . Forderivation ofNc and Nq, the center of the arc of the
log spiralcf is located at the edge of the foundation.However, for derivation of
N( , it is not so. In effect, two different surfaces are used in deriving Eq. (2.31).
However, it is on the safe side.
2.3 TERZAGHI’S BEARING CAPACITY THEORY
FOR LOCAL SHEAR FAILURE
It is obvious from Section 2.2 that Terzaghi’s bearing capacity theory was
obtained by assuming general shear failure in soil. However, the local shear
failure in soil, Terzaghi [1] suggested the following relationships:
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31.
FIGURE 2.6 Modified failure surface in soil supporting a shallow foundation at ultimate load
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32.
k D D Dr r r= + − ≤ ≤0 67 0 75 2
. . (for 0 0.67)
Strip foundation (B/L = 0; L = length of foundation)
qu = c´Nc´ + q´Nq´ + –1
2!BN!´ (2.39)
Square foundation (B = L)
qu = 1.3c´Nc´ + qNq´ + 0.4!BN!´ (2.40)
Circular foundation (B = diameter)
qu = 1.3c´Nc´ + qNq´ + 0.3!BN!´ (2.41)
where Nc´, Nq´, and N!´ = modified bearing capacity factors
c´ = 2c/3
The modified bearing capacity factors can be obtained by substituting "´
= tan!1
(0.67tan") for " in Eqs. (2.32), (2.33), and (2.34). The variations of Nc´,
Nq´, and N!´ with " are shown in Table 2.2.
TABLE 2.2 Terzaghi’s Modified Bearing Capacity Factors Nc´, Nq´, and N!!!!´
" Nc´ Nq´ N!´ " Nc´ Nq´ N!´ " Nc´ Nq´ N!´
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
5.70
5.90
6.10
6.30
6.51
6.74
6.97
7.22
7.47
7.74
8.02
8.32
8.63
8.96
9.31
9.67
10.06
1.00
1.07
1.14
1.22
1.30
1.39
1.49
1.59
1.70
1.82
1.94
2.08
2.22
2.38
2.55
2.73
2.92
0.00
0.005
0.02
0.04
0.055
0.074
0.10
0.128
0.16
0.20
0.24
0.30
0.35
0.42
0.48
0.57
0.67
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
10.47
10.90
11.36
11.85
12.37
12.92
13.51
14.14
14.80
15.53
16.03
17.13
18.03
18.99
20.03
21.16
22.39
3.13
3.36
3.61
3.88
4.17
4.48
4.82
5.20
5.60
6.05
6.54
7.07
7.66
8.31
9.03
9.82
10.69
0.76
0.88
1.03
1.12
1.35
1.55
1.74
1.97
2.25
2.59
2.88
3.29
3.76
4.39
4.83
5.51
6.32
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
23.72
25.18
26.77
28.51
30.43
32.53
34.87
37.45
40.33
43.54
47.13
51.17
55.73
60.91
66.80
73.55
81.31
11.67
12.75
13.97
15.32
16.85
18.56
20.50
22.70
25.21
28.06
31.34
35.11
39.48
44.54
50.46
57.41
65.60
7.22
8.35
9.41
10.90
12.75
14.71
17.22
19.75
22.50
26.25
30.40
36.00
41.70
49.30
59.25
71.45
85.75
Vesic [4] suggested a better mode to obtain "´ for estimatingNc´and Nq´for
foundations on sand in the form
"´ = tan!1
(k tan") (2.42)
(2.43)
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33.
FIGURE 2.7 Slip line fields for a rough continuous foundation
where Dr = relative density
2.4 MEYERHOF’S BEARING CAPACITY THEORY
In 1951, Meyerhof published a bearing capacity theory which could be applied
to rough shallow and deep foundations. The failure surface at ultimate load
under a continuous shallow foundation assumed by Meyerhof [5] is shown in
Fig. 2.7. In this figure, abc is the elastic triangular wedge shown in Fig. 2.6,
bcd is the radial shear zone with cd being an arc of a log spiral, and bde is a
mixed shear zone in which the shear varies between the limits of radial and
plane shear, depending on the depth and roughness of the foundation. The
plane be is called an equivalent free surface. The normal and shear stresses on
plane be are po and so , respectively. The superposition method is used to
determine the contribution of cohesion, c; po ; and ! and " on the ultimate
bearing capacity, qu , of the continuous foundation and is expressed as
qu = cN c + qN q + –
1
2
!BN ! (2.44)
where Nc´, Nq´, and N!´ = bearing capacity factors
B = width of the foundation
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34.
R
s
= 1
cosφ
s R
s
o = + =
+
cos( )
cos( )
cos
2
21
η φ
η φ
φ
cos( )
cos
tan
( tan ) cos
tan
2
1 1
η φ
φ
φ
φ φ
φ
+ =
+
=
+
+
s
c p
m c p
c p
o o
R
c p
=
+ 1 tan
cos
φ
φ
Derivation of Nc and Nq (""""""""0, !!!! = 0, po""""0, c""""0)
For this case, the center of the log spiral arc [Eq. (2.1)] is taken at b. Also it is
assumed that, along be,
so = m(c + po tan") (2.45)
where c = cohesion
" = soil friction angle
m = degree of mobilization of shear strength (0 # m #1)
Now, consider the linear zone bde (Fig. 2.8a). Plastic equilibriumrequires
that the shear strength s1 under the normal stress p1 is fully mobilized, or
s1 = c + p1 tan" (2.46)
Figure 2.8b shows the Mohr’s circle representing the stress conditions on
zone bde. Note that P is the pole. The traces of planes bd and be are also shown
in the figure. For the Mohr’s circle
(2.47)
where R = radius of the Mohr’s circle
Also
(2.48)
Combining Eqs. (2.45), (2.46), and (2.48)
(2.49)
Again, referring to the trace of plane de (Fig. 2.8c)
s1 = R cos"
(2.50)
Note that
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35.
FIGURE 2.8 Determination of Nq and Nc
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36.
FIGURE 2.8 (Continued)
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37.
FIGURE 2.8 (Continued)
p R p R
p R p
c p
p
o
o
o
1
1
1
2
2
2
+ = + +
= + − +
=
+
+ − +
sin sin( )
[sin( ) sin ]
tan
cos
[sin( ) sin ]
φ η φ
η φ φ
φ
φ
η φ φ
0
22
2
0
2
1
1
=+





′−





cp
M
r
p
r
p
r bc0 =
(2.51)
Figure 2.8d shows the free bodydiagramof zone bcd. Note that the normal
and shear stresses on the face bc are pp´ and sp´, or
sp´ = c + pp´tan"
or
pp´ = (sp´ ! c)cot" (2.52)
Taking the moment about b
(2.53)
where
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38.
r bd r eo1 = = θ φtan
M
c
r rc = −
2
1
2
0
2
tan
( )
φ
Bq
B
s
B
p pp
′=




 φ
+

















 φ
+
′+




 φ
+

















 φ
+
′
2
45sin
2
45cos
22
2
45cos
2
45cos
22





 φ
−′+′=′
2
45cotpp
spq
qoc
qN
o
cN
NpcN
e
p
e
cq
+=






φ+ηφ−
φ+
+


















φ+ηφ−
φ+
φ=′
φθφθ
444 3444 21444 3444 21
)2sin(sin1
)sin1(
)2sin(sin1
)sin1(
cot
tan2tan2
It can be shown that
(2.55)
Substituting Eqs. (2.54) and (2.55) into Eq. (2.53) yields
pp´ = p1 e2#tan"
+ c cot"(e2#tan"
! 1) (2.56)
Combining Eqs. (2.52) and (2.56)
sp´ = (c + p1 tan")e2#tan"
(2.57)
Figure 2.8e shows the free body diagram of wedge abc. Resolving the
forces in the vertical direction
where q´ = load per unit area of the foundation, or
(2.58)
Substituting Eqs. (2.51), (2.52), and (2.57) into Eq. (2.58), and further
simplifying yields
(2.59)
(2.55)
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39.
cos( )
( tan ) cos
tan
2
1
η φ
φ φ
φ
+ =
+
+
m c p
c p
o






φ−
φ+
= φπ
sin1
sin1tan
eNq
where Nc , Nq = bearing capacity factors
The bearing capacity factors will depend on the degree of mobilization, m,
of shear strength on the equivalent free surface. This is because m controls η.
From Eq. (2.49)
for m = 0, cos(2$ + ") = 0, or
$ = 45 ! –
"
2
(2.60)
For m = 1, cos(2$ + ") = cos", or
$ = 0 (2.61)
Also, the factors Nc and Nq are influenced by the angle of inclination of the
equivalent free surface %. From the geometry of Fig. 2.7
# = 135$ + % ! $ ! –
"
2
(2.62)
From Eq. (2.60), for m = 0, the value of $ is (45!"/2). So
# = 90 + % (for m = 0) (2.63)
Similarly, for m = 1, since $ = 0 [Eq. (2.61)]
# = 135$ + % ! –
"
2
(for m = 1) (2.64)
Figures 2.9 and 2.10 show the variation of Nc and Nq with ", %, and m. It
is of interest to note that, if we consider the surface foundation condition (as
done in Figs. 2.3 and 2.4 for Terzaghi’s bearing capacity equation derivation),
then % = 0 and m = 0. So, from Eq. (2.63)
# = –
&
2
(2.65)
Hence for m = 0, $ = 45 ! "/2, and # = &/2, the expressions for Nc and Nq are
as follows (surface foundation condition)
(2.66)
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40.
FIGURE 2.9 Meyerhof’s bearing capacity factor — variation of Nc with
␤, ␾, and m [Eq. (2.59)]
and
Nc = (Nq ! 1)cot␾ (2.67)
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41.
FIGURE 2.10 Meyerhof’s bearing capacity factor—variation of Nq
with ␤, ␾, and m [Eq. (2.59)]
Equations (2.66) and (2.67) are the same as those derived by Reissner [6]
for Nq and Prandtl [7] for Nc . For this condition, po = !Df = q. So, Eq. (2.59)
becomes
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42.
FIGURE 2.11 Determination of N␥
q´ = cNc + qNq (2.68)
% %
Eq. (2.66) Eq. (2.67)
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43.
P
W l P l
l
p
w p R R
p
γ =
+ ( )
γ
γ
γ=

















 φ
+−
γ





 φ
+
γ
=′′ BN
B
P
B
q
p
2
1
2
45tan
2
12
45sin4
2 2
Note that Ww is the weight of wedge abc in Fig. 2.11b.
Derivation of N!!!! ("""" """" 0, !!!! """" 0, po = 0, c = 0)
N! is determined by trial and error as in the case of the derivation of
Terzaghi’s bearing capacity factor N! (Section 2.2). Referring to Fig. 2.11a,
following is a step-by-step approach for the derivation of N! .
1. Choose values for " and the angle % (such as +30$, + 40$, !30$, ...).
2. Choose a value for m (such as, m = 0 or m = 1).
3. Determine the value of # from Eqs. (2.63) or (2.64) for m = 0 or m
= 1, as the case may be.
4. With known values of # and %, draw lines bd and be.
5. Select a trial center such as O and draw an arc of a log spiral
connecting points c and d. The log spiral follows the equation r =
r0e#tan"
.
6. Draw line de. Note that lines bd and de make angles of 90 ! " due to
the restrictions on slip lines in the linear zone bde. Hence the trial
failure surface is not, in general, continuous at d.
7. Consider the trial wedge bcdf. Determine the following forces per
unit length of the wedge at right angles to the cross section shown:
(a) weight of wedge bcdf —W, and (b) Rankine passive force on the
face df —Pp(R) .
8. Take the moment of the forces about the trial center of the log spiral
O, or
(2.69)
where Pp! = passive force due to !
Note that the line of action of Pp! acting on the face bc is located at
a distance of 2bc/3.
9. For given values of %, ", and m, and by changing the location of
point O (that is, the center of the log spiral), repeat Steps 5 through 8
to obtain the minimum value of Pp!.
10. Refer to Figure 2.11b. Resolve the forces acting on the triangular
wedge abc in the vertical direction, or
(2.70)
where q&& = force per unit area of the foundation
N ! = bearing capacity factor
and ␾ only.
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44.
FIGURE 2.12 Meyerhof’s bearing capacity factor—variation of with
␤, ␾, and m [Eq. (2.70)]
The variation of N␥ (determined in the above manner) with ␤, ␾, and m is given
in Fig. 2.12.
Combining Eqs. (2.59) and (2.70), the ultimate bearing capacity of a con-
tinuous foundation (for the condition c " 0, ! " 0, and " " 0) can be given as
N!
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45.
qu = q´ + q&& = cNc + po Nq + –1
2!BN!
The above equation is the same form as Eq. (2.44). Similarly, for surface
foundation conditions (that is, % = 0 and m =0), the ultimate bearing capacity
of a continuous foundation can be given as
qu = q´ + q! = cNc + qNq + –1
2!BN! (2.71)
% % % %
Eq. (2.68) Eq. (2.70) Eq. (2.67) Eq. (2.66)
For shallowfoundation design, the ultimate bearing capacityrelationship given
by Eq. (2.71) is presently used. The variation of N! for surface foundation
conditions (that is % = 0 and m = 0) is given in Fig. 2.12. In 1963, Meyerhof [8]
suggested that N! could be approximated as
N! = (Nq ! 1)tan(1.4") (2.72)
%
Eq. (2.66)
Table 2.3 gives the variation of Nc and Nq obtained from Eqs. (2.66) and (2.67)
and N! obtained from Eq. (2.72).
TABLE 2.3 Variation of Meyerhof’s Bearing Capacity Factors Nc , Nq , and N !!!!
[Eqs. (2.66), (2.67), and (2.72)]
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
5.14
5.38
5.63
5.90
6.19
6.49
6.81
7.16
7.53
7.92
8.35
8.80
9.28
9.81
10.37
10.98
11.63
1 00
1.09
1.20
1.31
1.43
1.57
1.72
1.88
2.06
2.25
2.47
2.71
2.97
3.26
3.59
3.94
4.34
0.00
0.002
0.01
0.02
0.04
0.07
0.11
0.15
0.21
0.28
0.37
0.47
0.60
0.74
0.92
1.13
1.38
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
12.34
13.10
13.93
14.83
15.82
16.88
18.05
19.32
20.72
22.25
23.94
25.80
27.86
30.14
32.67
35.49
38.64
4.77
5.26
5.80
6.40
7.07
7.82
8.66
9.60
10.66
11.85
13.20
14.72
16.44
18.40
20.63
23.18
26.09
1.66
2.00
2.40
2.87
3.42
4.07
4.82
5.72
6.77
8.00
9.46
11.19
13.24
15.67
18.56
22.02
26.17
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
42.16
46.12
50.59
55.63
61.35
67.87
75.31
83.86
93.71
105.11
118.37
133.88
152.10
173.64
199.26
229.93
266.89
29.44
33.30
37.75
42.92
48.93
55.96
64.20
73.90
85.38
99.02
115.31
134.88
158.51
187.21
222.31
265.51
319.07
31.15
37.15
44.43
53.27
64.07
77.33
93.69
113.99
139.32
171.14
211.41
262.74
328.73
414.32
526.44
674.91
873.84
" Nc Nq N! " Nc Nq N! " Nc Nq N!
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46.






φ−
φ+
= φπ
sin1
sin1tan
eNq
2.5 GENERAL DISCUSSION ON THE RELATIONSHIPS
OF BEARING CAPACITY FACTORS
At this time, the general trend among geotechnical engineers is to accept the
method of superposition as a proper means to estimate the ultimate bearing
capacity of shallow rough foundations. For rough continuous foundations, the
nature of failure surface in soil shown in Fig. 2.6 has also found acceptance,
as well as have Reissner’s [6] and Prandtl’s [7] solutions for Nc and Nq , which
are the same as Meyerhof’s [5] solution for surface foundations. Or
(2.66)
and
Nc = (N q ! 1)cot! (2.67)
There has been considerable controversyover the theoretical values of N".
Hansen [9] proposed an approximate relationship for N" in the form
N " = 1.5Nc tan2
! (2.73)
In the preceding equation, the relationship for Nc is that given by Prandtl’s
solution [Eq. (2.67)]. Caquot and Kerisel [10] assumed that the elastic tri-
angular soil wedge under a rough continuous foundation to be of the shape
shown inFig.2.6. Usingintegration of Boussinesq’s differential equation, they
presented numerical values of N" for various soil friction angles !. Vesic [4]
approximated their solution in the form
N" = 2(Nq + 1)tan! (2.74)
where Nq is given by Eq. (2.66)
Equation (2.74) has an error not exceeding 5% for 20" < ! < 40"
compared to the exact solution. Lundgren and Mortensen [11] developed
numerical methods (using the theory of plasticity) for exact determination of
rupture lines as well as the bearing capacity factor (N") for particular cases.
Figure 2.13 shows the nature of the rupture lines for this type of solution. Chen
[12] also gave a solution for N" in which he used the upper bound limit analysis
theorem suggested by Drucker and Prager [13]. Biarez et al. [14] also
recommended the following relationship for N"
N" = 1.8(Nq ! 1)tan! (2.75)
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47.
Soil
friction
angle, !
(deg)
N"
Terzaghi
[Eq. (2.34)]
Meyerhof
[Eq. (2.72)]
Vesic
[Eq. (2.74)]
Hansen
[Eq. (2.73)]
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
0.00
0.01
0.04
0.06
0.10
0.14
0.20
0.27
0.35
0.44
0.56
0.69
0.85
1.04
1.26
1.52
1.82
2.18
2.59
3.07
3.64
4.31
5.09
6.00
7.08
8.34
9.84
11.60
13.70
16.18
19.13
22.65
26.87
31.94
38.04
45.41
54.36
65.27
78.61
95.03
115.31
140.51
171.99
211.56
261.60
325.34
0.00
0.002
0.01
0.02
0.04
0.07
0.11
0.15
0.21
0.28
0.37
0.47
0.60
0.74
0.92
1.13
1.38
1.66
2.00
2.40
2.87
3.42
4.07
4.82
5.72
6.77
8.00
9.46
11.19
13.24
15.67
18.56
22.02
26.17
31.15
37.15
44.43
53.27
64.07
77.33
93.69
113.99
139.32
171.14
211.41
262.74
0.00
0.07
0.15
0.24
0.34
0.45
0.57
0.71
0.86
1.03
1.22
1.44
1.69
1.97
2.29
2.65
3.06
3.53
4.07
4.68
5.39
6.20
7.13
8.20
9.44
10.88
12.54
14.47
16.72
19.34
22.40
25.99
30.22
35.19
41.06
48.03
56.31
66.19
78.03
92.25
109.41
130.22
155.55
186.54
224.64
271.76
0.00
0.00
0.01
0.02
0.05
0.07
0.11
0.16
0.22
0.30
0.39
0.50
0.63
0.78
0.97
1.18
1.43
1.73
2.08
2.48
2.95
3.50
4.13
4.88
5.75
6.76
7.94
9.32
10.94
12.84
15.07
17.69
20.79
24.44
28.77
33.92
40.05
47.38
56.17
66.75
79.54
95.05
113.95
137.10
165.58
200.81
Table 2.4 gives a comparison of the N" values recommended by Meyerhof
[8], Terzaghi [1], Caquot and Kerisel [10], Vesic [4], and Hansen [9].
TABLE 2.4 Comparison of N"""" Values
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48.
FIGURE 2.13 Nature of rupture lines in soil under a continuous
foundation—plasticity solution to determine N"
FIGURE 2.14 Comparison of bearing capacity factor N" (Note: Curve 1–Chen
[12], Curve 2–Vesic [4], Curve 3–Terzaghi [1], Curve 4–Meyerhof
[8], Curve 5–Lundgren and Mortensen [11], Curve 6–Hansen [9])
Figure 2.14 shows a comparison of N" values obtained from various theories.
The primary reason several theories for N" were developed and their lack
of correlation with the experimental values lies in the difficulty in selecting a
representative value of the soil friction angle ! for computing the bearing
capacity. The parameter ! depends on many factors, such as intermediate prin-
cipal stress condition, friction angle anisotropy, and curvature of the Mohr-
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49.
t
L
B
φ











−=φ 1.01.1
φ α
φ
< < +min 45
2
Coulomb failure envelope. Ingra and Baecher [15] compared the theoretical
solutions of N" with the experimental results obtained by several investigators
for foundations with B/L = 1 and 6 (B = width and L = length of the founda-
tion). It was noticed that, when triaxial friction angles were used to deduce
experimental N" , their values were substantially higher than those obtained
theoretically. A regression analysis shows that the expected values of
variances can be given as
E(N")L/B = 1 = exp(!2.064 + 0.173!t) (2.76)
V(N")L/B = 1 = (0.0902)exp(!4.128 + 0.346!t) (2.77)
E(N")L/B = 6 = exp(!1.646 + 0.173!t) (2.78)
V(N")L/B = 6 = (0.0429)exp(!3.292 + 0.345!t) (2.79)
where !t = triaxial friction angle
It was previously suggested that the plane strain soil friction angle !p
instead of !t be used to estimate the bearing capacity [9]. To that effect, Vesic
[4] raised the issue that this type of assumption might help explain the dif-
ferences between the theoretical and experimental results for long rectangular
foundations. However, it does not help to interpret results of tests with square
or circular foundations. Ko and Davidson [16] also concluded that when plane
strain angles of internal friction are used in commonly accepted bearing capa-
city formulas, the bearing capacity for rough footings could be seriously over-
estimated for dense sands. To avoid the controversy, Meyerhof [8] suggested
the following:
(2.80)
2.6 OTHER BEARING CAPACITY THEORIES
Hu [17] proposed a theory according to which the base angle, #, of the
triangular wedge below the foundation (see Fig. 2.1) is a function of several
parameters, or
# = f (", !, q) (2.81)
The minimum and maximum values of # can be given as follows
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50.
FIGURE 2.15 Hu’s bearing capacity factors
α
φ
max = +45
2
and
The values of Nc , Nq , and N" determined by this procedure are shown in Fig.
2.15.
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51.
FIGURE 2.16 Nature of failure surface considered for Balla’s [18] bearing
capacity theory
Balla [18] proposed a bearing capacity theory which was developed for an as-
sumed failure surface in soil (Fig. 2.16). For this failure surface, the curve cd
was assumed to be an arc of a circle having a radius r. The bearing capacity
solution was obtained using Kötter’s equation to determine the distribution of
the normal and tangential stresses on the slip surface. Accordingto this solution
for a continuous foundation
qu = cNc + qNq + –
1
2
"Bn"
The bearing capacity factors can be determined as follows:
1. Obtain the magnitude of c/B" and Df /B.
2. With the values obtained in Step 1, go to Fig. 2.17 to obtain the
magnitude of $ = 2r/B.
3. With known values of $, go to Figs. 2.18, 2.19, and 2.20 respectively
to determine Nc , Nq , and N" .
2.7 SCALE EFFECTS ON BEARING CAPACITY
The problem of estimating the ultimate bearing capacity becomes complicated
if the scale effect is taken into consideration. The scale effect, which has come
to the limelight more recently, shows that the ultimate bearing capacity de-
creases with the increase in the size of the foundation. This condition is more
predominant in granular soils. Figure 2.21 shows the general nature of the
decrease in N" with the increase in foundation width B. The magnitude of N"
initially decreases with B and remains almost constant for larger values of B.
The reduction in N" for larger foundations may ultimately result in a sub-
stantial decrease in the ultimate bearing capacity which can primarily be attri-
buted to the following reasons.
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52.
FIGURE 2.17 Variation of $ with soil friction angle for determination
of Balla’s bearing capacity factors
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53.
FIGURE 2.18 Balla’s bearing capacity factor Nc
1. For larger-sized foundations, the rupture along the slip lines in soil is
progressive, and the average shear strength mobilized (and so !)
along a slip line decreases with the increase in B.
2. There are zones of weakness which exist in the soil under the
foundation.
3. The curvature of the Mohr-Coulomb envelope.
2.8 EFFECT OF WATER TABLE
The preceding sections assume that the water table is located below the failure
surface in the soil supporting the foundation. However, if the water table is
present close to the foundation, the terms q and " in Eqs. (2.31), (2.37), (2.38),
(2.39) to (2.41), and (2.71) need to be modified. This phenomenon can be ex-
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54.
FIGURE 2.19 Balla’s bearing capacity factor Nq
plained by referring to Fig. 2.22, in which the water table is located at a depth
d below the ground surface
Case I– d = 0
For d = 0, the term q = "Df associated with Nq should be changed to q = "´Df
("´ = effective unit weight of soil). Also, the term " associated with N" should
be changed to "´.
Case II– 0 < d #### Df
For this case, q will be equal to "d + (Df ! d)"´, and the term " associated with
N" should be changed to "´.
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55.
FIGURE 2.20 Balla’s bearing capacity factor N"
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56.
FIGURE 2.21 Nature of variation of N" with B
FIGURE 2.22 Effect of ground water table on ultimate bearing capacity
)( γ′−γ






 −
+γ′=γ
B
Dd f
Case III– Df #### d#### B
This condition is one in which the ground water table is located at or below the
bottom of the foundation. In such case, q = "Df and the last term " should be
replaced by an average effective unit weight of soil, !", or
(2.82)
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57.
Case IV– d > Df + B
For d > Df + B, q = !Df and the last term should remain !. This implies that
the ground water table has no effect on the ultimate capacity.
2.9 GENERAL BEARING CAPACITY EQUATION
The relationships to estimate the ultimate bearing capacity presented in the
preceding sections are for continuous (strip) foundations. They do not give (a)
the relationships for the ultimate bearing capacity for rectangular foundations
(that is, B/L > 0; B = width and L = length), and (b) the effect of the depth of
the foundation on the increase in the ultimate bearing capacity. Therefore, a
general bearing capacity may be written as
qu = cNc "cs "cd + qNq "qs "qd + –
1
2
!BN! "!s "!d (2.83)
where "cs , "qs , "!s = shape factors
"cd , "qd , "!d = depth factors
Most of the shape and depth factors available in literature are empirical and/or
semi-empirical, and they are given in Table 2.5.
It is recommended that, if Eqs.(2.67), (2.66), and (2.74) are used respec-
tively for Nc , Nq , and N! , then DeBeer’s shape factors and Hansen’s depth
factors should be used. However, if Eqs. (2.67), (2.66), and (2.72) are used for
bearing capacity factors Nc , Nq , and N! , then Meyerhof’s shape and depth
factors should be used.
EXAMPLE 2.1
A shallow foundation is 0.6 m wide and 1.2 m long. Given Df = 0.6 m. The soil
supporting the foundation has the following parameters: # = 25!, c = 48
kN/m2
, and ! = 18 kN/m3
. Determine the ultimate vertical load that the
foundation can carry by using
a. Prandtl’s value of Nc [Eq. (2.67)], Reissner’s value of Nq [Eq. (2.66)],
Vesic’s value of N ! [Eq. (2.74)], and the shape and depth factors
proposed by DeBeer and Hansen, respectively (Table 2.5).
b. Meyerhof’s values of Nc , Nq , and N! [Eqs. (2.67), (2.66), and (2.72)]
and the shape and depth factors proposed by Meyerhof [8] given in
Table 2.5.
Solution From Eq. (2.83):
qu = cNc "cs "cd + qNq "qs "qd + –
1
2
!BN! "!s "!d
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58.
TABLE 2.5 Summary of Shape and Depth Factors
Factor Relationship Reference
Shape For # = 0!:
1
1
2.01
=λ
=λ






+=λ
γs
qs
cs
L
B
For # "10!:





 φ
+





+=λ=λ





 φ
+





+=λ
γ
2
45tan1.01
2
45tan2.01
2
2
L
B
L
B
sqs
cs
Meyerhof [8]














+=λ
L
B
N
N
c
q
cs
1
[Note: Use Eq. (2.67) for Nc and Eq. (2.66) for Nq as given in
Table 2.3]






−=λ
φ





+=λ
γ
L
B
L
B
s
qs
4.01
tan1
DeBeer [19]
Depth For # = 0!:
1
2.01
=λ=λ








+=λ
γdqd
f
cd
B
D
For # " 10!:





 φ
+







+=λ=λ





 φ
+







+=λ
γ
2
45tan1.01
2
45tan2.01
B
D
B
D
f
dqd
f
cd
Meyerhof [8]
Factor Relationship Reference
For Df /B # 1:
1
)sin1(tan21
4.01
2
=λ








φ−φ+=λ








+=λ
γd
f
qd
f
cd
B
D
B
D
For Df /B > 1:
1
tan)sin1(tan21
tan4.01
12
1
=λ








φ−φ+=λ








+=λ
γ
−
−
d
f
qd
f
cd
B
D
B
D
















radiansinistan:Note 1-
B
Df
Hansen [9]
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59.
8.0
2.1
6.0
)4.0(14.01
233.125tan
2.1
6.0
1tan1
257.1
2.1
6.0
72.20
66.10
11
=





−=





−=λ
=





+=φ





+=λ
=











+=













+=λ
γ
L
B
L
B
L
B
N
N
s
qs
c
q
cs
1
155.1
6.0
6.0
)25sin1)(25(tan21
)sin1(tan21
4.1
6.0
6.0
)4.0(1)4.0(1
2
2
=λ
=





−+=








φ−φ+=λ
=





+=







+=λ
γd
f
qd
f
cd
B
D
B
D
246.1
2
25
45tan
2.1
6.0
2.01
2
45tan2.01 22
=





+





+=




 φ
+





+=λ
L
B
cs
Hansen’s depth factors are as follows:
So
qu = (48)(20.72)(1.257)(1.4) + (0.6)(18)(10.66)(1.233)(1.155)
+ –
1
2
(18)(0.6)(10.88)(0.8)(1)
a. From Table 2.3, for # = 25!, Nc = 20.72 and Nq = 10.66. Also, from
Table 2.4, for # = 25!, Vesic’s value of N! = 10.88. DeBeer’s shape
factors are as follows:
= 1750.2 + 163.96 + 47 $ 1961 kN/m2
b. From Table 2.3 for # = 25!, Nc = 20.72, Nq = 10.66, and N! = 6.77.
Now, referring to Table 2.5, Meyerhof’s shape and depth factors are
as follows:
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60.
123.1
2
25
45tan
2.1
6.0
1.01
2
45tan1.01 22
=






+





+=




 φ
+





+=λ=λ γ
L
B
sqs
157.1
2
25
45tan
6.0
6.0
1.01
2
45tan1.01
314.1
2
25
45tan
6.0
6.0
2.01
2
45tan2.01
=





+





+=





 φ
+







+=λ=λ
=





+





+=





 φ
+







+=λ
γ
B
D
B
D
f
dqd
f
cd
So
qu = (48)(20.72)(1.246)(1.314)
+ (0.6)(18)(10.66)(1.123)(1.157)
+ –
1
2
(18)(0.6)(6.77)(1.123)(1.157)
= 1628.3 + 149.6 + 47.7 = 1825.6 kN/m2
!!
2.10 EFFECT OF SOIL COMPRESSIBILITY
In Section 2.3, the ultimate bearing capacity equations proposed by Terzaghi
[1] for local shear failure were given [Eqs. (2.39)–(2.41)]. Also, suggestions by
Vesic [4] shown in Eqs. (2.42) and (2.43) address the problem of soil com-
pressibility and its effect on soil bearing capacity. In order to account for soil
compressibility, Vesic [4] proposed the following modifications to Eq. (2.83).
Or
qu = cNc "cs "cd "cc + qNq "qs "qd "qd + –
1
2
!BN! "!s "!d "!c (2.84)
where "cc , "qd , "!c = soil compressibility factors
The soil compressibility factors were derived by Vesic [4] from the
analogy of expansion of cavities [20]. According to this theory, in order to
calculate "cc , "qd , and "!c , the following steps should be taken.
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61.
I
G
c q
r =
+ tanφ



















 φ
−





−=
2
45cot45.03.3exp
2
1
(
L
B
Ir cr)














φ+
φ
+
φ











+−
=λ=λγ
sin1
)2)(logsin07.3(
tan6.04.4
exp
r
qcc
I
L
B
λ λ
λ
φ
cc qc
qc
qN
= −
−1
tan
1. Calculate the rigidity index, Ir , of the soil (approximately at a depth
of B/2 below the bottom of the foundation, or
(2.85)
where G = shear modulus of the soil
! = soil friction angle
q = effective overburden pressure at the level of the
foundation
2. The critical rigidity index of the soil, Ir(cr) , can be expressed as
(2.86)
3. If Ir ! Ir(cr) , then use λcc , λqc , and λγc equal to one. However if Ir < Ir(cr),
(2.87)
For ! = 0
"cc = 0.32 + 0.12 –
B
L
+0.6 logIr (2.88)
For other friction angles
(2.89)
Figures 2.23 and 2.24 show the variations of "#c = "qc [Eq. (2.87)] with ! and
Ir .
EXAMPLE 2.2
Refer to Example 2.1a. For the soil, the given modulus of elasticity, E = 620
kN/m2
; Poisson’s ratio, $ = 0.3. Determine the ultimate bearing capacity.
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62.
I
G
c q
E
c q
r =
+
=
+ +
=
+ + ×
=
tan ( )[ tan ]
( . )[( . ) tan ]
.
φ ν φ2 1
620
2 1 0 3 48 18 0 6 25
4 5
46.62
2
25
45cot
2.1
6.0
45.03.3exp
2
1
2
45cot45.03.3exp
2
1
=




















−





×−=



















 φ
−





−=
L
B
Ir(cr)
Solution
From Eq. (2.86)
FIGURE 2.23 Variation of "#c = "qc with ! and Ir for square foundation
(B/L = 1)
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63.
FIGURE 2.24 Variation of "#c = "qc with ! and Ir for foundations with
L/B > 5
353.0
25sin1
)5.42log()25sin07.3(
25tan
2.1
6.0
6.04.4
exp
sin1
)2)(logsin07.3(
tan6.04.4
exp
=














+
×
+





×+−
=














φ+
φ
+φ





+−
=λ=λ γ
r
cqc
I
L
B
Since Ir(cr) > Ir , use "cc , "qc , and "#c relationships from Eqs. (2.87) and (2.89)
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64.
λ λ
λ
φ
cc qc
qc
qN
= −
−
= −
−
=
1
0 353
1 0 353
10 66 25
0 228
tan
.
.
. tan
.






°
°
φ−φ−φ=φ
90
)( 211
i
Also
Equation (2.84):
qu = (48)(20.72)(1.257)(1.4)(0.228)
+ (0.6)(18) (10.66) (1.233) (1.155)(0.353)
+ –
1
2
(18)(0.6)(10.88)(0.8)(1)(0.353)
= 399.05 + 57.88 + 16.59 " 474 kN/m2
!!
2.11 BEARING CAPACITY OF FOUNDATIONS
ON ANISOTROPIC SOIL
Foundation on Sand (c = 0)
Most natural deposits of cohesionless soil have an inherent anisotropic
structure due to their nature of deposition in horizontal layers. Initial deposition
of the granular soil and subsequent compaction in the vertical direction causes
the soil particles to take a preferred orientation. For a granular soil of this type
Meyerhof suggested that, if the direction of application of deviator stress
makes an angle i with the direction of deposition of soil (Fig. 2.25), then the
soil friction angle ! can be approximated in a form
(2.90)
where!1 = soil friction angle with i = 0#
!2 = soil friction angle with i = 90#
Figure 2.26 shows a continuous (strip) rough foundation on an anisotropic
sand deposit. The failure zone in the soil at ultimate load is also shown in the
figure. In the triangular zone (Zone 1) the soil friction angle will be ! = !1 .
However, the magnitude of ! will vary between the limits of !1 and !2 in Zone
2. In Zone 3 the effective friction angle of the soil will be equal to !2 .
Meyerhof [21] suggested that the ultimate bearing capacity of a continuous
foundation on an anisotropic sand could be calculated by assuming an equiva-
lent friction angle ! = !eq , or
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65.
FIGURE 2.25 Aniostropy in sand deposit
FIGURE 2.26 Continuous rough foundation on anisotropic sand deposit
φ
φ φ φ
eq =
+
=
+( ) ( )2
3
2
3
1 2 1n
where friction ratio = 2
n =
φ
φ1
(2.91)
(2.92)
Once the equivalent friction angle is determined, the ultimate bearing capacity
for vertical loading conditions on the foundation can be expressed as
(neglecting the depth factors)
qu = q Nq(eq) "qs + –
1
2
#B N#(eq) "#s (2.93)
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66.
FIGURE 2.27 Variation of N#(eq) [Eq. (2.93)]
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67.
FIGURE 2.28 Variation of Nq(eq) [Eq. (2.93)]
where Nq(eq) , N#(eq) = equivalent bearing capacity factors corresponding to the
friction angle ! = !eq
In most cases the value of !1 will be known. Figures 2.27 and 2.28
present the plots of Nq(eq) and N#(eq) in terms of n and !1 . Note that the soil
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68.












−γ+





φ





+= γ
L
B
BN
L
B
qNq qu
4.01
2
1
tan1 (eq)eq(eq)
b
a
c
c c
u i
uV uH
=
= °( )
( )( )
45





 +
=
2)(
uHuV
icu
cc
Nq
friction angle ! = !eq was used in Eqs. (2.66) and (2.72) to prepare the graphs.
So combining the relationships for shape factors (Table 2.5) given by DeBeer
[19]
(2.94)
Foundations on Saturated Clay (!!!! = 0 concept)
As in the case of sand discussed above, saturated clay deposits also exhibit
anisotropic undrained shear strength properties. Figures 2.29a and 2.29b show
the nature of variation of the undrained shear strength of clays, cu , with respect
to the direction of principal stress application [22]. Note that the undrained
shear strength plot shown in Fig. 2.29b is elliptical. However, the center of the
ellipse does not match the origin. The geometry of the ellipse leads to the
equation
(2.95)
where cuV = undrained shear strength with i = 0#
cuH = undrained shear strength with i = 90#
A continuous foundation on a saturated clay layer (! = 0) whose
directional strength variation follows Eq. (2.95) is shown in Fig. 2.29c. The
failure surface in the soil at ultimate load is also shown in the figure. Note that,
in Zone I, the major principal stress direction is vertical. The direction of the
major principal stress is horizontal in Zone III; however, it gradually changes
from vertical to horizontal in Zone II. Using the stress characteristic solution,
Davis and Christian [22] determined the bearing capacity factor Nc(i) for the
foundation. For a surface foundation
(2.96)
The variation of Nc(i) with the ratio of a/b (Fig. 2.29b) is shown in Fig. 2.30.
Note that, when a = b, Nc(i) becomes equal to Nc = 5.14 [isotropic case; Eq.
(2.67)].
In manypractical conditions, the magnitudes of cuV and cuH maybe known,
but not the magnitude of cu(i = 45#) . If such is the case, the magnitude of a/b [Eq.
(2.95)]cannotbedetermined.Forsuchconditions,thefollowingapproximation
may be used
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69.
FIGURE 2.29 Bearing capacity of continuous foundation on anisotropic
saturated clay
14.5
2
9.0
=
↑





 +
≈ uHuV
cu
cc
Nq
(2.97)
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70.
FIGURE 2.30 Variation of Nc(i) with a/b based on the analysis of Davis and
Christian
qdqsqcdcs
iuHiuV
icu
qN
cc
Nq λλ+λλ




 +
= °=°=
2
)90()0(
)(
qdqsfcdcs
uHuV
icu
D
cc
Nq λλγ+λλ




 +
=
2)(
The preceding equation, which was suggested by Davis and Christian [22], is
based on the undrained shear strength results of several clays. So, in general,
for a rectangular foundation with vertical loading condition
(2.98)
For ! = 0 condition, Nq = 1 and q = #Df . So
(2.99)
The desired relationships for the shape and depth factors can be taken from
Table 2.5 and the magnitude of qu can be estimated.
Foundation on c–!!!! Soil
The ultimate bearing capacity of a continuous shallow foundation supported
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71.
FIGURE 2.31 Anisotropic clay soil—assumptions for bearing capacity evaluation
by anisotropic c–! soil was studied by Reddy and Srinivasan [23] using the
method of characteristics. Accordingto this analysis the shear strength of a soil
can be given as
s = %´tan! + c
However, it is assumed that the soil is anisotropic only with respect to
cohesion. As mentioned previously in this section, the direction of the major
principal stress (with respect to the vertical) along a slip surface located below
the foundation changes. In anisotropic soils, this will induce a change in the
shearing resistance to the bearing capacityfailure of the foundation. Reddy and
Srinivasan [23] assumed the directional variation of c at a given depth z below
the foundation as (Fig. 2.31a)
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72.
K
c
c
V z
H z
=
( )
( )
β
α
γ
c
V z
V z
l
c
l
c
=
′
=
=
=
( )
( )
0
0
where characteristic length =
ci(z) = cH(z) + [cV(z) $ cH(z)]cos2
i (2.100)
where ci(z) = cohesion at a depth z when the major principal stress is inclined
at an angle i to the vertical (Fig. 2.31b)
cV(z) = cohesion at depth z for i = 0#
cH(z) = cohesion at depth z for i = 90#
The preceding equation is of the form suggested by Casagrande and Carrillo
[24].
Figure 2.31b shows the nature of variation of ci(z) with i. The anisotropy
coefficient K is defined as the ratio of cV(z) to cH(z) .
(2.101)
In overconsolidated soils K is less than one and, for normally consolidated
soils the magnitude of K is greater than one.
For many consolidated soils, the cohesion increases linearly with depth
(Fig. 2.31c). Thus
cV(z) = cV(z=0) + &´s (2.102)
where cV(z) , cV(z=0) = cohesion in the vertical direction (that is, i = 0) at depths
of z and z = 0, respectively
&´ = the rate of variation with depth z
According to this analysis, the ultimate bearing capacity of a continuous foun-
dation may be given as
qu = cV(z=0) Nc(i´) + qNq(i´) + –
1
2
#BN#(i´) (2.103)
where Nc(i´) , Nq(i´) , N#(i´) = bearing capacity factors
q = #Df
This equation is similar to Terzaghi’s bearing capacity equation for continuous
foundations [Eq. (2.31)].
The bearing capacity factors are functions of the parameters 'c and K. The
term 'c can be defined as
(2.104)
(2.105)
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73.
FIGURE 2.32 Reddy and Srinivasan’s bearing capacity factor, Nc(i%) —
influence of K ('c = 0)
Furthermore, Nc(i´) is also a function of the nondimensional width of the
foundation, B´
B´ = –
B
l
(2.106)
The variations of the bearing capacity factors with 'c , B´, !, and K
determined using the method of analysis by Reddy and Srinivasan [23] are
shown in Figs. 2.32 to 2.37. This study shows that the rupture surface in soil
at ultimate load extends to a smaller distance below the bottom of the founda-
tion for the case where the anisotropic coefficient K is greater than one. Also,
when K changes from one to two with &´ = 0, the magnitude of Nc(i´) is reduced
by about 30% – 40%.
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74.
FIGURE 2.33 Reddy and Srinivasan’s bearing capacity factor Nc(i —
influence of K ('c = 0.2)
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75.
FIGURE 2.34 Reddy and Srinivasan’s bearing capacity factor, Nc(i%) —
influence of K ('c = 0.4)
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76.
FIGURE 2.35 Reddy and Srinivasan’s bearing capacity factors, N#(i%)
and Nq(i%) ( 'c = 0)
EXAMPLE 2.3
Estimate the ultimate bearing capacity qu of a continuous foundation with the
following: B = 9 ft, cV(z=0) = 250 lb/ft2
, &´ = 25 lb/ft2
/ft, Df = 3 ft, # = 110 lb/ft3
,
and ! = 20#. Assume K = 2.
Solution From Eq. (2.105)
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77.
Characteristic length,
Nondimensional width,
l
c
B
B
l
V z
= = =
′ = = =
=( )
.
.
.
0 250
110
227
9
227
396
γ
Also
FIGURE 2.36 Reddy and Srinivasan’s bearing capacity factors, N#(i%)
and Nq(i%) — influence of K ('c = 0)
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78.
FIGURE 2.37 Reddy and Srinivasan’s bearing capacity factors, N#(i%)
and Nq(i%) — influence of K ('c = 0.2)
β
α
c
V z
l
c
=
′
= =
=( )
( )( . )
.
0
25 2 27
250
0 227
Now, referring to Figs. 2.33, 2.34, 2.36, and 2.37, for ! = 20#, 'c = 0.227, K
= 2, and B´ = 3.96 (by interpolation)
Nc(i´) " 14.5; Nq(i´) " 6, and N#(i´) " 4
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79.
q
q
FS
u
all =
q
q
FS
u
all(net)
(net)
=
From Eq. (2.103)
qu = cV(z=0) Nc(i´) + qNq(i´) + –
1
2
!BN!(i´)
= (250)(14.5) + (3)(110)(6) + –
1
2
(110)(10)(4)
= 7,805 lb/ft2
!!
2.12 ALLOWABLE BEARING CAPACITY
WITH RESPECT TO FAILURE
Allowable bearing capacity for a given foundation may be (a) to protect the
foundation against a bearing capacity failure, or (b) to ensure that the foun-
dation does not undergo undesirable settlement. There are three definitions for
the allowable capacity with respect to a bearing capacity failure. They are:
Gross Allowable Bearing Capacity
The gross allowable bearing capacity is defined as
(2.107)
where qall = gross allowable bearing capacity
FS = factor of safety
In most cases a factor of safety, FS, of 3 to 4 is generally acceptable.
Net Allowable Bearing Capacity
The net ultimate bearing capacity is defined as the ultimate load per unit area
of the foundation that can be supported by the soil in excess of the pressure
caused by the surrounding soil at the foundation level. If the difference be-
tween the unit weight of concrete used in the foundation and the unit weight of
the surrounding soil is assumed to be negligible, then
qu(net) = qu ! q (2.108)
where q = !Df
qu(net) = net ultimate bearing capacity
The net allowable bearing capacity can now be defined as
(2.109)
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80.
A factor of safety of 3 to 4 in the preceding equation is generally considered







 φ
=φ
=
−
(shear)
(shear)
FS
FS
c
c
d
d
tan
tan 1
q
q
FS
u
all = = ≈
1961
4
490 kN / m2
satisfactory.
Allowable Bearing Capacity With Respect to Shear Failure, qall(shear)
For this case, a factor of safety with respect to shear failure, FS(shear) , which
may be in the range of 1.3 to 1.6 is adopted. In order to evaluate qall(shear) , the
following procedure may be used.
1. Determine the developed cohesion, cd , and the developed angle of
friction, "d , as
(2.110)
(2.111)
2. The gross and net ultimate allowable bearing capacities with respect
to shear failure can now be determined as [Eq. (2.83)]
qall(shear)—gross = cd Nc #cs #cd + qNq #qs #qd + –
1
2
!BN! #!s #!d (2.112)
qall(shear)—net = qall(shear)—gross ! q
= cd Nc #cs #cd + q(Nq ! 1)#qs #qd + –
1
2
!BN! #!s #!d (2.113)
where Nc , Nq , and N! = bearing capacity factors for friction angle "d
EXAMPLE 2.4
Refer to Example Problem 2.1a and determine
a. The gross allowable bearing capacity. Assume FS = 4.
b. The net allowable bearing capacity. Assume FS = 4.
c. The gross and net allowable bearing capacity with respect to shear
failure. Assume FS(shear) = 1.5.
Solution
a. From Example Problem 2.1a, qu = 1961 kN/m2
.
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81.
8.0
2.1
6.0
4.014.01
156.13.17tan
2.1
6.0
1tan1
192.1
2.1
6.0
5.12
8.4
11
=





−=





−=λ
=





+=φ





+=λ
=











+=













+=λ
γ
L
B
L
B
L
B
N
N
s
dqs
c
q
cs
1
308.1
6.0
6.0
)3.17sin1)(3.17)(tan2(1
)sin1(tan21
4.1
6.0
6.0
4.014.01
2
=λ
=





−+=








φ−φ+=λ
=





+=







+=λ
γd
f
qd
f
cd
B
D
B
D
b. q
q q
FS
u
all(net) =
−
=
−
≈
1961 0 6 18
4
( . )( )
488 kN / m2
c.
°=





=







 φ
=φ
===
−−
3.17
5.1
25tan
tan
tan
tan
32
5.1
48
11
(shear)
2
(shear)
kN/m
FS
FS
c
c
d
d
For " = 17.3", Nc = 12.5, Nq = 4.8 (Table 2.3), and N! = 3.6 (Table
2.4)
From Eq. (2.112)
qall(shear)—gross = cd Nc #cs #cd + qNq #qs #qd + –
1
2
!BN! #!s #!d
= (32)(12.5)(1.192)(1.4) + (0.6)(18)(4.8)(1.156)×
(1.308) + –
1
2
(18)(0.6)(3.6)(0.8)(1)
= 667.5 + 78.4 + 15.6 = 761.5 kN/m2
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82.
From Eq. (2.113)
qall(shear)—net = 761.5 ! q = 761.5 ! (0.6)(18) # 750.7 kN/m2
!!
2.13 INTERFERENCE OF CONTINUOUS FOUNDATIONS
IN GRANULAR SOIL
In earlier sections of this chapter, theories relating to the ultimate bearing
capacity of single rough continuous foundations supported by a homogeneous
soil medium extending to a great depth were discussed. However, if founda-
tions are placed close to each other with similar soil conditions, the ultimate
bearing capacity of each foundation maydecrease due to the interference effect
of the failure surface in the soil. This was theoretically investigated by Stuart
[25] for granular soils. The results of this study are summarized in this section.
Stuart [25] assumed the geometry of the rupture surface in the soil mass
to be the same as that assumed by Terzaghi (Fig. 2.1). According to Stuart, the
following conditions may arise (Fig. 2.38)
1. Case 1 (Fig. 2.38a): If the center-to-center spacing of the two
foundations is x $ x1 , the rupture surface in the soil under each
foundation will not overlap. So the ultimate bearing capacity of each
continuous foundation can be given by Terzaghi’s equation [Eq.
(2.31)]. For c = 0
qu = qNq + –
1
2
!BN! (2.114)
where Nq , N! = Terzaghi’s bearing capacity factors (Table 2.1)
2. Case 2 (Fig. 2.38b): If the center-to-center spacing of the two
foundations (x = x2 < x1) are such that the Rankine passive zones just
overlap, then the magnitude of qu will still be given by Eq. (2.114).
However, the foundation settlement at ultimate load will change
(compared to the case of an isolated foundation).
3. Case 3 (Fig. 2.38c): This is the case where the center-to-center
spacing of the two continuous foundations is x = x3 < x2 . Note that the
triangular wedges in the soil under the foundation make angles of
180" ! 2" at points d1 and d2 . The area of the logarithmic spirals d1
g1 and d1 e are tangent to each other at point d1 . Similarly, the arcs of
the logarithmic spirals d2 g2 and d2 e are tangent to each other at point
d2. For this case, the ultimate bearing capacity of each foundation can
be given as (c = 0)
qu = qNq $q + –
1
2
!BN! $! (2.115)
where $q , $! = efficiency ratios
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83.
FIGURE 2.38 Assumptionsforthefailuresurfaceingranularsoilundertwocloselyspacedroughcontinuousfoundations roughcontinuousfoundations
(Note: α1 = φ, α2 = 45 − φ/2, α3 = 180 − φ)
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84.
FIGURE 2.38 (Continued)
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85.
FIGURE 2.39 Stuart’s interference factor %q
FIGURE 2.40 Stuart’s interference factor %!
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86.
FIGURE 2.41 Comparison of experimental and theoretical %q
The efficiency ratios are functions of x/B and soil friction angle ".
The theoretical variations of $q and $! are given in Figs. 2.39 and
2.40.
4. Case 4 (Fig. 2.38d): If the spacing of the foundation is further
reduced such that x = x4 < x3 , blocking will occur, and the pair of
foundations will act as a single foundation. The soil between the
individual units will form an inverted arch which travels down with
the foundation as the load is applied. When the two foundations
touch, the zone of arching disappears and the system behaves as a
single foundation with a width equal to 2B. The ultimate bearing
capacity for this case can be given by Eq. (2.114), with B being
replaced by 2B in the third term.
Das and Larbi-Cherif [26] conducted laboratory model tests to determine
the interference efficiency ratios ($q and $!) of two rough continuous founda-
tions resting on sand extending to a great depth. The sand used in the model
tests was highly angular, and the tests were conducted at a relative density of
about 60%. The angle of friction " at this relative density of compaction was
39". Load-displacement curves obtained from the model tests were of local
shear type. The experimental variations of $q and $! obtained from these tests
are given in Figs. 2.41 and 2.42. From these figures it may be seen that,
although the general trend of the experimental efficiency ratio variations is
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87.
FIGURE 2.42 Comparison of experimental and theoretical %!
similar to those predicted by theory, there is a large variation in the magnitudes
between the theory and experimental results. Figure 2.43 shows the
experimental variations of Su /B with x/B (Su = settlement at ultimate load).
The elastic settlement of the foundation decreases with the increase in the
center-to-center spacing of the foundation and remains constant at x > about
4B.
REFERENCES
1. Terzaghi, K., Theoretical Soil Mechanics, John Wiley, New York, 1943.
2. Kumbhojkar, A. S., Numerical evaluation of Terzaghi’s N! , J. Geotech. Eng.,
ASCE, 119(3), 598, 1993.
3. Krizek, R. J., Approximation for Terzaghi’s bearing capacity, J. Soil Mech.
Found. Div., ASCE, 91(2), 146, 1965.
4. Vesi!, A. S., Analysis of ultimate loads of shallow foundations, J. Soil Mech.
Found. Div., ASCE, 99(1), 45, 1973.
5. Meyerhof, G. G., The ultimate bearing capacity of foundations, Geotechnique, 2,
301, 1951.
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88.
FIGURE 2.43 Variation of experimental elastic settlement (Si /B) with center-
to-center spacing of two continuous rough foundations
6. Reissner, H., Zum erddruckproblem, in Proc., First Intl. Conf. Appl. Mech.,
Delft, The Netherlands, 1924, 295.
7. Prandtl, L., Uber die eindringungs-festigkeit plastisher baustoffe und die
festigkeit von schneiden, Z. Ang. Math. Mech., 1(1), 15, 1921.
8. Meyerhof, G. G., Some recent research on the bearing capacity of foundations,
Canadian Geotech. J., 1(1), 16, 1963.
9. Hansen, J. B., A Revised and Extended Formula for Bearing Capacity, Bulletin
No. 28, Danish Geotechnical Institute, Copenhagen, 1970.
10. Caquot, A., and Kerisel, J., Sue le terme de surface dans le calcul des fondations
en milieu pulverulent, in Proc., III Intl. Conf. Soil Mech. Found. Eng., Zurich,
Switzerland, 1, 1953, 336.
11. Lundgren, H., and Mortensen, K., Determination by the theory of plasticity of the
bearing capacity of continuous footings on sand, in Proc., III Intl. Conf. Mech.
Found. Eng., Zurich, Switzerland, 1, 1953, 409.
12. Chen, W. F., Limit Analysis and Soil Plasticity, Elsevier Publishing Co., New
York, 1975.
13. Drucker, D. C., and Prager, W., Soil mechanics and plastic analysis of limit
design, Q. Appl. Math., 10, 157, 1952.
14. Biarez, J., Burel, M., and Wack, B., Contribution à l’étude de la force portante
des fondations, in Proc., V Intl. Conf. Soil Mech. Found. Eng., Paris, France, 1,
1961, 603.
15. Ingra, T. S., and Baecher, G. B., Uncertainty in bearing capacity of sand, J.
Geotech. Eng., ASCE, 109(7), 899, 1983.
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89.
16. Ko, H. Y., and Davidson, L. W., Bearing capacity of footings in plane strain, J.
Soil Mech. Found. Div., ASCE, 99(1), 1, 1973.
17. Hu, G. G. Y., Variable-factors theory of bearing capacity, J. Soil Mech. Found.
Div., ASCE, 90(4), 85, 1964.
18. Balla, A., Bearing capacity of foundations, J. Soil Mech. Found. Div., ASCE,
88(5), 13, 1962.
19. DeBeer, E. E., Experimental determination of the shape factors of sand,
Geotechnique, 20(4), 307, 1970.
20. Vesi!, A., Theoretical Studies of Cratering Mechanisms Affecting the Stability
of Cratered Slopes, Final Report, Project No. A-655, Engineering Experiment
Station, Georgia Institute of Technology, Atlanta, Ga., 1963.
21. Meyerhof, G. G., Bearing capacity of anisotropic cohesionless soils, Canadian
Geotech. J., 15(4), 593, 1978.
22. Davis, E., and Christian, J. T., Bearing capacity of anisotropic cohesive soil, J.
Soil Mech. Found. Div., ASCE, 97(5), 753, 1971.
23. Reddy, A. S., and Srinivasan, R. J., Bearing capacity of footings on anisotropic
soils, J.Soil Mech. Found. Div., ASCE, 96(6), 1967, 1970.
24. Casagrande, A., and Carrillo, N., Shear failure in anisotropic materials, in
Contribution to Soil Mechanics1941-53, Boston SocietyofCivil Engineers,122,
1944.
25. Stuart, J. G., Interference between foundations with special reference to surface
footing on sand, Geotechnique, 12(1), 15, 1962
26. Das, B. M., and Larbi-Cherif, S., Bearing capacity of two closely spaced shallow
foundations on sand, Soils and Foundations, 23(1), 1, 1983.
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90.
CHAPTER THREE
ULTIMATE BEARING CAPACITY UNDER INCLINED
AND ECCENTRIC LOADS
3.1 INTRODUCTION
Due to bending moments and horizontal thrusts transferred from the super-
structure, shallow foundations are many times subjected to eccentric and
inclinedloads.Undersuchcircumstances,theultimatebearingcapacitytheories
presented in Chapter 2 will need some modification, and this is the subject of
discussion in this chapter. The chapter is divided into two major parts. The first
part discusses the ultimate bearing capacity of shallow foundations subjected
to centric inclined load, and the second part is devoted to the ultimate bearing
capacity under eccentric loading.
FOUNDATIONS SUBJECTED TO INCLINED LOAD
3.2 MEYERHOF’S THEORY (CONTINUOUS FOUNDATION)
In 1953, Meyerhof [1] extended his theory for ultimate bearing capacity under
vertical loading (Section 2.4) to the case with inclined load. Figure 3.1 shows
the plastic zones in the soil near a rough continuous (strip) foundation with
small inclined load. The shear strength of the soil, s, is given as
s = c! !! tan" (3.1)
where c = cohesion
!! = effective vertical stress
" = angle of friction
The inclined load makes an angle # with the vertical. It needs to be pointed
out that Fig. 3.1 is an extension of Fig. 2.7. In Fig. 3.1, abc is an elastic zone,
bcd is a radial shear zone, and bde is a mixed shear zone. The normal and shear
stresses on plane ae are po and so , respectively. Also, the unit base adhesion is
c´a . The solution for the ultimate bearing capacity, qu , can be expressed as
qu(v) = qucos# = cNc+ po Nq+ –
1
2
$BN$ (3.2)
where Nc , Nq , N$ = bearing capacity factors for inclined loading condition
$ = unit weight of soil
Similar to Eqs. (2.71), (2.59), and (2.70), we can write
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91.
FIGURE 3.1 Plastic zones in soil near a foundation with inclined load
φθ
φθ
φ+ηφ−
φ−ψφ+
=












−
φ+ηφ−
φ−ψφ+
φ=
tan2
tan2
)2sin(sin1
)2sin(sin1
1
)2sin(sin1
)2sin(sin1
cot
eN
eN
q
c
qu(v) = qucos# = q´u(V) + q!u(V) (3.3)
where
q´u(v) = cNc + po Nq (for ""0, $"0, po"0, c"0) (3.4)
and
q!u(v) = –
1
2
$BN$ (for ""0, $"0, po"0, c"0) (3.5)
It was shown by Meyerhof [1] in Eq. (3.4) that
(3.6)
(3.7)
Note that the horizontal component of the inclined load per unit area on the
foundation, q´h , cannot exceed the shearing resistance at the base, or
q´u(h) # ca + q´u(v) tan% (3.8)
where ca = unit base adhesion
% = unit base friction angle
In order to determine the minimum passive force per unit length of the
foundation, Pp$(min) (see Fig. 2.11 for comparison), to obtain N$ , one can take
a numerical step-by-step approach as shown by Caquot and Kerisel [2] or a
semi-graphical approach based on the logarithmic spiral method as shown by
Meyerhof [3]. Note that the passive force Pp$ acts at an angle " with the
normal drawn to the face bc of the elastic wedge abc (Fig. 3.1). The
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92.
)(for δ≤α
φ
φ−ψψ
−





φ−ψ+
φ−ψ
ψ
γ
= γ
γ
cos
)cos(sin
)cos(
)cos(
sin2 2
2
(min)
B
P
N
p
FIGURE 3.2 Meyerhof’s [1] bearing capacity factor Ncq for purely
cohesive soil (" = 0)
relationship for N$ is
(3.9)
The ultimate bearing capacity expression given by Eq. (3.2) can also be
depicted as
qu(v) = qu cos# = cNcq + –
1
2
$BN$q (3.10)
where Ncq , N$q = bearing capacity factors which are functions of the soil
friction angle, ", and the depth of the foundation, Df
For a purely cohesive soil (" = 0)
qu(v) = qu cos# = cNcq (3.11)
Figure 3.2 shows the variation of Ncq for a purely cohesive soil (" = 0) for
various load inclinations (#).
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93.
FIGURE 3.3 Meyerhof’s [1] bearing capacity factor N$q for cohesionless
soil (c =0, % = ")
For cohesionless soils, c = 0 and, hence, Eq. (3.10) gives
qu(v) = qu cos# = –
1
2
$BN$q (3.12)
Figure 3.3 shows the variation of N$q with #.
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94.
2
2
1
90
1






°φ
°α
−=λ






°
°α
−=λ=λ
γi
qici
5
5
cotcos
sin7.0
1
1
1
cotcos
sin5.0
1








φ+α
α
−=λ
↑








−
λ−
−λ=λ








φ+α
α
−=λ
γ
BLcQ
Q
N
BLcQ
Q
u
u
i
q
qi
qici
u
u
qi
2.3Table
3.3 GENERAL BEARING CAPACITY EQUATION
The general ultimate bearing capacity equation for a rectangular foundation
given by Eq. (2.82) can be extended to account for inclined load and can be
expressed as
qu = cNc &cs &cd &ci + qNq &qs &qd &qi + –
1
2
$BN$ &$s &$d &$i (3.13)
where Nc , Nq , N$ = bearing capacity factors [for Nc and Nq , use Table 2.3;
for N$ , see Table 2.4 — Eqs. (2.72), (2.73), (2.74)]
&cs , &qs , &$s = shape factors (Table 2.5)
&cd , &qd , &$d = depth factors (Table 2.5)
&ci , &qi , &$i = inclination factors
Meyerhof [4] provided the following inclination factor relationships
(3.14)
(3.15)
Hansen [5] also suggested the following relationships for inclination factors
(3.16)
(3.17)
(3.18)
where, in Eqs. (3.14) to (3.18)
# = inclination of the load on the foundation with the vertical
Qu = ultimate load on the foundation = qu BL
B = width of the foundation
L = length of the foundation
3.4 OTHER RESULTS FOR FOUNDATIONS WITH
CENTRIC INCLINED LOAD
Based on the results of field tests, Muhs and Weiss [6] concluded that the ratio
of the vertical component Qu(v) of the ultimate load with the inclination # with
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95.
Q
Q
u v
u
( )
( )
( tan )
α
α
=
= −
0
2
1
Q
BL
Q
BL
q
q
u v
u
u v
u
( )
( )
( )
( )
( tan )
α α
α
= =
= = −
0 0
2
1
q
BN
u
q
=
γ
α
γ
2cos
D
B
f
= =
12
12
1
.
.
;
the vertical to the ultimate load Qu when the load is vertical (that is, # = 0) and
is approximately equal to (1$tan#)2
.
or
(3.19)
Dubrova [7] developed a theoretical solution for the ultimate bearing
capacity of a continuous foundation with centric inclined load and expressed
it in the following form
qu = c(Nq
*
$ 1)cot" +2qNq
*
+ B$N$
*
(3.20)
where Nq
*
, N$
*
= bearing capacity factors
q = $Df
The variations of Nq
*
and N$
*
are given in Figs. 3.4. and 3.5.
EXAMPLE 3.1
Consider a continuous foundation in a granular soil with the following: B = 1.2
m; Df = 1.2 m; unit weight of soil, $ = 17 kN/m3
; soil friction angle, " = 40%;
load inclination, # = 20%. Calculate the gross ultimate load bearing capacity
qu .
a. Use Eq. (3.12).
b. Use Eq. (3.13) and Meyerhof’s bearing capacity factors (Table 2.3),
his shape and depth factors (Table 2.5); and inclination factors [Eqs.
(3.14) and (3.15)].
Solution
a. From Eq. (3.12)
" = 40%; and #=20%. From Fig. 3.3, N$q & 100. So
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96.
FIGURE 3.4 Variation of Nq
*
qu = =
( )( . )( )
cos
17 12 100
2 20
1085.5 kN / m2
b. With c = 0 and B/L = 0, Eq. (3.13) becomes
qu = qNq &qd &qi + –
1
2
$BN$ &$d &$i
For "= 40%, from Table 2.3, Nq = 64.2 and N$ = 93.69. From Table
2.5,
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97.
214.1
2
40
45tan
2.1
2.1
1.01
2
45tan1.01
=





+





+=





 φ
+







+=λ=λ γ
B
Df
dqd
25.0
40
20
11
605.0
90
20
1
90
1
22
22
=





−=





φ
α
−=λ
=





−=




 α
−=λ
γi
qi
From Eqs. (3.14) and (3.15)
FIGURE 3.5 Variation of N$
*
© 1999 by CRC Press LLC

98.
q
BN
u
q
=
γ
α
γ
2cos
qu = ≈
( )( . )( )
cos
17 12 65
2 20
706 kN / m2
So
qu = (1.2 × 17)(64.2)(1.214)(0.65) + –
1
2
(17)(1.2)(93.69)(1.214)(0.25)
= 1323.5 kN / m2
!!
EXAMPLE 3.2
Consider the continuous foundation described in Example 3.1. Other quan-
tities remaining the same, let " = 35%.
a. Calculate qu using Eq. (3.12).
b. Calculate qu using Eq. (3.20).
Solution
a. From Eq. (3.12)
From Fig. 3.3, N$q & 65
b. For c = 0, Eq. (3.20) becomes
qu = 2qNq
*
+ B$N$
*
Using Figs. 3.4 and 3.5, for " = 35% and tan# = tan20=0.36, Nq
*
& 8.5
and N$
*
& 6.5 (extrapolation)
qu = (2)(17 × 1.2)(8.5) + (1.2)(17)(6.5) & 480 kN / m2
Note: Eq. (3.20) does not provide depth factors. !!
3.5 CONTINUOUS FOUNDATION WITH ECCENTRIC LOAD
When a shallow foundation is subjected to an eccentric load, it is assumed that
the contact pressure decreases linearly from the toe to the heel. However, at
ultimate load, the contact pressure is not linear. This problem was analyzed by
Meyerhof [1], who suggested the concept of effective width, B´. The effective
width is defined as (Fig. 3.6)
B´ = B $ 2e (3.21)
© 1999 by CRC Press LLC

99.
FIGURE 3.6 Effective width B!
where e = load eccentricity
According to this concept, the bearing capacityof a continuous foundation
can be determined by assuming that the load acts centrally along the effective
contact width as shown in Fig. 3.6. Thus, for a continuous foundation [from
Eq. (2.83)] with vertical loading
qu = cNc !cd + qNq !qd + –
1
2
" B´N" !"d (3.22)
Note that the shape factors for a continuous foundation are equal to one. The
ultimate load per unit length of the foundation, Qu , can now be calculated as
Qu = qu A´
where A´ = effective area = B´ × 1 = B´
Reduction Factor Method
Purkayastha and Char [8] carried out stability analysis of eccentrically loaded
continuous foundations using the method of slices proposed by Janbu [9].
Based on that analysis, they proposed that
© 1999 by CRC Press LLC

100.
R
q
q
k
u c
u
= −1
( )
( )
eccentri
centric
k
k
B
e
aR 





=














−=−=
k
ukuu
B
e
aqRqq 1)1( )()()( centriccentriceccentric
(3.23)
where Rk = reduction factor
qu(eccentric) = ultimate bearing capacity of eccentrically loaded
continuous foundations
qu(centric) = ultimate bearing capacity of centrally loaded continuous
foundations
The magnitude of Rk can be expressed as
where a and k are functions of the embedment ratio Df /B (Table 3.1).
TABLE 3.1 Variations of a
and k [Eq. (3.24)]
Df /B a k
0
0.25
0.5
1.0
1.862
1.811
1.754
1.820
0.73
0.785
0.80
0.888
Hence, combining Eqs. (3.23) and (3.24)
(3.25)
where
qu(centric) = cNc !dc + qNq !dq + –
1
2
" BN" !"d (3.26)
Theory of Prakash and Saran
Prakash and Saran [10] provided a comprehensive mathematical formulation
to estimate the ultimate bearing capacity for rough continuous foundations
under eccentric loading. According to this procedure, Fig. 3.7 shows the
assumed failure surface in a c–φ soil under a continuous foundation subjected
© 1999 by CRC Press LLC

101.
FIGURE 3.7 Derivation of the bearing capacity theory of Prakash and
Saran for eccentrically loaded rough continuous foundation
© 1999 by CRC Press LLC

102.
q
Q
B
BN D N cNu
u
e f q e c e=
×
= + +
( )
( ) ( ) ( )
1
1
2
γ γγ
32
83.563.263.10.1 





+





−





−=
B
e
B
e
B
e
S
S
o
e
to eccentric loading. Let Qu be the ultimate load per unit length of the foun-
dation of width B with an eccentricity e. In Fig. 3.7, Zone I is an elastic zone
with wedge angles of #1 and #2 . Zones II and III are similar to those assumed
by Terzaghi (that is, Zone II is a radial shear zone and Zone III is a Rankine
passive zone).
The bearing capacity expression can be developed by considering the
equilibrium of the elastic wedge abc located below the foundation (Fig. 3.7b).
Note that, in Fig. 3.7b, the contact width of the foundation with the soil is equal
to Bx1 . Neglecting the self-weight of the wedge
Qu = Pp cos(#1 " $) + Pm cos(#2 " $m) + Ca sin #1 + C´a sin#2 (3.27)
wherePp , Pm = passive forces per unit length of the wedge along the wedge
faces bc and ac, respectively
$ = soil friction angle
$m = mobilized soil friction angle (#$)
Ca = adhesion along wedge face bc =
cBx1 2
1 2
sin
sin( )
ψ
ψ ψ+
C´a = adhesion along wedge face ac =
mcBx1 1
1 2
sin
sin( )
ψ
ψ ψ+
m = mobilization factor (#1)
c = unit cohesion
Equation (3.27) can be expressed in the form
(3.28)
where N"(e) , Nq(e) , Nc(e) = bearing capacity factors for an eccentrically loaded
continuous foundation
The above-stated bearing capacity factors will be functions of e/B, $, and
also the foundation contact factor x1 . In obtaining the bearing capacity factors,
Prakash and Saran [10] assumed the variation of x1 as shown in Fig. 3.7c.
Figures 3.8, 3.9, and 3.10 show the variations of N"(e) , Nq(e) , and Nc(e) with $
and e/B. Note that for e/B = 0 the bearing capacity factors coincide with those
given by Terzaghi [11] for a centrically loaded foundation.
Prakash [12] also gave the relationships for settlement of a given founda-
tion under centric and eccentric loading conditions for a equal factor of safety,
FS. They are as follows (Fig.3.11)
(3.29)
© 1999 by CRC Press LLC

103.
FIGURE 3.8 Prakash and Saran’s bearing capacity
factors, Nc(e)
32
54.3161.2231.20.1 





+





−





−=
B
e
B
e
B
e
S
S
o
m
and
(3.30)
where So = settlement of a foundation under centric loading at
qall(centric) =
qu(centric)
FS
Se , Sm = settlements of the same foundation under eccentric loading at
qall(eccentric) =
qu(eccentric)
FS
© 1999 by CRC Press LLC