SHALLOW
FOUNDATIONS
Bearing Capacity and Settlement
Braja M. Das
CRC Press
Boca Raton London New
York Washington, D.C.
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
various theories developed during the past fifty years for estimating the ultimate
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 foundation settlement, it is desirable to know the principles for estimating the stress
increase in a soil mass supporting a foundation which carries the load transmitted from the superstructure. These principles are also discussed in this
Chapter 5. Recent developments regarding the ultimate bearing capacity of
shallow foundations 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 estimation 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 cameraready 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
To Janice and Valerie
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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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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 (! = 0)
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
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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 GeogridReinforced 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 GeogridReinforced 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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8.4
8.5
8.6
8.7
Theory of Vesic
Saeedy’s Theory
Discussion of Various Theories
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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5
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 categories: 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 foundations, are reinforced concrete slabs of considerable structural rigidity which
support a number of columns and wall loads. When the soil located immediately 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.
FIGURE 1.1 Individual footing
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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 . Theoretically, 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
FIGURE 1.2 General shear failure in soil
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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 foundation 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 = qú , which is usually referred to as the first failure load [2]. At this
FIGURE 1.3 Local shear failure in soil
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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-settlement curve becomes steeper and erratic with the gradual outward and
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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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
FIGURE 1.5 Nature of failure in soil with relative density of sand (Dr) and Df /R
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(1.1)
where A = area of the foundation = BL
P = perimeter of the foundation = 2(B + L)
Thus
R=
BL
2( B + L )
(1.2)
For a square foundation, B = L. So
R=
B
4
(1.3)
From Fig. 1.5 it can be seen that, when D f /R # about 18, punching shear
failure occurs in all cases, irrespective of the relative density of compaction of
sand.
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 by Vesic
[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.
Based on laboratory and field test results, the approximate ranges of values
of Su in various types of soil are given below.
Soil
Sand
Sand
Clay
Clay
Df
B
Su
(%)
B
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
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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
q all =
qu
FS
(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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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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∆=
S t (max) − S t (min)
L′
(for uniform tilt)
(1.5)
( for nonuniform settlement)
(1.6)
and
∆=
S t (max) − S t (min)
L1′
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.
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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 considered 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, q u , 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 sides ac and bc 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. The slip lines in this
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
r = r0 e θ tan φ
(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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FIGURE 2.1 Failure surface in soil at ultimate load for a continuous rough rigid foundation as assumed by Terzaghi
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The shear strength, s, of the soil can be given as
s = σ ′ tan φ + c
(2.2)
where F´ = effective normal stress
c = cohesion
The ultimate bearing capacity, q u , of the foundation can be determined if
we consider faces ac and bc of the triangular wedge abc 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 , cohesion c, 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 to bc. In order to obtain Ppq , Ppc , and
Pp( , the method of superposition can be used; however, it will not be an exact
solution.
Relationship for Ppq (N
N … 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
P p (1 ) = qK p H d = qH d tan
2
φ
45 +
2
(2.4)
where Hd = bb
fj
Kp = Rankine passive earth pressure coefficient = tan2(45 + N/2)
According to the property of a log spiral defined by the equation r =
2tanN
r0e
, 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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FIGURE 2.2 Passive force on the face bc of wedge abc shown in Fig. 2.1
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FIGURE 2.3 Determination of Ppq (N… 0, (= 0, q …0, c' 0)
bj
H
B
= q ( bj ) + P p ( 1) d
2
4
2
p pq
(2.5)
Let
B
sec φ
2
bc = r 0 =
From Eq. (2.1)
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(2.6)
3π φ
− tan φ
bf = r1 = r0 e 4 2
(2.7)
φ
bj = r1 cos 45 −
2
(2.8)
So
and
φ
H d = r1 sin 45 −
2
(2.9)
Combining Eqs. (2.4), (2.5), (2.8), and (2.9)
Ppq B
4
φ
φ
φ
qr12 cos 2 45 − qr12 sin 2 45 − tan 2 45 +
2
2
2
+
=
2
2
or
Ppq =
φ
4 2 2
qr1 cos 45 −
2
B
(2.10)
Now, combining Eqs.(2.6), (2.7), and (2.10)
2 3 π − φ tan φ
φ
Ppq = qB sec 2 φ e 4 2 cos 2 45 −
2
3π φ
2
− tan φ
qBe 4 2
=
φ
4 cos 2 45 +
2
(2.11)
Considering the stability of the elastic wedge abc under the foundation as
shown in Fig.2.3b
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q q (B × 1) = 2Ppq
where q q = load per unit area on the foundation, or
3π φ
2
− tan φ
4
2
2 Ppq
e
= qN
qq =
= q
q
B
φ
2
2
cos
45
+
2
144
42444
3
(2.12)
Nq
N … 0, ( = 0, q = 0, c … 0)
Relationship for Ppc (N
Figure 2.4 shows the free body diagram for the wedge bcfj (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,
P p ( 2 ) = 2 c K p H d = 2 cH d tan 45 +
φ
2
4. Cohesive force per unit area along arc cf, c.
Taking the moment of all the forces about point b
φ
r1 sin 45 −
2
+Mc
P pc = Pp ( 2 )
2
4
B
(2.13)
where Mc = moment due to cohesion c along arc cf
c
2
2
= 2 tan φ (r1 − r0 )
So
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(2.14)
FIGURE 2.4 Determination of Ppc (N… 0, ( = 0, q = 0, c … 0)
φ
r1 sin 45 −
B
φ
2
Ppc = 2 cH d tan 45 +
2
2
4
c 2
+
(r 1 − r 02 )
φ
2
tan
(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 sin 2(45 ! N/2) × tan(45 + N/2) = ½cosN
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P pc
2 3 π − φ tan φ cos φ
= Bc (sec 2 φ ) e 4 2
2
2 34π − φ2 tan φ
e
Bc
+
sec 2 φ
2 tan φ
(2.16)
Considering the equilibrium of the soil wedge abc (Fig. 2.4b)
q c (B × 1) = 2C sinN + 2Ppc
or
q c B = cB secN sinN + 2Ppc
(2.17)
where q c = load per unit area of the foundation
Combining Eqs. (2.16) and (2.17)
q c = c sec φe
−
3 π − φ tan φ
4 2
2
3π φ
φ 2 4 − 2 tan φ
+
e
tan φ
c sec
2
φ
+ c tan φ
tan φ
c sec
2
(2.18)
or
3π φ
4 − 2 tan φ
2
q c = ce
2
sec φ
sec 2 φ
sec
c
φ
+
−
− tan φ
tan φ
tan φ
(2.19)
However
sec φ +
sec
2
φ
tan φ
=
1
cos φ
+
1
cos φ sin φ
1 + sin φ
= cot φ
2
cos φ
1
= cot φ
φ
2
2 cos 45 + 2
Also
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(2.20)
sec 2 φ
tan φ
− tan φ = cot φ(sec φ − tan φ)
2
2
1
sin φ
= cot φ
−
cos φ cos φ
2
2
2
cos φ
= cot φ
= cot φ
cos φ
2
2
(2.21)
Substituting Eqs. (2.20) and (2.21) into Eq. (2.19)
3π φ
− tan φ
2
4 2
e
q c = c cot φ
− 1 = cN c = c cot φ( N q − 1 )
φ
2
2 cos 45 + 2
44 443
442
144
(2.22)
Nc
N … 0, ( … 0, q = 0, c = 0)
Relationship for Pp(( (N
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 which bf is an
arc is at a point O along line bf and not at b. This is because the minimum value
of Pp( has to be determined by several trials. Point O is 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
Pp(3) =
1
2
φ
γ H d2 tan 2 45 +
2
(2.23)
Also note that the line of action of force F will pass through O. Taking the
moment about O
Ppγ l p = Wl w + Pp (3) l R
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FIGURE 2.5 Determination of Pp( (N… 0, (… 0, q = 0, c = 0)
or
Ppγ =
1
Wlw + Pp(3) l R
lp
(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 determined.
Considering the stability of wedge abc as shown in Fig. 2.5, we can write
that
q ( B = 2Pp( !Ww
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(2.25)
where q ( = force per unit area of the foundation
Ww = weight of wedge abc
However,
Ww =
B
2
4
γ tan φ
(2.26)
So
qγ =
1
2
B
2 Pp γ − γ tan φ
4
B
(2.27)
The passive force Pp( can be expressed in the form
1
P pγ =
2
γ h 2 K pγ =
2
B tan φ
1
2
2
γ
K p γ = γ B K p γ tan φ
2
2
8
1
(2.28)
where Kp( = passive earth pressure coefficient
Substituting Eq. (2.28) into Eq. (2.27)
qγ =
=
1 1
2
B
γB 2 K p γ tan 2 φ −
γ tan φ
B 4
4
1
tan φ
1
γB K p γ tan 2 φ −
= γ BN γ
2
2
2
14 44424 4443
1
2
(2.29)
Nγ
Ultimate Bearing Capacity
The ultimate load per unit area of the foundation (that is, the ultimate bearing
capacity q u ) for a soil with cohesion, friction, and weight can now be given as
qu = qq + qc + q(
(2.30)
Substituting the relationships for q q , q c , and q ( given by Eqs. (2.12), (2.22), and
(2.29) into Eq. (2.30) yields
1
q u = cN c + qN q + γ BN γ
(2.31)
2
where Nc , Nq , and N( = bearing capacity factors, and
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3π φ
2
− tan φ
e 4 2
Nq =
φ
2 cos2 45 +
2
(2.32)
Nc = cot!(Nq!1)
Nγ =
(2.33)
tan φ
1
K pγ tan 2 φ −
2
2
(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:
Nc =
228 + 4.3φ
40 − φ
(2.35a)
Nq =
40 + 5φ
40 − φ
(2.35b)
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Nγ=
6φ
40 − φ
(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)
qu =
( 228 + 4.3 φ) c + ( 40 + 5 φ) q + 3 φγB
(for φ = 0° to 35° )
40 − φ
(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.
q u = 1.3cNc + qN q + 0.4(BN( (square foundation; plan B × B)
(2.37)
q u = 1.3cNc + qN q + 0.3(BN( (circular foundation; plan B × B)
(2.38)
and
Since Terzaghi’s founding work, numerous experimental studies to estimate
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( . For derivation of Nc and Nq , the center of the arc of the
log spiral cf 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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FIGURE 2.6 Modified failure surface in soil supporting a shallow foundation at ultimate load
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Strip foundation (B/L = 0; L = length of foundation)
1
qu = c´Nc´ + q´Nq´ + –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 5.70
1 5.90
2 6.10
3 6.30
4 6.51
5 6.74
6 6.97
7 7.22
8 7.47
9 7.74
10 8.02
11 8.32
12 8.63
13 8.96
14 9.31
15 9.67
16 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 3.13
10.90 3.36
11.36 3.61
11.85 3.88
12.37 4.17
12.92 4.48
13.51 4.82
14.14 5.20
14.80 5.60
15.53 6.05
16.03 6.54
17.13 7.07
18.03 7.66
18.99 8.31
20.03 9.03
21.16 9.82
22.39 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 estimating Nc´and Nq´ for
foundations on sand in the form
"´ = tan!1(k tan")
(2.42)
k = 0.67 + Dr − 0.75 Dr2 (for 0 ≤ Dr ≤ 0.67)
(2.43)
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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
1
qu = cN c + qN q + –2 !BN !
where Nc´, Nq´, and N!´ = bearing capacity factors
B = width of the foundation
FIGURE 2.7 Slip line fields for a rough continuous foundation
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(2.44)
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 equilibrium requires
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
R=
s1
cosφ
(2.47)
where R = radius of the Mohr’s circle
Also
so = R cos( 2 η + φ) =
s1 cos( 2 η + φ)
cos φ
(2.48)
Combining Eqs. (2.45), (2.46), and (2.48)
cos( 2 η + φ ) =
so cos φ
m( c + po tan φ ) cos φ
=
c + p1 tan φ
c + p1 tan φ
(2.49)
Again, referring to the trace of plane de (Fig. 2.8c)
s1 = R cos"
R=
c + p1 tan φ
cos φ
Note that
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(2.50)
FIGURE 2.8 Determination of Nq and Nc
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FIGURE 2.8 (Continued)
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FIGURE 2.8 (Continued)
p1 + R sin φ = po + R sin( 2 η + φ)
p1 = R[sin( 2 η + φ) − sin φ] + po
=
c + p1 tan φ
[sin( 2 η + φ) − sin φ] + po
cos φ
(2.51)
Figure 2.8d shows the free body diagram of 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
r2
r2
p1 1 − p′p 0 + M c = 0
2
2
where r0 = bc
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(2.53)
r1 = bd = ro e θ tan φ
(2.55)
It can be shown that
Mc =
c
( r12 − r02 )
2 tan φ
(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
B
B
φ
2
2
cos 45 + + 2 s′p
sin 45 + φ = q′B
2 p′p
φ
φ
2
2
cos 45 + 2
cos 45 + 2
where q´ = load per unit area of the foundation, or
φ
q ′ = p′p + s′p cot 45 −
2
(2.58)
Substituting Eqs. (2.51), (2.52), and (2.57) into Eq. (2.58), and further
simplifying yields
2 θ tan φ
(1 + sin φ)e
(1 + sin φ)e2 θ tan φ
q′ = c cot φ
+ po
− sin φ sin(2η + φ)
− sin φ sin(2η + φ)
14
14
1
4424443
1
4424443
Nc
Nq
= cN c + po N q
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(2.59)
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)
cos( 2 η + φ) =
m( c + p o tan φ) cos φ
c + p1 tan φ
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$ + % ! – (for m = 1)
2
(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)
1 + sin φ
N q = e π tan φ
1 − sin φ
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(2.66)
FIGURE 2.9
Meyerhof’s bearing capacity factor — variation of Nc with
b, f, and m [Eq. (2.59)]
and
Nc = (Nq ! 1)cotf
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(2.67)
FIGURE 2.10 Meyerhof’s bearing capacity factor—variation of Nq
with b, f, 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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FIGURE 2.11 Determination of Ng
q´ =
cNc +
qNq
%
%
Eq. (2.66) Eq. (2.67)
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(2.68)
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
Ppγ =
W lw + Pp ( R )lR
lp
(2.69)
where Pp! = passive force due to ! and f only.
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
φ
4 Ppγ sin 45 +
φ 1
γB
2 1
q′′ =
− tan 45 + = γBN γ
2
2
2
2 2
γB
where q&& = force per unit area of the foundation
N ! = bearing capacity factor
Note that Ww is the weight of wedge abc in Fig. 2.11b.
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(2.70)
FIGURE 2.12 Meyerhof’s bearing capacity factor—variation of N! with
b, f, and m [Eq. (2.70)]
The variation of Ng (determined in the above manner) with b, f, and m is given
in Fig. 2.12.
Combining Eqs. (2.59) and (2.70), the ultimate bearing capacity of a continuous foundation (for the condition c " 0, ! " 0, and " " 0) can be given as
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1
qu = q´ + q&& = cNc + po Nq + –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 + –12 !BN!
%
%
%
%
\
Eq. (2.68) Eq. (2.70) Eq. (2.67) Eq. (2.66)
(2.71)
For shallow foundation design, the ultimate bearing capacity relationship 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 , N q , and N !
[Eqs. (2.66), (2.67), and (2.72)]
Nq
N!
"
Nc
Nq
N!
"
Nc
Nq
N!
0 5.14 1 00
1 5.38 1.09
2 5.63 1.20
3 5.90 1.31
4 6.19 1.43
5 6.49 1.57
6 6.81 1.72
7 7.16 1.88
8 7.53 2.06
9 7.92 2.25
10 8.35 2.47
11 8.80 2.71
12 9.28 2.97
13 9.81 3.26
14 10.37 3.59
15 10.98 3.94
16 11.63 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
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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
1 + sin φ
N q = e π tan φ
1 − sin φ
(2.66)
Nc = (N q ! 1)cot!
(2.67)
and
There has been considerable controversy over 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 triangular soil wedge under a rough continuous foundation to be of the shape
shown in Fig. 2.6. Using integration 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!
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(2.75)
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
Soil
friction
angle, !
(deg)
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
N"
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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 principal stress condition, friction angle anisotropy, and curvature of the Mohr-
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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 foundation). 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 differences 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 capacity formulas, the bearing capacity for rough footings could be seriously overestimated for dense sands. To avoid the controversy, Meyerhof [8] suggested
the following:
B
φ = 1.1 − 0.1 φ t
L
(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
φ < α min < 45 +
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φ
2
and
α max = 45 +
φ
2
The values of Nc , Nq , and N" determined by this procedure are shown in Fig.
2.15.
FIGURE 2.15 Hu’s bearing capacity factors
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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 assumed 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. According to this solution
for a continuous foundation
1
qu = cNc + qNq + – "Bn"
2
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 decreases 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 substantial decrease in the ultimate bearing capacity which can primarily be attributed to the following reasons.
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FIGURE 2.17 Variation of $ with soil friction angle for determination
of Balla’s bearing capacity factors
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FIGURE 2.18 Balla’s bearing capacity factor Nc
1.
2.
3.
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.
There are zones of weakness which exist in the soil under the
foundation.
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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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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FIGURE 2.20 Balla’s bearing capacity factor N"
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FIGURE 2.21 Nature of variation of N" with B
FIGURE 2.22 Effect of ground water table on ultimate bearing capacity
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
", or
replaced by an average effective unit weight of soil, !
d − Df
( γ − γ′)
γ = γ′ +
B
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(2.82)
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
1
qu = cNc "cs "cd + qNq "qs "qd + – !BN! "!s "!d
2
(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 respectively 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):
1
qu = cNc "cs "cd + qNq "qs "qd + – !BN! "!s "!d
2
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TABLE 2.5 Summary of Shape and Depth Factors
Factor
Shape
Relationship
Reference
B
For # = 0!: λ cs = 1 + 0.2
L
λ qs = 1
Meyerhof [8]
λ γs = 1
φ
B 2
For # "10!: λ cs = 1 + 0.2 tan 45 +
2
L
φ
B
λ qs = λ γs = 1 + 0.1 tan 2 45 +
2
L
N B
λ cs = 1 + q
N c L
[Note: Use Eq. (2.67) for Nc and Eq. (2.66) for Nq as given in
Table 2.3]
B
λ qs = 1 + tan φ
L
DeBeer [19]
B
λ γs = 1 − 0.4
L
Depth
Df
For # = 0!: λ cd = 1 + 0 .2 B
λ qd = λ γd = 1
Meyerhof [8]
Df
φ
tan 45 +
For # " 10!: λ cd = 1 + 0.2
2
B
D
φ
λ qd = λ γd = 1 + 0.1 f tan 45 +
2
B
Factor
Relationship
Reference
Df
For Df /B # 1: λ cd = 1 + 0.4
B
Hansen [9]
D
λ qd = 1 + 2 tan φ(1 − sin φ) 2 f
B
λ γd = 1
Df
−1
For Df /B > 1: λ cd = 1 + 0.4 tan
B
D
λ qd = 1 + 2 tan φ(1 − sin φ ) 2 tan −1 f
B
λ γd = 1
Df
-1
is in radians
Note : tan
B
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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:
N B
10.66 0.6
λ cs = 1 + q = 1 +
= 1.257
N
L
20.72 1.2
c
0.6
B
λ qs = 1 + tan φ = 1 +
tan 25 = 1.233
1.2
L
0.6
B
λ γs = 1 − 0.4 = 1 − (0.4)
= 0.8
1.2
L
Hansen’s depth factors are as follows:
Df
0.6
= 1 + ( 0.4)
λ cd = 1 + ( 0.4)
= 1 .4
0.6
B
Df
λ qd = 1 + 2 tan φ(1 − sin φ) 2
B
0.6
= 1 + 2(tan 25)(1 − sin 25) 2
= 1.155
0.6
λ γd = 1
So
qu = (48)(20.72)(1.257)(1.4) + (0.6)(18)(10.66)(1.233)(1.155)
1
+ – (18)(0.6)(10.88)(0.8)(1)
2
= 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:
φ
25
B
0.6 2
λ cs = 1 + 0.2 tan 2 45 + = 1 + 0.2
tan 45 + = 1.246
2
2
L
2
1
.
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25
φ
B
0.6
λqs = λγs = 1+ 0.1 tan245+ = 1+ 0.1 tan245+
2
2
L
1.2
= 1.123
D
φ
λ cd = 1 + 0.2 f tan 45 +
2
B
25
0.6
= 1 + 0.2
= 1.314
tan 45 +
2
0
.
6
D
φ
λ qd = λ γd = 1 + 0.1 f tan 45 +
2
B
25
0.6
= 1 + 0.1
= 1.157
tan 45 +
2
0
.
6
So
qu = (48)(20.72)(1.246)(1.314)
+ (0.6)(18)(10.66)(1.123)(1.157)
1
+ – (18)(0.6)(6.77)(1.123)(1.157)
2
= 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 compressibility 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
1
qu = cNc "cs "cd "cc + qNq "qs "qd "qd + – !BN! "!s "!d "!c
2
(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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1.
Calculate the rigidity index, Ir , of the soil (approximately at a depth
of B/2 below the bottom of the foundation, or
Ir =
2.
G
c + q tan φ
where G = shear modulus of the soil
! = soil friction angle
q = effective overburden pressure at the level of the
foundation
The critical rigidity index of the soil, Ir(cr) , can be expressed as
I r ( cr) =
3.
(2.85)
1
φ
B
exp 3.3 − 0.45 cot 45 −
2
2
L
(2.86)
If Ir ! Ir(cr) , then use λcc , λqc , and λγc equal to one. However if Ir < Ir(cr),
B
− 4.4 + 0.6 tan φ
L
λ γc = λ qc = exp
(3.07 sin φ)(log 2 I r )
+
1 + sin φ
(2.87)
For ! = 0
B
"cc = 0.32 + 0.12 – +0.6 logIr
L
(2.88)
For other friction angles
λ cc = λ qc −
1 − λ qc
N q tan φ
(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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FIGURE 2.23 Variation of "#c = "qc with ! and Ir for square foundation
(B/L = 1)
Solution
Ir =
=
G
E
=
c + q tan φ 2(1 + ν)[c + q tan φ]
620
= 4.5
2(1 + 0.3)[(48 + 18 × 0.6) tan 25]
From Eq. (2.86)
I r (cr) =
=
φ
B
1
exp 3.3 − 0.45 cot 45 −
L
2
2
25
1
0.6
cot 45 − = 62.46
exp 3.3 − 0.45 ×
2
2
1.2
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FIGURE 2.24 Variation of "#c = "qc with ! and Ir for foundations with
L/B > 5
Since Ir(cr) > Ir , use "cc , "qc , and "#c relationships from Eqs. (2.87) and (2.89)
B
− 4 . 4 + 0 . 6 tan φ +
L
λ qc = λ γ c = exp
( 3 . 07 sin φ )(log 2 I r )
1 + sin φ
0 .6
− 4 . 4 + 0 . 6 × 1 . 2 tan 25 +
= exp
= 0 . 353
( 3 . 07 sin 25 ) log( 2 × 4 . 5 )
1 + sin 25
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Also
λ cc = λ qc −
1 − λ qc
N q tan φ
= 0.353 −
1 − 0.353
= 0.228
10.66 tan 25
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
+ – (18)(0.6)(10.88)(0.8)(1)(0.353)
2
= 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
i°
φ = φ 1 − (φ 1 − φ 2 )
90 °
(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 equivalent friction angle ! = !eq , or
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FIGURE 2.25 Aniostropy in sand deposit
FIGURE 2.26 Continuous rough foundation on anisotropic sand deposit
φ eq =
(2φ 1 + φ 2 ) (2 + n) φ1
=
3
3
where n = friction ratio =
φ2
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)
1
qu = q Nq(eq) "qs + – #B N#(eq) "#s
2
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(2.93)
FIGURE 2.27 Variation of N#(eq) [Eq. (2.93)]
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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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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]
B
1
B
q u = qN q (eq) 1 + tan φ eq + γBN γ (eq) 1 − 0.4
L
L
2
(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
b
=
a
cu (i = 45°)
( cuV )( cuH )
(2.95)
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
where
c + c uH
q u = N c ( i ) uV
2
(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 many practical conditions, the magnitudes of cuV and cuH may be known,
but not the magnitude of cu(i = 45#) . If such is the case, the magnitude of a/b [Eq.
(2.95)] cannot be determined. For such conditions, the following approximation
may be used
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FIGURE 2.29 Bearing capacity of continuous foundation on anisotropic
saturated clay
c + cuH
qu ≈ 0.9 N c uV
2
↑
= 5.14
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(2.97)
FIGURE 2.30 Variation of Nc(i) with a/b based on the analysis of Davis and
Christian
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
+ c uH ( i = 90° )
c
q u = N c ( i ) uV ( i = 0° )
λ cs λ cd + qN q λ qs λ qd
2
(2.98)
For ! = 0 condition, Nq = 1 and q = #Df . So
c + c uH
λ cs λ cd + γ D f λ qs λ qd
q u = N c ( i ) uV
2
(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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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. According to 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 capacity failure 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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ci(z) = cH(z) + [cV(z) $ cH(z)]cos2i
(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) .
K=
cV ( z )
(2.101)
cH ( z)
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 foundation may be given as
1
qu = cV(z=0) Nc(i´) + qNq(i´) + – #BN#(i´)
2
(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
βc =
α ′l
cV ( z = 0)
where l = characteristic length =
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(2.104)
cV ( z = 0)
γ
(2.105)
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 foundation 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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FIGURE 2.33 Reddy and Srinivasan’s bearing capacity factor Nc(i —
influence of K ('c = 0.2)
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FIGURE 2.34 Reddy and Srinivasan’s bearing capacity factor, Nc(i%) —
influence of K ('c = 0.4)
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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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FIGURE 2.36 Reddy and Srinivasan’s bearing capacity factors, N#(i%)
and Nq(i%) — influence of K ('c = 0)
cV ( z=0)
250
= 2.27
110
γ
B
9
= 396
Nondimensional width, B′ = =
.
l 2.27
Characteristic length, l =
Also
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=
FIGURE 2.37 Reddy and Srinivasan’s bearing capacity factors, N#(i%)
and Nq(i%) — influence of K ('c = 0.2)
βc =
α ′l
cV ( z = 0)
=
( 25)( 2.27)
= 0.227
250
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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From Eq. (2.103)
1
qu = cV(z=0) Nc(i´) + qNq(i´) + – !BN!(i´)
2
1
= (250)(14.5) + (3)(110)(6) + – (110)(10)(4)
2
= 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 foundation 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
q all =
qu
FS
(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 between 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)
q = !Df
qu(net) = net ultimate bearing capacity
The net allowable bearing capacity can now be defined as
where
q all(net) =
q u (net)
FS
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(2.109)
A factor of safety of 3 to 4 in the preceding equation is generally considered
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
cd =
(2.110)
c
FS (shear)
tan φ
φ d = tan −1
FS (shear)
2.
(2.111)
The gross and net ultimate allowable bearing capacities with respect
to shear failure can now be determined as [Eq. (2.83)]
1
qall(shear)—gross = cd Nc #cs #cd + qNq #qs #qd + – !BN! #!s #!d
2
(2.112)
qall(shear)—net = qall(shear)—gross ! q
1
= cd Nc #cs #cd + q(Nq ! 1)#qs #qd + – !BN! #!s #!d
2
(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.
q all =
qu
1961
=
≈ 490 kN / m 2
FS
4
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b.
q all(net) =
c.
cd =
q u − q 1961 − ( 0.6)(18)
=
≈ 488 kN / m 2
FS
4
c
FS (shear)
=
48
= 32 kN/m 2
1 .5
tan φ
−1 tan 25
φ d = tan −1
= 17.3°
= tan
FS
1. 5
(shear)
For " = 17.3", Nc = 12.5, Nq = 4.8 (Table 2.3), and N! = 3.6 (Table
2.4)
N B
4.8 0.6
λ cs = 1 + q = 1 +
= 1.192
N
L
12.5 1.2
c
B
0.6
λ qs = 1 + tan φ d = 1 +
tan17.3 = 1.156
L
1.2
0.6
B
λ γs = 1 − 0.4 = 1 − 0.4
= 0.8
L
1.2
D
0.6
λ cd = 1 + 0.4 f = 1 + 0.4
= 1.4
0.6
B
D
λ qd = 1 + 2 tan φ(1 − sin φ) 2 f
B
0.6
= 1 + (2)(tan 17.3)(1 − sin 17.3)
= 1.308
0.6
λ γd = 1
From Eq. (2.112)
1
qall(shear)—gross = cd Nc #cs #cd + qNq #qs #qd + – !BN! #!s #!d
2
= (32)(12.5)(1.192)(1.4) + (0.6)(18)(4.8)(1.156)×
1
(1.308) + – (18)(0.6)(3.6)(0.8)(1)
2
= 667.5 + 78.4 + 15.6 = 761.5 kN/m2
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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 foundations are placed close to each other with similar soil conditions, the ultimate
bearing capacity of each foundation may decrease 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
1
qu = qNq + – !BN!
2
2.
3.
(2.114)
where Nq , N! = Terzaghi’s bearing capacity factors (Table 2.1)
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).
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)
1
qu = qNq $q + – !BN! $!
(2.115)
2
where $q , $! = efficiency ratios
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FIGURE 2.38 Assumptions for the failure surface in granular soil under two closely spacedrough continuous foundations rough continuous foundations
(Note: α 1 = φ, α2 = 45 − φ/2, α3 = 180 − φ)
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FIGURE 2.38 (Continued)
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FIGURE 2.39 Stuart’s interference factor %q
FIGURE 2.40 Stuart’s interference factor %!
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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 foundations 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
FIGURE 2.41 Comparison of experimental and theoretical %q
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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
Terzaghi, K., Theoretical Soil Mechanics, John Wiley, New York, 1943.
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.
1.
2.
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FIGURE 2.43 Variation of experimental elastic settlement (Si /B) with centerto-center spacing of two continuous rough foundations
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
Reissner, H., Zum erddruckproblem, in Proc., First Intl. Conf. Appl. Mech.,
Delft, The Netherlands, 1924, 295.
Prandtl, L., Uber die eindringungs-festigkeit plastisher baustoffe und die
festigkeit von schneiden, Z. Ang. Math. Mech., 1(1), 15, 1921.
Meyerhof, G. G., Some recent research on the bearing capacity of foundations,
Canadian Geotech. J., 1(1), 16, 1963.
Hansen, J. B., A Revised and Extended Formula for Bearing Capacity, Bulletin
No. 28, Danish Geotechnical Institute, Copenhagen, 1970.
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.
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.
Chen, W. F., Limit Analysis and Soil Plasticity, Elsevier Publishing Co., New
York, 1975.
Drucker, D. C., and Prager, W., Soil mechanics and plastic analysis of limit
design, Q. Appl. Math., 10, 157, 1952.
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.
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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16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
Ko, H. Y., and Davidson, L. W., Bearing capacity of footings in plane strain, J.
Soil Mech. Found. Div., ASCE, 99(1), 1, 1973.
Hu, G. G. Y., Variable-factors theory of bearing capacity, J. Soil Mech. Found.
Div., ASCE, 90(4), 85, 1964.
Balla, A., Bearing capacity of foundations, J. Soil Mech. Found. Div., ASCE,
88(5), 13, 1962.
DeBeer, E. E., Experimental determination of the shape factors of sand,
Geotechnique, 20(4), 307, 1970.
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.
Meyerhof, G. G., Bearing capacity of anisotropic cohesionless soils, Canadian
Geotech. J., 15(4), 593, 1978.
Davis, E., and Christian, J. T., Bearing capacity of anisotropic cohesive soil, J.
Soil Mech. Found. Div., ASCE, 97(5), 753, 1971.
Reddy, A. S., and Srinivasan, R. J., Bearing capacity of footings on anisotropic
soils, J.Soil Mech. Found. Div., ASCE, 96(6), 1967, 1970.
Casagrande, A., and Carrillo, N., Shear failure in anisotropic materials, in
Contribution to Soil Mechanics 1941-53, Boston Society of Civil Engineers, 122,
1944.
Stuart, J. G., Interference between foundations with special reference to surface
footing on sand, Geotechnique, 12(1), 15, 1962
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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CHAPTER THREE
ULTIMATE BEARING CAPACITY UNDER INCLINED
AND ECCENTRIC LOADS
3.1 INTRODUCTION
Due to bending moments and horizontal thrusts transferred from the superstructure, shallow foundations are many times subjected to eccentric and
inclined loads. Under such circumstances, the ultimate bearing capacity theories
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
1
qu(v) = qucos# = cNc+ po Nq+ – $BN$
2
(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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FIGURE 3.1 Plastic zones in soil near a foundation with inclined load
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)
1
q!u(v) = – $BN$ (for ""0, $"0, po"0, c"0)
2
(3.5)
and
It was shown by Meyerhof [1] in Eq. (3.4) that
1 + sin φ sin(2ψ − φ) 2 θ tan φ
− 1
N c = cot φ
e
1 − sin φ sin(2η + φ)
Nq =
1 + sin φ sin(2ψ − φ) 2θ tan φ
e
1 − sin φ sin(2η + φ)
(3.6)
(3.7)
Note that the horizontal component of the inclined load per unit area on the
foundation, qh́ , cannot exceed the shearing resistance at the base, or
qú(h)
# ca + q´u(v) tan%
(3.8)
where ca = unit base adhesion
% = unit base friction angle
In order to determine the minimum passive force per unit length of the
foundation, Pp$(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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relationship for N$ is
Nγ =
2 Ppγ (min) sin 2 ψ
sin ψ cos(ψ − φ)
+ cos(ψ − φ) −
(for α ≤ δ)
2
γB cos(ψ − φ)
cos φ
(3.9)
The ultimate bearing capacity expression given by Eq. (3.2) can also be
depicted as
1
qu(v) = qu cos# = cNcq + – $BN$q
2
(3.10)
where Ncq , N$q = bearing capacity factors which are functions of the soil
friction angle, ", and the depth of the foundation, Df
For a purely cohesive soil (" = 0)
qu(v) = qu cos# = cNcq
(3.11)
Figure 3.2 shows the variation of Ncq for a purely cohesive soil (" = 0) for
various load inclinations (#).
FIGURE 3.2 Meyerhof’s [1] bearing capacity factor Ncq for purely
cohesive soil (" = 0)
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FIGURE 3.3 Meyerhof’s [1] bearing capacity factor N$q for cohesionless
soil (c =0, % = ")
For cohesionless soils, c = 0 and, hence, Eq. (3.10) gives
1
qu(v) = qu cos# = – $BN$q
2
Figure 3.3 shows the variation of N$q with #.
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(3.12)
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
1
qu = cNc &cs &cd &ci + qNq &qs &qd &qi + – $BN$ &$s &$d &$i
2
(3.13)
where Nc , Nq , N$ = 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 , &$i = inclination factors
Meyerhof [4] provided the following inclination factor relationships
α°
λ ci = λ qi = 1 −
90°
α°
λ γi = 1 −
φ°
2
(3.14)
2
(3.15)
Hansen [5] also suggested the following relationships for inclination factors
0.5Qu sin α
λ qi = 1 −
α
+
φ
cos
cot
Q
BLc
u
1 − λ qi
λ ci = λ qi −
N − 1
q
↑
Table 2.3
5
0.7Qu sin α
λ γi = 1 −
Qu cos α + BLc cot φ
5
(3.16)
(3.17)
(3.18)
where, in Eqs. (3.14) to (3.18)
# = inclination of the load on the foundation with the vertical
Qu = ultimate load on the foundation = qu BL
B = width of the foundation
L = length of the foundation
3.4 OTHER RESULTS FOR FOUNDATIONS WITH
CENTRIC INCLINED LOAD
Based on the results of field tests, Muhs and Weiss [6] concluded that the ratio
of the vertical component Qu(v) of the ultimate load with the inclination # with
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the vertical to the ultimate load Qu when the load is vertical (that is, # = 0) and
is approximately equal to (1$tan#)2.
Qu ( v )
Qu ( α = 0 )
= (1 − tan α ) 2
or
Qu ( v )
BL
Qu ( α = 0)
qu(v )
=
q u ( α = 0)
= (1 − tan α ) 2
(3.19)
BL
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)
Nq*, N$* = bearing capacity factors
q = $Df
The variations of Nq* and N$* are given in Figs. 3.4. and 3.5.
where
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)
qu =
Df
B
γBN γq
2 cos α
=
1.2
= 1;
1.2
" = 40%; and #=20%. From Fig. 3.3, N$q & 100. So
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FIGURE 3.4 Variation of Nq*
qu =
(17)(1.2)(100)
= 1085.5 kN / m 2
2 cos 20
b. With c = 0 and B/L = 0, Eq. (3.13) becomes
1
qu = qNq &qd &qi + – $BN$ &$d &$i
2
For "= 40%, from Table 2.3, Nq = 64.2 and N$ = 93.69. From Table
2.5,
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FIGURE 3.5 Variation of N$*
D
φ
λ qd = λ γd = 1 + 0.1 f tan 45 +
2
B
40
1.2
= 1 + 0.1
tan 45 +
= 1.214
2
1.2
From Eqs. (3.14) and (3.15)
α 20
λ qi = 1 − = 1 − = 0.605
90 90
2
2
2
α 20
λ γi = 1 − = 1 − = 0.25
φ 40
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2
So
1
qu = (1.2 × 17)(64.2)(1.214)(0.65) + – (17)(1.2)(93.69)(1.214)(0.25)
2
= 1323.5 kN / m2
!!
EXAMPLE 3.2
Consider the continuous foundation described in Example 3.1. Other quantities remaining the same, let " = 35%.
a. Calculate qu using Eq. (3.12).
b. Calculate qu using Eq. (3.20).
Solution
a.
From Eq. (3.12)
qu =
γBN γq
2 cos α
From Fig. 3.3, N$q & 65
qu =
(17)(1.2)( 65)
≈ 706 kN / m 2
2 cos 20
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
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(3.21)
FIGURE 3.6 Effective width B!
where e = load eccentricity
According to this concept, the bearing capacity of 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
1
qu = cNc !cd + qNq !qd + – " B´N" !"d
2
(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
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Rk = 1 −
q u ( eccentric )
q u ( centric )
(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
e
R k = a
B
k
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)
k
e
q u ( eccentric ) = q u ( centric ) (1 − Rk ) = q u ( centric ) 1 − a
B
(3.25)
where
1
qu(centric) = cNc !dc + qNq !dq + – " BN" !"d
2
(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
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FIGURE 3.7 Derivation of the bearing capacity theory of Prakash and
Saran for eccentrically loaded rough continuous foundation
© 1999 by CRC Press LLC
to eccentric loading. Let Qu be the ultimate load per unit length of the foundation 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 + Cá 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
$m = mobilized soil friction angle (#$)
cBx1 sin ψ 2
Ca = adhesion along wedge face bc = sin( ψ + ψ )
1
2
mcBx1 sin ψ 1
Cá = adhesion along wedge face ac =
sin( ψ 1 + ψ 2 )
m = mobilization factor (#1)
c = unit cohesion
Equation (3.27) can be expressed in the form
qu =
Qu
1
= γBN γ ( e ) + γD f N q ( e ) + cN c ( e )
( B × 1) 2
(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 foundation under centric and eccentric loading conditions for a equal factor of safety,
FS. They are as follows (Fig.3.11)
2
Se
e
e
e
= 1.0 − 1.63 − 2.63 + 5.83
So
B
B
B
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3
(3.29)
FIGURE 3.8 Prakash and Saran’s bearing capacity
factors, Nc(e)
and
2
Sm
e
e
e
= 1.0 − 2.31 − 22.61 + 31.54
So
B
B
B
where
3
(3.30)
So = settlement of a foundation under centric loading at
qu(centric)
qall(centric) =
FS
Se , Sm = settlements of the same foundation under eccentric loading at
qall(eccentric) =
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qu(eccentric)
FS
FIGURE 3.9 Prakash and Saran’s bearing capacity
factors, Nq(e)
EXAMPLE 3.3
Consider a continuous foundation having a width of 2 m. If e = 0.2 m and the
depth of the foundation Df = 1 m, determine the ultimate load per unit meter
length of the foundation. For the soil use $ = 40$, " = 17.5 kN / m3, and c =
0. Use Meyerhof’s bearing capacity and depth factors. Use the reduction factor
method.
Solution Since c = 0, B/L = 0. From Eq. (3.26)
1
qu(centric) = qNq !dq + – " BN" !"d
2
From Table 2.3, for $ = 40$, Nq = 64.2 and N" = 93.69. Again, from Table 2.5,
Meyerhof’s depth factors are as follows:
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FIGURE 3.10 Prakash and Saran’s bearing capacity
factors, N"(e)
D
φ
λ qd = λ γd = 1 + 0.1 f tan 45 +
2
B
40
1
= 1 + 0.1 tan 45 +
= 1.107
2
2
So
1
qu(centric) = (1)(17.5)(64.2)(1.107) + – (17.5)(2)(93.69)(1.107)
2
= 1243.7 + 1815.2 = 3058.9 kN / m2
According to Eq. (3.25)
e
q u ( eccentric ) = q u ( centric ) (1 − R k ) = q u ( centric ) 1 − a
B
k
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FIGURE 3.11 Notations for Eqs. (3.28) and (3.29)
For Df /B = 1/2 = 0.5, from Table 3.1, a = 1.754 and k = 0.80. So
0 .8
0.2
2
=
−
3058
.
9
1
1
.
754
q u ( eccentric )
≈ 2209 kN/m
2
The ultimate load per unit length
Q = (2209)(B)(1) = (2209)(2)(1) = 4418 kN / m
EXAMPLE 3.4
Solve Example Problem 3.3 using the method of Prakash and Saran.
Solution From Eq. (3.28)
1
Q u = ( B × 1) γBN γ ( e ) + γD f N q ( e ) + cN c ( e )
2
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!!
Given c = 0. For $ = 40$, e/B = 0.2/2 = 0.1. From Figs. 3.9 and 3.10, Nq(e) =
56.09 and N"(e) %55. So
1
Qu = ( 2 × 1) (17.5)(2)( 55) + (17.5)(1)(56.09)
2
= ( 2)( 962.5 + 981.6) = 3888 kN/m
!!
EXAMPLE 3.5
Solve Example Problem 3.3 using Eq. (3.22)..
Solution
For c = 0, from Eq. (3.22)
1
qu = qNq !qd + – " B´N" !"d
2
B´ = B " 2e = 2 " (2)(0.2) = 1.6 m
From Table 2.3, Nq = 64.2 and N" = 93.69. From Table 2.5, Meyerhof’s depth
factors are as follows:
D
φ
λ qd = λ γd = 1 + 0.1 f tan 45 +
2
B
40
1
= 1 + 0.1 tan 45 +
= 1.107
2
2
1
qu = (1 × 17.5)(64.2)(1.107) + – (17.5)(1.6)(93.69)(1.107)
2
= 2695.9 kN/m2
Qu = (B´ × 1)qu = (1.6)(2695.9) % 4313 kN
!!
3.6 ULTIMATE LOAD ON RECTANGULAR FOUNDATION
Meyerhof’s effect area method [1] described in the preceding section can be
extended to determine the ultimate load on rectangular foundations. Eccentric
loading of shallow foundations occurs when a vertical load Q is applied at
a location other than the centroid of the foundation (Fig. 3.12a), or when a
foundation is subjected to a centric load of magnitude Q and momentum M
(Fig. 3.12b). In such cases, the load eccentricities may be given as
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FIGURE 3.12 Eccentric load on rectangular foundation
eL =
MB
Q
(3.31)
ML
Q
(3.32)
and
eB =
where eL , eB = load eccentricities, respectively, in the direction of long and
short axes of the foundation
MB , ML = moment components about the short and long axes of the
foundation, respectively
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FIGURE 3.13 One-way eccentricity of load on foundation
According to Meyerhof [1], the ultimate bearing capacity qu and the
ultimate load Qu of an eccentrically loaded foundation (vertical load) can be
given as
1
qu = cNc !cs !cd + qNq !qs !qd + – "B´N" !"s !"d
2
(3.33)
Qu = (qu)A´
(3.34)
and
where
A´ = effective area = B´L´
B´ = effective width
L´ = effective length
The effective area A´ is a minimum contact area of the foundation such that
its centroid coincides with that of the load. For one-way eccentricity, that is, if
eL =0 (Fig. 3.13a), then
B´ = B ! 2eB; L´ = L; A´ = B´L
(3.35)
However, if eB = 0 (Fig. 3.13b), calculate L ! 2eL . The effective area is
A´ = B(L ! 2eL)
(3.36)
The effective width B´ is the smaller of the two values, that is, B or L ! 2eL .
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FIGURE 3.14 Rectangular foundation
with one-way eccentricity
Based on their model test results Prakash and Saran [10] suggested that, for
rectangular foundations with one-way eccentricity in the width direction (Fig.
3.14), the ultimate load may be expressed as
1
2 γ BN γ ( e ) λ γs ( e ) + γ D f N q ( e ) λ qs ( e )
Q u = q u ( BL ) = ( BL )
+ cN λ
c( e)
cs ( e )
(3.37)
where !"s(e) ,
!qs(e) , !cs(e) = shape factors
The shape factors may be expressed by the following relationships
3e B
B
2e
λ γs ( e ) = 1 .0 + B − 0 .68 + 0 .43 − B
2 B L
B
L
2
(3.38)
where L = length of the foundation
!qs(e) = 1
and
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(3.39)
B
λ cs ( e ) = 1 + 0.2
L
(3.40)
Note that Eq. (3.37) does not contain the depth factors.
For two-way eccentricities (that is, eL " 0 and eB " 0), five possible cases
may arise as discussed by Highter and Anders [13]. They are as follows:
Case I (eL /L # 1/6 and eB /B # 1/6)
This case is shown in Fig. 3.15. For this, calculate
3e
B1 = B1.5 − B
B
(3.41)
3e
L1 = L1.5 − L
L
(3.42)
and
So, the effective area
1
A´ = – B1 L1
2
(3.43)
The effective width B´ is equal to the smaller of B1 or L1 .
Case II (eL /L < 0.5 and 0 < eB /B < 1/6)
This case is shown in Fig. 3.16. Knowing the magnitudes of eL /L and eB /B,
the values of L1 /L and L2 /L (and thus L1 and L2) can be obtained from Figs.
3.17 and 3.18. The effective area is given as
1
A´ = – (L1 + L2)B
2
(3.44)
The effective length L´ is the larger of the two values L1 or L2 . The effective
width is equal to
A´
B´ = —
L´
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(3.45)
FIGURE 3.15 Effective area for the case of
eL /L # 1/6 and eB /B # 1/6
FIGURE 3.16 Effective area for the case of
eL/L < 0.5 and 0 < e B /B < 1/6
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FIGURE 3.17 Plot of eL /L versus L1 /L for eL /L < 0.5 and 0 < eB /B < 1/6
(redrawn after Highter and Anders [13])
FIGURE 3.18 Plot of eL /L versus L2 /L for eL /L < 0.5 and 0 < eB /B < 1/6
(redrawn after Highter and Anders [13])
© 1999 by CRC Press LLC
FIGURE 3.19 Effective area for the case of
eL /L < 1/6 and 0 < eB /B < 0.5
Case III (eL /L < 1/6 and 0 < eB /B < 0.5)
Figure 3.19 shows the case under consideration. Knowing the magnitudes of
eL /L and eB /B, the magnitudes of B1 and B2 can be obtained from Figs. 3.20 and
3.21. So the effective area can be obtained as
1
A´ = – (B1 + B2)L
2
(3.46)
In this case, the effective length is equal to
L´ = L
(3.47)
The effective width can be given as
A´
B´ = —
L
(3.48)
Case IV (eL /L < 1/6 and eB /B < 1/6)
The eccentrically loaded plan of the foundation for this condition is shown in
Fig. 3.22. For this case, the eL /L curves sloping upward in Fig. 3.23 represent
the values of B2 /B on the abscissa. Similarly, in Fig. 3.24 the family of eL /L
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FIGURE 3.20 Plot of eB /B versus B1 /B for eL /L < 1/6 and 0 < eB /B < 0.5
(redrawn after Highter and Anders [13])
FIGURE 3.21 Plot of eB /B versus B2 /B for eL /L < 1/6 and 0 < eB /B < 0.5
(redrawn after Highter and Anders [13])
© 1999 by CRC Press LLC
FIGURE 3.22 Effective area for the case of
eL /L < 1/6 and eB /B < 1/6
FIGURE 3.23 Plot of eB /B versus B2 /B for eL /L < 1/6 and eB /B < 1/6
(redrawn after Highter and Anders [13])
© 1999 by CRC Press LLC
FIGURE 3.24 Plot of eB /B versus L2 /L for eL /L < 1/6 and eB /B < 1/6
(redrawn after Highter and Anders [13])
curves which slope downward represent the values of L2 /L on the abscissa.
Knowing B2 and L2 , the effective area A´ can be calculated. For this case, L´
= L and B´ = A´/L´.
Case V (Circular Foundation)
In the case of circular foundations under eccentric loading (Fig. 3.25a), the
eccentricity is always one way. The effective area A´ and the effective width B´
for a circular foundation are given in a nondimensional form in Fig. 3.25b.
Depending on the nature of the load eccentricity and the shape of the foundation, once the magnitudes of the effective area and the effective width are
determined, they can be used in Eqs. (3.33) and (3.34) to determine the ultimate load for the foundation. In using Eq. (3.33), one needs to remember that
1. The bearing capacity factors for a given friction angle are to be
determined from those presented in Tables 2.3 and 2.4.
2. The shape and depth factors are determined by using the relationships
given in Table 2.5 by replacing B´ for B and L´ for L whenever they
appear.
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FIGURE 3.25 Normalized effective dimension of circular
foundations (after Highter and Anders [13])
3.
The depth factors are determined from the relationships given in Table
2.5. However, for calculating the depth factor, the term B is not
replaced by B´.
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EXAMPLE 3.5
A shallow foundation measuring 4 ft × 6 ft in plan is subjected to a centric load
and a moment. If eB = 0.45 ft, eL = 1.2 ft, and the depth of the foundation is 3
ft, determine the allowable load the foundation can carry. Use a factor of safety
of 4. For the soil given unit weight, ! = 115 lb / ft3 ; friction angle, " = 35!; and
cohesion, c = 0. Use Vesic’s N! (Table 2.4), DeBeer’s shape factors (Table
2.5), and Hansen’s depth factors (Table 2.5).
Solution For this case
e B 0.4
=
= 0.1;
B
4
e L 1.2
=
= 0.2
L
6
For this type of condition, Case II as shown in Fig. 3.16 applies. Referring to
Figs. 3.17 and 3.18
L1
— = 0.865, or L1 = (0.865)(6) = 5.19 ft
L
L2
— = 0.22, or L2 = (0.22)(6) = 1.32 ft
L
From Eq. (3.44)
1
1
A´ = – (L1 + L2)B = – (5.19 + 1.32)(4) = 13.02 ft2
2
2
So
B′ =
A ′ A ′ 13.02
=
=
= 2.51 ft
519
L ′ L1
.
Since c = 0
1
qu = qNq #qs #qd + – !B´N! #!s #!d
2
From Table 2.3 for " = 35!, Nq = 33.30. Also from Table 2.4 for " = 35!,
Vesic’s N! = 48.03.
The shape factors given by DeBeer are as follows (Table 2.5)
B′
2.51
λ qs = 1 + tan φ = 1 +
tan 35 = 1.339
L′
5.19
B′
2.51
λ γs = 1 − 0.4 = 1 − (0.4)
= 0.806
L′
5.19
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The depth factors given by Hansen are as follows:
D
λ qd = 1 + 2 tan φ(1 − sin φ) 2 f
B
3
= 1 + (2)(tan 35)(1 − sin 35) 2 = 1.191
4
λ γd = 1
So
1
qu = (115)(3)(33.3)(1.339)(1.191) + – (115)(2.51)(48.03)(0.806)(1)
2
= 18,321 + 5,587 = 23,908 lb / ft2
So the allowable load on the foundation is
Q=
q u A ′ ( 23,908)(13.02)
=
= 77,820 lb
FS
4
!!
REFERENCES
1.
2.
3.
4.
5.
6.
7.
8.
9.
Meyerhof, G. G., The bearing capacity of foundations under eccentric and
inclined loads, in Proc., III Intl. Conf. on Soil Mech. Found. Eng., Zurich,
Switzerland, 1, 1953, 440.
Caquot, A., and Kerisel, J., Tables for the Calculation of Passive Pressure,
Active Pressure, and the Bearing Capacity of Foundations, Gauthier-Villars,
Paris, 1949.
Meyerhof, G. G., The ultimate bearing capacity of foundations, Geotechnique,
2, 301, 1951.
Meyerhof, G. G., Some recent research on the bearing capacity of foundations,
Canadian Geotech. J., 1(1), 16, 1963.
Hansen, J. B., A Revised and Extended Formula for Bearing Capacity, Bulletin
No. 28, Danish Geotechnical Institute, Copenhagen, 1970.
Muhs, H., and Weiss, K., Inclined load tests on shallow strip footing, in Proc.,
VIII Int. Conf. Soil Mech. Found. Eng., Moscow, 1.3, 1973.
Dubrova, G. A., Interaction of Soils and Structures, Rechnoy Transport,
Moscow, 1973.
Purkayastha, R. D., and Char, R. A. N., Stability analysis for eccentrically
loaded footings, J. Geotech. Eng. Div., ASCE, 103(6), 647, 1977.
Janbu, N., Earth pressures and bearing capacity calculations by generalized
procedure of slices, in Proc., IV Int. Conf. Soil Mech. Found. Eng., London, 2,
1957, 207.
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10.
11.
12.
13.
Prakash, S., and Saran, S., Bearing capacity of eccentrically loaded footings,
J. Soil Mech. Found. Div., ASCE, 97(1), 95, 1971.
Terzaghi, K., Theoretical Soil Mechanics, John Wiley, New York, 1943.
Prakash, S., Soil Dynamics, McGraw-Hill, New York, 1981.
Highter, W. H., and Anders, J. C., Dimensioning footings subjected to
eccentric loads., J. Geotech. Eng., ASCE, 111(5), 659, 1985.
© 1999 by CRC Press LLC
CHAPTER FOUR
SPECIAL CASES OF SHALLOW FOUNDATIONS
4.1 INTRODUCTION
The bearing capacity problems described in Chapters 2 and 3 assume that the
soil supporting the foundation is homogeneous and extends to a great depth
below the bottom of the foundation. They also assume that the ground surface
is horizontal; however, this is not true in all cases. It is possible to encounter a
rigid layer at a shallow depth, or the soil may be layered and have different
shear strength parameters. It may be necessary to construct foundations on or
near a slope. Bearing capacity problems related to these special cases will be
described in this chapter.
4.2 FOUNDATION SUPPORTED BY A SOIL WITH
A RIGID ROUGH BASE AT A LIMITED DEPTH
Figure 4.1a shows a shallow rigid rough continuous foundation supported by
a soil which extends to a great depth. The ultimate bearing capacity of this
foundation can be expressed (neglecting the depth factors) as (Chapter 2)
1
qu= cNc + qNq + – !BN!
2
(4.1)
The procedure for determining the bearing capacity factors Nc , Nq , and N!
in homogeneous and isotropic soils was outlined in Chapter 2. The extent of the
failure zone in soil at ultimate load qu is equal to D. The magnitude of D
obtained during the evaluation of the bearing capacity factor Nc by Prandtl [1]
and Nq by Reissner [2] is given in a nondimensional form in Fig. 4.2. Similarly,
the magnitude of D obtained by Lundgren and Mortensen [3] during the
evaluation of N! is given in Fig. 4.3.
Now if a rigid rough base is located at a depth of H < D below the bottom
of the foundation, full development of the failure surface in soil will be
restricted. In such a case, the soil failure zone and the development of slip lines
at ultimate load will be as shown in Fig. 4.1b. Mandel and Salencon [4]
determined the bearing capacity factors for such a case by numerical integration
using the theory of plasticity. According to Mandel and Salencon’s theory, the
ultimate bearing capacity of a rough continuous foundation with a rigid rough
base located at a shallow depth can be given by the relation
1
qu! cNc* " qNq* " – ! BN!*
2
© 1999 by CRC Press LLC
(4.2)
FIGURE 4.1 Failure surface under a rigid rough continuous foundation:
(a) Homogeneous soil extending to a great depth; (b) With
a rough rigid base located at a shallow depth
where Nc*, Nq*, N!* = modified bearing capacity factors
B = width of foundation
! = unit weight of soil
Note that, for H ! D, Nc* = Nc , Nq* = Nq , and N!* = N! (Lundgren and
Mortensen). The variations of Nc*, Nq*, and N!*with H/B and soil friction angle
" are given in Figs. 4.4, 4.5, and 4.6, respectively.
Neglecting the depth factors, the ultimate bearing capacity of rough
circular and rectangular foundations on a sand layer (c = 0) with a rough rigid
base located at a shallow depth can be given as
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FIGURE 4.2 Variation of D/B with soil friction angle (for Nc and Nq )
FIGURE 4.3 Variation of D/B with soil friction angle (for N! )
1
qu = qNq*#*qs + – !BN!*#*!s
2
(4.3)
where #*qs , #*!s = modified shape factors
The above-mentioned shape factors are functions of H/B and ". Based on
the work of Meyerhof and Chaplin [5] and with simplifying the assumption
that, in radial planes, the stresses and shear zones are identical to those in
transverse planes, Meyerhof [6] evaluated the approximate values of #*qs and
#*!s as
© 1999 by CRC Press LLC
FIGURE 4.4 Mandel and Salencon’s bearing capacity factor
N*c [Eq. (4.2)]
B
λ*qs = 1− m1
L
(4.4)
B
λ*γs = 1− m2
L
(4.5)
and
where L = length of the foundation
© 1999 by CRC Press LLC
FIGURE 4.5 Mandel and Salencon’s bearing capacity factor
N*q [Eq. (4.2)]
The variations of m1 and m2 with H/B and " are given in Figs. 4.7 and 4.8.
Milovic and Tournier [7] and Pfeifle and Das [8] conducted laboratory
tests to verify the theory of Mandel and Salencon [4]. Figure 4.9 shows the
comparison of the experimental evaluation of N!* for a rough surface
foundation (Df = 0) on a sand layer with theory. The angle of friction of the
sand used for these tests was 35". From Fig. 4.9 the following conclusions can
be drawn:
1 The value of N!* for a given foundation increases with the decrease
in H/B.
2. The magnitude of H/B = D/B beyond which the presence of a rigid
rough base has no influence on the N!* value of a foundation is about
50-75% more than that predicted by the theory.
© 1999 by CRC Press LLC
FIGURE 4.6 Mandel and Salencon’s bearing capacity factor N*! [Eq. (4.2)]
For H/B between 0.6 to about 1.9, the experimental values of N!* are
higher than those predicted theoretically.
4. For H/B < about 0.6, the experimental values of N!* are substantially
lower than those predicted by theory. This may be due to two factors:
(a) the crushing of sand grains at such high values of ultimate load,
and (b) the curvilinear nature of the actual failure envelope of soil at
high normal stress levels.
For saturated clay (that is, " = 0), Eq. (4.2) will simplify to the form
3.
q u = cu Nc* + q
(4.6)
Mandel and Salencon [9] performed calculations to evaluate Nc* for continuous
foundations. Similarly, Buisman [10] gave the following relationship for
© 1999 by CRC Press LLC
FIGURE 4.7 Variation of m1 (Meyerhof’s values) for use in the
modified shape factor equation [Eq. (4.4)]
obtaining the ultimate bearing capacity of square foundations, or
2
B
qu( square) = π + 2 +
c +q
−
u
2
2
H
(4.7)
for B − 2 ≥ 0
2H
2
where cu = undrained shear strength
Equation 4.7 can be rewritten as
B
0.5 − 0.707
H
cu + q
qu (square ) = 5.14 1 +
5.14
1444442444443
N c*( square )
© 1999 by CRC Press LLC
(4.8)
FIGURE 4.8 Variation of m2 (Meyerhof’s values) for use in Eq. (4.5)
FIGURE 4.9 Comparison of theory with the experimental results of N*!
(Note: " ! 43"; c ! 0)
© 1999 by CRC Press LLC
Table 4.1 gives the values of Nc*for continuous and square foundations.
TABLE 4.1 Values of Nc*for Continuous
"!0 Condition)
and Square Foundations ("
B
H
2
3
4
5
6
8
10
a
Nc*
Squarea
Continuousb
5.43
5.93
6.44
6.94
7.43
8.43
9.43
5.24
5.71
6.22
6.68
7.20
8.17
9.05
Buisman’s analysis [10]
Mandel and Salencon’s analysis [9]
b
FIGURE 4.10 Foundation on a weaker clay layer underlain by
a stronger clay layer [Note: c u (1) < cu (2) ]
Equations 4.6 and 4.7 assume the existence of a rough rigid layer at a
limited depth. However, if a soft saturated clay layer of limited thickness
[undrained shear strength = cu (1) ] is located over another saturated clay with a
somewhat larger shear strength cu (2) [Note: cu (1) < cu (2) ; see Fig. 4.10], then the
following relationship suggested by Vesic [11] and DeBeer [12] may be used
to estimate the ultimate bearing capacity
© 1999 by CRC Press LLC
B
− 2
c
B
qu = 1 + 0.2 5.14 + 1 − u (1) H
cu(1) + q
B
L
c
2 + 1
u (2 )
L
(4.9)
where L = length of the foundation
4.3 FOUNDATION ON LAYERED SATURATED
ANISOTROPIC CLAY ("
" = 0)
Figure 4.11 shows a shallow continuous foundation supported by layered saturated anisotropic clay. The width of the foundation is B, and the interface
between the clay layers is located at a depth H measured from the bottom of the
foundation. It is assumed that the clays are anisotropic with respect to strength
following the Casagrande-Carillo relationship [13], or
cu(i) = cu(h) + [cu(v) # cu(H) ] sin2i
(4.10)
where
cu(i) = undrained shear strength at a given depth where the
major principal stress is inclined at an angle i with the
horizontal
cu(v) , cu(h) = undrained shear strength for i = 90" and 0", respectively
The ultimate bearing capacity of the continuous foundation can be given as
qu = cu(v)#1 Nc(L) + q
(4.11)
where cu(v)#1 = undrained shear strength of the top soil layer when the
major principal stress is vertical
q = !1 Df
Df = depth of foundation
!1 = unit weight of the top soil layer
Nc(L) = bearing capacity factor
However, the bearing capacity factor, Nc(L) , will be a function of H/B and
cu(v)#2/cu(v)#1, or
H c
N c ( L ) = f , u ( v )− 2
B c u ( v ) − 1
(4.12)
where cu(v)#2 = undrained shear strength of the bottom clay layer when the
major principal stress is vertical
© 1999 by CRC Press LLC
FIGURE 4.11 Shallow continuous foundation on layered
anisotropic saturated clay
Reddy and Srinivasan [14] developed a procedure to determine the variation of Nc(L) . In developing their theory, they assumed that the failure surface
was cylindrical as shown in Fig. 4.12 when the center of the trial failure surface
was at O. They also assumed that the magnitude of cu(v) for the top clay layer
[cu(v)-1 ], and the bottom clay layer [cu(v)-2 ] remained constant with depth z as
shown in Fig. 4.12b.
In Fig. 4.12a, for equilibrium of the foundation, considering forces per unit
length, and taking the moment about point O
θ1
θ
∫
2bqu (r sinθ − b) = 2 r [cu(i)−1 ]
θ1
2
∫
dα + 2 r2[cu(i)−2 ]
dα
(4.13)
0
where
b = half-width of the foundation = B/2
r = radius of the trial failure circle
cu(i)#1 , cu(i)#2 = directional undrained shear strength for layers 1 and 2,
respectively
As shown in Fig. 4.12, let $ be the angle between the failure plane and the
direction of the major principal stress. Referring to Eq. (4.10)
Along arc AC
cu (i)#1 = cu (h)#1 + [cu (v)#1#cu (h)#1]sin2(%"$)
© 1999 by CRC Press LLC
(4.14)
FIGURE 4.12 Assumptions in deriving Nc (L) for a continuous foundation
on anisotropic layered clay
Along arc CE
cu (i)#2 = cu (h)#2 + [cu (v)#2#cu (h)#2]sin2(%"$)
(4.15)
Similarly, along arc DB
cu (i)#1 = cu (h)#1 + [cu (v)#1#cu (h)#1]sin2(%#$)
(4.16)
and, along arc ED
cu (i)#2 = cu (h)#2 + [cu (v)#2#cu (h)#2]sin2(%#$)
© 1999 by CRC Press LLC
(4.17)
Note that for the portion of the arc AE, i = %"$ and, for the portion BE, i =
%#$. Let the anisotropy coefficient be defined as
K=
cu ( v ) − 1
cu ( h ) − 1
=
cu ( v ) − 2
cu ( h ) − 2
(4.18)
The magnitude of the anisotropy coefficient K is less than one for overconsolidated clays and K > 1 for normally consolidated clays. Also, let
c
c
n = u ( v ) − 2 − 1 = u ( h )− 2 − 1
cu ( v ) −1
cu ( h )−1
(4.19)
where n = a factor representing the relative strength of two clay layers
Combining Eqs. (4.13), (4.14), (4.15), (4.16), (4.17), and (4.18)
θ
∫
2bqu (r sin θ − b) = r 2 {cu ( h )−1 + [cu ( v )−1 − cu ( h )−1 ]sin 2 (α + ψ )}dα
θ1
θ
∫
+ r 2 {cu ( h )−1 + [cu ( v )−1 − cu ( h )−1 ] sin 2 (α − ψ )}dα
θ1
θ1
∫
+ r 2 ( n + 1){cu ( h )−1 + [cu ( v )−1 − cu ( h )−1 ]sin 2 (α + ψ )}dα
0
θ1
∫
+ r 2 ( n + 1){cu ( h )−1 + [cu ( v )−1 − cu ( h )−1 ]sin 2 (α − ψ )}dα
(4.20)
0
Or, combining Eqs. (4.18) and (4.20)
2θ + 2 nθ 1 + ( K − 1)θ + n( K − 1)θ 1
2
r
2
qu
K − 1 sin 2( θ + ψ ) sin 2( θ − ψ )
b
=
+
−
2
2
2
c u ( v ) −1
r
2 K sin θ − 1
b
n( K − 1) sin 2(θ + ψ ) sin 2(θ − ψ )
1
1
+
−
2
2
2
(4.21)
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H
where θ 1 = cos −1 cos θ +
r
From Eq. (4.11) note that, with q = 0 (surface foundation),
N c( L ) =
qu
cu ( v ) −1
(4.22)
In order to obtain the minimum value of Nc(L) = qu /cu (v)-1 , the theorem of
maxima and minima need to be used, or
∂N c ( L )
∂θ
=0
(4.23)
=0
(4.24)
and
∂N c ( L )
∂r
Equations (4.21), (4.23), and (4.24) will yield two relationships in terms
of the variables ! and r/b. So, for given values of H/b, K, n, and ", the above
relationships may be solved to obtain values of ! and r/b. These can then be
used in Eq. (4.21) to obtain the desired value of Nc LL) (for given values of H/b,
K, n, and "). Lo [15] showed that the angle " between the failure plane and the
major principal stress for anisotropic soils can be taken to be approximately
equal to 35!. The variations of the bearing capacity factor Nc (L) obtained in this
manner for K = 0.8, 1 (isotropic case), 1.2, 1.4, 1.6, and 1.8 are shown in Fig.
4.13.
If a shallow rectangular foundation B × L in plan is located at a depth Df,
the general ultimate bearing capacity equation [see Eq. (2.83 )] will be of the
form (# = 0 condition)
qu = cu (v)-1 Nc (L) $cs $cd + q $qs $qd
where $cs , $qs = shape factors
$cd , $qd = depth factors
The proper shape and depth factors can be selected from Table 2.5.
© 1999 by CRC Press LLC
(4.25)
FIGURE 4.13
Bearing capacity factor Nc(L) (After Reddy and Srinivasan [15])
© 1999 by CRC Press LLC
FIGURE 4.13
(Continued)
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FIGURE 4.13
(Continued)
© 1999 by CRC Press LLC
EXAMPLE 4.1
Refer to Fig. 4.11. For the foundation, given Df = 0.8 m, B = 1 m, L = 1.6 m,
H = 0.5 m, %1 = 17.8 kN / m3, %2 = 17.0 kN / m3, cu (v)-1 = 45 kN / m2, cu (v)-2 =
30 kN / m2, and anisotropy coefficient, K = 1.4. Estimate the allowable load
bearing capacity of the foundation with a factor of safety, FS = 4.
Solution From Eq. (4.25)
qu = cu (v)-1 Nc (L) $cs $cd + q $qs $qd
0 .5
H
H
=
=
= 1;K = 1 .4
b
B 0 .5
2
cu ( v )− 2
=
c u ( v ) −1
30
= 0 . 67
45
n = 1 " 0.67 = 0.33
So, from Fig. 4.13(d), the value of Nc (L) = 4.75
Using Meyerhof’s shape and depth factors given in Table 2.5
B
1
λ cs = 1 + 0.2 = 1 + (0.2)
= 1.125
L
1.6
λ qs = 1
D
0.8
λ cd = 1 + 0.2 f = 1 + (0.2)
= 1.16
1.0
B
λ cd = 1
So
qu = (45)(4.75)(1.125)(1.16) + (17.8)(0.8)(1.0)(1.0)
= 278.9 + 14.24 = 293.14 kN / m2
q all =
qu
29314
.
=
= 73.29 kN / m 2
FS
4
© 1999 by CRC Press LLC
!!
FIGURE 4.14 Rough continuous foundation on layered soil — stronger
over weaker
4.4 FOUNDATION ON LAYERED c – # SOIL —
STRONGER SOIL UNDERLAIN BY WEAKER SOIL
Meyerhof and Hanna [16] developed a theory to estimate the ultimate bearing
capacity of a shallow rough continuous foundation supported by a strong soil
layer underlain by a weaker soil layer as shown in Fig. 4.14. According to their
theory, at ultimate load per unit area, qu , the failure surface in soil will be as
shown in Fig. 4.14. If the ratio H/B is relatively small, a punching shear failure
will occur in the top (stronger) soil layer followed by a general shear failure in
the bottom (weaker) layer. Considering the unit length of the continuous
foundation, the ultimate bearing capacity can be given as
q u = qb +
2 ( Ca + Pp sin δ )
B
− γ 1H
where B = width of the foundation
%1 = unit weight of the stronger soil layer
Ca = adhesive force along aa# and bb#
Pp = passive force on faces aa# and bb#
qb = bearing capacity of the bottom soil layer
& = inclination of the passive force Pp with the horizontal
© 1999 by CRC Press LLC
(4.26)
Note that, in Eq. (4.26)
Ca = ca H
(4.27)
where ca = unit adhesion
Pp =
=
K pH
K
1
+ ( γ 1 D f )( H ) pH
γ 1 H 2
cos δ
2
cos δ
2 D f K pH
1
γ 1 H 2 1 +
2
H cos δ
(4.28)
where KpH = horizontal component of the passive earth pressure coefficient
Also
1
qb = c2 Nc (2) + %1 (Df + H)Nq (2) + – %2 BN%(2)
2
(4.29)
c2 = cohesion of the bottom (weaker) layer of soil
%2 = unit weight of bottom soil layer
Nc (2) , Nq (2) , N% (2) = bearing capacity factors for the bottom soil layer
(that is, with respect to the soil friction angle of
the bottom soil layer, #2)
Combining Eqs. (4.26), (4.27), and (4.28)
where
qu = q b +
= qb +
1
2 D f K pH sin δ
2c a H
+ 2 γ 1 H 2 1 +
−γ H
B
H cos δ B 1
2
2 D f K pH tan δ
2c a H
+ γ 1 H 2 1 +
− γ 1H
B
H
B
(4.30)
Let
KpH tan& = Ks tan#1
(4.31)
where Ks = punching shear coefficient
So
qu = qb +
2 D f K s tan φ1
2ca H
− γ1H
+ γ 1 H 2 1 +
B
H B
© 1999 by CRC Press LLC
(4.32)
FIGURE 4.15 Meyerhof and Hanna’s theory — variation of Ks with #1
and q2 /q1
The punching shear coefficient can be determined using the passive earth pressure coefficient charts proposed by Caquot and Kerisel [17]. Figure 4.15 gives
the variation of Ks with q2 /q1 and #1 . Note that q1 and q2 are the ultimate
bearing capacities of a continuous surface foundation of width B under vertical
load on homogenous beds of upper and lower soils, respectively, or
1
q1 = c1 Nc (1) + – %1 BN%(1)
2
(4.33)
where Nc (1) , N% (1) = bearing capacity factors corresponding to soil friction
angle #1
1
q2 = c2 Nc (2) + – %2 BN%(2)
2
© 1999 by CRC Press LLC
(4.34)
FIGURE 4.16 Continuous rough foundation on layered soil —
H/B is relatively small
If the height H (Fig. 4.14) is large compared to the width B, then the failure
surface will be completely located in the upper stronger soil layer as shown in
Fig. 4.16. In such case, the upper limit for qu will be of the following form
1
qu = qt = c1 Nc (1) + qNq (1) + – %1 BN%(1)
2
(4.35)
Hence, combining Eqs. (4.32) and (4.35)
qu = qb +
2 D f K s tan φ 1
2c a H
+ γ 1 H 2 1 +
− γ 1 H ≤ qt
B
H
B
(4.36)
For rectangular foundations, the preceding equation can be modified as
B 2c H
qu = qb + 1 + a λ a
L B
2 D f K s tan φ 1
B
λ s − γ 1 H ≤ qt
+ 1 + γ 1 H 2 1 +
L
H
B
where
(4.37)
$a , $s = shape factors
1
qb = c2 Nc(2)$cs(2) + %1(Df + H)Nq(2) $qs(2) + – %2BN%(2) $%s(2) (4.38)
2
© 1999 by CRC Press LLC
1
qt = c1 Nc(1) !cs(1) + "1Df Nq(1) !qs(1) + – "1BN"(1) !"s(1)
2
(4.39)
!cs(1) , !qs(1) , !"s(1) = shape factors for the top soil layer (friction angle = #1;
see Table 2.5)
!cs(2) , !qs(2) , !"s(2) = shape factors for the bottom soil layer (friction angle =
#2; see Table 2.5)
Based on the general equations [Eqs. (4.37), (4.38), and (4.39)], some
special cases may be developed. They are as follow:
Case I: Stronger Sand Layer Over Weaker Saturated Clay (#
#2 = 0)
For this case c1 = 0 and, hence, ca = 0. Also for #2 = 0, Nc (2) = 5.14, N" (2) = 0,
Nq (2) = 1, !cs = 1 + 0.2(B/L), !qs = 1 (shape factors are Meyerhof’s values as
given in Table 2.5). So
2 D f K s tan φ1
B
B
qu = 5.14c2 1 + 0.2 + 1 + γ 1 H 2 1 +
λs
L
H
B
L
+ γ 1 D f ≤ qt
( 4.40)
φ
B
where q t = γ 1 D f N q ( 1 ) 1 + 0 . 1 tan 2 45 + 1
2
L
+
φ
1
B
γ 1 BN γ ( 1 ) 1 + 0 . 1 tan 2 45 + 1
2
2
L
(4.41)
In Eq. (4.41) the relationships for the shape factors !qs and !"s are those
given by Meyerhof [18] as shown in Table 2.5. Note that Ks is a function of
q2/q1 [Eqs. (4.33) and (4.34) ]. For this case
c2 N c ( 2 )
q2
514
. c2
=
=
1
q1
BN
γ 1 BN γ (1) 0.5γ 1 γ (1)
2
(4.42)
Once q2 /q1 is known, the magnitude of Ks can be obtained from Fig. 4.15
which, in turn, can be used in Eq. (4.40) to determine the ultimate bearing
capacity of the foundation qu . The value of the shape factor !s for a strip
foundation can be taken as one. For square or circular foundations, as per the
© 1999 by CRC Press LLC
experimental work for Hanna and Meyerhof [19], the magnitude of !s appears
to vary between 1.1 and 1.27. For conservative design it may be taken as one.
Based on this concept, Hanna and Meyerhof [19] developed some alternative design charts to determine the punching shear coefficient Ks , and these
charts are shown in Figs. 4.17 and 4.18. In order to use these charts, the
ensuing steps need to be followed.
1. Determine q2 /q1 .
2. With known values of #1 and q2 /q1 , determine the magnitude of $/#1
from Fig. 4.17.
3. With known values of #1 , $/#1 , and c2 , determine Ks from Fig. 4.18.
FIGURE 4.17 Hanna and Meyerhof’s analysis — variation of $/#1
with q2 /q1 and #1 (for stronger sand over weaker clay)
EXAMPLE 4.2
Figure 4.19 shows a continuous foundation. If H = 1.5 m, determine the
ultimate bearing capacity qu . Use the results shown in Figs. 4.17 and 4.18.
Solution For a continuous foundation B/L = 0 and with !s = 1, Eq. (4.40)
takes the form
© 1999 by CRC Press LLC
FIGURE 4.18 Hanna and Meyerhof’s analysis for coefficient of punching shear—
stronger sand over weaker clay: (a) #1=50!; (b) #1=45!; (c) #1=40!
© 1999 by CRC Press LLC
FIGURE 4.19
2 D f K s tan φ1
qu = 5.14c2 + γ 1 H 2 1 +
+ γ1Df
H
B
(2)(1.2) K s tan 40
= (5.14)(30) + (17.5)( H 2 )1 +
+ (17.5)(1.2)
H
2
2.4
= 175.2 + 7.342H 2 K s 1 +
H
(a)
To determine Ks , we need to obtain q2/q1 . From Eq. (4.42)
q2
514
. c2
=
q 1 0.5γBN γ (1)
From Table 2.4 for #1 = 40!, Meyerhof’s value of N" (1) is equal to 93.7. so
q2
(5.14)( 30)
=
= 0.094
q 1 ( 0.5)(17.5)( 2)( 93.7)
Referring to Fig. 4.17 for q2 /q1 = 0.094 and #1 = 40!, the value of $/#1 = 0.42.
With $/#1 = 0.42 and c2 = 30 kN/m2, Figure 4.18c gives the value of Ks = 3.89.
Substituting this value of Eq. (a) gives
© 1999 by CRC Press LLC
2.4
qu = 175.2 + 28.56 H 2 1 +
≤ qt
H
(b)
From Eq. (4.41)
φ
B
q t = γ 1 D f N q (1) 1 + 0.1 tan 2 45 + 1
2
L
+
φ
1
B
γ 1 BN γ (1) 1 + 0.1 tan 2 45 + 1
2
2
L
For a continuous foundation B/L = 0. So
1
qt = "1 Df Nq (1) + – "1BN"(1)
2
For #1 = 40!, Meyerhof’s value of N" (1) = 93.7 and Nq (1) = 62.4 (Table 2.4).
Hence
1
qt = (17.5)(1.2)(62.4) + – (17.5)(2)(93.7)
2
= 1348.2 + 1639.75 = 2987.95 kN / m2
(c)
If H = 1.5 m is substituted into Eq. (b)
2.4
2
qu = 175.2 + (28.56)(1.5) 2 1 +
= 342.3 kN/m
1
.
5
Since qu = 342.3 < qt , the ultimate bearing capacity is 342.3 kN / m 2.
!!
Case II: Stronger Sand Layer Over Weaker Sand Layer
For this case, c1 = 0 and ca = 0. Hence, referring to Eq. (4.37)
2 D f K s tan φ 1
B
λ s − γ 1 H ≤ q t
q u = q b + 1 + γ 1 H 2 1 +
L
H
B
where
1
qb = "1(Df + H)Nq(2) !qs(2) + – "2BN"(2) !"s(2)
2
© 1999 by CRC Press LLC
(4.43)
(4.44)
1
qt = "1Df Nq(1) !qs(1) + – "1BN"(1) !"s(1)
2
(4.45)
Using Meyerhof’s shape factors given in Table 2.5
φ
B
λ qs(1) = λ γs(1) = 1 + 0.1 tan2 45 + 1
2
L
(4.46)
φ
B
λ qs( 2 ) = λ γs( 2 ) = 1 + 0.1 tan2 45 + 2
2
L
(4.47)
and
For conservative design, for all B/L ratios, the magnitude of !s can be taken
as one. For this case
q 2 0.5γ 2 BN γ ( 2 ) γ 2 N γ ( 2 )
=
=
q1 0.5γ 1 BN γ (1)
γ 1 N γ (1)
(4.48)
Once the magnitude of q2/q1 is determined, the value of the punching shear
coefficient Ks can be obtained from Fig. 4.15. Hanna [20] suggested that the
friction angles obtained from direct shear tests should be used.
Hanna [20] also provided an improved design chart for estimating the
punching shear coefficient Ks in Eq. (4.43). In this development he assumed
that the variation of $ for the assumed failure surface in the top stronger sand
layer will be of the nature shown in Fig. 4.20, or
$z" = %#2 + az"2
(4.49)
where
η=
q2
q1
q
φ 1 − 2 φ 2
q1
a=
H2
(4.50)
(4.51)
So
q
φ 1 − 2 φ 2
q
q1 z ′ 2
δ z ′ = 2 φ 2 +
H2
q1
© 1999 by CRC Press LLC
(4.52)
FIGURE 4.20 Hanna’s assumption for variation of $
with depth for determination of Ks
The preceding relationship means that at z"! 0 (that is, at the interface of the
two soil layers)
© 1999 by CRC Press LLC
q
δ = 2 φ 2
q1
(4.53)
and at the level of the foundation, that is z! = H
! = "1
(4.54)
Equation (4.49) can also be rewritten as
q
φ 1 − 2 φ 2
q
q1 ( H − z ) 2
δ z = 2 φ 2 +
H2
q1
(4.55)
where ! is the angle of inclination of the passive pressure with respect to the
horizontal at a depth z measured from the bottom of the foundation. So, the
passive force per unit length of the vertical surface aa! (or bb!) is
H
Pp =
γ 1 K pH ( z )
∫ cos δ ( z + D ) dz
f
0
(4.56)
z
where KpH (z) = horizontal component of the passive earth pressure coefficient
at a depth z measured from the bottom of the foundation
The magnitude of Pp expressed by Eq. (4.56), in combination with the
expression !z given in Eq. (4.55), can be determined. In order to determine the
magnitude of the punching shear coefficient Ks given in Eq. (4.31), we need to
know an average value of !. In order to achieve that, the following steps are
taken:
1. Assume an average value of ! and obtain KpH as given in the tables by
Caquot and Kerisel [17].
2. Calculate Pp from Eq. (4.28) using the average values of ! and KpH
obtained from Step 1.
3. Repeat Steps 1 and 2 until the magnitude of Pp obtained from Eq.
(4.28) is the same as that calculated from Eq. (4.56).
4. The average value of !, for which Pp calculated from Eqs. (4.28) and
(4.56) is the same, is the value which needs to be used in Eq. (4.31) to
calculate Ks .
Figure 4.21 gives the relationship for !/"1 versus "2 for various values of
"1 obtained by the above procedure. Using Fig. 4.21, Hanna [20] gave a design
chart for Ks , and this design chart is shown in Fig. 4.22.
© 1999 by CRC Press LLC
FIGURE 4.21 Hanna’s analysis — variation of !/"1
Case III: Stronger Clay Layer ("
"1 = 0) Over Weaker Clay ("
"2 = 0)
For this case Nq (1) and Nq (2) are both equal to one, and N# (1) = N# (2) = 0. Also,
Nc (1) = Nc (2) = 5.14. So, from Eq. (4.37)
B 2c H
B
q u = 1 + 0.2 c 2 N c ( 2 ) + 1 + a λ a + γ 1 D f ≤ q t
L B
L
B
where qt = 1 + 0.2 c1 N c (1) + γ 1 D f
L
(4.57)
(4.58)
For conservative design the magnitude of the shape factor $a may be taken as
one. The magnitude of the adhesion ca is a function of q2 /q1 . For this condition
q 2 c2 N c ( 2 ) 514
. c2 c2
=
=
=
q1 c1 N c (1)
514
. c1 c1
Figure 4.23 shows the theoretical variation of ca with q2 /q1 [16].
© 1999 by CRC Press LLC
(4.59)
FIGURE 4.22 Hanna’s analysis — variation of Ks for stronger
sand over weaker sand
FIGURE 4.23 Analysis of Meyerhof and Hanna for the variation
of ca /c1 with c2 /c1
© 1999 by CRC Press LLC
EXAMPLE 4.3
Figure 4.24a shows a shallow foundation. Given: undrained shear strength c1
(for "1 = 0 condition) = 80 kN / m 2; undrained shear strength c2 (for "2 = 0
condition) = 32 kN / m 2; #1 = 18 kN / m 3; Df = 1 m; B = 1.5 m; and L = 3 m.
Plot the variation of qu with H/B.
Solution From Eq. (4.59)
q 2 c2 32
=
=
= 0.4
q1 c1 80
From Fig. 4.23 for q2 /q1 = 0.4, ca /c1 = 0.9. So
ca = (0.9)(80) = 72 kN / m 2.
From Eq. (4.58)
B
q1 = 1 + 0.2 c1 N c (1) + γ 1 D f
L
1.5
= 1 + 0.2
(80)(5.14) + (18)(1) = 470.32 kN/m 2
3
With $s = 1, Eq. (4.57) yields
B 2c H
B
qu = 1 + 0.2 c2 N c ( 2 ) + 1 + a + γ 1 D f
L
L B
H
= (1 + 0.1)(32)(5.14) + (1.5)(2)( 72) + (18)(1)
B
H
= 198.93 + 216
B
Note that qu = 198.93 + 96(H/B) gives a straight line variation. In order to
determine the value of H/B at which qu is equal to qt , we set
H
qt = 470.32 kN/m 2 = 198.93 + 216
B
So
© 1999 by CRC Press LLC
FIGURE 4.24
© 1999 by CRC Press LLC
H
— = 1.26
B
Figure 4.24b shows the variation of qu with H/B.
!!
4.5 FOUNDATION ON LAYERED c – " SOIL —
WEAKER SOIL UNDERLAIN BY A STRONGER SOIL
In general when a foundation is supported by a weaker soil layer underlain by
a stronger soil at a shallow depth, as shown in the left-hand side of Fig. 4.25a,
the failure surface at ultimate load will pass through both soil layers. However,
when the magnitude of H is relatively large compared to the width of the
foundation B, the failure surface at ultimate load will be fully located in the
weaker soil layer (see the right-hand side of Fig. 4.25a). For estimating the
ultimate bearing capacity of such foundations, Meyerhof [6] and Meyerhof and
Hanna [16] proposed the following semiempirical relationship.
2
H
qu = qt + (qb − qt ) 1 − ≥ qt
D
where
(4.60)
D = depth of failure surface beneath the foundation in the thick bed
of the upper weaker soil layer
qt = ultimate bearing capacity in a thick bed of the upper soil layer
qb = ultimate bearing capacity in a thick bed of the lower soil layer
So
1
qt = c1 Nc (1) $cs (1) + #1 Df Nq (1) $qs (1) + – #1 N# (1) $#s (1)
2
(4.61)
1
qb = c2 Nc (2) $cs (2) + #2 Df Nq (2) $qs (2) + – #2 N# (2) $#s (2)
2
(4.62)
and
where Nc (1) , Nq (1) , N# (1) = bearing capacity factors corresponding to the soil
friction angle "1
Nc (2) , Nq (2) , N# (2) = bearing capacity factors corresponding to the soil
friction angle "2
$cs (1) , $qs (1) , $#s (1) = shape factors corresponding to the soil friction
angle "1
$cs (2) , $qs (2) , $#s (2) = shape factors corresponding to the soil friction
angle "2
© 1999 by CRC Press LLC
FIGURE 4.25 (a) Foundation on weaker soil layer underlain by stronger sand
layer; (b) Nature of variation of qu with H/B [Eq. (4.60)]
© 1999 by CRC Press LLC
Equations (4.60), (4.61), and (4.62) imply that the maximum and minimum
limits of qu are qb and qt , respectively (Fig. 4.25). The magnitude of D/B varies
from 1 for loose sand and clay to about 2 for dense sands [16]. The author
conducted several laboratory model tests and found a better approximation is
obtained by modifying Eq. (4.60) as
H
q u = q t + ( q b − q t ) 1 −
2B
1 .8
≥ qt
(4.63)
Based on several laboratory model tests, Hanna [21] proposed the following relationship for estimating the ultimate bearing capacity qu for a founda-tion
resting on a weak sand layer underlain by a strong sand layer.
1
qu = – #1 B$*#s N# (m) + #1 Df $*qs Nq (m)
2
1
1
" – #2 B$#s(2) N# (2) + – #2 Df $qs(2) Nq (2)
2
2
where
(4.64)
N# (2) , Nq (2) = Meyerhof’s bearing capacity factors with reference to
the soil friction angle "2 (Table 2.4)
$#s(2) , $qs(2) = Meyerhof’s shape factors (Table 2.5) with reference to
soil friction angle "2
φ
B
= 1 + 0.1 tan 2 45 + 2
2
L
N# (m) , Nq (m) = modified bearing capacity factors
$*#s , $*qs = modified shape factors
The modified bearing capacity factors can be obtained as follows
H
N γ(m ) = N γ(2) −
N γ ( 2 ) − N γ (1)
D ( γ )
]
(4.65)
H
N q( m ) = N q (2 ) −
N q ( 2 ) − N q ( 1)
D ( γ )
]
(4.66)
[
[
where N# (1) , Nq (1) = Meyerhof’s bearing capacity factors with reference to soil
friction angle "1 (Table 2.4)
The variations of D(#) and D(q) with "1 are shown in Figs. 4.2 and 4.3. The
relationships for the modified shape factors are the same as those given in Eqs.
(4.4) and (4.5). The term m1 [Eq. (4.4)] can be determined from Fig. 4.7 by
substituting D(q) for H and "1 for ". Similarly, the term m2 [Eq. (4.5)] can be
determined from Fig. 4.8 by substituting D(#) for H and "1 for ".
© 1999 by CRC Press LLC
FIGURE 4.26
EXAMPLE 4.4
A shallow foundation is shown in Fig. 4.26. Use Eq. (4.64) and determine the
ultimate bearing capacity qu .
Solution
35
D( γ )
B
H = 0.5 m, !1 = 35 , !2 = 45 . From Figs. 4.2 and 4.3 for !1 =
= 1.0;
D( q )
B
= 1.9
So D(") = 2.0 m and D(q) = 3.8 m. From Table 2.4 for !1 = 35 and !2 = 45 , Nq
(1) = 33.30, Nq (2) = 134.88, N" (1) = 37.1, and N" (2) = 262.7. Using Eqs. (4.65)
and (4.66)
0.5
N γ ( m ) = 262.7 −
(262.7 − 37.1) = 206.3
2
0.5
N q( m ) = 134.88 −
(134.88 − 33.3) = 121.5
3.8
From Eq. (4.64)
© 1999 by CRC Press LLC
1
qu = – "1 B#*"s N" (m) + "1 Df #*qs Nq (m)
2
H 0. 5
= 0.25 and φ1 = 35°
From Eqs. (4.4) and (4.5) Note : =
2
B
B
2
λ*qs = 1 − m1 ≈ 1 − 0.73 = 0.27
L
2
and
B
2
λ*γs = 1 − m2 ≈ 1 − 0.72 = 0.28
L
2
So
qu = (0.5)(16.5)(2)(0.28)(206.3) + (16.5)(0.8)(0.27)(121.5)
= 953.1 + 433 1386 kN / m 2
CHECK
1
1
qu = qb = – "2 B#"s(2) N" (2) + – "2 Df #qs(2) Nq (2)
2
2
φ
B
λ γs ( 2 ) = λ qs ( 2 ) = 1 + 0.1 tan 2 45 + 2
2
L
45
2
= 1 + 0.1 tan 2 45 +
= 1.583
2
2
qu = (0.5)(18.5)(2)(1.583)(262.7) + (16.5)(0.8)(1.53)(134.88)
= 7693.3 + 2818.4 = 10,511.7 kN / m 2
Hence, qu = 1386 kN / m 2
!!
4.6 CONTINUOUS FOUNDATION ON WEAK CLAY
WITH A GRANULAR TRENCH
In practice, there are several techniques to improve the load bearing capacity
and settlement of shallow foundations on weak compressible soil layers. One
© 1999 by CRC Press LLC
FIGURE 4.27 Continuous rough foundation on a weak soil with a granular trench
of those techniques is the use of a granular trench under the foundation. Figure
4.27 shows a continuous rough foundation on a granular trench made in a
weak soil extending to a great depth. The width of the trench is W, the width
of the foundation is B, and the depth of the trench is H. The width of the
trench, W, can be smaller or larger than B. The parameters of the stronger
trench material and the weak soil for bearing capacity calculation are as
follow:
Angle of friction
Cohesion
Unit weight
Trench material
Weak soil
!1
c1
"1
!2
c2
"2
Madhav and Vitkar [22] assumed a general shear failure mechanism in the soil
under the foundation to analyze the ultimate bearing capacity of the foundation
using the upper bound limit analysis suggested by Drucker and Prager [23], and
this is shown in Fig. 4.27. The failure zone in soil can be divided into subzones,
and they are as follow:
1. An active Rankine zone ABC with a wedge angle of $.
2. A mixed transition zone such as BCD bounded by angle %1 . CD is an
arc of a log spiral defined by the equation
r = r0 e%tan!1
© 1999 by CRC Press LLC
3.
where !1 = angle of friction of the trench material.
A transition zone such as BDF with an central angle %2 . DF is an arc
of a log spiral defined by the equation
r = r0 e%tan!2
4. A Rankine passive zone like BFH.
Note that %1 and %2 are functions of $, &, W/B, and !1 .
By using the upper bound limit analysis theorem, Madhav and Vitkar [22]
expressed the ultimate bearing capacity of the foundation as
γ B
q u = c 2 N c ( T ) + D f γ 2 N q ( T ) + 2 N γ ( T )
2
(4.67)
where Nc (T) , Nq (T) , N" (T) = bearing capacity factors with the presence of the
trench
The variations of the bearing capacity factors [that is, Nc (T) , Nq (T) , and N" (T)]
for purely granular trench soil (c1 = 0) and soft saturated clay (with !2 = 0 and
c2 = cu ) determined by Madhav and Vitkar [22] are given in Figs. 4.28, 4.29
and 4.30. The values of N" (T) given in Fig. 4.30 are for "1 /"2 = 1. In an actual
case, the ratio "1 /"2 may be different than one. The error for this assumption,
however, is less than 10%.
Sufficient experimental results are not available in the literature to verify
the above theory. Hamed, Das, and Echelberger [24] conducted several
laboratory model tests to determine the variation of the ultimate bearing capacity of a strip foundation resting on a granular trench (sand, c1 = 0) made in a
saturated soft clay medium (c2 = cu ; !2 =0). For these tests the width of the
foundation B was kept equal to the width of the trench W, and the ratio of H/B
= H/W was varied. Figures 4.31 and 4.32 show the variation of the experimental qu with H/B. In both cases the experimental ultimate bearing capacity
increased with H/B up to a maximum value and remained practically constant
thereafter. The theoretical values of qu calculated using Eq. (4.67) and Figs.
4.28 and 4.30 are also shown in Figs. 4.31 and 4.32. From this comparison,
the following major conclusions can be drawn:
1. Madhav and Vitkar’s theory [22] yields a higher and unsafe value of
qu when compared to the experimental maximum value of qu .
2. The minimum height H of the granular trench necessary to obtain the
maximum value of qu is about 2.5B to 3B.
Since Eq. (4.67) yields unsafe values of the ultimate bearing capacity,
Hamed, Das and Echelberger [24] suggested an approximate procedure for
estimating the maximum qu (for c1 = 0, c2 = cu , !2 = 0, and B = W). This procedure can be explained by referring to Fig. 4.33, in which A and B are two soil
elements. For soil element A the major principal stress is 'I , and the minor
principal stress is 'III . So
© 1999 by CRC Press LLC
FIGURE 4.28 Madhav and Vitkar’s bearing capacity factor, Nc (T)
© 1999 by CRC Press LLC
FIGURE 4.29 Madhav and Vitkar’s bearing capacity factor, Nq (T)
'I = qu = Kp (1) 'III
(4.68)
where K p(1) = Rankine passive earth pressure coefficient
φ
= tan2 45 + 1
2
For soil element B the major principal stress is '1 and the minor principal
stress is '3 . However,
'3 = "2 Df
and
σ 1 = σ 3 K p ( 2 ) + 2 cu
© 1999 by CRC Press LLC
K p(2)
FIGURE 4.30
Madhav and Vitkar’s bearing capacity factor, N " (T)
where K p( 2) = Rankine passive earth pressure coefficient
φ
= tan2 45 + 2
2
Since ! 2 = 0, the value of Kp (2) = 1. Hence
'1 = '3 + 2cu = "2 Df + 2cu
(4.69)
Again referring to Figure 4.33, it can be seen that, at failure (which is of the
bulging type)
© 1999 by CRC Press LLC
FIGURE 4.31 Ultimate bearing capacity of a continuous foundation on soft clay with
a granular trench (Note: B =W, !1 = 40!. c2 = 210 lb / ft2, !2 = 0)
FIGURE 4.32 Ultimate bearing capacity of a continuous foundation on soft clay with
a granular trench (Note: B =W, !1 = 43!. c2 = 210 lb / ft2, !2 = 0, Df = 0)
© 1999 by CRC Press LLC
FIGURE 4.33 Derivation of Eq. (4.71)
'1 = 'III
(4.70)
Now, combining Eq.s (4.68), (4.69), and (4.70)
φ
q u = K p (1) ( γ 2 D f + 2 c u ) = ( γ 2 D f + 2 c u ) tan 2 45 + 1
2
(4.71)
With proper parameters, the ultimate bearing capacity calculated using Eq.
(4.71) is also shown in Figs. 4.31 and 4.32. This comparison shows that the
theoretical qu [Eq. (4.71)] is equal to or somewhat less than the maximum
experimental value of qu .
4.7 SHALLOW FOUNDATIONS ABOVE A VOID
Mining operations may leave underground voids at relatively shallow depths.
Additionally, in some instances, soluble bedrock dissolves at the interface of
the soil and the bedrock leaving void spaces. Estimating the ultimate bearing
capacity of shallow foundations constructed over these voids, as well as the
stability of the foundations, is gradually becoming an important issue. Only a
few studies have been reported in published literature so far. Baus and Wang
[25] reported some experimental results for the ultimate bearing capacity of a
shallow rough continuous foundation located above voids as shown in Fig.
4.34. It is assumed that the top of the rectangular void is located at a depth H
below the bottom of the foundation. The void is continuous and has cross-
© 1999 by CRC Press LLC
FIGURE 4.34 Shallow continuous rough foundation over a void
sectional dimensions of W! × H!. The laboratory tests of Baus and Wang [25]
were conducted with a soil having the following properties.
Friction angle of soil, ! = 13.5"
Cohesion = 65.6 kN / m 2
Modulus in compression = 4,670 kN / m 2
Modulus in tension = 10,380 kN / m 2
Poisson’s ratio = 0.28
Unit weight of compacted soil, " = 18.42 kN / m 3
The results of Baus and Wang [25] are shown in a nondimensional form in
Fig. 4.35. Note that the results of the tests which constitute Fig. 4.35 are for the
case of Df = 0. From this figure the following conclusions can be drawn:
1. For a given H/B, the ultimate bearing capacity decreases with the increase in the void width, W!.
2. For any given W!/B, there is a critical H/B ratio beyond which the
void has no effect on the ultimate bearing capacity. For W!/B = 10,
the value of the critical H/B is about 12.
Baus and Wang [25] conducted finite analysis to compare the validity of
their experimental findings. In the finite element analysis, the soil was treated
as an elastic, perfectly plastic material. They also assumed that Hooke’s law
is void in the elastic range, and the soil follows the von Mises yield criterion in
the perfectly plastic range, or
f = α J1 +
© 1999 by CRC Press LLC
J2 = k ′
(4.72)
FIGURE 4.35 Experimental bearing capacity of a continuous foundation
as a function of void size and location [1]
f& = 0
(4.73)
where f = yield function
α=
tanφ
(9 + 12tanφ) 0.5
(4.74)
k′ =
3c
(9 + 12tanφ) 0.5
(4.75)
J1 = first stress invariant
J2 = second stress invariant
The relationships as shown in Eq. (4.74) and (4.75) are based on the study of
Drucker and Prager [23]. The results of the finite element analysis have shown
good agreement with experiments.
4.8 FOUNDATION ON A SLOPE
In 1957, Meyerhof [26] proposed a theoretical solution to determine the
ultimate bearing capacity of a shallow foundation located on the face of a
slope. Figure 4.36 shows the nature of the plastic zone developed in the soil
© 1999 by CRC Press LLC
FIGURE 4.36 Nature of plastic zone under a rough continuous foundation
on the face of a slope
under a rough continuous foundation (width = B) located on the face of a slope.
In Fig. 4.36, abc is the elastic zone, acd is a radial shear zone, and ade is a
mixed shear zone. The normal and shear stresses on the plan ae are po and so ,
respectively. Note that the slope makes an angle # with the horizontal. The
shear strength parameters of the soil are c and !, and its unit weight is equal to
". As in Eq. (2.71), the ultimate bearing capacity can be expressed as
1
qu = cNc + po Nc + – "BN"
2
(4.76)
The preceding relationship can also be expressed as
1
qu = cNcq + – "BN"q
2
(4.77)
where Ncq , N"q = bearing capacity factors
For purely cohesive soil (that is, ! = 0)
qu = cNcq
(4.78)
Figure 4.37 shows the variation of Ncq with slope angle # and the slope
stability number, Ns . Note that
"H
Ns = –—
c
(4.79)
where H = height of the slope
In a similar manner, for a granular soil (c = 0)
1
qu = – "BN"q
2
© 1999 by CRC Press LLC
(4.80)
FIGURE 4.37 Variation of Meyerhof’s bearing capacity
factor, Ncq , for a purely cohesive soil
(foundation on a slope)
The variation of N"q (for c = 0) applicable to Eq. (4.80) is shown in Fig. 4.38.
4.9 FOUNDATION ON TOP OF A SLOPE
Meyerhof’s Solution
Figure 4.39 shows a rough continuous foundation of width B located on top of
a slope of height H. It is located at a distance b from the edge of the slope. The
ultimate bearing capacity of the foundation can be expressed by Eq. (4.77), or
1
qu = cNcq + – "BN"q
2
© 1999 by CRC Press LLC
(4.81)
FIGURE 4.38
Variation of Meyerhof’s bearing capacity factor,
N"q , for a granular soil (foundation on slope)
FIGURE 4.39 Continuous foundation on a slope
© 1999 by CRC Press LLC
FIGURE 4.40 Meyerhof’s bearing capacity factor, Ncq , for a purely
cohesive soil (foundation on top of a slope)
Meyehof [26] developed the theoretical variations of Ncq for a purely cohesive
soil (! = 0) and N"q for a granular soil (c = 0), and these variations are shown
in Figs. 4.40 and 4.41. Note that, for purely cohesive soil (Fig. 4.40)
qu = cNcq
and, for granular soil (Fig. 4.41)
1
qu = – "BN"q
2
© 1999 by CRC Press LLC
FIGURE 4.41 Meyerhof’s bearing capacity factor, N"q , for a granular
soil (foundation on top of a slope)
It is important to note that, in using Fig. 4.40, the stability number Ns should be
taken as zero when B < H. If B # H the curve for the actual stability number
should be used.
Typical plots of load per unit area q versus settlement s obtained by the
author from laboratory model tests in saturated clay soil (with b/B = 0, Df /B =
0, cu = 27.5 kN / m 2, and B = 76.2 mm) are shown in Fig. 4.42. It can be seen
that, for similar foundation conditions, the settlement at ultimate load decreases
with the increase in the slope angle #.
Solutions of Hansen [27] and Vesic [11]
Referring to the condition of b = 0 in Fig. 4.39 (that is, the foundation is
© 1999 by CRC Press LLC
FIGURE 4.42 Typical load per unit area versus settlement plots
for a continuous foundation on top of a slope —
model test results (B = 76.2 mm, cu = 27.5 kN / m2,
b/B = 0, Df /B)
located at the edge of the slope), Hansen [27] proposed the following
relationship for the ultimate bearing capacity of a continuous foundation
1
qu = cNc $c# + qNq $q# + – "BN" $"#
2
(4.82)
where Nc , Nq , N" = bearing capacity factors (see Table 2.3 for Nc and Nq and
Table 2.4 for N" )
$c# , $q# , $"# = slope factors
q = "Df
According to Hansen [27]
λ qβ = λ γβ = (1 − tan β ) 2
© 1999 by CRC Press LLC
(4.83)
λ cβ =
N q λ qβ − 1
Nq − 1
λ cβ = 1 −
(for φ > 0)
2β
(for φ = 0)
π+2
(4.84)
(4.85)
For ! = 0 condition, Vesic [11] pointed out that, with the absence of
weight due to the slope, the bearing capacity factor N" has a negative value and
can be given as
N" = $2sin#
(4.86)
Thus for ! = 0 condition, with Nc = 5.14 and Nq = 1, Eq. (4.82) takes the form
2β
qu = c(5.14) 1 −
+ γD f (1 − tan β) 2 − γB sin β(1 − tan β) 2
5.14
or
qu = (5.14 $ 2#)c + "Df (1 $ tan#)2 $ "B sin#(1 $ tan#)2
(4.87)
Solution by Limit Equilibrium and Limit Analysis
Saran, Sud, and Handa [28] provided a solution to determine the ultimate
bearing capacity of shallow continuous foundations on the top of a slope (Fig.
4.39) using the limit equilibrium and limit analysis approach. According to
this theory, for a strip foundation
1
qu = cNc + qNc + – "BN"
2
(4.88)
where Nc , Nq , N" = bearing capacity factors
q = "Df
Referring to the notations used in Fig. 4.39, the numerical values of Nc , Nq , and
N" are given in Table 4.2.
Stress Characteristics Solution
As shown in Eq. (4.81), for granular soils (that is, c = 0)
1
qu = – "BN"q
2
© 1999 by CRC Press LLC
(4.89)
TABLE 4.2 Bearing Capacity Factors Based on Saran, Seed, and Handa’s Analysis
Factor
N"
Nq
Nc
#
(deg)
Df
B
b
B
30
20
10
0
0
30
20
10
0
Soil friction angle, ! (deg)
40
35
0
25.37
53.48
101.74
165.39
0
1111
30
25
20
%15
1111
30
25
%20
30
25
20
15
10
12.41 6.14
24.54 11.62
43.35 19.65
66.59 28.98
3.20
5.61
9.19
13.12
1.26
4.27
4.35
6.05
0.70
1.79
1.96
2.74
0.10
0.45
0.77
1.14
60.06
85.98
125.32
165.39
34.03
42.49
55.15
66.59
18.95
21.93
25.86
28.89
10.33
11.42
12.26
13.12
5.45
5.89
6.05
6.05
0.00
1.35
2.74
2.74
0
91.87
115.65
143.77
165.39
49.43
59.12
66.00
66.59
26.39
28.80
28.89
28.89
111
111
131.34
151.37
166.39
64.37 28.89
66.59 28.89
66.59 28.89
30
20
%10
111
0
12.13
12.67
81.30
16.42 8.98
19.48 16.80
41.40 22.50
7.04
12.70
12.70
5.00
7.40
7.40
3.60
4.40
4.40
30
20
%10
111
111
28.31
42.25
81.30
24.14
41.4
41.4
22.5
22.5
22.5
50
40
30
20
%10
0
0
21.68
31.80
44.80
63.20
88.96
16.52
22.44
28.72
41.20
55.36
12.60
16.64
22.00
28.32
36.50
10.00
12.80
16.20
20.60
24.72
8.60
10.04
12.20
15.00
17.36
7.10
8.00
8.60
11.30
12.61
50
40
30
20
%10
0
11111
38.80
48.00
59.64
75.12
95.20
30.40
35.40
41.07
50.00
57.25
24.20
27.42
30.92
35.16
36.69
19.70
21.52
23.60
27.72
24.72
16.42
17.28
17.36
17.36
17.36
50
40
30
20
%10
11111
0
35.97
51.16
70.59
93.79
95.20
28.11
37.95
50.37
57.20
57.20
22.38
29.42
36.20
36.20
36.20
18.38
22.75
24.72
24.72
24.72
15.66
17.32
17.36
17.36
17.36
50
40
30
%20
1111
1111
53.65
67.98
85.38
95.20
42.47
51.61
57.25
57.25
35.00
36.69
36.69
36.69
24.72
24.72
24.72
24.72
5.50
6.25
6.70
8.76
9.44
10.00
12.16
12.16
12.16
12.16
Graham, Andrews, and Shields [29] provided a solution for the bearing capacity factor, N"q , for a shallow continuous foundation on the top of a slope in
© 1999 by CRC Press LLC
FIGURE 4.43 Schematic diagram of failure zones for embedment and setback:
(a) Df /B > 0; (b) b/B > 0
granular soil based on the method of stress characteristics. Figure 4.43 shows
the schematics of the failure zone in the soil for embedment (Df /B) and setback
(b/B) assumed for this analysis. The variations of N!q obtained by this method
are shown in Figs. 4.44, 4.45, and 4.46.
Empirical Relationship Based on Centrifuge Testing
In 1988 Gemperline [30] reported the results of 215 centrifuge tests on continuous foundations located at the top of a slope of cohesionless sand. Based
on these 215 tests, Gemperline proposed that the ultimate bearing capacity of
a continuous foundation can be expressed as
© 1999 by CRC Press LLC
FIGURE 4.44 Graham et al.’s theoretical values of N!q (Df /B = 0)
© 1999 by CRC Press LLC
FIGURE 4.45 Graham et al.’s theoretical values of N!q (Df /B = 0.5)
© 1999 by CRC Press LLC
FIGURE 4.46 Graham et al.’s theoretical values of N!q (Df /B = 1)
© 1999 by CRC Press LLC
1
qu = – !BN!q
2
Shields, Chandler, and Garnier [31] normalized the N!q values proposed by
Gemperline in the following form
N γq
D f
= 1 + 0.65 1 − 0.8
N γqR
B
2
2
1 −( 1 − tanβ)
2
b
2 + tanβ
B
[
D
2
×1 − 0.33 f tanβ
2
B
b
β
+
2
tan
B
]
(4.90)
where N!qR = the value of N!q for a reference continuous foundation at the
surface of level ground (that is, Df /B = 0 and b/B = !)
The magnitude of N!qR can be given by the following relationship
N!qR = (100.1159""2.386)(100.34"0.2 logB )
(4.91)
where " is in degrees and B is in inches.
EXAMPLE 4.5
Figure 4.47 shows a continuous foundation on the top of a saturated clay slope.
Estimate the ultimate bearing capacity by
a. Meyerhof’s method [Eq. (4.81)]
b. Hansen and Vesic’s method [Eq. (4.87)]
Solution
a.
qu = cNcq
Given
Df
© 1999 by CRC Press LLC
B
=
15
.
= 1,
15
.
b
H
0
=
= 0. Since
> 1, use N s = 0.
B 15
B
.
FIGURE 4.47
From Fig. 4.40, for
Df
B
= 1,
b
= 0, β = 30° , and N s = 0, the
B
value of N cq is about 5.85. So
qu = (49)(5.85) = 286.7 kN / m 2
b. From Eq. (4.87)
qu = (5.14− 2β)c + γDf (1− tanβ)2 − γBsinβ(1− tanβ)2
= 5.14− (2)
π
×30
180
(49) + (18.5)(1.5)(1− tan30)2
−(18.5)(1.5)(sin30)(1− tan30)2
= 203 kN/m2
!!
EXAMPLE 4.6
Figure 4.48 shows a continuous foundation on a slope of a granular soil. Estimate the ultimate bearing capacity by
a. Meyerhof’s method [Eq. (4.81)]
b. Saran and Handa’s method [Eq. (4.88)]
c. the stress characteristic solution [Eq. (4.91)]
© 1999 by CRC Press LLC
FIGURE 4.48
Solution
a.
For granular soil (c = 0) from Eq. (4.81)
1
qu = – !BN!q
2
Given
Df
b 15
.
15
.
=
= 1,
=
= 1, φ = 40° , and β = 30° .
B 15
B
.
.
15
From Fig. 4.41, N!q # 120. So
1
qu = – (16.8)(1.5)(120) = 1512 kN / m 2
2
b. For c = 0, from Eq. (4.88)
1
qu = qNq + – !BN!
2
For
Df
b
= 1,
= 1, φ = 40° , and β = 30° , the value of N γ =
B
B
131.34 and the value of N q = 28.31 (Table 4.2)
1
qu = (16.8)(1.5)(28.31) + – (16.8)(1.5)(131.34) # 2368 kN / m 2
2
© 1999 by CRC Press LLC
c.
From Eq. (4.89)
1
qu = – !Bn!q
2
From Fig. 4.46b, N!q # 110
1
qu = – (16.8)(1.5)(110) = 1386 kN / m 2
2
!!
REFERENCES
1. Prandtl, L., Uber die eindringungsfestigkeit plastisher baustoffe und die
festigkeit von schneiden, Z. Ang. Math. Mech., 1(1), 15, 1921.
2. Reissner, H., Zum erddruckproblem, in Proc., I Intl. Conf. Appl. Mech., Delft,
The Netherlands, 1924, 295.
3. Lundgren, H., and Mortensen, K., Determination by the theory of plasticity of
the bearing capacity of continuous footings on sand, in Proc., III Int. Conf. Soil
Mech. Found. Engg., Zurich, Switzerland, 1, 409, 1953.
4. Mandel, J., and Salencon, J., Force portante d’un sol sur une assise rigide (étude
theorizue), Geotechnique, 22(1), 79, 1972.
5. Meyerhof, G. G., and Chaplin, T. K., The compression and bearing capacity of
cohesive soils, Br. J. Appl. Phys., 4, 20, 1953.
6. Meyerhof, G. G., Ultimate bearing capacity of footings on sand layer overlying
clay. Canadian Geotech. J., 11(2), 224, 1974.
7. Milovic, D. M., and Tournier, J. P., Comportement de foundations reposant sur
une coche compressible d´épaisseur limitée, in Proc., Conf. Comportement des
Sols Avant la Rupture, Paris, France, 1971, 303.
8. Pfeifle, T. W., and Das, B. M., Bearing capacity of surface footings on sand layer
resting on rigid rough base. Soils and Foundations, 19(1), 1979, 1.
9. Mandel, J., and Salencon, J., Force portante d’un sol sur une assise rigide, in
Proc., VII Int. Conf. Soil Mech. Found Engg., Mexico City, 2, 1969, 157.
10. Buisman, A. S. K., Grondmechanica, Waltman, Delft, 1940.
11. Vesic, A. S., Bearing capacity of shallow foundations, in Foundation
Engineering Hand-book, Winterkorn, H. F., and Fang, H. Y., Eds., Van
Nostrant Reinhold Co., 1975, 121.
12. DeBeer, E. E., Analysis of shallow foundations, in Geotechnical Modeling and
Applications, Sayed, S. M., Ed., Gulf Publishing Co., 1975, 212.
13. Casagrande, A., and Carrillo, N., Shear failure in anisotropic materials, in
Contribution to Soil Mechanics 1941-53, Boston Society of Civil Engineers.,
1954, 122.
14. Reddy, A. S., and Srinivasan, R. J., Bearing capacity of footings on layered
clays, J. Soil Mech. Found. Div., ASCE, 93(SM2), 83, 1967.
15. Lo, K. Y., Stability of slopes in anisotropic soil, J. Soil Mech. Found. Div.,
ASCE, 91(SM4), 85, 1965.
16. Meyerhof, G. G., and Hanna, A. M., Ultimate bearing capacity of foundations on
layered soils under inclined load, Canadian Geotech. J., 15(4), 565,1978.
© 1999 by CRC Press LLC
17. Caquot, A., and Kerisel, J., Tables for the Calculation of Passive Pressure,
Active Pressure, and Bearing Capacity of Foundations. Gauthier-Villars, Paris,
1949.
18. Meyerhof, G. G., Some recent research on the bearing capacity of foundations,
Canadian Geotech. J., 1(1), 16, 1963.
19. Hanna, A. M., and Meyerhof, G. G., Design charts for ultimate bearing capacity
for sands overlying clays, Canadian Geotech. J., 17(2), 300, 1980.
20. Hanna, A. M., Foundations on strong sand overlying weak sand, J. Geotech.
Eng,, ASCE, 107(GT7), 915, 1981.
21. Hanna, A. M., Bearing capacity of foundations on a weak sand layer overlying
a strong deposit, Canadian Geotech. J., 19(3), 392, 1982.
22. Madhav, M. R., and Vitkar, P. P., Strip footing on weak clay stabilized with a
granular trench or pile, Canadian Geotech. J., 15(4), 605, 1978.
23. Drucker, D. C., and Prager, W., Soil mechanics and plastic analysis of limit
design, Q. Appl. Math., 10, 157, 1952.
24. Hamed, J. T., Das, B. M., and Echelberger, W. F., Bearing capacity of a strip
foundation on granular trench in soft clay. Civil Engineering for Practicing and
Design Engineers, Pergamon Press, 5(5), 359, 1986.
25. Baus, R. L., and Wang, M. C., Bearing capacity of strip footing above void. J.
Geotech. Eng., ASCE, 109(GT1), 1, 1983.
26. Meyerhof, G. G., The ultimate bearing capacity of foundations on slopes, in
Proc., IV Int. Conf. Soil Mech. Found. Eng., London England, 1, 1957, 384.
27. Hansen, J. B., A revised and extended formula for bearing capacity, Bulletin 28,
Danish Geotechnical Institute, Copenhagen, 1970.
28. Saran, S., Sud, V. K., and Handa, S. C., Bearing capacity of footings adjacent to
slopes, J. Geotech. Eng., ASCE, 115(4), 553, 1989.
29. Graham, J., Andrews, M., and Shields, D. H., Stress characteristics for shallow
footings in cohesionless slopes, Canadian Geotech. J., 25(2), 238, 1988.
30. Gemperline, M. C., Centrifuge modelling of shallow foundations, in Proc.,
ASCE Spring Convention, 1988.
31. Shields, D., Chandler, N., and Garnier, J., Bearing capacity of foundations in
slopes, J. Geotech. Eng., ASCE, 116(3), 528, 1990.
© 1999 by CRC Press LLC
CHAPTER FIVE
SETTLEMENT AND ALLOWABLE BEARING CAPACITY
5.1 INTRODUCTION
Various theories relating to the ultimate bearing capacity of shallow foundations were presented in Chapters 2, 3, and 4. In Section 2.12, a number of
definitions for the allowable bearing capacity were discussed. In the design of
any foundation, one must consider the safety against bearing capacity failure
as well as against excessive settlement of the foundation. In the design of most
foundations, there are specifications for allowable levels of settlement. Refer
to Fig. 5.1 which is a plot of load per unit area q versus settlement S for a foundation. The ultimate bearing capacity is realized at a settlement level of Su . Let
Sall be the allowable level of settlement for the foundation and qall(S) be the
corresponding allowable bearing capacity. If FS is the factor of safety against
bearing capacity failure, then the allowable bearing capacity is qall (b) = qu /FS.
FIGURE 5.1 Load-settlement curve for shallow foundation
© 1999 by CRC Press LLC
The settlement corresponding to qall (b) is S´. For foundations with smaller
widths of B, S´ may be less than Sall ; however, for larger values of B, Sall < S´.
Hence, for smaller foundation widths, the bearing capacity controls and, for
larger foundation widths, the allowable settlement controls. This chapter
describes the procedures for estimating the settlement of foundations under
load and thus estimating the allowable bearing capacity.
The settlement of a foundation can have three components: (a) elastic
settlement, Se ; (b) primary consolidation settlement, Sc ; (c) secondary consolidation settlement, Ss .
Thus the total settlement St can be expressed as
St = Se + Sc + Ss
For any given foundation, one or more of the components may be zero or
negligible.
Elastic settlement is caused by deformation of dry soil, as well as moist and
saturated soils, without any change in moisture content. Primary consoli-dation
settlement is a time-dependent process which occurs in clayey soils located
below the ground water table as a result of the volume change in soil because
of the expulsion of water that occupies the void spaces. Secondary consolidation settlement follows the primary consolidation process in saturated clayey
soils and is a result of the plastic adjustment of soil fabrics. The pro-cedures for
estimating the above three types of settlement are discussed in this chapter.
Any type of settlement is a function of the additional stress imposed on the
soil by the foundation. Hence, it is desirable to know the relationships for
calculating the stress increase in the soil caused by application of load to the
foundation. These relationships are given in Section 5.2 and are derived
assuming that the soil is a semi-infinite, elastic and homogeneous medium.
5.2 STRESS INCREASE IN SOIL DUE TO APPLIED LOAD
Point Load
Boussinesq [1] developed a mathematic relationship for the stress increase due
to a point load Q acting on the surface of a semi-infinite mass. In Fig. 5.2 the
stress increase at a point A is shown in the Cartesian coordinate system, and the
stress increase in the cylindrical coordinate system is shown in Fig. 5.3. The
components of the stress increase can be given by the following relationships.
Cartesian coordinate system (Fig. 5.2)
© 1999 by CRC Press LLC
FIGURE 5.2
Boussinesq’s problem—Stress increase at a point in the Cartesian
coordinate system due to a point load on the surface
σz =
3Qz3
2πR5
(5.1)
σx =
3Q x 2 z 1 − 2ν 1
(2R + z)x 2 z
−
−
+
2π R5
3 R(R + z) R3 (R + z)2 R3
(5.2)
σy =
3Q y2 z 1 − 2ν 1
(2R + z)y 2
z
+
−
− 3
5
3
2
2π R
3 R( R + z) R ( R + z)
R
(5.3)
τ xy =
3Q xyz 1 − 2ν (2R + z)xy
−
2π R 5
3 R 3 ( R + z)2
(5.4)
τ xz =
3Q xz 2
2π R 5
(5.5)
© 1999 by CRC Press LLC
FIGURE 5.3 Boussinesq’s problem—Stress increase at a point in the cylindrical
coordinate system due to a point load on the surface
τ yz =
where
3Q yz 2
2π R 5
! = normal stress
" = shear stress
R=
z2 + r2
r=
x2 + y2
# = Poisson’s ratio
Cylindrical coordinate system (Fig. 5.3):
© 1999 by CRC Press LLC
(5.6)
σz =
3Qz3
2πR5
(5.7)
σr =
1 − 2ν
Q 3zr2
−
5
2π R
R(R + z)
(5.8)
σθ =
1
Q
z
(1 − 2ν)
− 3
2π
R( R + z) R
(5.9)
τrz =
3Qrz2
2πR 5
(5.10)
Uniformly Loaded Flexible Circular Area
Boussinesq’s solution for a point load can be extended to determine the stress
increase due to a uniformly loaded flexible circular area on the surface of a
semi-infinite mass (Fig. 5.4). In Fig. 5.4 the circular area has a radius R, and the
uniformly distributed load per unit area is q. If the components of stress
increase at a point A below the center are to be determined, then we consider
an elemental area dA = rd$dr. The load on the elemental area is dQ = qrd$dr.
This can be treated as a point load. Now the vertical stress increase, d!z , at A
due to dQ can be obtained by substituting dQ for Q and
5.7. Thus
dσ z =
r 2 + z 2 for R in Eq.
3z 3qrdθdr
2 π ( r 2 + z 2 ) 5/2
The vertical stress increase, !z , due to the entire loaded area is then
σz =
∫
R 2π
dσ z =
3 z 3 qrd θ dr
z3
= q 1 −
2
2 3/2
2
2 3/2
π
+
+
2
(
)
(
)
r
z
R
z
r = 0θ = 0
∫∫
(5.11)
Similarly, the magnitudes of !$ and !r below the center can be obtained as
2(1 + ν)z
q
z3
+ 2
σr = σ θ = 1 + 2ν − 2
2 1/ 2
2 3/ 2
2
(R + z )
(R + z )
© 1999 by CRC Press LLC
(5.12)
188 CHAPTER FIVE
FIGURE 5.4
Stress increase below the center of a
uniformly-loaded flexible circular area
FIGURE 5.5
Stress increase below any point under a
uniformly-loaded flexible circular area
© 1999 by CRC Press LLC
Table 5.1 gives the variation of !z /q at any point A below a circularly loaded
flexible area (Figure 5.5). A more detailed tabulation of the stress increase
(that is, !z , !$ , !r , and "rz ) below a uniformly loaded flexible area is given by
Ahlvin and Ulery [2].
TABLE 5.1 Variation of !z /q at a point A (Figure 5.5)
r/R !
!z /q
"z/R
0
0.2
0.4
0.6
0.8
1.0
1.5
2.0
2.5
3.0
4.0
5.0
0
0.2
0.4
0.6
0.8
1.0
1.000
0.992
0.979
0.864
0.756
0.646
0.424
0.284
0.200
0.146
0.087
0.057
1.000
0.991
0.943
0.852
0.742
0.633
0.416
0.281
0.197
0.145
0.086
0.057
1.000
0.987
0.920
0.814
0.699
0.591
0.392
0.268
0.196
0.141
0.085
0.056
1.000
0.970
0.860
0.732
0.619
0.525
0.355
0.248
0.188
0.135
0.082
0.054
1.000
0.890
0.713
0.591
0.504
0.434
0.308
0.224
0.167
0.127
0.080
0.053
0.500
0.468
0.435
0.400
0.366
0.332
0.288
0.196
0.151
0.118
0.075
0.052
Uniformly Loaded Flexible Rectangular Area
Figure 5.6 shows a flexible rectangular area of length L and width B subjected
to a uniform vertical load of q per unit area. The load on the elemental area dA
is equal to dQ = qdxdy. This can be treated as an elemental point load. The
vertical stress increase d!z due to this at A, which is located at a depth z below
the corner of the rectangular area, can be obtained by using Eq. (5.7), or
dσ z =
3qz 3dxdy
2 π ( x 2 + + y 2 + z 2 ) 5/ 2
(5.13)
Hence, the vertical stress increase at A due to the entire loaded area is
σz =
∫
B
dσ z =
© 1999 by CRC Press LLC
L
3 qz 3 dxdy
= qI
2 π( x 2 + y 2 + z 2 )5/2
y=0x =0
∫∫
(5.14)
FIGURE 5.6 Stress increase below the corner of a uniformly-loaded flexible
rectangular area
2mn(m 2 + n 2 + 1) 0.5 m 2 + n 2 + 2
× 2
2
2
2 2
2
1 m + n + m n + 1 m + n + 1
where I =
4π
2mn(m 2 + n 2 + 1) 0.5
+ tan −1 2
m + n 2 − m2n 2 + 1
m=
B
z
n=
L
z
(5.15)
Table 5.2 shows the variation of I with m and n. The stress below any other
point C below the rectangular area (Fig. 5.7) can be obtained by dividing it into
four rectangles as shown. For rectangular area No. 1, m1 = B1 /z; n1 = L1 /z.
Similarly for rectangles No. 2, 3, and 4, m2 = B1 /z; n2 = L2 /z, m3 = B2 /z; n3 =
L2 /z, and m4 = B2 /z; n4 = L1 /z.
© 1999 by CRC Press LLC
TABLE 5.2 Variation of I with m and n
m
n
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
0.1
0.0047 0.0092
0.0132
0.0168
0.0198
0.0222
0.0242
0.0258
0.0270
0.0279
0.2
0.0092
0.0179
0.0259
0.0328
0.0387
0.0435
0.0474
0.0504
0.0528
0.0547
0.3
0.0132
0.0259
0.0374
0.0474
0.0559
0.0629
0.0686
0.0731
0.0766
0.0794
0.4
0.0168
0.0328
0.0474
0.0602
0.0711
0.0801
0.0873
0.0931
0.0977
0.1013
0.5
0.0198
0.0387
0.0559
0.0711
0.0840
0.0947
0.1034
0.1104
0.1158
0.1202
0.6
0.0222
0.0435
0.0629
0.0801
0.0947
0.1069
0.1168
0.1247
0.1311
0.1361
0.7
0.0242
0.0474
0.0686
0.0873
0.1034
0.1169
0.1277
0.1365
0.1436
0.1491
0.8
0.0258
0.0504
0.0731
0.0931
0.1104
0.1247
0.1365
0.1461
0.1537
0.1598
0.9
0.0270
0.0528
0.0766
0.0977
0.1158
0.1311
0.1436
0.1537
0.1619
0.1684
1.0
0.0279
0.0547
0.0794
0.1013
0.1202
0.1361
0.1491
0.1598
0.1684
0.1752
1.2
0.0293
0.0573
0.0832
0.1063
0.1263
0.1431
0.1570
0.1684
0.1777
0.1851
1.4
0.0301
0.0589
0.0856
0.1094
0.1300
0.1475
0.1620
0.1739
0.1836
0.1914
1.6
0.0306
0.0599
0.0871
0.1114
0.1324
0.1503
0.1652
0.1774
0.1874
0.1955
1.8
0.0309
0.0606
0.0880
0.1126
0.1340
0.1521
0.1672
0.1797
0.1899
0.1981
2.0
0.0311
0.0610
0.0887
0.1134
0.1350
0.1533
0.1686
0.1812
0.1915
0.1999
2.5
0.0314
0.0616
0.0895
0.1145
0.1363
0.1548
0.1704
0.1832
0.1938
0.2024
3.0
0.0315
0.0618
0.0898
0.1150
0.1368
0.1555
0.1711
0.1841
0.1947
0.2034
4.0
0.0316
0.0619
0.0901
0.1153
0.1372
0.1560
0.1717
0.1847
0.1954
0.2042
5.0
0.0316
0.0620
0.0901
0.1154
0.1374
0.1561
0.1719
0.1849
0.1956
0.2044
6.0
0.0316
0.0620
0.0902
0.1154
0.1374
0.1562
0.1719
0.1850
0.1957
0.2045
Continued
FIGURE 5.7
Stress increase below any point of a uniformly
loaded flexible rectangular area
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TABLE 5.2
Continued
m
n
1.2
1.4
1.6
1.8
2.0
2.5
3.0
4.0
5.0
6.0
0.1
0.0293
0.0301
0.0306
0.0309
0.0311
0.0314
0.0315
0.0316
0.0316
0.0316
0.2
0.0573
0.0589
0.0599
0.0606
0.0610
0.0616
0.0618
0.0619
0.0620
0.0620
0.3
0.0832
0.0856
0.0871
0.0880
0.0887
0.0895
0.0898
0.0901
0.0901
0.0902
0.4
0.1063
0.1094
0.1114
0.1126
0.1134
0.1145
0.1150
0.1153
0.1154
0.1154
0.5
0.1263
0.1300
0.1324
0.1340
0.1350
0.1363
0.1368
0.1372
0.1374
0.1374
0.6
0.1431
0.1475
0.1503
0.1521
0.1533
0.1548
0.1555
0.1560
0.1561
0.1562
0.7
0.1570
0.1620
0.1652
0.1672
0.1686
0.1704
0.1711
0.1717
0.1719
0.1719
0.8
0.1684
0.1739
0.1774
0.1797
0.1812
0.1832
0.1841
0.1847
0.1849
0.1850
0.9
0.1777
0.1836
0.1874
0.1899
0.1915
0.1938
0.1947
0.1954
0.1956
0.1957
1.0
0.1851
0.1914
0.1955
0.1981
0.1999
0.2024
0.2034
0.2042
0.2044
0.2045
1.2
0.1958
0.2028
0.2073
0.2103
0.2124
0.2151
0.2163
0.2172
0.2175
0.2176
1.4
0.2028
0.2102
0.2151
0.2184
0.2206
0.2236
0.2250
0.2260
0.2263
0.2264
1.6
0.2073
0.2151
0.2203
0.2237
0.2261
0.2294
0.2309
0.2320
0.2323
0.2325
1.8
0.2103
0.2183
0.2237
0.2274
0.2299
0.2333
0.2350
0.2362
0.2366
0.2367
2.0
0.2124
0.2206
0.2261
0.2299
0.2325
0.2361
0.2378
0.2391
0.2395
0.2397
2.5
0.2151
0.2236
0.2294
0.2333
0.2361
0.2401
0.2420
0.2434
0.2439
0.2441
3.0
0.2163
0.2250
0.2309
0.2350
0.2378
0.2420
0.2439
0.2455
0.2461
0.2463
4.0
0.2172
0.2260
0.2320
0.2362
0.2391
0.2434
0.2455
0.2472
0.2479
0.2481
5.0
0.2175
0.2263
0.2324
0.2366
0.2395
0.2439
0.2460
0.2479
0.2486
0.2489
6.0
0.2176
0.2264
0.2325
0.2367
0.2397
0.2441
0.2463
0.2482
0.2489
0.2492
Now, using Table 5.2, the magnitudes of I (= I1 , I2 , I3 , I4 ) for the four
rectangles can be determined. The total stress increase below point C at depth
z can thus be determined as
!z = q (I1 + I2 + I3 + I4)
(5.16)
In any practical problem the stress increase below the center of a loaded
rectangular area is of primary importance. The vertical stress increase below the
center of a uniformly-loaded flexible rectangular area can be calculated as
© 1999 by CRC Press LLC
1 + m12 + 2n12
m1n1
2
2
2
1 + m12 + n12 (1 + n1 )(m1 + n1 )
2q
σz ( c ) =
π
m1
−1
+ sin
2
2
2
m1 + n1 1 + n1
wherem1 =
(5.17)
L
B
n1 =
(5.18)
z
B
2
(5.19)
Table 5.3 gives the variation of !z (c) /q with L/B and z/B based on Eq. (5.17).
TABLE 5.3 Variation of !z (c) /q [Eq. 5.17)]
L/B
z/B
3
4
5
6
7
8
9
10
0.1
0.994 0.997
1
2
0.997
0.997
0.997
0.997
0.997
0.997
0.997
0.997
0.2
0.960 0.976
0.977
0.977
0.977
0.977
0.977
0.977
0.977
0.977
0.3
0.892 0.932
0.936
0.936
0.937
0.937
0.937
0.937
0.937
0.937
0.4 0.800 0.870
0.878
0.880
0.881
0.881
0.881
0.881
0.881
0.881
0.5
0.701 0.800
0.814
0.817
0.818
0.818
0.818
0.818
0.818
0.818
0.6
0.606 0.727
0.748
0.753
0.754
0.755
0.755
0.755
0.755
0.755
0.7
0.522 0.658
0.685
0.692
0.694
0.695
0.695
0.696
0.696
0.696
0.8
0.449 0.593
0.627
0.636
0.639
0.640
0.641
0.641
0.641
0.642
0.9
0.388 0.534
0.573
0.585
0.590
0.591
0.592
0.592
0.593
0.593
1.0
0.336 0.481
0.525
0.540
0.545
0.547
0.548
0.549
0.549
0.549
1.5
0.179 0.293
0.348
0.373
0.384
0.389
0.392
0.393
0.394
0.395
2.0
0.108 0.190
0.241
0.269
0.285
0.293
0.298
0.301
0.302
0.303
2.5
0.072 0.131
0.174
0.202
0.219
0.229
0.236
0.240
0.242
0.244
3.0
0.051 0.095
0.130
0.155
0.172
0.184
0.192
0.197
0.200
0.202
3.5
0.038 0.072
0.100
0.122
0.139
0.150
0.158
0.164
0.168
0.171
4.0
0.029 0.056
0.079
0.098
0.113
0.125
0.133
0.139
0.144
0.147
4.5
0.023 0.045
0.064
0.081
0.094
0.105
0.113
0.119
0.124
0.128
5.0
0.019 0.037
0.053
0.067
0.079
0.089
0.097
0.103
0.108
0.112
© 1999 by CRC Press LLC
FIGURE 5.8
EXAMPLE 5.1
Figure 5.8 shows the plan of a flexible loaded area located at the ground
surface. The uniformly distributed load q on the area is 150 kN / m2. Determine the stress increase !z below points A and C at a depth of 10 m below the
ground surface. Note C is at the center of the area.
Solution
Stress increase below point A: The following table can now be prepared.
Area
No.
B
(m)
L
(m)
z
(m)
m=B/z
n=L/z
I
(Table 5.2)
1
2
3
4
2
2
2
2
2
4
4
2
10
10
10
10
0.2
0.2
0.2
0.2
0.2
0.4
0.4
0.2
0.0179
0.0328
0.0328
0.0179
!0.1014
From Eq. (5.14)
!z = qI = (150)(0.1014) = 15.21 kN / m2
Stress increase below point C:
L 6
z 10
= = 15
=
= 2.5
.;
4
B 4
B
© 1999 by CRC Press LLC
From Table 5.3
σz
≈ 0.104
q
σ z = (0.104)(150) = 15.6 kN / m 2
!!
ELASTIC FOUNDATIONS
5.3 FLEXIBLE AND RIGID FOUNDATIONS
Before discussing the relationships for elastic settlement of shallow foundations, it is important to understand the fundamental concepts and the differences between a flexible foundation and a rigid foundation. When a flexible
foundation on an elastic medium is subjected to a uniformly distributed load,
the contact pressure will be uniform as shown in Fig. 5.9a. Figure 5.9a also
shows the settlement profile of the foundation. If a similar foundation is placed
on a granular soil it will undergo larger elastic settlement at the edges rather
than at the center (Fig. 5.9b); however, the contact pressure will be uniform.
The larger settlement at the edges is due to the lack of confinement in the soil.
If a fully rigid foundation is placed on the surface of an elastic medium,
the settlement will remain the same at all points; however, the contact distribution will be as shown in Fig. 5.10a. If this rigid foundation is placed on a
granular soil, the contact pressure distribution will be as shown in Fig. 5.10b,
although the settlement at all points below the foundation will be the same.
Theoretically, for an infinitely rigid foundation supported by a perfectly
elastic material, the contact pressure can be expressed as (Fig. 5.11)
σ z=0 =
σ z=0 =
2q
2x
π 1−
B
q
2x
2 1−
B
2
2
(continuous foundation)
(5.20)
(circular foundation )
(5.21)
where q = applied load per unit area of the foundation
B = foundation width (or diameter)
Borowicka [3] developed solutions for the distribution of contact pressure
beneath a continuous foundation supported by a perfectly elastic material.
According to this theory
© 1999 by CRC Press LLC
FIGURE 5.9
Contact pressure and settlement for flexible foundation:
(a) Elastic material; (b) Granular soil
!z=0 = f (K)
(5.22)
where K = relative stiffness factor
1 1 − ν 2s E f t
=
6 1 − ν 2f Es B
2
3
(5.23)
"s = Poisson’s ratio of the elastic material
"f = Poisson’s ratio of the foundation material
t = thickness of the foundation
Es , Ef = modulus of elasticity of the elastic material and foundation
material, respectively
Although soil is not perfectly elastic and homogeneous, the theory of
elasticity may be used to estimate the settlement of shallow foundations at
allowable loads. Judicious use of these results have done well in the design,
construction, and maintenance of structures.
© 1999 by CRC Press LLC
FIGURE 5.10 Contact pressure and settlement for rigid foundation:
(a) Elastic material; (b) Granular soil
5.4 SETTLEMENT UNDER A CIRCULAR AREA
Figure 5.12 shows a uniformly loaded flexible circular area (q / unit area) on
the surface of an elastic material (in this case soil which is assumed to be
elastic). The radius of the circular area is R. The vertical strain #z at a point
A due to the loading can be given as
∈z =
1
[σ − ν(σr + σθ )]
Es z
(5.24)
Es = modulus of elasticity of soil
" = Poisson’s ratio of soil
The elastic settlement at A can be evaluated as
where
∞
∫
z
∞
∈ z dz =
∫ E [σ − ν (σ + σ )]dz
© 1999 by CRC Press LLC
1
z
z
s
r
θ
(5.25)
FIGURE 5.11 Contact pressure distribution under an infinitely rigid
foundation supported by a perfectly elastic material
FIGURE 5.12 Settlement under a uniformly
loaded flexible circular area
© 1999 by CRC Press LLC
By substituting the proper relations for !z , !$ , and !r into Eq. (5.25), we
obtain
Se = q
1+ ν
R
Es
z
R I1 + (1 − ν)I2
(5.26)
r
z
where I 1 and I 2 = f and
R
R
The variations of I1 and I2 at r/R = 0 and 1 (that is, below the center and
edge of the loaded area) are given in Table 5.4 and 5.5.
At the surface, that is, z/R = 0
Se = q
1 − ν2
RI 2
Es
(5.27)
Using Tables 5.4 and 5.5 and Eq. (5.27), at the surface
S e (center ) =
2q (1 − ν 2 ) R qB (1 − ν 2 )
=
Es
Es
(5.28)
S e ( edge) =
1.273qB (1 − ν 2 ) 0.636qB (1 − ν 2 )
=
2Es
Es
(5.29)
where B = 2R = diameter of the loaded area
TABLE 5.4 Variations
of I1 and I2 for r/R = 0
TABLE 5.5 Variations
of I1 and I2 for r/R =1
z/R
I1
I2
z/R
I1
I2
0
0.2
0.4
0.6
0.8
1.0
1.5
2.0
2.5
3.0
4.0
5.0
1.000
0.804
0.629
0.486
0.375
0.293
0.168
0.106
0.072
0.051
0.030
0.019
2.000
1.640
1.350
1.130
0.961
0.828
0.606
0.472
0.385
0.325
0.246
0.198
0
0.2
0.4
0.6
0.8
1.0
1.5
2.0
2.5
3.0
4.0
5.0
0.500
0.383
0.310
0.256
0.213
0.179
0.119
0.083
0.060
0.045
0.027
0.018
1.273
1.100
0.962
0.849
0.756
0.678
0.531
0.432
0.362
0.310
0.239
0.195
© 1999 by CRC Press LLC
FIGURE 5.13 Average settlement under a uniformly loaded flexible circular
area (Note: Diameter = B = 2R)
The average settlement at the surface can now be determined as (Fig.
5.13)
B/2
∫ S dx
e
S e (average ) =
−B / 2
B
For most practical purposes, Se (average) ! 0.85Se (center). Hence, at the
surface
S e (average) =
0.85qB (1 − ν 2 )
Es
(5.30)
For rigid foundations, the surface settlement (z/R = 0) is
S e ( rigid ) ≈ 0.93S e (average flexible) ≈
0.79qB(1 − ν 2 )
Es
(5.31)
EXAMPLE 5.2
A flexible circularly loaded area located on the ground surface has a radius of
1.5 m. The uniformly distributed load on the area is 250 kN / m2. Determine
the settlement at the ground surface for the following conditions:
a. Below the center
b. At the edge
c. Average
Assume Poisson’s ratio to be 0.3 and Es = 9500 kN / m2.
© 1999 by CRC Press LLC
Solution
a.
From Eq. (5.28)
Se =
. )(1 − 0.32 )
qB (1 − ν 2 ) (250)( 2 × 15
=
= 0.0718 m = 71.8 mm
9500
Es
b. From Eq. (5.29)
Se =
c.
0.636qB (1 − ν 2 )
= (0.636)( 718
. ) = 45.66 mm
Es
From Eq. (5.30)
Se =
0.85qB (1 − ν 2 )
. ) = 61.03 mm
= ( 0.85)(718
Es
!!
EXAMPLE 5.3
Refer to Example 5.2. Assume the loaded area to be rigid and determine the
elastic surface settlement.
Solution From Eq. (5.31)
Se =
0.79qB (1 − ν 2 ) ( 0.79)( 250)( 2 × 15
. )(1 − 0.32 )
=
9500
Es
= 0.0568 m = 56.8 mm
!!
5.5 SETTLEMENT UNDER A RECTANGULAR AREA
The elastic settlement at any depth below the corner of a flexible rectangular
area of dimension L × B (Fig. 5.6) can be obtained by proper integration of the
expression for vertical strain as [4]
Se (corner ) =
© 1999 by CRC Press LLC
qB
1 − 2ν
(1 − ν 2 ) I 3 −
I 4
2Es
1− ν
(5.32)
where I 3 =
1 + m′2 + n′2 + 1
1 1 + m′2 + n′2 + m′
ln
+ m′ ln
1 + m′2 + n′2 − 1
π 1 + m′2 + n′2 − m′
(5.33)
I4 =
n′
m′
tan −1
2
2
π
n′ 1 + m′ + n′
(5.34)
m′ =
L
B
(5.35)
n′ =
z
B
(5.36)
The variations of I3 and I4 with m! and n! are given in Table 5.6 and 5.7,
respectively. At the surface n! = z/B = 0, so I4 = 0. Hence from Eq. (5.37) at the
surface
S e ( corner ) =
qB
(1 − ν 2 ) I 5
2E
(5.37)
where
1 + m′2 + 1
1 1 + m′2 + m′
+ m′ ln
I5 = ln
1 + m′2 − 1
π 1 + m′2 − m′
(5.38)
Using the method of superposition it can be shown that, at the surface,
S e (center ) =
qB
(1 − ν 2 ) I 5
Es
(5.39)
The variations of I5 with L/B are given in Table 5.8.
As in the case of a circularly loaded area [Eqs. (5.30) and (5.31)] at the
surface
S e ( average) =
© 1999 by CRC Press LLC
qB
0.85qB
(1 − ν 2 ) I 5 =
(1 − ν 2 ) I 6
Es
Es
(5.40)
TABLE 5.6 Variation of I3
m!
n!
1
2
3
4
5
6
7
8
9
10
0.00
0.25
0.50
0.75
1.00
1.25
1.50
1.75
2.00
2.25
2.50
2.75
3.00
3.25
3.50
3.75
4.00
4.25
4.50
4.75
5.00
5.25
5.50
5.75
6.00
6.25
6.50
6.75
7.00
7.25
7.50
7.75
8.00
8.25
8.50
8.75
9.00
9.25
9.50
9.75
10.00
1.122
1.095
1.025
0.933
0.838
0.751
0.674
0.608
0.552
0.504
0.463
0.427
0.396
0.369
0.346
0.325
0.306
0.289
0.274
0.260
0.248
0.237
0.227
0.217
0.208
0.200
0.193
0.186
0.179
0.173
0.168
0.162
0.158
0.153
0.148
0.144
0.140
0.137
0.133
0.130
0.126
1.532
1.510
1.452
1.371
1.282
1.192
1.106
1.026
0.954
0.888
0.829
0.776
0.728
0.686
0.647
0.612
0.580
0.551
0.525
0.501
0.479
0.458
0.440
0.422
0.406
0.391
0.377
0.364
0.352
0.341
0.330
0.320
0.310
0.301
0.293
0.285
0.277
0.270
0.263
0.257
0.251
1.783
1.763
1.708
1.632
1.547
1.461
1.378
1.299
1.226
1.158
1.095
1.037
0.984
0.935
0.889
0.848
0.809
0.774
0.741
0.710
0.682
0.655
0.631
0.608
0.586
0.566
0.547
0.529
0.513
0.497
0.482
0.468
0.455
0.442
0.430
0.419
0.408
0.398
0.388
0.379
0.370
1.964
1.944
1.89
1.816
1.734
1.650
1.570
1.493
1.421
1.354
1.291
1.233
1.179
1.128
1.081
1.037
0.995
0.957
0.921
0.887
0.855
0.825
0.797
0.770
0.745
0.722
0.699
0.678
0.658
0.639
0.621
0.604
0.588
0.573
0.558
0.544
0.531
0.518
0.506
0.494
0.483
2.105
2.085
2.032
1.959
1.878
1.796
1.717
1.641
1.571
1.505
1.444
1.386
1.332
1.281
1.234
1.189
1.147
1.107
1.070
1.034
1.001
0.969
0.939
0.911
0.884
0.858
0.834
0.810
0.788
0.767
0.747
0.728
0.710
0.692
0.676
0.660
0.644
0.630
0.616
0.602
0.589
2.220
2.200
2.148
2.076
1.995
1.914
1.836
1.762
1.692
1.627
1.567
1.510
1.457
1.406
1.359
1.315
1.273
1.233
1.195
1.159
1.125
1.093
1.062
1.032
1.004
0.977
0.952
0.927
0.904
0.881
0.860
0.839
0.820
0.801
0.783
0.765
0.748
0.732
0.717
0.702
0.688
2.318
2.298
2.246
2.174
2.094
2.013
1.936
1.862
1.794
1.730
1.670
1.613
1.561
1.511
1.465
1.421
1.379
1.339
1.301
1.265
1.231
1.199
1.167
1.137
1.109
1.082
1.055
1.030
1.006
0.983
0.960
0.939
0.918
0.899
0.879
0.761
0.843
0.826
0.810
0.794
0.778
2.403
2.383
2.331
2.259
2.179
2.099
2.022
1.949
1.881
1.817
1.758
1.702
1.650
1.601
1.555
1.511
1.470
1.431
1.393
1.358
1.323
1.291
1.260
1.230
1.201
1.173
1.147
1.121
1.097
1.073
1.050
1.028
1.007
0.987
0.967
0.948
0.930
0.912
0.895
0.878
0.862
2.477
2.458
2.406
2.334
2.255
2.175
2.098
2.025
1.958
1.894
1.836
1.780
1.729
1.680
1.634
1.591
1.550
1.511
1.476
1.438
1.404
1.372
1.341
1.311
1.282
1.255
1.228
1.203
1.178
1.154
1.131
1.109
1.087
1.066
1.046
1.027
1.008
0.990
0.972
0.955
0.938
2.544
2.525
2.473
2.401
2.322
2.242
2.166
2.093
2.026
1.963
1.904
1.850
1.798
1.750
1.705
1.662
1.621
1.582
1.545
1.510
1.477
1.444
1.413
1.384
1.355
1.328
1.301
1.275
1.251
1.227
1.204
1.811
1.160
1.139
1.118
1.099
1.080
1.061
1.043
1.026
1.009
S e ( rigid ) =
qB
0.79 qB
(1 − ν 2 ) I 5 =
(1 − ν 2 ) I 7
Es
Es
The variations of I6 and I7 with L/B are given in Table 5.9.
© 1999 by CRC Press LLC
(5.41)
TABLE 5.7 Variation of I4
m!
n!
1
2
3
4
5
6
7
8
9
10
0.25
0.50
0.75
1.00
1.25
1.50
1.75
2.00
2.25
2.50
2.75
3.00
3.25
3.50
3.75
4.00
4.25
4.50
4.75
5.00
5.25
5.50
5.75
6.00
6.25
6.50
6.75
7.00
7.25
7.50
7.75
8.00
8.25
8.50
8.75
9.00
9.25
9.50
9.75
10.00
0.098
0.248
0.166
0.167
0.160
0.149
0.139
0.128
0.119
0.110
0.102
0.096
0.090
0.084
0.079
0.075
0.071
0.067
0.064
0.061
0.059
0.056
0.054
0.052
0.050
0.048
0.046
0.045
0.043
0.042
0.040
0.039
0.038
0.037
0.036
0.035
0.034
0.033
0.032
0.032
0.103
0.167
0.202
0.218
0.222
0.220
0.213
0.205
0.196
0.186
0.177
0.168
0.160
0.152
0.145
0.138
0.132
0.126
0.121
0.116
0.111
0.107
0.103
0.099
0.096
0.093
0.089
0.087
0.084
0.081
0.079
0.077
0.074
0.072
0.070
0.069
0.067
0.065
0.064
0.062
0.104
0.172
0.212
0.234
0.245
0.248
0.247
0.243
0.237
0.230
0.223
0.215
0.208
0.200
0.193
0.186
0.179
0.173
0.167
0.161
0.155
0.150
0.145
0.141
0.136
0.132
0.128
0.124
0.121
0.117
0.114
0.111
0.108
0.105
0.103
0.100
0.098
0.095
0.093
0.091
0.105
0.174
0.216
0.241
0.254
0.261
0.263
0.262
0.259
0.255
0.250
0.244
0.238
0.232
0.226
0.219
0.213
0.207
0.201
0.195
0.190
0.185
0.179
0.174
0.170
0.165
0.161
0.156
0.152
0.149
0.145
0.141
0.138
0.135
0.132
0.129
0.126
0.123
0.120
0.118
0.105
0.175
0.218
0.244
0.259
0.267
0.271
0.273
0.272
0.269
0.266
0.262
0.258
0.253
0.248
0.243
0.237
0.232
0.227
0.221
0.216
0.211
0.206
0.201
0.197
0.192
0.188
0.183
0.179
0.175
0.171
0.168
0.164
0.160
0.157
0.154
0.151
0.147
0.145
0.142
0.105
0.175
0.219
0.246
0.262
0.271
0.277
0.279
0.279
0.278
0.277
0.274
0271
0.267
0.263
0.259
0.254
0.250
0.245
0.241
0.236
0.232
0.227
0.223
0.218
0.214
0.210
0.205
0.201
0.197
0.193
0.190
0.186
0.182
0.179
0.176
0.172
0.169
0.166
0.163
0.105
0.175
0.220
0.247
0.264
0.274
0.280
0.283
0.284
0.284
0.283
0.282
0.279
0.277
0.273
0.270
0.267
0.623
0.259
0.255
0.251
0.247
0.243
0.239
0.235
0.231
0.227
0.223
0.219
0.216
0.212
0.208
0.205
0.201
0.198
0.194
0.191
0.188
0.185
0.182
0.105
0.176
0.220
0.248
0.265
0.275
0.282
0.286
0.288
0.288
0.288
0.287
0.285
0.283
0.281
0.278
0.276
0.272
0.269
0.266
0.263
0.259
0.255
0.252
0.248
0.245
0.241
0.238
0.234
0.231
0.227
0.224
0.220
0.217
0.214
0.210
0.207
0.204
0.201
0.198
0.105
0.176
0.220
0.248
0.265
0.276
0.283
0.288
0.290
0.291
0.291
0.291
0.290
0.288
0.287
0.285
0.282
0.280
0.277
0.274
0.271
0.268
0.265
0.262
0.259
0.256
0.252
0.249
0.246
0.243
0.240
0.236
0.233
0.230
0.227
0.224
0.221
0.218
0.215
0.212
0.105
0.176
0.220
0.248
0.266
0.277
0.284
0.289
0.292
0.293
0.294
0.294
0.293
0.292
0.291
0.289
0.287
0.285
0.283
0.281
0.278
0.276
0.273
0.270
0.267
0.265
0.262
0.259
0.256
0.253
0.250
0.247
0.244
0.241
0.238
0.235
0.233
0.230
0.227
0.224
© 1999 by CRC Press LLC
TABLE 5.8 Variation of I5 for Surface Center
Settlement Calculation
L/B
I5
L/B
I5
L/B
I5
1
2
3
4
5
1.122
1.532
1.783
1.964
2.105
6
7
8
9
10
2.220
2.318
2.403
2.447
2.544
15
20
30
50
100
2.802
2.985
3.243
3.568
4.010
TABLE 5.9 Variations of I6 and I7 with L/B for Surface Settlement
Calculation
L/B
I6
I7
L/B
I6
I7
L/B
I6
I7
1
2
3
4
5
0.954
1.302
1.516
1.699
1.789
0.886
1.210
1.409
1.552
1.663
6
7
8
9
10
1.887
1.970
2.043
2.105
2.162
1.754
1.831
1.898
1.957
2.010
15
20
30
50
100
2.382
2.537
2.757
3.033
3.409
2.214
2.358
2.562
2.819
3.168
EXAMPLE 5.4
Figure 5.14 shows a flexible rectangular area measuring 10 ft × 5 ft in plan on
the ground surface. The area is subjected to a uniformly distributed load of
4000 lb / ft2. For the soil, given í = 0.35; Es = 2500 lb / in.2. Determine:
a. The settlement at a depth of 5 ft below a corner point
b. The surface settlement below the center of the loaded area
Solution
a.
From Eq. (5.32)
FIGURE 5.14
© 1999 by CRC Press LLC
Se =
qB
1 − 2ν
(1 − ν 2 ) I 3 −
I4
2Es
1− ν
Se =
qB
1 − 2ν
(1 − ν 2 ) I 3 −
I 4
2Es
1− ν
m′ =
L 10
=
=2
5
B
n′ =
z 5
= =1
B 5
From Tables 5.6 and 5.7 for m! = 2 and n! = 1, I3 = 1.282 and I4 =
0.218. Hence,
Se =
( 4000)(5)
1 − 2 × 0.35
(1 − 0.352 )1.282 −
0.218
(2)(2500 × 144)
1 − 0.35
= 0.0288 ft = 0.346 in.
b. From Eq. (5.39)
Se =
qB
(1 − ν 2 ) I 5
Es
L 10
=
= 2. From Table 5.8, I 5 = 1532
.
B
5
Se =
(4000)(5)
(1 − 0.352 )(1532
. ) = 0.075 ft = 0.9 in.
2500 × 144
!!
5.6 EFFECT OF A RIGID BASE
AT A LIMITED DEPTH
The relationships for elastic settlement given in the preceding section assume
that the elastic soil layer extends to an infinite depth. However, if the soil layer
is underlain by a rigid base at a limited depth h as shown in Fig. 5.15, the
elastic settlement of the compressible soil layer can be calculated as
© 1999 by CRC Press LLC
FIGURE 5.15 Effect of a rigid base located at a limited depth
on elastic settlement of a foundation
∞
Se =
∞
∫ ∈ dz − ∫ ∈ dz
z
0
z
h
where !z = vertical compressive strain at a depth z, or
Se = Se (z=0) " Se (z=h)
(5.42)
Egorov [5] calculated the elastic settlement below the center of a uniformly loaded circular area of radius R and a rectangular area with dimensions
of L × B. He assumed that no friction exists between the soil layer and the
rigid base. According to this method [4]
Circular area:
S e (center, flexible) =
S e (center, rigid) =
Rq
(1 − ν 2 ) α 1
Es
Rq
(1 − ν 2 )α 2
Es
where R = radius of the circular area
Rectangular Area (B × L)
© 1999 by CRC Press LLC
(5.43)
(5.44)
S e (center, flexible) =
S e (center, rigid) =
Bq
(1 − ν 2 )α 3
Es
(5.45)
Bq
(1 − ν 2 )α 4
Es
(5.46)
The values of "1 and "2 for various values of h/R are given in Table 5.10.
Similarly, the variations of "3 and "4 are given in Table 5.11.
TABLE 5.10 Variations
of "1 and "2
h/R
"1
"2
0
0.5
1.0
2.0
3.0
5.0
10.0
0
0.52
1.00
1.44
1.62
1.78
1.88
0
0.45
0.79
1.16
1.32
1.48
1.64
Table 5.11 Variations of "3 and "4
L/B
1
h/B
2
3
4
5
"3
"4
"3
"4
"3
"4
"3
"4
"3
"4
0
0
0
0
0
0
0
0
0
0
0
0.25
0.25
0.226
0.25
0.23
0.25
0.23
0.25
0.24
0.25
0.238
0.5
0.51
0.403
0.51
0.43
0.51
0.44
0.51
0.44
0.51
0.446
1
0.77
0.609
0.87
0.7
0.88
0.73
0.88
0.75
0.88
0.764
1.5
0.88
0.711
1.07
0.86
1.12
0.91
1.13
0.95
1.13
0.982
2.5
0.98
0.8
1.24
1.01
1.36
1.12
1.44
1.2
1.45
1.256
5
1.05
0.873
1.39
1.16
1.56
1.31
1.75
1.48
1.87
1.619
5.7 EFFECT OF DEPTH OF EMBEDMENT
The theories presented in Sections 5.4, 5.5, and 5.6 are for the conditions
where the load is at the surface of the soil layer (Fig. 5.16a). In actuality, foundations are always placed at a certain depth below the ground surface (Fig.
5.16b). The elastic settlement of an embedded foundation is always less than
© 1999 by CRC Press LLC
FIGURE 5.16 Effect of embedment on the elastic settlement of a foundation
(Note: Foundation length = L; foundation width = B)
when the foundation is placed at the surface. There are limited theoretical
studies available on this subject. Fox [6] evaluated the effect of embedment on
the average settlement of a uniformly loaded flexible rectangular area (L × B),
and this was presented in a graphical form. The interpolated values from this
graph are given in Table 5.12. Note that Table 5.12 is valid only for ! = 0.5.
TABLE 5.12 Variation of
Se (average, Df )/Se (average, Df =0)
L/B
Df
LB
1
5
10
0
0.1
0.2
0.3
0.4
0.5
0.6
0.8
1.0
1
0.981
0.959
0.928
0.894
0.863
0.825
0.763
0.732
1
0.975
0.947
0.906
0.875
0.844
0.819
0.775
0.737
1
0.969
0.931
0.893
0.856
0.831
0.813
0.781
0.738
© 1999 by CRC Press LLC
TABLE 5.13 Suggested Values for Poisson’s Ratio
Soil type
Poisson’s ratio, !
Coarse sand
Medium loose sand
Fine sand
Sandy silt and silt
Saturated clay (undrained)
Saturated clay--lightly overconsolidated (drained)
0.15 – 0.20
0.20 – 0.25
0.25 – 0.30
0.30 – 0.35
0.50
0.2 – 0.4
5.8 ELASTIC PARAMETERS
Parameters such as the modulus of elasticity Es and Poisson’s ratio ! for a
given soil must be known to calculate the elastic settlement of a foundation. In
most cases, if laboratory test results are not available, they are estimated from
empirical correlations. Table 5.13 provides some suggested values for
Poisson’s ratio.
Trautman and Kulhawy [7] used the following relationship for Poisson’s
ratio (drained state)
! = 0.1 + 0.3"rel
(5.47)
φ rel = relative friction angle =
φ tc − 25°
(0 ≤ φ rel ≤ 1)
45°−25°
(5.48)
where "tc = friction angle from drained triaxial compression test
A general range of the modulus of elasticity of sand, Es , is given in Table
5.14.
TABLE 5.14 General Range of Modulus of Elasticity
of Sand
Type
Coarse and medium coarse sand
Loose
Medium dense
Dense
Fine sand
Loose
Medium dense
Dense
Sandy silt
Loose
Medium dense
Dense
© 1999 by CRC Press LLC
Es (kN / m2)
25,000 – 35,000
30,000 – 40,000
40,000 – 45,000
20,000 – 25,000
25,000 – 35,000
35,000 – 40,000
8,000 – 12,000
10,000 – 12,000
12,000 – 15,000
A number of correlations for the modulus of elasticity of sand with the
field standard penetration resistance N and cone penetration resistance qc were
made in the past. Schmertmann [8] proposed that
Es (kN / m2) = 766N
(5.49a)
Es (U.S. ton/ft2) = 8N
(5.49b)
Schmertmann and Hartman [9] made the following recommendations for
estimating the Es of sand from cone penetration resistance, or
Es = 2.5qc (for square and circular foundations)
(5.50)
Es = 3.5qc (for strip foundation; L/B ! 10)
(5.51)
In many cases, the modulus of elasticity of saturated clay soils (undrained)
was correlated with the undrained shear strength, cu . D’Appolonia et al. [10]
compiled several field test results and concluded that
Es
= 1000 to 1500 for lean inorganic clays from
moderate to high plasticity
cu
(5.52)
Duncan and Buchignani [11] correlated Es /cu with the overconsolidation
ratio OCR and plasticity index PI of several clay soils. This broadly generalized
correlation is shown in Fig. 5.17.
5.9 SETTLEMENT OF FOUNDATIONS
ON SATURATED CLAYS
Method of Janbu et al.
Janbu et al. [12] proposed a generalized equation for estimating the average
elastic settlement of a uniformly loaded flexible foundation located on a saturated clay (! = 0.5) which is similar to those presented in Sections 5.4 and 5.5.
This relationship incorporates (a) the effect of embedment Df and (b) the
possible existence of a rigid layer at a shallow depth under the foundation as
shown in Fig. 5.18, or
S e = µ 1µ 2
qB
Es
© 1999 by CRC Press LLC
(5.53)
FIGURE 5.17
Correlation of Duncan and Buchignani for the
modulus of elasticity of clay in an undrained state
FIGURE 5.18 Settlement of foundation on saturated clay
© 1999 by CRC Press LLC
Df
µ 1 = f
B
where
h L
,
µ2 =
B B
L = foundation length
B = foundation width
Christian and Carrier [13] made a critical evaluation of the factors µ 1 and
µ 2 , and the results were presented in a graphical form. The interpolated values
of µ 1 and µ 2 from these graphs are given in Tables 5.15 and 5.16.
TABLE 5.15 Variation
of µ 1 with Df /0 [Eq. (5.53)]
Df /B
µ1
0
2
4
6
8
10
12
14
16
18
20
1.0
0.9
0.88
0.875
0.87
0.865
0.863
0.860
0.856
0.854
0.850
TABLE 5.16 Variation of µ2 with h/B
L/B
h/B
Circle
1
2
5
10
"
1
2
4
6
8
10
20
30
0.36
0.47
0.58
0.61
0.62
0.63
0.64
0.66
0.36
0.53
0.63
0.67
0.68
0.70
0.71
0.73
0.36
0.63
0.82
0.88
0.90
0.92
0.93
0.95
0.36
0.64
0.94
1.08
1.13
1.18
1.26
1.29
0.36
0.64
0.94
1.14
1.22
1.30
1.47
1.54
0.36
0.64
0.94
1.16
1.26
1.42
1.74
1.84
Method of D’Appolonia et al.
D’Appolonia et al. [10] proposed an improved method for estimating the
© 1999 by CRC Press LLC
FIGURE 5.19 Idealized plot of load per unit area versus settlement
of foundation
FIGURE 5.20 Estimation of initial settlement for a foundation supported
by a saturated clay layer
initial settlement, Si , for foundations supported by saturated clay. The purpose
of this theory was to account for the limitation of the elastic theory and to account for the stress redistribution and the strains occurring after local yielding.
© 1999 by CRC Press LLC
This hypothesis can be explained by referring to Fig. 5.19, which shows an
idealized plot of applied load per unit area on the foundation q versus
settlement Se plot. In this figure OA is a linear plot. The first local yield occurs
at A. It is followed by nonlinear segment AB. When elastic theories are used
to calculate settlement the nonlinearity of segment AB is not taken into
account.
According to this approach, the steps to estimate the initial settlement
follow (refer to Fig. 5.20).
1. Using any procedure, estimate the ultimate bearing capacity, qu , of the
foundation.
2. Determine the allowable load per unit area, q, and then the applied
stress ratio, q/qu .
3. Calculate the initial shear stress ratio, f, as
f =
4.
5.
6.
1 − Ko
2 cu
σ ′v
(5.54)
where Ko = coefficient of lateral at-rest earth pressure
cu = undrained shear strength
##v = initial vertical effective stress
The initial shear stress ratio f can be estimated from Fig. 5.21, which
is the average of tests conducted on several clays [10].
Estimate the magnitude of the modulus of elasticity, Es (Section 5.8).
Calculate the elastic settlement, Se , by using one of the applicable
equations given in Sections 5.4 and 5.5.
For a given applied stress ratio, q/qu , f, and h/B (h = thickness of
compressible layer; B = width of the rectangular foundation or
diameter of the circular foundation), determine the settlement ratio,
$, using Fig. 5.22
ρ=
Se
Si
(5.55)
Se
ρ
(5.56)
So
Si =
From Fig. 5.22 it is clear than h/B has minor influence on the settlement
ratio, $. The result for h/B = 1.5 will apply for all values of h/B > 1.5.
© 1999 by CRC Press LLC
FIGURE 5.21 Relationship between f and OCR (after D’Appolonia et al. [10])
5.10 SETTLEMENT OF FOUNDATIONS ON SAND
Correlation of Standard Penetration Resistance
and Cone Penetration Resistance
Geotechnical engineers have used empirical approaches based on a large
number of case studies to estimate the elastic settlement of foundations
constructed on sand. The two most widely used empirical approaches utilize
the corrected standard penetration resistance N´ and the cone penetration resistance number qc . Meyerhof [14] proposed a correlation for the net allowable
bearing capacity for foundations for one inch (25.4 mm) of estimated
maximum settlement with the corrected standard penetration resistance N´ as
qnet(all) (kip/ft2 ) =
N′
(for B ≤ 4 ft)
4
(5.57)
and
N′ B + 1
qnet(all) (kip/ft ) =
(for B > 4 ft)
6 B
2
2
© 1999 by CRC Press LLC
(5.58)
FIGURE 5.22 Relationship between $ and q/qu for continuous
foundation (after D’Appolonia et al. [10])
where B = width of the foundation
N# = corrected standard penetration resistance
qnet(all) = qall $ %Df
© 1999 by CRC Press LLC
(5.59)
The N# value to be used in the above equation should be the average N# value
in the influence zone of the foundation, that is, 0.5Df above the foundation and
2B below the foundation.
Again, based on several additional field observations, Meyerhof [15]
suggested that the qnet(all) predicted by Eqs. (5.57) and (5.58) can be safely
increased by 50%. Thus, these two equations can be modified as
q net(all) (kip/ft 2 ) =
N′
F S (for B ≤ 4 ft)
2.5 d e
q net(all) (kip/ft 2 ) =
N′ B + 1
F S (for B > 4 ft)
4 B d e
(5.60a)
2
where Se = estimate maximum elastic settlement, in inches
B = width, in ft
Df
Fd = depth factor = 1 + 0.33
B
(5.60b)
(5.61)
Df = depth of foundation
In SI units, Eqs. (5.60a) and (5.60b) can be expressed as
S
q net(all) (kN/m 2 ) = 19.16 N ′Fd e (for B ≤ 1.22 m)
25.4
(5.62)
2
3.28B + 1 Se
(for B > 1.22 m)
qnet(all) (kN/m ) = 11.98N ′
Fd
3.28B 25.4
2
(5.63)
where
B = width, in meters
Se = estimated maximum elastic settlement, in mm
Meyerhof [14] also developed empirical relationships for the net allowable
bearing capacity of foundations based on cone penetration resistance, qc , or
q net(all) (kip / ft 2 ) =
q c ( kip / ft 2 )
15
(for B ≤ 4 ft and settlement of 1 in. )
and
© 1999 by CRC Press LLC
(5.64)
q net(all) (kip/ft 2 ) =
q c ( kip/ft 2 ) B + 1
25
B
2
(5.65)
(for B > 4 ft and 1 in. of settlement )
In SI units, the preceding two equations can be written as
q net(all ) ( kN/m
2
) =
for B ≤ 1.22 m and
settlement of 25.4 mm
(5.66)
for B > 1.22 m and
settlement
of 25.4 mm
(5.67)
q c ( kN/m 2 )
15
and
q net(all ) =
q c 3 . 28 B + 1
25 3 . 28 B
2
Based on the observation of the settlement of 48 buildings all constructed on
sand, Schultze and Sherif [16] gave the following expression for predicting
elastic settlement
qall f ′
1
B
Se =
0
.
4
D
B
f
1 +
1.71N ′ 0.87
B
B1
(5.68)
where qall is in kip/cm2
f´ = influence factor of the settlement
B1 = 1 cm
Df = depth of foundation
Based on the study of Schultze and Sherif [16], Meyerhof [17] suggested the
following relations for estimating elastic settlement
Se =
q B
(for sand and gravel)
2N ′
(5.69)
Se =
q B
(for silty sand)
N′
(5.70)
where Se = settlement, in inches, q is in U.S. ton / ft2, and B is in inches.
© 1999 by CRC Press LLC
Use of Strain Influence Factor
The equation for vertical strain, !z , below the center of a flexible circular load
of radius R was in given in Eq. (5.24) as
1
[σ − ν(σr + σθ )]
Es z
∈z =
(5.24)
After proper substitution for "z , "r , and "# in the preceding equation, one
obtains
∈z =
q(1 + ν)
[(1 − 2ν)A′ + B′]
Es
(5.71)
where A!, B! = nondimensional factors and functions of z/R
The variations of A! and B! below the center of a loaded area as estimated
by Ahlvin and Ulery [2] are given in Table 5.17.
TABLE 5.17 Variations of A!! and B!! (Below
the Center of a Flexible Loaded Area)
z/R
A!
B!
z/R
A!
B!
0
0.2
0.4
0.6
0.8
1.0
1.5
2.0
1.0
0.804
0.629
0.486
0.375
0.293
0.168
0.106
0
0.189
0.320
0.378
0.381
0.354
0.256
0.179
2.5
3.0
4.0
5.0
6.0
7.0
8.0
9.0
0.072
0.051
0.030
0.019
0.104
0.010
0.008
0.006
0.128
0.095
0.057
0.038
0.027
0.020
0.015
0.012
From Eq. (5.71) we can write
Iz =
∈z E s
q
= (1 + ν )[(1 − 2ν ) A′ + B ′]
(5.72)
Figure 5.23 shows plots of Iz versus z/R obtained from the experimental
results of Eggstad [18] along with the theoretical values calculated from Eq.
(5.72). Based on Fig. 5.23, Schmertmann [8] proposed a practical variation of
Iz and z/B (B = foundation width) for calculating the elastic settlement of foundations. This model was later modified by Schmertmann and Hartman [9], and
the variation is shown in Fig. 5.24 for L/B = 1 and L/B " 10. Interpolations can
be used to obtain the Iz –z/B variation for other L/B values. Using the simplified
strain influence factor, elastic settlement can be calculated as
© 1999 by CRC Press LLC
FIGURE 5.23 Comparison of experimental and theoretical variation of
Iz below the center of a flexible circularly loaded area
(Note: R = radius of circular area; Dr = relative density)
S e = c1 c 2 ( q − q )
Iz
∑ E ∆z
s
c1 = a correction factor for depth of foundation
q
= 1 - 0.5
q−q
c2 = a correction factor for creep in soil
time in years
= 1 + 0.2log
0.1
© 1999 by CRC Press LLC
(5.73)
FIGURE 5.24 Variation of Iz versus z/B
q = $Df
!q = stress at the level of the foundation
The use of Eq. (5.73) can be explained by the following example. Figure 5.25a
shows a continuous foundation for which B = 8 ft; Df = 4 ft; unit weight of
sand, $ = 110 lb / ft3; !q = 25 lb / in.2.
For this case, L/B is greater than 10. Accordingly, the plot of Iz with depth
is shown in Fig. 5.25a. Note that
Iz = 0.2 at z = 0
Iz = 0.5 at z = 8 ft (= B)
Iz = 0 at z = 32 ft (= 4B)
© 1999 by CRC Press LLC
FIGURE 5.25
Based on the results of the standard penetration test or cone penetration test, the
variation of Es can be calculated by using Eq. (5.49) or (5.51) (or similar
relationships). The variation is shown by the broken line in Fig. 5.25b. The
© 1999 by CRC Press LLC
actual variation of Es can be approximated by several linear plots, and this is
also shown in Fig. 5.25b (solid lines). For elastic settlement, Table 5.18 can
now be prepared.
TABLE 5.18 Elastic Settlement Calculation (Fig. 5.25)
Iz at the
middle of
the layer
Iz
%z
(in.)
Es
(lb / in.2)
z to the
middle of
the layer
(in.)
1
48
750
24
0.275
0.0176
2
48
1250
72
0.425
0.016
3
96
1250
144
0.417
0.031
4
48
1000
216
0.292
0.014
5
144
2000
312
0.125
0.0009
Layer
No.
Es
Δz
(in.3 / lb)
#384 in. = 4B
#0.0886 in.3/lb
Since $ = 110 lb / ft3, q = $Df = (4)(110) = 440 lb / ft2 = 3.06 lb / in.2. Given
!q = 25 lb / in.2. Thus !q $ q = 25 $ 3.06 = 21.94 lb / in.2. Also
q
3.06
= 1 − .05
c1 = 1 − 0.5
= 0.93
21.94
q −q
Assume the time for creep is 10 years. Hence,
10
c2 = 1 + 0.2 log
= 1.4
0.1
Thus,
S e = c1 c 2 ( q − q )
Iz
∑ E ∆z
= ( 0.93)(1.4 )( 21.94 )( 0.0886 ) = 2.53 in .
5.11 FIELD PLATE LOAD TESTS
In some cases it is desirable to conduct field load tests to determine the
allowable and ultimate bearing capacities of foundations and associated elastic
© 1999 by CRC Press LLC
settlement. The standard method for field load tests is given by the American
Society for Testing and Materials under Designation D-1194. Circular steel
bearing plates of 6 in. to 30 in. (152.4 mm to 762 mm) in diameter and 1 ft ×
1 ft (304.8 mm × 304.8 mm) square plates are available for this type of test.
Based on the plate load tests conducted in the field, an estimation of the
bearing capacity and associated elastic settlement of full-scale foundations can
be made.
Figure 5.26 shows the general nature of the plots of load per unit area, q,
versus settlement, Se , for two plates with widths (or diameters) of B1 and B2 in
a clay soil (& = 0 condition). In this case B2 > B1 . Similarly, in Fig. 5.27 are
the plots for two plates supported by a sand extending to a great depth. From
these two figures we can see that, in clay
(5.74)
qu ( B1 ) = q u ( B2 )
where q u ( B1 ) , q u( B2 ) = ultimate bearing capacity of foundations with
widths B1 and B2 , respectively
However
B
S eu ( B1 ) ≈ S eu ( B2 ) 2
B1
(5.75)
where S eu ( B1 ) , S eu( B1 ) = elastic settlement of foundations with widths
B1 and B2 , respectively, at ultimate load
In a similar manner, for a given load intensity q (< qu ) on the foundation
B
S e ( B 2 ) ≈ S e ( B1 ) 2
B1
(5.76)
For foundations on sand
B
q u ( B2 ) ≈ q u ( B1 ) 2
B1
(5.77)
2
B 3 . 28 B 2 + 1
S eu ( B2 ) ≈ S eu ( B1 ) 2
B
B
+
3
.
28
1
1
1
2
where B is in meters
Also, for a given load intensity q (< qu ) on the foundation
© 1999 by CRC Press LLC
(5.78)
FIGURE 5.26 Nature of elastic settlement variation of two foundations in clay
FIGURE 5.27 Nature of elastic settlement variation of two foundations in sand
© 1999 by CRC Press LLC
2
B 3 .28 B2 + 1
S e ( B2 ) ≈ S e ( B1 ) 2
B1 3 .28 B1 + 1
(5.79)
where B is in meters
The above relationships for load intensity q and elastic settlement Se show
that, for a given allowable load intensity in any soil, the level of settlement increases with the increase in the foundation width. For smaller foundations
(that is, smaller B), the ultimate bearing capacity may be the controlling factor.
On the other hand, design of foundations having a larger B may be controlled
by settlement criteria.
CONSOLIDATION SETTLEMENT
5.12 GENERAL PRINCIPLES OF
CONSOLIDATION SETTLEMENT
As explained in Section 5.1, consolidation settlement is a time-dependent
process which occurs due to the expulsion of excess pore water pressure in
saturated clayey soils below the ground water table and is created by the increase in stress created by the foundation load. For a normally consolidated
clay, the nature of the variation of void ratio e with vertical effective stress "!
is shown in Fig. 5.28a. A similar plot for an overconsolidated clay is also
shown in Fig. 5.28b. In this figure the preconsolidation pressure is "!c . The
slope of the e vs. log "! plot for the normally consolidated portion of the soil
is referred to as compression index Cc , or
Cc =
e1 − e2
(for σ′1 ≥ σ′c )
σ ′2
log
σ′1
(5.80)
Similarly, the slope of the e vs. log "! plot for the overconsolidated portion of
the clay is called the swell index Cs , or
Cs =
e3 − e 4
σ′
log 4
σ ′3
(for σ′4 ≤ σ′c )
(5.81)
For normally consolidated clays, Terzaghi and Peck [19] gave a correlation for
the compression index as
Cc = 0.009(LL $ 10)
© 1999 by CRC Press LLC
(5.82a)
FIGURE 5.28 Nature of variation of void ratio with effective
stress: (a) Normally consolidated clay; (b) Overconsolidated clay
where LL = liquid limit
The preceding relation has a reliability in the range of ±30% and should not be
used for clays with sensitivity ratios greater than 4.
© 1999 by CRC Press LLC
Terzaghi and Peck [19] also gave a similar correlation for remolded clays
Cc = 0.007(LL $ 10)
(5.82b)
Several other correlations for the compression index with the basic index
properties of soils have been made, and some of these are given below [20]
Cc = 0.01wN (for Chicago clays)
(5.83)
Cc = 0.0046(LL $ 9) (for Brazilian clays)
(5.84)
Cc = 1.21 + 1.055(eo $ 1.87) (for Motley clays, São Paulo city)
(5.85)
Cc = 0.208eo + 0.0083 (for Chicago clays)
(5.86)
Cc = 0.0115 wN
(5.87)
where wN = natural moisture content, in percent
eo = in situ void ratio
The swell index, Cs , for a given soil is about 1/4 to 1/5 Cc .
5.13 RELATIONSHIPS FOR PRIMARY CONSOLIDATION
SETTLEMENT CALCULATION
Figure 5.29 shows a clay layer of thickness Hc . Let the initial void ratio before
the construction of the foundation be eo , and let the average effective vertical
stress on the clay layer be "!o . The foundation located at a depth Df is subjected
to a net average pressure increase of q. This will result in an increase in the
vertical stress in the soil. If the vertical stress increase at any point below the
center line of the foundation is %", the average vertical stress increase %"av
in the clay layer can thus be given as
z = H2
∫
1
∆σ av =
( ∆σ)dz
H 2 − H1 z=H
(5.88)
1
The consolidation settlement Sc, due to this average stress increase can be
calculated as follows
Sc =
σ′ + ∆ σ av
CH
∆e
= c c log o
σ′o
1 + eo 1 + eo
(for normally consolidated clay, that is, "!o = "!c )
© 1999 by CRC Press LLC
(5.89)
FIGURE 5.29 Primary consolidation settlement calculation
Sc =
σ′ + ∆ σ av
CH
∆e
= s c log o
σ′o
1 + eo 1 + eo
(5.90)
(for overconsolidated clay and !!o + "!av " !!c )
Sc =
σ′ C H
σ′ + ∆σ av
CH
∆e
= s c log c + c c log o
σ′c
1 + eo 1 + eo
σ′o 1 + eo
(5.91)
(for overconsolidated clay and !!o < !!c < !!o < "!av)
where "e = change of void ratio due to primary consolidation
Equations (5.89), (5.90), and (5.91) can be used in two ways to calculate
the primary consolidation settlement. They are as follows:
© 1999 by CRC Press LLC
FIGURE 5.30 Average stress increase, "!av
Method A
According to this method, !!o is the in situ average effective stress (that is, the
effective stress at the middle of the clay layer). The magnitude of "!av can be
calculated as (Fig. 5.29)
1
"!av = – ("!t + 4"!m + "!b )
6
(5.92)
where "!t , "!m , "!b = increase in stress, respectively, at the top, middle,
and bottom of the clay layer
The stress increase can be calculated by using the principles given previously
in this chapter.
The average stress increase, "!av , from z = 0 to z = H below the center of
a uniformly loaded flexible rectangular area (Fig. 5.30) was obtained using the
integration method by Griffiths [21], or
© 1999 by CRC Press LLC
FIGURE 5.31 Variation of Iav with a/h and b/h
"!av = qIav
(5.93)
a b
where I av = f ,
h h
(5.94)
a, b = half-length and half-width of the foundation
The variation of Iav is given in Fig. 5.31 as a function of a/h and b/h. It is
important to realize that Iav calculated by using this figure is for the case of
average stress increase from z = 0 to z = h (Fig. 5.30). For calculating the
average stress increase in a clay layer as shown in Fig. 5.32
I av ( h1 / h2 ) =
h2 I av ( h2 ) − h1 I av ( h1 )
© 1999 by CRC Press LLC
h2 − h1
FIGURE 5.32 Average stress increase in a clay layer
a b
where I av ( h2 ) = f ,
h2 h2
a b
I av ( h1 ) = f ,
h1 h 1
h 2 − h1 = H c
So
h 2 I av ( h 2 ) − h 1 I av ( h1 )
∆ σ av = q
Hc
(5.95)
Method B
In this method, a given clay layer can be divided into several thin layers
having thicknesses of Hc(1) , Hc(2) , . . . , Hc(n) (Fig. 5.33). The in situ effective
stresses at the middle of each layer are !!o(1) , !!o(2) , . . ., !!o(n) . The average
stress increase for each layer can be approximated to be equal to the vertical
© 1999 by CRC Press LLC
FIGURE 5.33
Consolidation settlement calculation using Method B
© 1999 by CRC Press LLC
stress increase at the middle of each soil layer [that is, "!av(1) # "!1 , "!av(2)
# "!2 , . . . , "!av(n) # "!n]. Hence, the consolidation settlement of the entire
layer can be calculated as
∆ei
Hi
i =1 1 + eo ( i )
i =n
Sc = ∑
(5.96)
EXAMPLE 5.5
Refer to Fig. 5.34. Using Method A, determine the primary consolidation
settlement of a foundation measuring 1.5 m × 3 m (B × L) in plan.
Solution From Eq. (5.89)
Sc =
σ′ + ∆ σ av
Cc H c
log o
σ′
1 + eo
o
Given Cc = 0.27, Hc = 3 m, eo = 0.92
3
!!o = (1 + 1.5)(16.5) + (1.5)(17.8 $ 9.81) + – (18.2 $ 9.81)
2
2
= 65.82 kN / m
a=
L 3
. m
= = 15
2 2
b=
B 15
.
=
= 0.75 m
2
2
FIGURE 5.34
© 1999 by CRC Press LLC
h1 = 1.5 + 1.5 = 3 m
h2 = 1.5 + 1.5 + 3 = 6 m
a 15
b 0.75
.
=
= 0.5;
=
= 0.25
h1
h1
3
3
From Fig. 5.31, I av( h1 ) = 0.54
Similarly,
a 15
b
.
0.75
=
= 0.25;
=
= 0.125
h2
h2
6
6
From Fig. 5.31, I av( h2 ) = 0.34
From Eq. (5.95)
h2 I av ( h2 ) − h1 I av ( h1 )
∆σ av = q
Hc
(6)( 0.34) − (3)( 0.54 )
2
= 170
= 23.8 kN/m
3
( 0.27)(3) 65.82 + 23.8
Sc =
log
= 0.057 m = 57 mm
65.82
1 + 0.9
!!
EXAMPLE 5.6
Solve Example 5.5 by Method B. (Note: Divide the clay layer into three
layers, each 1 m thick).
Solution The following tables can now be prepared.
Calculation of !!o :
Layer
No.
Layer
thickness,
Hi
(m)
1
1
1.0 + 1.5 + 1.5 + 0.5 = 4.5
(1 + 1.5)16.5 + (1.5)(17.8 $ 9.81)
+ (0.5)(18.2 $ 9.81) = 57.43
2
1
4.5 + 1 = 5.5
57.43 + (1)(18.2 $ 9.81) = 65.82
3
1
5.5 + 1 = 6.5
65.82 + (1)(18.2 $ 9.81) = 74.21
© 1999 by CRC Press LLC
Depth to the middle
of clay layer
(m)
!!o
(kN / m2)
Calculation of !"av :
Layer
No.
Layer
thickness
Hi
(m)
Depth to middle of
layer from bottom
of foundation, z
(m)
1
1
2
1
3
1
a
b
Δσ(av)
!"avc
L/Ba
z/B
3.5
2
2.33
0.16
27.2
4.5
2
3.0
0.095
16.15
5.5
2
3.67
0.07
11.9
q
B = 1.5 m; L = 3 m; b Table 5.3; c q = 170 kN / m2
Sc =
σ′o( i ) + ∆σ av ( i )
σ′o ( i )
∑ 1 + e log
Cc H i
o
57.43 + 27.2
65.82 + 16.15
+ log
log
65.82
57.43
(0.27)(1)
=
1 + 0.9
+ log 74.21 + 11.9
74.21
= (0.142)(0.168 + 0.096 + 0.065) = 0.047 m = 47 mm
5.14
!!
THREE-DIMENSIONAL EFFECT ON
PRIMARY CONSOLIDATION SETTLEMENT
The procedure described in the preceding section is for one-dimensional consolidation and will provide a good estimation for a field case where the width
of the foundation is large relative to the thickness of the compressible stratum,
Hc , and also when the compressible material lies between two stiffer soil
layers. This is because the magnitude of horizontal strains is relatively less in
the above cases.
In order to account for the three-dimensional effect, Skempton and
Bjerrum [22] proposed a correction to the one-dimensional consolidation settlement for normally consolidated clays. This can be explained by referring to
Fig. 5.35, which shows a circularly loaded area (diameter = B) on a layer of
normally consolidated clay of thickness Hc . Let the stress increases at a depth
z under the center line of the loaded area be !"1 (vertical) and !"3 (lateral).
The increase in pore water pressure due to the increase in stress, !u, can be
given as
!u = !"3 + A(!"1 ! !"3)
© 1999 by CRC Press LLC
(5.97)
FIGURE 5.35 Three-dimensional effect on primary consolidation
settlement (circular foundation of diameter B)
where A = pore water pressure parameter
The consolidation settlement dSc of an elemental soil layer of thickness dz is
∆e
dSc = m v ⋅ ∆u ⋅ dz =
( ∆u)( dz)
(1 + eo )∆σ 1
(5.98)
where mv = volume coefficient of compressibility
!e = change in void ratio
eo = initial void ratio
Hence,
Hc
∫
Sc = dSc =
∆e
∫ (1 + e )∆σ [∆σ + A(∆σ − ∆σ )]dz
3
1
3
1
0
o
∆σ 3
or
Hc
Sc =
∫ m ∆σ A + ∆σ (1 − A)dz
v
0
1
1
For conventional one-dimensional consolidation (Section 5.12)
© 1999 by CRC Press LLC
(5.99)
Hc
Hc
∆e
S c ( oed ) =
dz =
1
+ eo
0
Hc
∆e
∆ σ 1 dz =
(
1 + eo )
∆
σ
1
0
∫
∫
∫ m ∆ σ dz
v
1
(5.100)
0
From Eqs. (5.99) and (5.100)
Hc
Sc
=
S c ( oed )
µ c ( NC ) =
∆σ 3
∫ m ∆ σ A + ∆ σ (1 − A ) dz
v
1
1
0
Hc
∫ m ∆ σ dz
v
1
0
Hc
∫ ∆ σ dz
3
= A + (1 − A ) H0c
∫ ∆ σ dz
= A + (1 − A ) M 1
(5.101)
1
0
Hc
∫ ∆ σ dz
3
where M 1 = H0c
(5.102)
∫ ∆ σ dz
1
0
The variation of µ c(NC) with A and Hc /B is shown in Fig. 5.36.
In a similar manner, we can derive an expression for a uniformly loaded
strip foundation of width B supported by a normally consolidated clay layer
(Fig. 5.37). Let !"1 , !"2 , and !"3 be increase in stress at a depth z below the
center line of the foundation. For this condition it can be shown that
3
1 1
∆u = ∆σ 3 +
A − + ( ∆ σ 1 − ∆σ 3 )
3 2
2
(for ν = 0.5)
(5.103)
In a similar manner as in Eq. (5.99)
Hc
Sc =
∆σ 3
∫ m ∆ σ N + (1 − N ) ∆ σ ∆ z
v
1
where N =
(5.104)
1
0
3
1 1
A− +
2
3 2
Thus
© 1999 by CRC Press LLC
(5.105)
FIGURE 5.36 Variation of µc (NC) with A and Hc /B [Eq. (5.101)]
FIGURE 5.37 Three-dimensional effect on primary consolidation
settlement (continuous foundation of width B)
© 1999 by CRC Press LLC
Hc
µ s ( NC ) =
∆σ 3
∫ m ∆σ N + (1 − N ) ∆σ dz
v
Sc
Sc ( oed )
= 0
1
1
Hc
∫ m ∆σ dz
v
1
0
= N + (1 − N ) M 2
(5.106)
Hc
∫ ∆ σ dz
3
where M 2 = H0c
(5.107)
∫ ∆ σ dz
1
0
The plot of µs(NC) with A for varying values of Hc /B is shown in Fig. 5.38.
Leonards [23] considered the correction factor, µc(OC) , for three-dimensional consolidation effect in the field for a circular foundation located over
overconsolidated clays. Referring to Fig. 5.39
Sc = µc(OC) Sc(oed)
FIGURE 5.38 Variation of µs (NC) with A and Hc /B [Eq. (5.105)]
© 1999 by CRC Press LLC
(5.108)
FIGURE 5.39 Three-dimensional effect on primary consolidation
settlement of overconsolidated clay (circular foundation)
B
where µ c (OC ) = f OCR,
Hc
OCR =
σ′c
σ′o
(5.109)
(5.110)
""c = preconsolidation pressure
""o = present effective consolidation pressure
The interpolated values of µ c(OC) from the work of Leonard [24] are given in
Table 5.19.
EXAMPLE 5.7
Refer to Example 5.5. Assume that the pore water pressure parameter A for the
clay is 0.6. Considering the three-dimensional effect, estimate the consolidation settlement.
Solution
Note that Eq. (5.101) and Fig. 5.36 are valid for only an axisymmetrical case;
however, an approximate procedure can be adopted. Refer to Fig. 5.40. If we
assume that the load from the foundation spreads out along planes having
slopes of 2V:1H, then the dimension of the loaded area on the top of the clay
layer is
© 1999 by CRC Press LLC
TABLE 5.19 Variation of µc(OC) with OCR
and B/Hc
µ c(OC)
OCR
B/Hc = 4.0
B/Hc = 1.0
B/Hc = 0.2
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
1
0.986
0.972
0.964
0.950
0.943
0.929
0.914
0.900
0.886
0.871
0.864
0.857
0.850
0.843
0.843
1
0.957
0.914
0.871
0.829
0.800
0.757
0.729
0.700
0.671
0.643
0.629
0.614
0.607
0.600
0.600
1
0.929
0.842
0.771
0.707
0.643
0.586
0.529
0.493
0.457
0.429
0.414
0.400
0.386
0.371
0.357
FIGURE 5.40
1
B" = 1.5 + – (3) = 3 m
2
1
L" = 3 + – (3) = 4.5 m
2
The diameter of an equivalent circular area, Beq , can be given as
© 1999 by CRC Press LLC
π 2
B = B′L ′
4 eq
or
Beq =
π
π
B′L ′ = (3)(4.5) = 4.15 m
4
4
Hc
3
=
= 0.723
B
4.15
From Fig. 5.36, for A = 0.6 and Hc /B = 0.723, the magnitude of µ c(NC) !
0.76. So
Sc = Sc(oed) µ c(NC) = (57)(0.76) = 43.3 mm
!!
5.15 SECONDARY CONSOLIDATION SETTLEMENT
Secondary consolidation follows the primary consolidation process and takes
place under essentially constant effective stress as shown in Fig. 5.41. The
slope of the void ratio versus log-of-time plot is equal to C! , or
C α = secondary compressio n index =
∆e
t
log 2
t1
(5.111)
The secondary consolidation settlement, Ss , can be calculated as
Ss =
t
Cα H c
log 2
1 + ep
t1
(5.112)
ep = void ratio at the end of primary consolidation
t2 , t1 = time
The magnitude of the secondary compression index can vary widely, and some
general ranges are as follow:
where
Overconsolidated clays (OCR > 2 to 3) . . . . . . . . >0.001
Organic soils . . . . . . . . . . . . . . . . . . . . . . 0.025 or more
Normally consolidated clays . . . . . . . . . . . . 0.004–0.025
© 1999 by CRC Press LLC
FIGURE 5.41 Secondary consolidation settlement
In a majority of cases, secondary consolidation is small compared to primary
consolidation settlement. It can, however, be substantial for highly plastic
clays and organic soils.
DIFFERENTIAL SETTLEMENT
5.16 GENERAL CONCEPT OF
DIFFERENTIAL SETTLEMENT
In most instances the subsoil is not homogeneous and the load carried by
various shallow foundations of a given structure can vary widely. As a result,
it is reasonable to expect varying degrees of settlement in different parts of a
given building. The differential settlement of various parts of a building can
lead to damage of the superstructure. Hence, it is important to define certain
parameters to quantify differential settlement and develop limiting values for
these parameters for desired safe performance of structures. Burland and
Worth [24] summarized the important parameters relating to differential settlement. Figure 5.42 shows a structure in which various foundations at A, B, C,
D, and E have gone through some settlement. The settlement at A is AA", at B
© 1999 by CRC Press LLC
FIGURE 5.42 Definition of parameters for differential settlement
is BB", . . . . Based on this figure the definitions of the various parameters
follow:
ST = total settlement of a given point
"ST = difference between total settlement between any two parts
! = gradient between two successive points
∆S T (ij )
β = angular distortion =
lij
( Note: lij = distance between points i and j )
# = tilt
" = relative deflection (that is, movement from a straight line joining
two reference points
∆
= deflection ratio
L
Since the 1950's, attempts have been made by various researchers and
building codes to recommend allowable values for the above parameters. A
summary of some of these recommendations is given in the following section.
© 1999 by CRC Press LLC
5.17 LIMITING VALUE OF DIFFERENTIAL
SETTLEMENT PARAMETERS
In 1956, Skempton and MacDonald [25] proposed the following limiting
values for maximum settlement, maximum differential settlement, and maximum angular distortion to be used for building purposes
Maximum settlement, ST(max)
In sand . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 mm
In clay . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 mm
Maximum differential settlement, " ST(max)
Isolated foundations in sand . . . . . . . . . . . . . . . . 51 mm
Isolated foundations in clay . . . . . . . . . . . . . . . . . 76 mm
Raft in sand . . . . . . . . . . . . . . . . . . . . . . . . . . 51–76 mm
Raft in clay . . . . . . . . . . . . . . . . . . . . . . . . . . 76–127 mm
Maximum angular distortion, $max . . . . . . . . . . . . . . . . 1/300
Based on experience, Polschin and Tokar [26] provided the allowable deflection ratios for buildings as a function of L/H (L = length; H = height of
building), which are as follows:
"/L = 0.0003 for L/H # 2
"/L = 0.001 for L/H = 8
The 1955 Soviet Code of Practice gives the following allowable values.
Building type
Multistory buildings and
civil dwellings
One-story mills
"/L
L/H
#3
0.0003 (for sand)
0.0004 (for clay)
$5
0.0005 ( for sand)
0.0007 (for clay)
0.001 (for sand and clay
Bjerrum [27] recommended the following limiting angular distortion ($max)
for various structures
Category of potential damage
Safe limit for flexible brick wall (L/H > 4)
Danger of structural damage to most buildings
Cracking of panel and brick walls
Visible tilting of high rigid buildings
First cracking of panel walls
Safe limit for no cracking of building
Danger to frames with diagonals
© 1999 by CRC Press LLC
$max
1/150
1/150
1/150
1/250
1/300
1/500
1/600
Grant et al.[28] correlated ST(max) and $max for several buildings with the
following results.
Soil type
Clay
Clay
Sand
Sand
Foundation type
Isolated shallow foundation
Raft
Isolated shallow foundation
Raft
Correlation
ST(max) (mm) = 30,000 $max
ST(max) (mm) = 35,000 $max
ST(max) (mm) = 15,000 $max
ST(max) (mm) = 18,000 $max
Using the above correlations, if the maximum allowable value of $max is
known, the magnitude of the allowable ST(mzx) can be calculated.
The European Committee for Standardization more recently provided
values for limiting values for serviceability limit states [29] and the maximum
accepted foundation movements [30], and these are given in Table 5.20.
TABLE 5.20 Recommendation of European Committee for Standardization on
Differential Settlement Parameters
Item
Parameter
Magnitude
Comments
Limiting values for
serviceability [29]
ST
$
25 mm
50 mm
5 mm
10 mm
20 mm
1/500
Isolated shallow foundation
Raft foundation
Frames with rigid cladding
Frames with flexible cladding
Open frames
—
ST
"ST
$
50
20
!1/500
Isolated shallow foundation
Isolated shallow foundation
—
"ST
Maximum acceptable
foundation movement [30]
REFERENCES
1.
2.
3.
4.
5.
Boussinesq, J., Application des Potentials a L’Etude de L’Equilibre et due
Mouvement des Solides Elastiques, Gauthier-Villars, Paris, 1883.
Ahlvin, R. G., and Ulery, H. H., Tabulated values for determining the
complete pattern of stresses, strains, and deflections beneath a uniform load on
a homogeneous half space, High-way Res. Rec., Bulletin 342, 1, 1962.
Borowicka, H., Influence of rigidity of a circular foundation slab on the
distribution of pressures over the contact surface, In. Proc. I Int. Conf. Soil
Mech. Found. Eng., 2, 1936, 144.
Harr, M. E., Fundamentals of Theoretical Soil Mechanics, McGraw-Hill, New
York, 1966.
Egorov, K. E., Concerning the question of the deformation of bases of finite
thickness, Mekhanika Gruntov, Sb. Tr. 34, Gosstroiizdat, Moscow, 1958.
© 1999 by CRC Press LLC
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
Fox, E. N., The mean elastic settlement of a uniformly loaded area at a depth
below the ground surface, in Proc., II Int. Conf. Soil Mech. Found. Eng., 1,
1948, 129.
Trautmann, C. H., and Kulhawy, F. H., CUFAD—A Computer Program for
Compression and Uplift Foundation Analysis and Design, Report EL-4540CCM, Electric Power Research Institute, 16, Palo Alto, 1987.
Schmertmann, J. H., Static cone to compute settlement over sand, J. Soil
Mech. Found. Div., ASCE, 96(8), 1011, 1970.
Schmertmann, J. H., and Hartman, J. P., Improved strain influence factor
diagrams, J. Geotech. Eng. Div., ASCE, 104(8), 1131, 1978.
D’Appolonia, D. T., Poulos, H. G., and Ladd, C. C., Initial settlement of
structures on clay, J. Soil Mech. Found. Div., ASCE, 97(10), 1359, 1971.
Duncan, J. M., and Buchignani, A. L., An Engineering Manual for Settlement
Studies, Department of Civil Engineering, University of California, Berkeley,
1976.
Janbu, N., Bjerrum, L., and Kjaernsli, B., Veiledning ved losning av
Fundamenteringsoppgaver, Norwegian Geotechnical Institute Publication 16,
Oslo, 1956.
Christian, J. T., and Carrier III, W. D., Janbu, Bjerrum and Kjaernsli’s chart
reinterpreted, Canadian Geotech. J., 15(1), 124, 1978.
Meyerhof, G. G., Penetration tests and bearing capacity of cohesionless soils,
J. Soil Mech. Found.. Div., ASCE, 82(1), 1, 1956.
Meyerhof, G. G., Shallow foundations, J. Soil Mech. Found. Div., ASCE,
91(2), 21, 1965.
Schultze, E., and Sherif, G., Prediction of settlement from evaluation of
settlement observations for sand, In Proc., VIII Int. Conf. Soil Mech. Found.
Eng., Moscow, U.S.S.R., 1(3), 1973, 225.
Meyerhof, G. G., General report: state-of-the-art of penetration testing in
countries outside Europe, in Proc. Eur. Symp. Pen. Test., 1974.
Eggstad, A., Deformation measurements below a model footing on the surface
of dry sand, in Proc. Eur Conf. Soil Mech. Found. Eng., Weisbaden, W.
Germany, 1, 1963, 223.
Terzaghi, K., and Peck, R. B., Soil mechanics in engineering practice, 2nd Ed.,
Wiley, New York, 1967.
Azzouz, A. S., Krizek, R. T., and Corotis, R. B., Regression analysis of soil
compressibility, Soils and Foundations, 16(2), 19, 1976.
Griffiths, D. V., A chart for estimating the average vertical stress increase in
an elastic foundation below a uniformly loaded rectangular area, Canadian
Geotech. J., 21(4), 710, 1984.
Skempton, A. W., and Bjerrum, L., A contribution to settlement analysis of
foundations in clay, Geotechnique, 7, 168, 1957.
Leonards, G. A., Estimating Consolidation Settlement of Shallow Foundations
on Over-consolidated Clay, Transportation Research Board, Special Report
163, Washington, D.C., 13, 1976.
Burland, J. B., and Worth, C. P., Allowable and differential settlement of
structures, including damage and soil-structure interaction, in Proc. Conf. On
Settlement of Structures, Cambridge University, U.K., 1970, 611.
© 1999 by CRC Press LLC
25.
26.
27.
28.
29.
30.
Skempton, A. W., and MacDonald, D. H., The allowable settlement of
buildings, in Proc. of Institute of Civil Engineers, 5, Part III, 1956, 727.
Polshin, D. E., and Tokar, R. A., Maximum allowable non-uniform settlement
of structures, in Proc., IV Int. Conf. Soil Mech. Found. Eng., London, 1, 1957,
402.
Bjerrum, L., Allowable settlement of structures, in Proc. European Conf. Soil
Mech. Found. Eng., Weisbaden Germany, 3, 1963, 135.
Grant, R., Christian, J. T., and Vanmarcke, E. H., Differential settlement of
buildings, J. Geotech. Eng. Div., ASCE, 100(9), 1974, 973.
European Committee for Standardization, Basis of design and actions on
structures, Eurocode 1, Brussels, Belgium, 1994.
European Committee for Standardization, Geotechnical design, general
rules—Part I, Eurocode 7, Brussels, Belgium, 1994.
© 1999 by CRC Press LLC
CHAPTER 6
DYNAMIC BEARING CAPACITY AND SETTLEMENT
6.1 INTRODUCTION
Depending on the type of superstructure and the type of loading, a shallow
foundation may be subjected to dynamic loading. The dynamic loading may be
of various types, such as (a) monotonic loading with varying velocities; (b)
earthquake loading; (c) cyclic loading; and (d) transient loading. The ultimate
bearing capacity and settlement of shallow foundations subjected to dynamic
loading is the subject of discussion of this chapter.
6.2 EFFECT OF LOAD VELOCITY ON
ULTIMATE BEARING CAPACITY
The static ultimate bearing capacity of shallow foundations was discussed in
Chapters 2, 3, and 4. Vesic, Banks, and Woodward [1] conducted laboratory
model tests to study the effect of the velocity of loading on the ultimate bearing
capacity. These tests were conducted on a rigid rough circular model foundation having a diameter of 101.6 mm. The model foundation was placed on the
surface of a dense sand layer. The velocity of loading to cause failure was
varied from 10-5 in. / sec to 10 in. / sec. The tests were conducted in dry and
submerged sand. From Eq. (2.83), for a surface foundation in sand subjected
to vertical loading
1
qu = – !BN! "!s
2
or
N γ λ γs =
qu
1
γB
2
(6.1)
where qu = ultimate bearing capacity
! = effective unit weight of sand
B = diameter of foundation
N! = bearing capacity factor
"!s = shape factor
The variation of N! "!swith the velocity of loading obtained in the study of
Vesic et al. [1] is shown in Fig. 6.1. It can be seen from this figure that, when
the loading velocity is between 10-3 in. / sec to 10-2 in. / sec, the ultimate bearing
© 1999 by CRC Press LLC
FIGURE 6.1
Variation of N! λ!s with loading velocity (after Vesic et al. [1])
© 1999 by CRC Press LLC
capacity reaches a minimum value. Vesic [2] suggested that the minimum
value of qu in granular soil can be obtained by using a soil friction angle of #dy
instead of # in the bearing capacity equation [Eq. (2.83)], which is conventionally obtained from laboratory tests, or
#dy = #!2" (6.2)
The above relationship is consistent with the findings of Whitman and Healy
[3]. The increase in the ultimate bearing capacity when the loading velocity is
very high is due to the fact that the soil particles in the failure zone do not
always follow the path of least resistance, resulting in high shear strength of
soil and thus ultimate bearing capacity.
Unlike in the case of sand, the undrained shear strength of saturated clay
increases with the increase in the strain rate of loading. An excellent example
can be obtained from the unconsolidated undrained triaxial tests conducted by
Carroll [4] on Buckshot clay. The tests were conducted with a chamber
confining pressure of 2000 lb / ft2 (#96 kN / m2), and the moisture contents
of the specimens were 33.5 ± 0.2%. A summary of the test results follows:
Strain rate
(% / sec)
Undrained cohesion, cu
(lb / ft2)
0.033
4.76
14.4
53.6
128
314 and 426
1660
1850
2170
2430
2550
2620
From the above data, it can be seen that cu (dynamic) /cu (static) may be about 1.5. For
a given foundation the strain rate, $!, can be approximated as (Fig. 6.2)
•
∈
=
1 ∆Se
∆ t 2 B
(6.2)
where t = time
Se = settlement
So, if the undrained cohesion cu (# = 0 condition) for a given soil at a given
strain rate is known, this value can be used in Eq. (2.83) to calculate the
ultimate bearing capacity.
© 1999 by CRC Press LLC
FIGURE 6.2
Strain rate definition under a foundation
6.3 ULTIMATE BEARING CAPACITY
UNDER EARTHQUAKE LOADING
Richards et al. [5] more recently proposed a bearing capacity theory for a
continuous foundation supported by a granular soil under earthquake loading.
This theory assumes a simplified failure surface in soil at ultimate load. Figure
6.3a shows this failure surface under static conditions based on Coulomb’s
active and passive pressure wedges. Note that, in Zone I, %A is the angle that
Coulomb’s active wedge makes with the horizontal at failure
[tanφ(tanφ + cotφ)(1 + tanδ cotφ]0.5 − tanφ
α A = φ + tan
1 + tanδ(tanφ + cotφ)
−1
(6.3)
Similarly, in Zone II, %P is the angle that Coulomb’s passive wedge makes with
the horizontal at failure, or
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FIGURE 6.3 Bearing capacity of a continuous foundation on sand--static condition
0.5
+ tanφ (6.4)
−1 [tanφ(tanφ + cotφ)(1 + tanδ cotφ]
tan
αP = −φ +
1 + tanδ(tanφ + cotφ)
where # = soil friction angle
& = wall friction angle (AB in Fig. 6.3a)
Considering a unit length of the foundation, Fig. 6.3b shows the equilibrium analysis of wedges I and II. In this figure the following notations are
used.
PA = Coulomb’s active pressure
PP = Coulomb’s passive pressure
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RA = resultant of shear and normal force along AC
RP = resultant of shear and normal force along CD
WI , WII = weight of wedges ABC and BCD, respectively
Now, if # $ 0, ! = 0, and q $ 0, then
qu = q%u
and
PA cos & = PP cos &
(6.5)
However
PA cos & =q%u KA H
where
(6.6)
H = !!
AB
KA = horizontal component of Coulomb’s active earth pressure
coefficient, or
KA =
cos2 φ
sin(φ + δ) sin φ
cos δ1 +
cos δ
2
(6.7)
Similarly
PP cos& = qKP H
(6.8)
where KP = horizontal component of Coulomb’s passive earth pressure coefficient, or
KP =
cos2 φ
sin(φ − δ) sin φ
cos δ1 −
cos δ
2
(6.9)
Combining Eqs. (6.5), (6.6), and (6.8)
q u′ = q
KP
= qN q
KA
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(6.10)
where Nq = bearing capacity factor
Again, if # $ 0, ! $ 0, and q = 0, then qu = q&u .
1
PA cos& = q&u HKA + – !H 2KA
2
(6.11)
Also
1
PP cos& = – !H 2KP
2
(6.12)
Equating the right-hand sides of Eqs. (6.11) and (6.12)
1
1
q&u HKA + – !H 2KA = – !H 2KP
2
2
1
1
qu′′ = γH 2 (K P − K A )
2
HK A
or
qu′′ =
1 KP
γH
− 1
2 KA
(6.13)
However,
H = Btan%A
(6.14)
Combining Eqs. (6.13) and (6.14)
q u′′ =
K
1
1
γ B tan α A P − 1 = γBN γ
2
KA
2
K
where N γ = bearing capacity factor = tan α A P − 1
KA
(6.15)
(6.16)
If # $ 0, ! $ 0, and q$ 0, using the superposition we can write
1
(6.17)
qu = q%u + q&u = qNq + – !BN!
2
Richards et al. [5] suggested that, in calculating the bearing capacity factors
Nq and N! which are functions of # and &, we may assume & = #/2. With this
assumption, the variations of Nq and N! are given in Table 6.1.
© 1999 by CRC Press LLC
TABLE 6.1 Variation of Nq , N! , and Nc
(Assumption: & = #/2)
Soil friction
angle, #
(deg)
10203040
&
(deg)
Nq
N!
Nc
0
5
10
15
20
1
2.37
5.9
16.51
59.04
0
1.38
6.06
23.76
111.9
6
7.77
13.46
26.86
58.43
It can also be shown that for # = 0 condition if Coulomb’s wedge analysis
is performed, it will give a value of 6 for the bearing capacity factor, Nc . For
brevity, we can assume
Nc = (Nq ! 1)cot#
(2.67)
Using Eq. (2.67) and the Nq values given in Table 6.1, the Nc values can be
calculated, and these values are shown in Table 6.1. Figures 6.4, 6.5, and 6.6
show the variations of the bearing capacity factors with soil friction angle, #.
FIGURE 6.4 Variation of Nc with soil friction angle #
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FIGURE 6.5 Variation of Nq with soil friction angle #
Thus, the ultimate bearing capacity qu for a continuous foundation supported
by a c–# soil can be given as
1
(6.18)
qu = cNc + qNq + – !BN!
2
The ultimate bearing capacity of a continuous foundation under earthquake loading can be evaluated in a manner similar to that for the static condition shown above. Figure 6.7 shows the wedge analysis for this condition for
a foundation supported by a granular soil. In Fig. 6.7a note that %AE and %PE
are, respectively, the angles that the Coulomb’s failure wedges would make for
active and passive conditions, or
(1 + tan 2 a)[1 + tan(δ + θ) cot a] − tan a
α AE = a + tan −1
(6.19)
1
tan(
)(tan
cot
)
a
a
+
δ
+
θ
+
and
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FIGURE 6.6 Variation of N! with soil friction angle #
© 1999 by CRC Press LLC
FIGURE 6.7 Bearing capacity of a continuous foundation on sand--earthquake
loading
(1 + tan2 a)[1 + tan(δ − θ) cot a] + tan a
α PE = −a + tan−1
(6.20)
1 + tan(δ + θ)(tana + cot a)
where a = !! "
θ = tan −1
(6.21)
kh
1 − kv
kh = horizontal coefficient of acceleration
kv = vertical coefficient of acceleration
© 1999 by CRC Press LLC
(6.22)
Figure 6.7b shows the equilibrium analysis of wedges I and II as shown
in Fig. 6.7a. As in the static analysis [similar to Eq. (6.17)]
1
quE = qNqE + – #BN#E
2
(6.23)
quE = ultimate bearing capacity,
NqE , N#E = bearing capacity factors
Similar to Eqs. (6.10) and (6.16)
where
N qE =
K PE
K AE
(6.24)
K
N γ E = tan α AE PE − 1
K AE
(6.25)
where KAE , KPE = horizontal coefficients of active and passive earth pressure
(under earthquake conditions), respectively, or
K AE =
cos2 (φ − θ)
sin(φ + δ) sin(φ − θ)
cos θ cos(δ + θ)1 +
cos(δ + θ)
2
(6.26)
2
(6.27)
and
K PE =
cos2 (φ − θ)
sin(φ + δ) sin(φ − θ)
cos θ cos(δ + θ)1 −
cos(δ + θ)
Using $ = !/2 as before, the variations of KAE and KPE for various values
of " can be calculated. They can then be used to calculate the bearing capacity
factors NqE and N#E . Again, for a continuous foundation supported by a c–!
soil
1
quE = cNcE + qNqE + – #BN#E
2
(6.28)
where NcE = bearing capacity factor
The magnitude of NcE can be approximated as
NcE " (NqE !1)cot!
© 1999 by CRC Press LLC
(6.29)
FIGURE 6.8 Variation of N#e /N# with tan" and ! (after Richards et al. [5])
Figures 6.8, 6.9, and 6.10 show the variations of N#E /N#, NqE /Nq, and NcE /Nc.
These plots in combination with those given in Figs. 6.4, 6.5, and 6.6 can be
used to estimate the ultimate bearing capacity of a continuous foundation quE.
EXAMPLE 6.1
Consider a shallow continuous foundation. Given: B = 1.5 m, Df = 1 m, # =
17 kN / m3, ! = 25#, c = 30 kN / m2, kh = 0.25; kv = 0. Estimate the ultimate
bearing capacity quE .
Solution From Eq. (6.28)
1
quE = cNcE + qNqE + – #BN#E
2
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FIGURE 6.9 Variation of Nqe /Nq with tan" and ! (after Richards et al. [5])
For ! = 25#, from Figs. 6.4, 6.5, and 6.6, Nc " 20, Nq " 10, N# " 14. From
Figs. 6.8, 6.9, and 6.10, for tan" = kh /(1 ! kv) = 0.25/(1 ! 0) = 0.25,
N cE
= 0.44;
Nc
N qE
Nq
N γE
Nγ
N cE = (0.44)(20) = 8.8
= 0.38;
N qE = (0.38)(10) = 38
.
= 013
. ;
N γE = (013
. )(14) = 182
.
So
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FIGURE 6.10 Variation of Nce /Nc with tan" and ! (after Richards et al. [5])
1
quE = (30)(8.8) + (1 × 17)(3.8) + – (17)(1.5)(1.82) = 351.8 kN / m2
2
!!
6.4 SETTLEMENT OF FOUNDATION ON GRANULAR
SOIL DUE TO EARTHQUAKE LOADING
Bearing capacity settlement of a foundation (supported by a granular soil)
during an earthquake takes place only when the critical acceleration ratio
kh /(1!kv) reaches a certain critical value. Thus, if kv " 0, then
kh
kh
*
1 − k ≈ 1 − 0 ≈ k h
v cr
cr
© 1999 by CRC Press LLC
(6.30)
FIGURE 6.11
Critical acceleration kh*for incipient foundation settlement (after Richards et al. [5])
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FIGURE 6.11
(Continued)
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FIGURE 6.12 Variation of tan %AE with kh and ! (after Richards et al. [5])
The critical value k*h is a function of the factor of safety (FS) taken over the
ultimate static bearing capacity, embedment ratio (Df /B), and the soil friction
angle (!). Richards et al. [5] developed this relationship, and it is shown in a
graphical form in Fig. 6.11. According to Richards et al. [5], the settlement
of a foundation during an earthquake can be given as
V 2 k h*
S e = 0.174
Ag A
−4
tan α AE
(6.31)
where Se = settlement
V = peak velocity of the design earthquake
A = peak acceleration coefficient of the design earthquake
The variations of tan%AE with kh and ! are given in Fig. 6.12.
EXAMPLE 6.2
Consider a shallow foundation on a granular soil with B = 1.5 m, Df = 1 m, #
= 16.5 kN / m3, and ! = 35#. If the allowable bearing capacity is 304 kN / m2,
A = 0.32, and V = 0.35 m / s, determine the settlement the foundation may
undergo.
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Solution From Eq. (6.17)
1
qu = qNq + – #Bn#
2
From Figs. 6.5 and 6.6, for ! = 35#, Nq " 30; N# " 42. So
1
qu = (1 × 16.5)(30) + – (16.5)(1.5)(42) " 1015 kN / m2
2
Given qall = 340 kN / m2, so
FS =
qu
1015
=
= 2.98
q all
340
From Fig. 6.11, for FS = 2.98 and Df /B = 1/1.5 = 0.67, the magnitude of k*h is
about 0.28. From Eq. (6.31)
V 2 k h*
S e = 0.174
Ag A
−4
tan α AE
From Fig. 6.12 for ! = 35# and k*h = 0.28, tan%AE " 0.95. So
S e = ( 0.174)
( 0.35 m / s) 2
0.28
2
0
( 0.32)( 9.81 m / s ) .32
−4
( 0.95) = 0.011 m = 11 mm
!!
6.5 FOUNDATION SETTLEMENT DUE TO CYCLIC
LOADING—GRANULAR SOIL
Raymond and Komos [6] reported laboratory model test results on surface (Df
= 0) continuous foundations supported by granular soil and subjected to a lowfrequency (1 cps) cyclic loading of the type shown in Fig. 6.13. In this figure,
&d is the amplitude of the intensity of the cyclic load. The laboratory tests were
conducted for foundation widths (B) of 75 mm and 228 mm. The unit weight
of sand was 16.97 kN / m3. Since the settlement of the foundation Se after the
first cycle of load application was primarily due to the placement of the foundation rather than the foundation behavior, it was taken to be zero (that is, Se =
0 after the first cycle load application). Figures 6.14 and 6.15 show the
© 1999 by CRC Press LLC
FIGURE 6.13 Cyclic load on a foundation
FIGURE 6.14 Variation of Se (after first load cycle) with &d /qu and N —
B = 75 mm (after Raymond and Komos [6])
variation of Se (after the first cycle) with the number of load cycles, N, and
&d/qu (qu = ultimate static bearing capacity). Note that: (a) for a given number
of load cycles, the settlement increased with the increase in &d /qu and, (b) for
a given &d /qu , Se increased with N. These load-settlement curves can be approximated by the relation (for N = 2 to 105)
© 1999 by CRC Press LLC
FIGURE 6.15 Variation of Se (after first load cycle) with &d /qu and N —
B = 228 mm (after Raymond and Komos [6])
Se =
a
1
−b
log N
(6.32)
σ
where a = − 0.15125 + 0.0000693 B 1.18 d + 6.09
qu
(6.33)
σ
b = 0.153579 + 0.0000363 B 0.821 d − 23.1
qu
(6.34)
In Eqs. (6.33) and (6.34), B is in mm and &d /qu is in percent.
Figures 6.16 and 6.17 show the contours of the variation of Se with &d and
N for B = 75 mm and 228 mm. Studies of this type are useful in designing
railroad ties.
Settlement of Machine Foundations
Machine foundations subjected to sinusoidal vertical vibration (Fig. 6.18) may
undergo permanent settlement (Se). In Fig. 6.18, the weight of the machine and
the foundation is W, and the diameter of the foundation is B. The impressed
cyclic force Q is given by the relationship
© 1999 by CRC Press LLC
FIGURE 6.16
Contours of variation of Se with σd and N --B = 75 mm (after Raymond and Komos [6])
FIGURE 6.17
© 1999 by CRC Press LLC
Contours of variation of Se with σd and N --B = 228 mm (after Raymond and Komos [6])
FIGURE 6.18 Sinusoidal vertical vibration of machine foundations
FIGURE 6.19 Settlement, Se , with time due to cyclic load application
Q = Qo sin!t
(6.35)
where Qo = amplitude of the force
! = angular velocity
t = time
Many investigators believe that the peak acceleration is the primary controlling parameter for the settlement. Depending on the degree of compaction
of the granular soil, the solid particles come to an equilibrium condition for a
given peak acceleration resulting in a settlement Se (max) as shown in Fig. 6.19.
This threshold acceleration must be acceded before additional settlement can
take place.
Brumund and Leonards [7] evaluated the settlement of circular foundations subjected to vertical sinusoidal loading by laboratory model tests. For this
study, the model foundation had a diameter of 4 in., and 20-30 Ottawa sand
© 1999 by CRC Press LLC
FIGURE 6.20 Variation of Se (max) with peak
acceleration and weight of foundation
(after Brumund and Leonards [7])
compacted at a relative density of 70% was used. Based on their study, it
appears that energy per cycle of vibration can be used to determine Se (max) .
Figure 6.20 shows the variation of Se (max) versus peak acceleration for
weights of foundation, W = 48.8 lb, 73.4 lb, and 98 lb. The frequency of vibration was kept constant at 20 Hz for all tests. It is obvious that, for a given value
of W, the magnitude of Se increases linearly with the peak acceleration level.
The maximum energy transmitted to the foundation per cycle of vibration
can be theorized as follows. Figure 6.21 shows the schematic diagram of a
lumped-parameter one degree-of-freedom vibrating system for the machine
foundation. The soil supporting the foundation has been taken to be equivalent
to a spring and a dashpot. Let the spring constant be equal to k and the viscous
damping constant of the dashpot be c. The spring constant k and the viscous
damping constant c can be given by the following relationships (for further
details see any soil dynamics text, for example, Das [8]).
k=
2GB
1 − νs
© 1999 by CRC Press LLC
(6.36)
FIGURE 6.21 Lumped–parameter one degree –
of–freedom vibrating system
c=
0.85 2 Gγ
B
1 − νs
g
(6.37)
where G = shear modulus of the soil
"s = Poisson’s ratio of the soil
B = diameter of the foundation
# = unit weight of soil
g = acceleration due to gravity
The vertical motion of the foundation can be expressed as
z = Z cos(!t + $)
(6.38)
where
Z = amplitude of the steady-state vibration of the foundation
α = phase angle by which the motion lags the impressed force
The dynamic force transmitted by the foundation can be given as
Fdynamic = kz + c
dz
dt
Substituting Eq. (6.38) into Eq. (6.39) we obtain
Fdynamic = kZ cos(!t + $) ! c!Z sin(!t + $)
Let kZ = Acos% and c!Z = Asin%. So
Fdynamic = Acos% cos(!t + $) ! Asin% sin(!t + $)
© 1999 by CRC Press LLC
(6.39)
or
Fdynamic = Acos(!t + $ + %)
(6.40)
where A = magnitude of maximum dyanmic force = Fdynamic(max)
=
( A cos β) 2 + ( A sin β) 2 = Z k 2 + ( cω ) 2
(6.41)
The energy transmitted to the soil per cycle of vibration, Etr , is
Etr = "F dz = Fav Z
where
(6.42)
F = total contact force on soil
Fav = average contact force on the soil
However
1
Fav = – (Fmax + Fmin )
2
(6.43)
Fmax = W + Fdynamic(max)
(6.44)
Fmin = W ! Fdynamic(max)
(6.45)
Combining Eqs. (6.43), (6.44), and (6.45)
Fav = W
(6.46)
Hence, from Eqs. (6.42) and (6.46)
Etr = WZ
(6.47)
Figure 6.22 shows the experimental results of Brumund and Leonards [7],
which is a plot of Se (max) versus Etr . The data includes
a. a frequency range of 14–59.3 Hz,
b. a range of W varying from 0.27W to 0.55W, and
c. a maximum downward dynamic force of 0.3W to W.
The results show that Se (max) increases linearly with Etr . Figure 6.23 shows a
plot of the experimental results of Se (max) against peak acceleration for different
ranges of Etr . This clearly demonstrates that, if the value of the transmitted
energy is constant, the magnitude of Se (max) remains constant irrespective of the
level of peak acceleration.
© 1999 by CRC Press LLC
FIGURE 6.22 Plot of Se (max) versus Etr (after Brumund and Leonards [7])
FIGURE 6.23 Se (max) versus peak acceleration for three levels of
transmitted enery (after Brumund and Leonards [7])
6.6 FOUNDATION SETTLEMENT DUE TO
CYCLIC LOADING IN SATURATED CLAY
Das and Shin[9] provided small-scale model test results for the settlement of
a continuous surface foundation (Df = 0) supported by a saturated clay and
subjected to cyclic loading. For these tests, the width of the model foundation,
B, was 76.2 mm, and the average undrained shear strength of the clay was 11.9
kN / m2. The load to the foundation was applied in two stages (Fig. 6.24):
© 1999 by CRC Press LLC
FIGURE 6.24 Load application sequence to observe foundation settlement in
saturated clay due to cyclic loading based on the laboratory
model tests of Das and Shin [9]
Stage I—application of a static load per unit area of qs = qu /FS (where qu =
ultimate bearing capacity; FS = factor of safety) as shown in Fig. 6.24a, and
Stage II— application of a cyclic load, the intensity of which has an amplitude
of &d as shown in Fig. 6.24b. The frequency of the cyclic load was 1 Hz. Figure
6.24c shows the variation of the total load intensity on the foundation. Typical
experimental plots obtained from these laboratory tests are shown by the
© 1999 by CRC Press LLC
FIGURE 6.25
Typical plot of Se/B versus N for FS 3.33 and &d /qu 4.38%, 9.38%, and 18.75%
© 1999 by CRC Press LLC
FIGURE 6.26 General nature of plot of Se versus N for given values
of FS and &d /qu
broken lines in Fig. 6.25 (FS = 3.33; &d /qu = 4.38%, 9.38%, and 18.75%). It
is important to note that Se in this figure refers to the settlement obtained due
to cyclic load only (that is, after application of Stage II load; Fig 6.24b). The
general nature of these plots is shown in Fig. 6.26. They consist of
approximately three linear segments, and they are
1. An initial rapid settlement Se(r) (branch Oa).
2. A secondary settlement at a slower rate Se(s) (branch ab). The settlement practically ceases after application of N = Ncr cycles of load.
3. For N > Ncr cycles of loading, the settlement of the foundation due to
cyclic load practically ceases (branch bc).
The linear approximations of Se with number of load cycles N are shown
in Fig. 6.25 (solid lines). Hence, the total settlement of the foundation is
Se(max) = Se(r) + Se(s)
(6.48)
The tests of Das and Shin [9] had a range of FS =3.33 to 6.67 and &d /qu
= 4.38% to 18.75%. Based on these test results, the following general conclusions were drawn:
1. The initial rapid settlement is completed within the first 10 cycles of
loading.
2. The magnitude of Ncr varied between 15,000 to 20,000 cycles. This is
independent of FS and &d /qu .
3. For a given FS, the magnitude of Se increased with an increase of
&d /qu.
4. For a given &d /qu , the magnitude of Se increased with a decrease in
FS.
Figure 6.27 shows a plot of Se(max) /Se(u) versus &d /qu for various values of
FS. Note that Se(u) is the settlement of the foundation corresponding to static
© 1999 by CRC Press LLC
FIGURE 6.27 Results of laboratory model tests of Das and
Shin [9] — Plot of Se (max) /Se (u) versus &d /qu
FIGURE 6.28 Results of laboratory model tests of Das and
Shin [9] — Plot of Se (r) /Se (max) versus &d /qu
© 1999 by CRC Press LLC
ultimate bearing capacity. Similarly, Fig. 6.28 is the plot of Se(r) /Se(max) versus
&d /qu for various values of FS. From, these plots it can be seen that
1.
S e (max)
Se(u )
n1
σ
= m 1 d , and
q u!
$!#
"
( Fig. 6.27)
2. for any FS and &d /qu (Fig.6.28), the limiting value of Se(r) may be
about 0.8Se(max) .
6.7 SETTLEMENT DUE TO TRANSIENT LOAD
ON FOUNDATION
A limited number of test results are available in the literature which relate to
the evaluation of settlement of shallow foundations (supported by sand and
clay) subjected to transient loading. The findings of these tests will be discussed in this section.
Cunny and Sloan [10] conducted several model tests on square surface
foundations (Df = 0) to observe the settlement when the foundations were
subjected to transient loading. The nature of variation of the transient load
with time used for this study in shown in Fig. 6.29. Tables 6.2 and 6.3 show
the results of these tests conducted in sand and clay, respectively. Other details
of the tests are as follows:
FIGURE 6.29 Nature of transient load in the laboratory tests of Cunny and Sloan [10]
© 1999 by CRC Press LLC
TABLE 6.2 Load-Settlement Relationship of Square Surface Model Foundation
on Sand Due to Transient Loading (compiled from Cunny and Sloan [10])
Parameter
Test 1
Test 2
Test 3
Test 4
6
8
8
9
Ultimate static load-carrying capacity, Qu
(lb)
770
1820
1820
2590
Qd (max) (lb)
800
3140
2275
3500
Q#d (lb)
800
2800
2175
3250
Qd (max) /Qu
Width of model foundation, B (in.)
1.04
1.73
1.25
1.35
tr (ms)
18
8
90
11
tdw (ms)
122
420
280
0
tde (ms)
110
255
290
350
Se (Pot. 1) (in.)
0.280
—
0.830
0.400
Se (Pot. 2) (in.)
0.050
—
0.930
0.420
Se (Pot. 3) (in.)
0.110
—
0.950
0.400
Average Se (in.)
0.147
—
0.903
0.407
TABLE 6.3 Load-Settlement Relationship of Square Surface Model Foundation
on Clay Due to Transient Loading (compiled from Cunny and Sloan [10])
Parameter
Test 1
Test 2
Test 3
Test 4
4.5
4.5
4.5
5.0
Ultimate static load-carrying capacity, Qu
(lb)
2460
2460
2460
3040
Qd (max) (lb)
2850
3100
3460
3580
Q#d (lb)
2275
2820
2970
2950
Qd (max) /Qu
Width of model foundation, B (in.)
1.16
1.26
1.41
1.18
tr (ms)
9
9
10
9
tdw (ms)
170
0
0
0
tde (ms)
350
380
365
360
Se (Pot. 1) (in.)
0.500
0.660
1.700
0.580
Se (Pot. 2) (in.)
0.500
0.720
1.680
0.550
Se (Pot. 3) (in.)
0.480
0.700
1.700
0.550
Average Se (in.)
0.493
0.693
1.693
0.560
© 1999 by CRC Press LLC
Tests in sand (Table 6.2)
Dry unit weight, # = 103.4 lb / ft3
Relative density of compaction = 96%
Triaxial angle of friction = 32$
Tests in clay (Table 6.3)
Compacted moist unit weight = 94.1 to 98.4 lb / ft3
Moisture content = 22.5 ± 1.7%
Angle of friction (undrained triaxial test) = 4$
Cohesion (undrained triaxial test) = 2400 lb / ft2
For all tests, the settlement of the model foundation was measured at three
corners by linear potentiometers. Based on the results of these tests, the following general conclusions can be drawn.
1. The settlement of the foundation under transient loading is generally
uniform.
2. Failure in soil below the foundation may be in punching mode.
3. Settlement under transient loading may be substantially less than that
observed under static loading. As an example, for Test No. 4 in Table
6.2, the settlement at ultimate load, Qu (static bearing capacity test)
was about 2.620 in. However, when subjected to a transient load with
Qd (max) = 1.35Qu , the observed settlement was about 0.410 in. Similarly for Test No. 2 in Table 6.3, the settlement at ultimate load was
about 2 in. Under transient load with Qd (max) = 1.26Qu , the observed
settlement was only about 0.7 in.
Jackson and Hadala [11] reported several laboratory model test results on
square surface foundation with width B varying from 4.5 in. to 8 in. which
were supported by saturated Buckshot clay. For these tests, the nature of the
transient load applied to the foundation is shown in Fig. 6.30. The rise time,
tr , varied from 2 to 16 ms and the decay time from 240 to 425 ms. Based on
these tests, it showed that there is a unique relationship between Qd(max)/(B2cu)
and Se /B. This relationship can be found in the following manner.
1. From the plate load test (square plate, B × B) in the field, determine
the relationship between load Q and Se /B.
2. Plot a graph of Q/B2cu versus Se /B as shown by the broken lines in
Fig. 6.31.
3. Since the strain-rate factor in clays is about 1.5 (Section 6.2), determine 1.5Q/B2cu and develop a plot of 1.5Q/B2cu versus Se /B as shown
by the solid lines in Fig. 6.31. This will be the relationship between
Qd(max)/(B2cu) versus Se /B.
© 1999 by CRC Press LLC
FIGURE 6.30 Nature of transient load in the laboratory tests of Johnson and
Hadala [11]
FIGURE 6.31 Relationship of Qd (max) /B 2cu versus Se /B from plate
load tests (plate size B × B)
© 1999 by CRC Press LLC
REFERENCES
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
Vesic, A. S., Banks, D. C., and Woodward, J. M., An experimental study of
dynamic bearing capacity of footings on sand, in Proceedings, VI Int. Conf.
Soil Mech. Found. Eng., Montreal, Canada, 2, 1965, 209.
Vesic, A. S., Analysis of ultimate loads of shallow foundations, J. Soil Mech.
Found. Engg. Div., ASCE, 99(1), 45, 1973.
Whitman, R. V., and Healy, K. A., Shear strength of sands during rapid
loading, J. Soil Mech. Found. Engg. Div., ASCE, 88(2), 99, 1962.
Carroll, W. F., Dynamic Bearing Capacity of Soils: Vertical Displacement of
Spread Footing on Clay: Static and Impulsive Loadings, Technical Report 3599, Report 5, U.S. Army Corps of Engineers, Waterways Experiment Station,
Mississippi, 1963.
Richards, R., Jr., Elms, D. G., and Budhu, M., Seismic bearing capacity and
settlement of foundations, J. Geotech. Eng., ASCE, 119(4), 622, 1993.
Raymond, G. P., and Komos, F. E., Repeated load testing of a model plane
strain footing, Canadian Geotech. J., 15(2), 190, 1978.
Brumund, W. F., and Leonards, G. A., Subsidence of sand due to surface
vibration, J. Soil Mech. Found. Eng. Div., ASCE, 98(1), 27, 1972.
Das, B. M., Principles of Soil Dynamics, PWS Publishers, Boston,
Massachusetts, 1993.
Das, B. M., and Shin, E. C., Laboratory model tests for cyclic load-induced
settlement of a strip foundation on a clayey soil, Geotech. Geol. Eng., London,
14, 213, 1996.
Cunny, R. W., and Sloan, R. C., Dynamic Loading Machine and Results of
Preliminary Small-Scale Footing Tests, Spec. Tech. Pub. 305, ASTM, 65,
1961.
Jackson, J. G., Jr., and Hadala, P. F., Dynamic Bearing Capacity of Soils.
Report 3: The Application Similitude to Small-Scale Footing Tests, U. S.
Army Corps of Engineers, Waterways Experiment Station, Mississippi, 1964.
© 1999 by CRC Press LLC
CHAPTER SEVEN
SHALLOW FOUNDATIONS ON REINFORCED SOIL
7.1 INTRODUCTION
Reinforced soil, or mechanically stabilized soil, is a construction technique that
consists of soil that has been strengthened by tensile elements such as metal
strips, geotextiles, or geogrids. In the 1960's the French Road Research
Laboratory conducted extensive research to evaluate the beneficial effects of
using reinforced soil as a construction technique. Results of the early work
were well documented by Vidal [1]. During the last thirty years many retaining
walls and embankments were constructed all over the world using reinforced
soil, and they have performed very well.
The metallic strips that are used for reinforced soil are usually galvanized
steel strips. However, the galvanized steel strips are subject to corrosion at the
rate of about 0.025 to 0.05 mm/year. Hence, depending on the projected service
life of a given structure, allowances must be made for the rate of corrosion.
Geotextiles and geogrids are non-biodegradable materials. They are made
from petroleum products such as polyester, polyethylene, and polypropylene.
Geotextiles perform four major functions: (a) allow drainage from the soil;
(b) keep the soil layers separated; (c) provide reinforcement to the soil; and
(d) allow free seepage from one layer of soil to the other; however, they protect
fine-grained soil from being washed into the coarse-grained soil.
Geogrids are made by tensile drawing of polymer materials such as polyethylene and polypropylene. They are relatively stiff material compared to geotextiles. They have large apertures which allow interlocking with surrounding
soil to perform the functions of reinforcement and/or segregation. Commercially available geogrids are generally of two types—uniaxial and biaxial.
Figures 7.1a and 7.1b show these two types of geogrids. Geogrids are manufactured so that the open areas of the grids are greater than 50% of the total
area. They develop reinforcing strength at low strain levels such as 2%.
FOUNDATIONS ON METALLIC STRIP-REINFORCED
GRANULAR SOIL
7.2 FAILURE MODE
Binquet and Lee [2,3] conducted several laboratory tests and proposed a theory
for designing a continuous foundation on sand reinforced with metallic strips.
Figure 7.2 defines the general parameters in this design procedure. In Fig. 7.2,
the width of the continuous foundation is B. The first layer of reinforcement
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FIGURE 7.1
Geogrids: (a) Uniaxial; (b) Biaxial
FIGURE 7.2 Foundation on metallic strip reinforced granular soil
© 1999 by CRC Press LLC
© 1999 by CRC Press LLC
FIGURE 7.3 Failure in reinforced earth by tie break (u/B < 2/3 and N ! 4)
FIGURE 7.4 Failure surface in reinforced earth at ultimate load
is placed at a distance u measured from the bottom of the foundation. The
distance between each layer of reinforcement is h. It was experimentally
shown [2, 3] that the most beneficial effect of reinforced earth is obtained
when u/B is less than about bB and the number of layers of reinforcement (N)
is greater than 4 but no more than 6 to 7. If the length of the ties (that is,
reinforcement strips) is sufficiently long, failure occurs when the upper ties
break. This phenomenon is shown in Fig. 7.3.
Figure 7.4 shows an idealized condition for the development of a failure
© 1999 by CRC Press LLC
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FIGURE 7.5 Variation of x"/B with z/B
surface in reinforced earth which consists of two zones. Zone I is immediately
below the foundation which settles with the foundation during the application
of load. In zone II the soil is pushed outward and upward. Points A1 , A2 , A3 ,
. . . , and B1 , B2 , B3 , . . . , which define the limits of zones I and II, are points
at which maximum shear stress, !max , occurs in the xz plane. The distance x!x"
of the points measured from the center line of the foundation where maximum
shear stress occurs is a function of z/B. This is shown in a nondimensional
form in Fig. 7.5.
7.3 FORCE IN REINFORCEMENT TIES
In order to obtain the forces in the reinforcement ties, Binquet and Lee [3]
made the following assumptions.
1. Under the application of bearing pressure by the foundation, the reinforcing ties at points A1 , A2 , A3 , . . . , and B1 , B2 , B3 , . . . , (Fig. 7.4)
take the shape shown in Fig. 7.6a. That is, the tie takes two right
angle turns on each side of zone I around two frictionless rollers.
2. For N reinforcing layers, the ratio of the load per unit area on the
foundation supported by reinforced earth qR to the load per unit area
on the foundation supported by unreinforced earth qo is constant,
irrespective of the settlement level, Se (see Fig. 7.6b). Binquet and Lee
[2] proved this relation by laboratory experimental results.
© 1999 by CRC Press LLC
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FIGURE 7.6 Assumptions to calculate the force in reinforcement ties
With the above assumptions it can be seen that
T =
1 qR
− 1 (α B − β h )
q o
N q o
(7.1)
where T = tie force per unit length of the foundation at a depth z (lb / ft or
kN / m)
© 1999 by CRC Press LLC
© 1999 by CRC Press LLC
N = number of reinforcement layers
qo = load per unit area of the foundation on unreinforced soil for a foundation settlement level of Se = S"e
qR = load per unit area of the foundation on reinforced soil for a foundation settlement level of Se = S"e
", # = parameters which are functions of z/B
The variations of " and # with z/B are shown in Figs. 7.7 and 7.8 respectively.
FIGURE 7.7 Variation of " with z/B
FIGURE 7.8 Variation of # with z/B
© 1999 by CRC Press LLC
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7.4 FACTOR OF SAFETY AGAINST
TIE BREAKING AND TIE PULLOUT
In designing a foundation it is essential to determine if the reinforcement ties
will fail either by breaking or by pullout. Let the width of a single tie (at right
angles to the cross section shown in Fig. 7.2) be w and its thickness be t. If the
number of ties per unit length of the foundation placed at any depth z is equal
to n, then the factor of safety against the possibility of tie break (FSB) is
FS B =
where
wtnf y
T
=
tf y ( LDR )
T
(7.2)
fy = yield or breaking strength of tie material
LDR = linear density ratio = wn
(7.3)
Figure 7.9 shows a layer of reinforcement located at a depth z. The
frictional resistance against tie pullout at that depth can be calculated as
x=X
F P = 2 tan φ µ wn σ dx + wn γ ( X − x ′)( z + D f )
x = x′
∫
FIGURE 7.9 Frictional resistance against tie pullout
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(7.4)
where
%µ = soil-tie interface friction angle
& = effective normal stress at a depth z due to the uniform load per
unit area, qR , on the foundation
X # distance at which & = 0.1qR
Df = depth of the foundation
' = unit weight of soil
Note that the second term in the right-hand side of Eq. (7.4) is due to the fact
that frictional resistance is derived from the top and bottom of the ties. Thus,
from Eq. (7.4)
q
FP = 2 tan φ µ ( LDR ) δBq o R + γ ( X − x ′)( z + D f )
qo
(7.5)
The term $ is a function of z/B and is shown in Fig. 7.10. Figure 7.11 shows a
plot of X/B versus z/B. Hence, at any given depth z, the factor of safety against
tie pullout, FSP , can be given as
FS P =
FP
T
FIGURE 7.10 Variation of $ with z/B
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(7.6)
FIGURE 7.11 Variation of X/B with z/B
7.5 DESIGN PROCEDURE FOR A
CONTINUOUS FOUNDATION
Following is a step-by-step procedure for designing a continuous foundation
on a granular soil reinforced with metallic strips.
Step 1. Establish the following parameters
a. Foundation
$ Net load per unit length, Q
$ Depth, Df
$ Factor of safety, FS, against bearing capacity failure on unreinforced soil
$ Allowable settlement, Se
b. Soil
$ Unit weight, '
$ Friction angle, %
$ Modulus of elasticity, Es
$ Poisson’s ratio, µ s
c. Reinforcement ties
$ Width, w
© 1999 by CRC Press LLC
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$ Soil-tie friction angle, %µ
$ Factor of safety against tie pullout, FSP
$ Factor of safety against tie break, FSB
Step 2. Assume values of B, u, h, and number of reinforcement layers N.
Note that the depth of reinforcement, d, from the bottom of the
foundation
d = u + (N % 1)h & 2B
(7.7)
Step 3. Assume a value of LDR = wn
Step 4. Determine the allowable bearing capacity, q"all , on unreinforced sand,
or
1
qN q + γBN γ
qu
2
′ ≈
=
qall
FS
FS
(7.8)
where qu = ultimate bearing capacity on unreinforced soil
q = 'Df
Nq , N' = bearing capacity factors (Table 2.3)
Step 5. Determine the allowable bearing capacity, q'all , based on allowable
settlement. From Eq. (5.41),
qall
′′ =
E s Se
B (1 − µ 2s ) I 7
(7.9)
The magnitude of I7 for continuous foundations can be taken to be
approximately 2 for this calculation.
Step 6. The smaller of the two allowable bearing capacities (that is, q"all or q'all)
is equal to qo .
Step 7. Calculate qR (load per unit area of the foundation on reinforced soil)
as
qR =
Q
B
Step 8. Calculate T for all layers of reinforcement using Eq. (7.1).
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(7.10)
Step 9. Calculate the magnitude of FP /T for each layer to see if FP /T ! FSP .
If FP /T < FSP , the length of reinforcing strips may have to be increased by substituting X´ (>X) in Eq. (7.5) so that FP /T is equal to
FSP .
Step 10. Use Eq. (7.2) to obtain the thickness of the reinforcement strips.
Step 11. If the design is unsatisfactory, repeat Steps 2 through 10.
EXAMPLE 7.1
Design a continuous foundation with the following:
Foundation:
Net load to be carried, Q = 1.5 MN / m
Df = 1.2 m
Factor of safety against bearing capacity failure in
unreinforced soil, Fs = 3.5
Tolerable settlement, Se = 25 mm
Soil: Unit weight, ! = 16.5 kN / m3
Friction angle, " = 36"
Es = 3.4 × 104 kN / m2
µ s = 0.3
Reinforcement Ties: Width, w = 70 mm
"µ = 25"
FSB = 3
FSP = 2
fy = 2.5 × 105 kN / m2
Solution Let B = 1.2 m, u = 0.5 m, h = 0.5 m, N = 4, and LDR = 60%. With
LDR = 60%
Number of strips, n =
LDR
0.6
=
= 8.57 / m
w
0.007
From Eq. (7.8)
q all
′ =
1
γBN γ
2
FS
qN q +
From Table 2.3, for " = 36" the magnitudes of Nq and N! are 37.75 and 44.43
respectively. So
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q all
′ =
(1.2 × 16.5)( 37.75) + ( 0.5)(16.5)(1.2)( 44.43)
= 339.23 kN / m 2
3.5
From Eq. (7.9)
qall
′′ =
E s Se
B (1 − µ 2s ) I 7
=
( 3.4 × 10 4 )( 0.025)
2
(1.2)[1 − ( 0.3) ]( 2)
= 389.2 kN / m 2
Since q!all < qáll , qo = qáll = 339.23 kN / m2. Thus
qR =
Q 15
. × 10 3 kN
=
= 1250 kN / m 2
B
1.2
Now the tie forces can be calculated using Eq. (7.1)
qR
−1(αB − βh)
q
o
q
T = o
N
Layer
No.
1
2
3
4
qR
−1
q
o
qo
N
227.7
227.7
227.7
227.7
z
(m)
z/B
#B#$h
T
(kN/m)
0.5
1.0
1.5
2.0
0.47
0.83
1.25
1.67
0.285
0.300
0.325
0.330
64.89
68.31
74.00
75.14
Note: B = 1.2 m; # from Fig. 7.7; $ from Fig. 7.8; h = 0.5 m
The magnitudes of FP /T for each layer are calculated in the following table.
From Eqs. (7.5) and (7.6)
FP 2 tan φ µ ( LDR)
=
T
T
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q
δBqo R + γ ( X − x ′)(z + D f )
qo
Layer
Parameter
1
2
3
4
0.0086
0.0082
0.0076
0.0075
z/B
0.47
0.83
1.25
1.67
%
0.12
0.14
0.15
0.16
q
δ Bq o R (kN/m)
qo
180
210
225
240
X/B
1.4
2.3
3.2
3.6
X (m)
1.68
2.76
3.84
4.32
x´/B
0.7
0.8
1.0
1.3
x´ (m)
0.84
0.96
1.2
1.56
γ ( X − x ′)( z + D f ) (kN/m)
23.56
65.34
117.6
145.7
FP /T
1.75
2.26
2.6
2.89
2 tan φ µ ( LDR )
T
(m / kN)
The minimum factor of safety FSP required is 2. In all layers except Layer 1,
FP /T is greater than 2. So we need to find a new value of x = X´ so that FP /T
is equal to 2. So for Layer 1
2 tan φ µ ( LDR )
q
FP
=
δ Bq o R + γ ( X ′ − x ′)( z + D f )
T
T
qo
or
2 = 0.0086[180 + 16.5(X´ # 0.84)(0.5 + 1.2)]; X´ = 2.71 m
Toe thickness, t: From Eq. (7.2)
FS B =
t=
tf y ( LDR )
T
( FS B )( T )
( 3)( T )
=
= 2 × 10 −5 T
( f y )( LDR ) ( 2.5 × 105 )( 0.6)
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The following table can now be prepared.
Layer
No.
T
(kN/m)
t
(mm)
1
2
3
4
64.89
68.31
74.00
75.14
$1.3
$1.4
$1.5
$1.503
.
A tie thickness of 1.6 mm will be sufficient for all layers. Figure 7.12 shows
a diagram of the foundation with the ties.
FIGURE 7.12
!!
FOUNDATIONS ON
GEOTEXTILE-REINFORCED SOIL
7.6 LABORATORY MODEL TEST RESULTS
Results of a limited number of model tests conducted in the early to mid-1980's
to determine the bearing capacity of surface foundations (that is, Df =0) resting
on geotextile-reinforced soils can be found in the literature. Guido et al. [4] reported results of several laboratory model tests for a square surface foundation
measuring 0.31 m × 0.31 m (B × B) and supported by a loose to medium sand
reinforced with multiple layers of nonwoven melt-bonded geotextile (size b ×
b). Figure 7.13 shows the geometric parameters of the problem under consideration, and Fig. 7.14 shows some of the results of these tests. For the tests
reported in Fig. 7.14, the following parameters apply: relative density of sand,
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FIGURE 7.13 Foundation on geotextile-reinforced soil
FIGURE 7.14 Model test results of Guido et al. [4] on geotextilereinforced sand for a square surface foundation
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Dr = 50%; width of geogrid layers, b = 0.62 m; b/B = 0.2; u/B = 0.5; and h/B
= 0.25.
Similar model test results on a continuous surface foundation supported by
a saturated clay (! = 0 condition) reinforced by heat-bonded nonwoven geotextile were reported by Sakti and Das [5]. The load-settlement curves for some
of these tests are given in Fig. 7.15. For these tests the following parameters
apply: width, B = 76.2 mm; undrained cohesion of clay, cu = 22.5 kN/m2, and
u/B = h/B = 0.33. The tests clearly show that the ultimate bearing capacity of
foundations increases when geotextile reinforcement is used.
7.7 COMMENTS ON GEOTEXTILE REINFORCEMENT
Figures 7.14 and 7.15 show that geotextile reinforcement contributes to the
increase in ultimate bearing capacity of foundations on sand and saturated clay.
However, at low settlement level of the foundation, geotextile reinforcement
hardly contributes to the load bearing capacity. This is primarily because geotextiles are made of flexible material. Sufficient settlement of the foundation
will be necessary to give the layers of geotextile a catenary shape and to develop
tension to resist the stress transmitted from the foundation. The design of most
foundations with a width B greater than about 1 m will be controlled by the
allowable level of settlement rather than the ultimate bearing capacity. For that
reason, geotextiles may not be a suitable material for soil reinforcement in the
improvement of bearing capacity.
FIGURE 7.15 Model test results of Sakti and Das [5] on geotextile-reinforced
saturated clay for a continuous surface foundation
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FOUNDATIONS ON
GEOGRID-REINFORCED SOIL
7.8 GENERAL PARAMETERS
Geogrids are stiffer material than geotextiles. Since the mid 1980's, a number
of laboratory model studies have been reported relating to the evaluation of the
ultimate and allowable bearing capacities of shallow foundations supported by
soil reinforced with multiple layers of geogrids. The results obtained so far seem
promising. In this section the general parameters of the problem are defined.
FIGURE 7.16 Geometric parameters of a rectangular foundation
supported by geogrid-reinforced soil
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Figure 7.16 shows the general parameters of a rectangular surface
foundation on a soil layer reinforced with several layers of geogrid. The size
of the foundation is B × L (width × length) and the size of the geogrid layers
is b × l (width × length). The first layer of geogrid is located at a depth u below
the foundation, and the vertical distance between consecutive layers of geogrid
is h. The total depth of reinforcement d can be given as
d = u + (N ! 1)h
(7.11)
where N = number of reinforcement layers
The beneficial effects of reinforcement to increase the bearing capacity can
be expressed in terms of a nondimensional parameter called the bearing
capacity ratio (BCR). The bearing capacity ratio can be expressed with respect
to the ultimate bearing capacity or the allowable bearing capacity (at a given
settlement level of the foundation). Figure 7.17 shows the general nature of the
load-settlement curve of a foundation both with and without geogrid reinforcement. Based on this concept the bearing capacity ratio can be defined as
BCRu =
qu( R)
qu
FIGURE 7.17 General nature of the load-settlement curves
for unreinforced and geogrid-reinforced soil
supporting a foundation
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(7.12)
and
BCRs =
qR
q
(7.13)
where
BCRu = bearing capacity ratio with respect to the ultimate load
BCRs = bearing capacity ratio at a given settlement level Se for the
foundation
For a given foundation and given values of b/B, l/B, u/B, and h/B, the
magnitude of BCRu increases with d/B and reaches a maximum value at (d/B)cr
beyond which the bearing capacity remains practically constant. The term
(d/B)cr is the critical-reinforcement-depth ratio. For given values of l/B, u/B,
h/B, and d/B, BCRu attains a maximum value at (b/B)cr , which is called the
critical-width ratio. Similarly, a critical-length ratio (l/B)cr can be established
(for given values of b/B, u/B, h/B, and d/B) for a maximum value of BCRu . This
concept is schematically illustrated in Fig. 7.18. As an example, Fig. 7.19
shows the variation of BCRu with d/B for four model foundations (B/L = 0, 1/3,
FIGURE 7.18 Definition of critical nondimensional parameters —
(d/B)cr , (b/B)cr , and (l /B)cr
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FIGURE 7.19 Variation of BCRu with d/B (after Omar et al. [6])
1/2, and 1) as reported by Omar et al. [6]. It was also shown from laboratory
model tests [6,7] that for a given foundation, if b/B, l/B, d/B, and h/B are kept
constant, the nature of variation of BCRu with u/B will be as shown in Fig. 7.20.
Initially (Zone 1) BCRu increases with u/B to a maximum value at (u/B)cr . For
u/B > (u/B)cr the magnitude of BCRu decreases (Zone 2). For u/B > (u/B)max the
plot of BCRu versus u/B generally flattens out (Zone 3). The present understanding (in general) among investigators is that, in Zones 1 and 2, the nature
of the failure surface in soil will be as shown in Fig. 7.21a. In Zone 1 the initial
increase in BCRu with u/B is due to the increase in confining pressure on the
geogrid layers. In Zone 3 [that is, u/B " (u/B)max ] the failure surface in soil
below the foundation is located fully above the first layer of geogrid, which
acts as a semi-rigid rough base (Fig. 7.21b).
7.9 RELATIONSHIPS FOR CRITICAL NONDIMENSIONAL
PARAMETERS FOR FOUNDATIONS ON GEOGRIDREINFORCED SAND
Based on the results of their model tests and other existing results, Omar et al.
[6] developed the following empirical relationships for the nondimensional
parameters (d/B)cr , (b/B)cr , and (l/B)cr described in the preceding section.
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FIGURE 7.20 Nature of variation of BCRu with u/B
FIGURE 7.21 Failure surface in geogrid-reinforced soil under a
foundation (a) u/B < (u/B)max ; (b) u/B = (u/B)max
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Critical Reinforcement-Depth Ratio
B
B
for0 ≤ ≤ 0.5
L
L
d
= 2 − 1.4
b cr
B
B
for0.5≤ ≤ 1
L
L
d
= 1.43− 0.26
b cr
(7.14)
(7.15)
The preceding relationships suggest that the bearing capacity increase is
realized only when the reinforcement is located within a depth of 2B for a
continuous foundation and a depth of 1.2B for a square foundation.
Critical Reinforcement-Width Ratio
b
= 8 − 3.5
Bcr
0.51
B
L
(7.16)
According to Eq. (7.16), (b/B)cr is about 8 for a continuous foundation and
about 4.5 for a square foundation. It needs to be realized that generally, with
other parameters remaining constant, about 70% or more of BCRu is realized
with b/B # 2. The remaining 30% of BCRu is realized when b/B increased from
about 2 to (b/B)cr .
Critical Reinforcement-Length Ratio
l
= 3.5
Bcr
B L
+
L B
(7.17)
The magnitude of (u/B)max was recommended by Binquet and Lee [2,3] to be
about 0.67. However, model test results of Guido et al. [8] on a square foundation and Omar et al. [9] on a continuous foundation show that (u/B)max is about
0.9 to 1. Based on the existing studies it appears that (u/B)cr is approximately
equal to 0.25 to 0.4.
7.10 RELATIONSHIP BETWEEN
BCRu AND BCRs IN SAND
It was pointed out in Section 7.7 that the design of most foundations with B
greater than about 1 m will be generally controlled by the allowable level of
settlement Se , not by the ultimate bearing capacity. For that reason, it is
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important to determine BCRs at various levels of settlement [see definition in
Eq. (7.13)] and its relationship with BCRu .
Omar et al. [9] conducted several laboratory model tests on square and
continuous surface foundations (that is, Df = 0) supported by unreinforced and
geogrid-reinforced sand to determine the relationship between BCRs and BCRu.
For these tests, the parameters of the sand and the model foundations were:
B = 76.2 mm
Relative density of sand, Dr = 70%
Soil friction angle, ! = 40.3$
Size of square foundation = 76.2 mm × 76.2 mm
Width of continuous foundation, B = 76.2 mm
Geogrid used: TENSAR BX1000
The nondimensional parameters u/B, b/B, and d/B were varied during the tests.
Figure 7.22 shows typical plots of BCRu and BCRs versus d/B. The variation
of BCRs shown in Fig. 7.22 is the average plot for Se /Se (u) = 0.25, 0.50, and
0.75 [Se (u) = settlement at ultimate load for unreinforced sand] since the scatter
FIGURE 7.22 Plot of BCRu and BCRs with d/B (after Omar et al. [9])
(Note: The plots for BCRs are average for Se /Se(u) =
0.25, 0.5, and 0.75)
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of experimental results was relatively small. Based on the results of this study,
it was shown that (for u/B = 0.25 to 0.4)
S
B
BCRu = 1.7 to 1.8BCRs for = 0 and e ≤ 0.75
L
Se ( u )
(7.18)
S
B
BCRs = 1.4 to 1.45BCRu for = 1 and e ≤ 0.75
L
Se ( u )
(7.19)
and
7.11 CRITICAL NONDIMENSIONAL PARAMETERS
FOR FOUNDATIONS ON GEOGRID-REINFORCED
CLAY (!
! = 0 CONDITION)
Large-scale field and small-scale laboratory model tests results relating to the
determination of bearing capacity of foundations supported by saturated clay
are relatively scarce. Shin et al. [10] reported some laboratory model test results
which were intended to determine the nondimensional parameters as defined
by Section 7.8. These tests were conducted with a continuous surface foundation (Df = 0; B/L = 0) supported by a saturated clay having a liquid limit of 44%
and a plasticity index of 20%. TENSAR BX1100 geogrid was used for
reinforcement. Based on this study, the following conclusions were drawn:
1.
For all cases
BCRu = BCRs = BCR (for Se /Se (u) % 1)
(7.20)
2.
(u/B)cr # 0.4. This fact is illustrated in Fig. 7.23. Note that, for all b/B
values, the magnitude of BCR increases from u/B = 0.25 to u/B = 0.4
and decreases thereafter.
3.
(u/B)max # 0.9 to 1.0.
4.
Critical-width ratio: (b/B)cr # 4.0 to 4.5. This can clearly be seen from
Fig. 7.23.
5.
Critical reinforcement-depth ratio: (d/B)cr # 1.8. This is illustrated in
Fig. 7.24. Note that (d/B)cr is independent of the undrained cohesion
of the clay, cu .
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FIGURE 7.23 Variation of BCR with B/B (after Shin et al. [4])
( Note: h/B = 1/3 and N = 4)
FIGURE 7.24 Variation of BCR with N (that is d/B) (after Shin
et al. [4]) ( Note: b/B = 4, u/B = 0.4, h/B = 1/3)
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7.12 BEARING CAPACITY THEORY
Most of the existing data available relating to the ultimate and allowable bearing capacities for shallow foundations supported by geogrid-reinforced soil is
based on small-scale laboratory model tests. More recently, Adams and Collin
[11] provided large-scale model test results in sand which could be used to
verify laboratory test results. Also Huang and Meng [12] recently proposed an
ultimate bearing capacity theory for foundations on geogrid-reinforced sand
based on the wide-slab effect. However, more study is necessary to develop a
bearing capacity theory that will take into account the failure mode in soil, the
strength and stiffness of geogrid reinforcement, and scale effects.
7.13 SETTLEMENT OF FOUNDATIONS ON GEOGRIDREINFORCED SOIL DUE TO CYCLIC LOADING
In many cases shallow foundations supported by geogrid-reinforced soil may
be subjected to cyclic loading. This problem will primarily be encountered by
vibratory machine foundations. Das [13] and Das and Shin [14] reported
laboratory model test results on settlement caused by cyclic loading on surface
foundations supported, respectively, by reinforced sand and saturated clay.
The model tests of Das [13] were conducted with a square model foundation on unreinforced and geogrid-reinforced sand. Details of the sand and geogrid parameters were:
Model foundation: Square; B = 76.2 mm
Sand: Relative density of compaction, Dr = 76%
Angle of friction, ! = 42$
Reinforcement: Geogrid; TENSAR BX1000
Reinforcement-width ratio:
b b
≈ [see Eq. (7.16)]
B B cr
u u
≈ = 0.33
B B cr
h
= 0.33
B
d d
Reinforcement-depth ratio: ≈ = 1.33 [see Eq. (7.15)]
B B cr
Number of layers of reinforcement: N = 4
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FIGURE 7.25 Nature of load application—cyclic load test
FIGURE 7.26 Plot of Sec /B versus n (after Das [13]) ( Note: For
reinforced sand u/B = h/B = 1/3; b/B = 4; d/B = 1-1/3)
The laboratory tests were conducted by first applying a static load of
intensity qs (= qu (R) /FS; FS = factor of safety) followed by a cyclic load of low
frequency (1 cps). The amplitude of the intensity of cyclic load was qdc (max) . The
nature of load application described is shown in Fig. 7.25. Figure 7.26 shows
the nature of variation of foundation settlement due to cyclic load application
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Sec with qdc (max) /qu (R) and number of load cycles n. This is for the case of FS =
3. Note that, for any given test, Sec increases with n and reaches practically a
maximum value Sec (max) at n = ncr . Based on these tests the following
conclusions can be drawn.
1. For given values of FS and n, the magnitude of Sec /B increases with
the increase in qdc (max) /qu (R) .
2. If the magnitude of qdc (max) /qu (R) and n remain constant, the value of Sec
/B increases with a decrease in FS.
3. The magnitude of ncr for all tests in reinforced soil is approximately
the same, varying between 1.75 × 105 and 2.5 × 105 cycles. Similarly,
the magnitude of ncr for all tests in unreinforced soil varies between
1.5 × 105 and 2.0 × 105 cycles.
The variations of Sec (max) /B obtained from these tests for various values of
qdc (max) /qu (R) and FS are shown in Fig. 7.27. This figure clearly demonstrates
FIGURE 7.27
Plot of Sec(max) /B versus qdc(max) /qu(R) . (after
Das [10]) (Note: For reinforced sand, u/B
= h/B = 1/3, b/B = 4, d/B = 1-1/3)
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FIGURE 7.28 Variation of qdc(max) /qu(R) with ! (after
Das [10]) ( Note: For reinforced sand,
u/B = h/B = 1/3, b/B = 4, d/B = 1-1/3)
the reduction of the level of permanent settlement caused by geogrid reinforcement due to cyclic loading. Using the results of Sec (max) given in Fig. 7.27, the
variation of settlement ratio ! for various combinations of qdc (max) /qu (R) and FS
are plotted in Fig. 7.28. The settlement ratio is defined as
ρ=
S ec(max) − reinforced
Sec(max) − unreinforced
(7.21)
From Fig. 7.28 it can be seen that, although some scattering exists, the settlement ratio is only a function of qdc (max) /qu (R) and not the factor of safety, FS.
Laboratory model test results on continuous foundations with similar
loading conditions as those described above (Fig. 7.25) in reinforced saturated
clay were provided by Das and Shin [14]. General parameters of the test
program were as follows:
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Model foundation: Continuous; B = 76.2 mm
Clay: Moisture content = 34%
Degree of saturation = 96%
Undrained shear strength, cu = 12 kN/m2
Reinforcement: Geogrid; TENSAR BX1100
u u
= = 0. 4
B B cr
h
= 0.33
B
b b
≈ =5
B B cr
N =5
d d
= = 1.73
B B cr
Range of test:
q d (max)
qu( R )
FS =
= 3.4 to 14.6%
qu ( R )
qs
= 315
. to 6.85
FIGURE 7.29 Nature of variation of foundation settlement in clay
due to cyclic load application
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FIGURE 7.30
Plot of Sec(max) /B versus qdc(max) /qu(R) ( Note:
cu = 12 kN / m2; for reinforced clay, u/B = 0.4,
h/B = 0.33, b/B = 5, d/B = 1.73)
The general nature of the foundation settlement curve [for a given FS and
qdc (max) /qu (R) ] obtained is shown in Fig. 7.29, which can be divided into three
major zones. Zone 1 (for n = 1 to n = nr ) is a rapid settlement zone during
which about 70% of the maximum settlement [Sec (max) ] takes place. The magnitude of nr is about 10. Zone 2 (n = nr to n = ncr ) is a zone in which the settlement continues at a retarding rate reaching a maximum at n = ncr . For n! ncr,
the settlement of the foundation due to cyclic loading is negligible. The magnitude of ncr for reinforced soil varied from 1.8 × 104 to 2.5 × 104 cycles.
Figure 7.30 shows the summary of the tests conducted, and it is a plot of
Sec(max)/B for various combinations of qdc (max) /qu (R) and FS. It is important to
note that, for FS = 4.27, Sec (max) for reinforced soil was about 20% to 30%
smaller than that in unreinforced soil.
7.14 SETTLEMENT DUE TO IMPACT LOADING
Geogrid reinforcement can reduce the settlement of shallow foundations that
are likely to be subjected to impact loading. This is shown in the results of
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FIGURE 7.31 Nature of transient load
laboratory model tests in sand reported by Das [13]. The tests were conducted
with a square surface foundation (Df = 0; B = 76.2 mm). TENSAR BX1000
geogrid was used as reinforcement. Following are the physical parameters of
the soil and reinforcement:
Sand: Relative density of compaction = 76%
Angle of friction, " = 42"
Reinforcement:
u
= 0.33;
B
b
= 4;
B
h
= 0.33
B
Number of reinforcement layers, N = 0,1, 2, 3, and 4
The idealized shape of the impact load applied to the model foundation is
shown in Fig. 7.31, in which tr and td are rise and decay times, and qt (max) is the
maximum intensity of the impact load. For these tests the average values of tr
and td were approximately 1.75 s and 1.4 s, respectively. The maximum settlements observed due to the impact loading Set (max) are shown in a nondimensional form in Fig. 7.32. In this figure, qu and Se (u) , respectively, are the
ultimate bearing capacity and the corresponding foundation settlement on
unreinforced sand. From this figure it is obvious that
1. For a given value of qt (max) /qu , the foundation settlement decreases
with an increase in the number of geogrid layers.
2. For a given number of reinforcement layers, the magnitude of Set (max)
increases with the increase in qt (max) /qu .
The effectiveness with which geogrid reinforcement helps reduce the
settlement can be expressed by a quantity called the settlement reduction factor
R or
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FIGURE 7.32
R=
where
Variation of Set(max) /Se(u) with qt(max) /qu
and d/B (after Das [13])
S et (max)-d
S et (max)-d = 0
Set (max)-d = maximum settlement due to impact load with reinforcement depth of d
Set (max)-d=0 = maximum settlement with no reinforcement (that is, d =
0 or N = 0)
Based on the results given in Fig. 7.32, the variation of R with qt (max) /qu and
d/dcr is shown in Fig. 7.33. From the plot it is obvious that the geogrid
reinforcement acts as an excellent settlement retardant under impact loading.
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FIGURE 7.33 Plot of settlement reduction
factor with qt(max) /qu and d/dcr
(after Das [14])
REFERENCES
1. Vidal, H., La terre Armée, Anales de l’institut Technique du Bâtiment et des
Travaus Publiques, France, July-August, 888, 1966.
2. Binquet, J., and Lee, K. L., Bearing capacity tests on reinforced earth mass, J.
Geotech. Eng. Div., ASCE, 101(12), 1241, 1975.
3. Binquet, J., and Lee, K. L. Bearing capacity analysis of reinforced earth slabs,
J. Geotech. Eng. Div., ASCE, 101(12), 1257, 1975.
4. Guido, V. A., Biesiadecki, G. L., and Sullivan, M. J., Bearing capacity of a
geotextile rein-forced foundation, in Proc, XI Int. Conf. Soil Mech. Found.
Eng., San Francisco, 3, 1985, 1777.
5. Sakti, J., and Das, B. M., Model tests for strip foundation on clay reinforced
with geotextile layers, Trans. Res. Rec. No. 1153, National Academy of
Sciences, Washington, DC, 40, 1987.
6. Omar, M. T., Das, B. M., Yen, S. C., Puri, V. K., and Cook, E. E., Ultimate
bearing capacity of rectangular foundations on geogrid-reinforced sand,
Geotech. Testing J., ASTM, 16(2), 246, 1993.
7. Yetimoglu, T., Wu, J. T. H., and Saglamer, A., Bearing capacity of rectangular
footings on geogrid-reinforced sand, J. Geotech. Eng., ASCE, 120(12), 2083,
1994.
8. Guido, V. A., Knueppel, J. D., and Sweeney, M.. A., Plate load tests on
geogrid-reinforced earth slabs, in Proc., Geosynthetics ‘87, 1987, 216.
© 1999 by CRC Press LLC
© 1999 by CRC Press LLC
9. Omar, M. T., Das, B. M., Yen, S. C., Puri, V. K., and Cook, E. E., Shallow
foundations on geogrid-reinforced sand, Trans. Res. Rec. No. 1414, National
Academy of Sciences, Washington, DC, 59, 1993.
10. Shin, E. C., Das, B. M., Puri, V. K., Yen, S. C., and Cook, E. E., Bearing
capacity of strip foundation on geogrid-reinforced clay, Geotech. Testing J.,
ASTM, 17(4), 534, 1993.
11. Adams, M.. T., and Collin, J. G., Large model spread footing load tests on
geosynthetic rein-forced soil foundation, J. Geotech. Geoenviron. Eng., ASCE,
123(1), 66, 1997.
12. Huang, C. C., and Meng, F. Y., Deep footing and wide-slab effects on
reinforced sandy ground, J. Geotech. Geoenviron. Eng., ASCE, 123(1), 30,
1997.
13. Das, B. M., Dynamic loading on foundation on reinforced soil, in Geosynthetics
in Foundation Reinforcement and Erosion Control Systems, Bowders, J. J.,
Scranton, H. B., and Broderick, G. P., Eds., Geotechnical Special Publication
No. 76, ASCE, 1998, 19.
14. Das, B. M., and Shin, E. C., Strip foundation on geogrid-reinforced clay:
behavior under cyclic loading, Geotextiles and Geomembranes, 13(10), 657,
1993.
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CHAPTER EIGHT
UPLIFT CAPACITY OF SHALLOW FOUNDATIONS
8.1 INTRODUCTION
Foundations and other structures may be subjected to uplift forces under
special circumstances. For those foundations, during the design process it is
desirable to apply sufficient factor of safety against failure by uplift. During
the last thirty or so years, several theories have been developed to estimate the
ultimate uplift capacity of foundations embedded in sand and clay soils, and
some of those theories are detailed in this chapter. The chapter is divided into
two major parts: foundations in granular soil and foundations in saturated clay
soil (! = 0).
Figure 8.1 shows a shallow foundation of width B. The depth of embedment is Df . The ultimate uplift capacity of the foundation Qu can be expressed
as
Qu = frictional resistance of soil along the failure surface
+ weight of soil in the failure zone and the foundation
(8.1)
If the foundation is subjected to an uplift load of Qu , the failure surface in
the soil for relatively small Df /B values will be of the type shown in Fig. 8.1.
The intersection of the failure surface at the ground level will make an angle
" with the horizontal. However, the magnitude of " will vary with the relative
density of compaction in the case of sand and with the consistency in the case
of clay soils.
FIGURE 8.1 Shallow foundation subjected to uplift
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When the failure surface in soil extends up to the ground surface at
ultimate load, it is defined as a shallow foundation under uplift. For larger
values of Df /B, failure takes place around the foundation and the failure surface does not extend to the ground surface. These are called deep foundations
under uplift. The embedment ratio, Df /B, at which a foundation changes from
shallow to deep condition is referred to as the critical embedment ratio,
(Df /B)cr. In sand the magnitude of (Df /B)cr can vary from 3 to about 11 and, in
saturated clay, it can vary from 3 to about 7.
FOUNDATIONS IN SAND
During the last thirty years, several theoretical and semiempirical methods
have been developed to predict the net ultimate uplifting load of continuous,
circular, and rectangular foundations embedded in sand. Some of these
theories are briefly described in the following sections.
8.2 BALLA’S THEORY
Based on results of several model and field tests conducted in dense soil, Balla
[1] established that, for shallow circular foundations, the failure surface in soil
will be as shown in Fig. 8.2. Note from the figure that aa! and bb! are arcs of
a circle. The angle " is equal to 45! !/2. The radius of the circle, of which aa!
and bb! are arcs, is equal to
FIGURE 8.2 Balla’s theory for shallow circular foundations
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r=
Df
φ
sin 45 +
2
(8.2)
As mentioned before, the ultimate uplift capacity of the foundation is the sum
of two components: (a) the weight of the soil and the foundation in the failure
zone and (b) the shearing resistance developed along the failure surface. Thus,
assuming that the unit weight of soil and the foundation material are approximately the same
Df
D
+ F3 φ, f
Q u = D 3f γ F1 φ,
B
B
(8.3)
where # = unit weight of soil
! = soil friction angle
B = diameter of the circular foundation
The sums of the functions F1(!, Df /B) and F3(!, Df /B) developed by Balla [1]
are plotted in Fig. 8.3 for various values of the soil friction angle ! and the
embedment ratio, Df /B.
In general, Balla’s theory is in good agreement with the uplift capacity of
shallow foundations embedded in dense sand at an embedment ratio of Df /B
" 5. However for foundations located in loose and medium sand, the theory
overestimates the ultimate uplift capacity. The main reason Balla’s theory
overestimates the ultimate uplift capacity for Df /B > about 5 even in dense
sand is because it is essentially deep foundation condition, and the failure
surface does not extend to the ground surface.
The simplest procedure to determine the embedment ratio at which deep
foundation condition is reached may be determined by plotting the nondimensional breakout factor Fq against Df /B as shown in Fig. 8.4. The breakout
factor is derived as
Fq =
Qu
γAD f
(8.4)
where A = area of the foundation.
The breakout factor increases with Df /B up to a maximum value of Fq" Fq* at
Df /B" (Df /B)cr . For Df /B > (Df /B)cr the breakout factor remains practically
constant (that is, Fq*).
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FIGURE 8.3 Variation of F1 + F3 [Eq. (8.3)]
© 1999 by CRC Press LLC
FIGURE 8.4 Nature of variation of Fq with Df /B
FIGURE 8.5 Continuous foundation subjected to uplift
8.3 THEORY OF MEYERHOF AND ADAMS
One of the most rational methods for estimating the ultimate uplift capacity of
a shallow foundation was proposed by Meyerhof and Adams [2], and it is described in detail in this section. Figure 8.5 shows a continuous foundation of
width B subjected to an uplifting force. The ultimate uplift capacity per unit
length of the foundation is equal to Qu . At ultimate load the failure surface in
© 1999 by CRC Press LLC
soil makes an angle " with the horizontal. The magnitude of " depends on
several factors, such as the relative density of compaction and the angle of
friction of the soil, and it varies between 90#! a! to 90#! b!. Let us
consider the free body diagram of the zone abcd. For stability consideration, the
following forces per unit length of the foundation need to be considered.
a. the weight of the soil and concrete, W, and
b. the passive force Pṕ per unit length along the faces ad and bc. The
force Pṕ is inclined at an angle $ to the horizontal. For an average
value of "" 90! !/2, the magnitude of $ is about b!.
If we assume that the unit weights of soil and concrete are approximately
the same, then
W = #Df B
Pp′ =
Ph′
1 1
2
=
( K γ D )
cos δ 2 cos δ ph f
(8.5)
where Ph́ = horizontal component of the passive force
Kph = horizontal component of the passive earth pressure coefficient
Now, for equilibrium, summing the vertical components of all forces
$Fv = 0
Qu = W + 2P´p sin$
Qu = W + 2(P´p cos$)tan$
Qu = W + 2P´h tan$
or
1
Q u = W + 2 K ph γD 2f tan δ = W + K ph γD 2f tan δ
2
(8.6)
The passive earth pressure coefficient based on the curved failure surface
for $ = b! can be obtained from Caquot and Kerisel [3]. Furthermore, it is
convenient to express Kphtan$ in the form
Ku tan! = Kphtan$
(8.7)
Combining Eqs. (8.6) and (8.7)
Qu = W + Ku #Df² tan!
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(8.8)
FIGURE 8.6 Variation of Ku
where Ku = nominal uplift coefficient
The variation of the nominal uplift coefficient Ku with the soil friction
angle ! is shown in Fig. 8.6. It falls within a narrow range and may be taken
as equal to 0.95 for all values of ! varying from 30# to about 48#. The ultimate uplift capacity can now be expressed in a nondimensional form (that is,
the breakout factor, Fq ) as defined in Eq. (8.4) [4]. Thus, for a continuous
foundation, the breakout factor per unit length is
Fs =
Qu
γBD f
or
Fq =
W + K u γD 2f tan φ
W
D
= 1 + K u f tan φ
B
(8.9)
For circular foundations, Eq. (8.8) can be modified to the form
%
Qu = W + – SF #BDf²Ku tan!
2
(8.10)
%
W% – B2Df #
4
(8.11)
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where SF = shape factor
B = diameter of the foundation
The shape factor can be expressed as
D
S F = 1 + m f
B
(8.12)
where m = coefficient which is a function of the soil friction angle !
Thus, combining Eqs. (8.10), (8.11), and (8.12), we obtain
Qu =
D f
π 2
π
γ BD 2f K u tan φ
B D f γ + 1 + m
4
2
B
(8.13)
The breakout factor Fq can be given as
Fq =
Qu
=
γ AD f
D
π 2
π
B D f γ + 1 + m f γ BD 2f K u tan φ
4
2
B
π
γ B 2 D f
4
D D
= 1 + 2 1 + m f f K u tan φ
B B
(8.14)
For rectangular foundations having dimensions of B × L, the ultimate
capacity can also be expressed as
Qu = W + #D2f (2SFB + L & B)Ku tan!
(8.15)
The preceding equation was derived with the assumption that the two end portions of length B/2 are governed by the shape factor SF , while the passive
pressure along the central portion of length L & B is the same as the continuous
foundation. In Eq. (8.15)
W % #BLDf
(8.16)
D
S F = 1 + m f
B
(8.17)
and
© 1999 by CRC Press LLC
Thus
D f
B + L − B K u tan φ
Q u = γBLD f + γD 2f 2 1 + m
B
(8.18)
The breakout factor Fq can now be determined as
Fq =
Qu
γBLD f
(8.19)
Combining Eqs. (8.18) and (8.19), we obtain [4]
D B D
Fq = 1 + 1 + 2 m f + 1 f K u tan φ
B L B
(8.20)
The coefficient m given in Eq. (8.12) was determined from experimental
observations [2] and its values are given in Table 8.1. In Fig. 8.7, m is also
plotted as a function of the soil friction angle !.
TABLE 8.1 Variation
of m [Eq. (8.12)]
Soil friction
angle, !
m
20
25
30
35
40
45
48
0.05
0.1
0.15
0.25
0.35
0.5
0.6
As shown in Fig. 8.4, the breakout factor Fq increases with Df /B to a
maximum value of Fq* at (Df /B)cr and remains constant thereafter. Based on
experimental observations, Meyerhof and Adams [2] recommended the
variation of (Df /B)cr for square and circular foundations with soil friction angle
! and this is shown in Fig. 8.8.
Thus, for a given value of ! for square (B = L) and circular (diameter =
B) foundations, we can substitute m (Table 1) into Eqs. (8.14) and (8.20) and
© 1999 by CRC Press LLC
FIGURE 8.7 Variation of m
FIGURE 8.8 Variation of (Df /B)cr for square and circular
foundations
© 1999 by CRC Press LLC
calculate the breakout factor (Fq) variation with embedment ratio (Df /B). The
maximum value of Fq = F q* will be attained at Df /B = (Df /B)cr . For Df /B >
(Df /B)cr , the breakout factor will remain constant as F q* . The variation of Fq
with Df /B for various values of ! made in this manner is shown in Fig. 8.9.
The variation of the maximum breakout factor F q* for deep square and circular
foundations with the soil friction angle ! is shown in Fig. 8.10.
Laboratory experimental observations have shown that the critical embedment ratio (for a given soil friction angle !) increases with the L/B ratio.
Meyerhof [5] indicated that, for a given value of !,
FIGURE 8.9 Plot of Fq [Eqs. (8.14) and (8.20)] for square and circular foundations
© 1999 by CRC Press LLC
FIGURE 8.10 Fq* for deep square and circular foundations
Df
B
cr -continuous ≈ 1.5
Df
B
cr -square
(8.21)
Based on laboratory model test results, Das and Jones [6] gave an empirical
relationship for the critical embedment ratio of rectangular foundations in the
form
Df
D
D
L
= f 0.133 + 0. 867 ≤ 1.4 f
B
B
cr - R B cr -S
B cr -S
© 1999 by CRC Press LLC
(8.22)
D
where f = criticalembedmentratio of a rectangular foundation
B cr-R
having dimensions of L × B
Df
= criticalembedmentratio of a square foundation having
B
cr-S
dimensions of B × B
Using Eq. (8.22) and the (Df /B)cr-S values given in Fig. 8.8, the magnitude of
(Df /B)cr-R for a rectangular foundation can be estimated. These values of
(Df /B)cr-R can be substituted into Eq. (8.20) to determine the variation of Fq =
F *with the soil friction angle !.
8.4 THEORY OF VESIC
Vesic [7] studied the problem of an explosive point charge expanding a
spherical cavity close to the surface of a semi-infinite, homogeneous, isotropic
solid (in this case, the soil). Now, referring to Fig. 8.11, it can be seen that, if
the distance Df is small enough, there will be an ultimate pressure po that will
shear away the soil located above the cavity. At that time, the diameter of the
spherical cavity is equal to B. The slip surfaces ab and cd will be tangent to
the spherical cavity at a and c. At points b and d they make an angle " =
45!!/2. For equilibrium, summing the components of forces in the vertical
direction we can determine the ultimate pressure po in the cavity. Forces that
will be involved are:
1. Vertical component of the force inside the cavity, PV ;
2. Effective self-weight of the soil, W = W1 + W2 ; and
3. Vertical component of the resultant of internal forces, FV .
FIGURE 8.11 Vesic’s theory of expansion of cavities
© 1999 by CRC Press LLC
For a c–! soil, we can thus determine that
!q
Fc + #Df F
po = c!
B
D
D
2 2
where Fq = 1.0 − + A1 f + A2 f
3 Df
B
B
2
2
(8.23)
2
D
D
Fc = A3 f + A4 f
B
B
2
2
(8.24)
(8.25)
where A1 , A2 , A3 , A4 = functions of the soil friction !
For granular soils c = 0, so
!q
po = #Df F
(8.26)
Vesic [8] applied the preceding concept to determine the ultimate uplift
capacity of shallow circular foundations. In Fig. 8.12 consider that the circular
foundation ab having a diameter B is located at a depth Df below the ground
FIGURE 8.12 Cavity expansion theory applied to
circular foundation uplift
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surface. Assuming that the unit weight of the soil and the unit weight of the
foundation are approximately the same, if the hemispherical cavity above the
foundation (that is, ab) is filled with soil, it will have a weight of
3
2 B
W3 = π γ
3 2
(8.27)
This weight of soil will increase the pressure by p1 , or
3
2 B
π γ
W3
3 2 = 2 γ B
=
p1 =
2
2
3 2
B
B
π
π
2
2
If the foundation is embedded in a cohesionless soil (c = 0), then the pressure
p1 should be added to Eq. (8.26) to obtain the force per unit area of the anchor,
qu , needed for complete pullout. Thus
qu =
Qu
Qu
2 B
=
= po + p1 = γD f Fq + γ
π
A
3 2
( B) 2
2
2 B
3 2
= γD f Fq +
Df
(8.28)
or
2
D
D
Qu
f
+ A2 f = γD f Fq
qu =
= γD f 1 + A1
A
B
B
2
2
↑
Breakout factor
(8.29)
The variation of the breakout factor Fq for shallow circular foundations
is given in Table 8.2 and Fig. 8.13. In a similar manner, using the analogy of
© 1999 by CRC Press LLC
TABLE 8.2 Vesic’s Breakout Factor, Fq , for
Circular Foundations
Soil friction
angle, !
(deg)
0.5
1.0
1.5
2.5
5.0
0
10
20
30
40
50
1.0
1.18
1.36
1.52
1.65
1.73
1.0
1.37
1.75
2.11
2.41
2.61
1.0
1.59
2.20
2.79
3.30
3.56
1.0
2.08
3.25
4.41
5.43
6.27
1.0
3.67
6.71
9.89
13.0
15.7
Df /B
FIGURE 8.13 Vesic’s breakout factor, Fq , for shallow circular foundations
© 1999 by CRC Press LLC
expansion of long cylindrical cavities, Vesic determined the variation of the
breakout factor Fq for shallow continuous foundations. These values are given
in Table 8.3 and are also plotted in Fig. 8.14.
TABLE 8.3 Vesic’s Breakout Factor, Fq , for
Continuous Foundations
Soil friction
angle, !
(deg)
0.5
1.0
1.5
2.5
5.0
0
10
20
30
40
50
1.0
1.09
1.17
1.24
1.30
1.32
1.0
1.16
1.33
1.47
1.58
1.64
1.0
1.25
1.49
1.71
1.87
2.04
1.0
1.42
1.83
2.19
2.46
2.60
1.0
1.83
2.65
3.38
3.91
4.20
Df /B
FIGURE 8.14 Vesic’s breakout factor, Fq , for shallow
continuous foundation
© 1999 by CRC Press LLC
8.5 SAEEDY’S THEORY
A theory for the ultimate uplift capacity of circular foundations embedded in
sand was proposed by Saeedy [9] in which the trace of the failure surface was
assumed to be an arc of a logarithmic spiral. According to this solution, for
shallow foundations the failure surface extends to the ground surface. However, for deep foundations [that is, Df > Df (cr)] the failure surface extends only
to a distance of Df (cr) above the foundation. Based on this analysis, Saeedy [9]
proposed the ultimate uplift capacity in a nondimensional form (Qu /!B2Df ) for
various values of " and the Df /B ratio. The author converted the solution into
a plot of breakout factor Fq = Qu /!ADf (A = area of the foundation) versus the
soil friction angle " as shown in Fig. 8.15. According to Saeedy, during the
foundation uplift the soil located above the anchor gradually becomes compacted, in turn increasing the shear strength of the soil and, hence, the ultimate
uplift capacity. For that reason, he introduced an empirical compaction factor
µ, which is given in the form
µ = 1.044Dr + 0.44
(8.30)
where Dr = relative density of sand
Thus, the actual ultimate capacity can be expressed as
Qu (actual) = (Fq !ADf ) µ
(8.31)
8.6 DISCUSSION OF VARIOUS THEORIES
Based on the various theories presented in the preceding sections, we can make
some general observations:
1. The only theory that addresses the problem of rectangular foundations
is that given by Meyerhof and Adams [3].
2. Most theories assume that shallow foundation conditions exist for
Df /B ! 5. Meyerhof and Adams’ theory provides a critical embedment
ratio (Df /B)cr for square and circular foundations as a function of the
soil friction angle.
3. Experimental observations generally tend to show that, for shallow
foundation in loose sand, Balla’s theory [1] overestimates the ultimate
uplift capacity. Better agreement, however, is obtained for foundations
in dense soil.
4. Vesic’s theory [8] is, in general, fairly accurate in estimating the ultimate uplift capacity of shallow foundations in loose sand. However,
laboratory experimental observations have shown that, for shallow
foundations in dense sand, this theory can underestimate the actual
uplift capacity by as much as 100% or more.
© 1999 by CRC Press LLC
FIGURE 8.15 Plot of Fq based on Saeedy’s theory
Figure 8.16 shows a comparison of some published laboratory experimental results for the ultimate uplift capacity of circular foundations with the
theories of Balla, Vesic, and Meyerhof and Adams. Table 8.4 gives the references to the laboratory experimental curves shown in Fig. 8.16. In developing
the theoretical plots for " = 30" (loose sand condition) and " = 45" (dense
sand condition) the following procedures were used.
1. According to Balla’s theory [1], from Eq. (8.3) for circular foundations
Qu = D 3f γ ( F1 + F3 )
!#"#
$
Fig. 8.3
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FIGURE 8.16
© 1999 by CRC Press LLC
Comparison of theories with laboratory experimental results for circular foundations
TABLE 8.4 References to Laboratory Experimental Curves Shown in Fig. 8.16
Curv
e
Reference
Circular
foundation
diameter, B
1
2
3
4
5
6
7
8
9
Baker and Kondner [10]
Baker and Kondner [10]
Baker and Kondner [10]
Baker and Kondner [10]
Sutherland [11]
Sutherland [11]
Esquivel-Diaz [12]
Esquivel-Diaz [12]
Balla [1]
25.4
38.1
50.8
76.2
38.1–152.4
38.1–152.4
76.2
76.2
61–119.4
Soil properties
" = 42"; ! = 17.61 kN / m3
" = 42"; ! = 17.61 kN / m3
" = 42"; ! = 17.61 kN / m3
" = 42"; ! = 17.61 kN / m3
" = 45"
" = 31"
" # 43"; ! = 14.81 kN / m3–15.14 kN / m3
" = 33"; ! = 12.73 kN / m3–12.89 kN / m3
Dense sand
So
π 2
B Qu
Q
4
=
F1 + F3 = u3 =
γD f
3 π
2
γD f B
4
2
B
π
Q
4 D f u
γD f A
or
Fq =
Qu
F1 + F3
=
2
γ AD f
π B
4 D f
(8.32)
So, for a given soil friction angle the sum of F1 + F3 was obtained
from Fig. 8.3, and the breakout factor was calculated for various values
of Df /B. These values are plotted in Fig. 8.16.
2. For Vesic’s theory [8] the variations of Fq versus Df /B for circular
foundations are given in Table 8.2. These values of Fq are also plotted
in Fig. 8.16.
3. The breakout factor relationship for circular foundations based on
Meyerhof and Adams’ theory [3] is given in Eq. (8.14). Using Ku #
0.95, the variations of Fq with Df /B were calculated, and they are also
plotted in Fig. 8.16.
© 1999 by CRC Press LLC
Based on the comparison between the theories and the laboratory experimental results shown in Fig. 8.16, it appears that Meyerhof and Adams’ theory
[3] is more applicable to a wide range of foundations and provides as good an
estimate as any for the ultimate uplift capacity. So this theory is recommended
for use. However, it needs to be kept in mind that the majority of the experimental results presently available in the literature for comparison with the
theory are from laboratory model tests. When applying these results to the
design of an actual foundation, the scale effect needs to be taken into
consideration. For that reason, a judicious choice is necessary in selecting the
value of the soil friction angle ".
EXAMPLE 8.1
Consider a circular foundation in sand. Given for the foundation: diameter, B
= 1.5 m; depth of embedment, Df = 1.5 m. Given for the sand: unit weight, !
= 17.4 kN / m3; friction angle, " = 35". Using Balla’s theory, calculate the
ultimate uplift capacity.
Solution From Eq. (8.3)
Qu = D3f !(F1 + F3)
From Fig. 8.3, for " = 35" and Df /B = 1.5/1.5 = 1, the magnitude of F1 + F3
# 2.4. so
Qu = (1.5)3(17.4)(2.4) = 140.9 kN
!!
EXAMPLE 8.2
Redo Example Problem 8.1 using Vesic’s theory.
Solution From Eq. (8.29)
Qu = A! D3f Fq
From Fig. 8.13, for " = 35" and Df /B = 1, Fq is about 2.2. So
π
Qu = (1.5) 2 (17.4)(1.5)(2.2) = 101.5 kN
4
!!
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EXAMPLE 8.3
Redo Example Problem 8.1 using Meyerhof and Adams’ theory.
Solution From Eq. (8.14)
D f D f
K u tan φ
Fq = 1 + 2 1 + m
B B
For " = 35", m = 0.25 (Table 8.1). So
Fq = 1+2[1 + (0.25)(1)](1)(0.95)(tan35) = 2.66
So
π
Qu = Fq γAD f = (2.66 )(17.4) (1.5) 2 (1.5) = 122.7 kN
4
!!
8.7 EFFECT OF BACKFILL ON UPLIFT CAPACITY
Spread foundations constructed for electric transmission towers are subjected
to uplifting force. The uplift capacity of such foundations can be estimated by
using the same relationship described in the preceding sections. During the
construction of such foundations, the embedment ratio Df /B is usually kept at
3 or less. For foundation construction, the native soil is first excavated. Once
the foundation construction is finished, the excavation is backfilled and compacted. The degree of compaction of the backfill material plays an important
role in the actual ultimate uplift capacity of the foundation. Kulhawy,
Trautman, and Nicolaides [13] conducted several laboratory model tests to observe the effect of the degree of compaction of the backfill compared to the
native soil. According to their observations, failure in soil in most cases takes
place by side shear as shown in Fig. 8.17. However, wedge or combined shear
failure occurs for foundations with Df /B < about 2 in medium to dense native
soil where the backfill is at least 85 percent as dense as the native soil (Fig.
8.18). Figure 8.19 shows the effect of backfill compaction on the breakout
factor Fq when the native soil is loose. Similarly, Fig. 8.20 is for the case
where the native soil is dense. Based on the observations of Kulhawy et al.
[13], this study shows that the compaction of the backfill has a great influence
on the breakout factor of the foundation, and the ultimate uplift capacity
greatly increases with the degree of backfill compaction.
© 1999 by CRC Press LLC
FIGURE 8.17 Failure by side shear
FIGURE 8.18 Wedge or combined shear failure
© 1999 by CRC Press LLC
FIGURE 8.19 Effect of backfill on breakout factor —
square foundation with loose native soil
(after Kulhawy et al. [13])
FIGURE 8.20 Effect of backfill on breakout factor —
square foundation with dense native soil
(after Kulhawy et al. [13])
© 1999 by CRC Press LLC
FIGURE 8.21 Shallow foundation in saturated clay subjected to uplift
FOUNDATIONS IN SATURATED CLAY
(!
! = 0 CONDITION)
8.8 ULTIMATE UPLIFT CAPACITY—GENERAL
Theoretical and experimental research results presently available for determining the ultimate uplift capacity of foundations embedded in saturated clay
soil are rather limited. In the following sections, the results of some of the
existing studies are reviewed.
Figure 8.21 shows a shallow foundation in a saturated clay. The depth of
the foundation is Df , and the width of the foundation is B. The undrained
shear strength and the unit weight of the soil are cu and γ, respectively. If we
assume that the unit weight of the foundation material and the clay are approximately the same, then the ultimate uplift capacity can be expressed as [8]
Qu = A("Df + cu Fc)
(8.33)
where A = area of the foundation
Fc = breakout factor
" = saturated unit weight of the soil
8.9 VESIC’S THEORY
Using the analogy of the expansion of cavities, Vesic [8] presented the
theoretical variation of the breakout factor Fc (for ! = 0 condition) with
© 1999 by CRC Press LLC
! = 0 Condition)
TABLE 8.5 Variation of Fc (!
Df /B
Foundation type
Circular (diameter = B)
Continuous (width = B)
0.5
1.0
1.5
2.5
5.0
1.76 3.80
0.81 1.61
6.12
2.42
11.6
4.04
30.3
8.07
FIGURE 8.22 Vesic’s breakout factor Fc
the embedment ratio Df /B, and these values are given in Table 8.5. A plot of
these same values of Fc against Df /B is also shown in Fig. 8.22. Based on the
laboratory model test results available at the present time, it appears that
Vesic’s theory gives a closer estimate only for shallow foundations embedded
in softer clay.
In general, the breakout factor increases with embedment ratio up to a
maximum value and remains constant thereafter as shown in Fig. 8.23. The
maximum value of Fc = Fc* is reached at Df /B = (Df /B)cr . Foundations located
at Df /B > (Df /B)cr are referred to as deep foundations for uplift capacity consideration. For these foundations at ultimate uplift load, local shear failure in soil
© 1999 by CRC Press LLC
FIGURE 8.23 Nature of variation of Fc with Df /B
located around the foundation takes place. Foundations located at Df /B!
(Df /B)cr are shallow foundations for uplift capacity consideration.
8.10 MEYERHOF’S THEORY
Based on several experimental results, Meyerhof [5] proposed the following
relationship
Qu = A("Df + Fc cu)
For circular and square foundations
Df
≤9
Fc = 1.2
B
(8.34)
and, for strip foundations
Df
≤8
Fc = 0.6
B
(8.35)
The preceding two equations imply that the critical embedment ratio (Df /B)cr
is about 7.5 for square and circular foundations and about 13.5 for strip
foundations.
© 1999 by CRC Press LLC
8.11 MODIFICATIONS TO MEYERHOF’S THEORY
Das [14] compiled a number of laboratory model test results on circular
foundations in saturated clay with cu varying from 5.18 kN / m2 to about 172.5
kN / m2. Figure 8.24 shows the average plots of Fc versus Df /B obtained from
these studies, along with the critical embedment ratios.
From Fig. 8.24 it can be seen that, for shallow foundations
D
Fc ≈ n f ≤ 8 to 9
B
(8.36)
where n = a constant
The magnitude of n varies between 5.9 to 2.0 and is a function of the undrained cohesion. Since n is a function of cu and Fc = Fc* is about 8 to 9 in all
cases, it is obvious that the critical embedment ratio (Df /B)cr will be a function
of cu .
Das [14] also reported some model test results with square and rectangular
foundations. Based on these tests, it was proposed that
Df
= 0.107 c u + 2.5 ≤ 7
B
cr −S
(8.37)
D
where f = critical embedment ratio of square foundations
B cr −S (or circular foundation)
cu = undrained cohesion, in kN/m 2
It was also observed by Das [19] that
Df
D
D
L
= f 0 .73 + 0 . 27 ≤ 1 . 55 f
B
B
cr − R B cr − S
B cr −S
(8.38)
D
where f
= critical embedment ratio of rectangula r foundation s
B cr − R
L = length of foundation
© 1999 by CRC Press LLC
FIGURE 8.24 Variation of Fc with Df/B from various experimental observations—circular foundations; diameter = B
© 1999 by CRC Press LLC
Based on the above findings, Das [19] proposed an empirical procedure
to obtain the breakout factors for shallow and deep foundations. According to
this procedure, #" and $" are two nondimensional factors defined as
Df
α′ =
B
Df
B
cr
β′ =
Fc
(8.39)
and
Fc*
(8.40)
For a given foundation, the critical embedment ratio can be calculated by using
Eqs. (8.37) and (8.38). The magnitude of Fc* can be given by the following
empirical relationship
B
Fc*−R = 7.56 + 1.44
L
(8.41)
*
= breakout factor for deep rectangular foundations
where Fc-R
Figure 8.25 shows the experimentally derived plots (upper limit, lower
limit, and average of $" and #". Following is a step-by-step procedure to estimate the ultimate uplift capacity.
1. Determine the representative value of the undrained cohesion, cu .
2. Determine the critical embedment ratio using Eqs. (8.37) and (8.38).
3. Determine the Df /B ratio for the foundation.
4. If Df /B > (Df /B)cr as determined in Step 2, it is a deep foundation.
However, if Df /B ! (Df /B)cr , it is a shallow foundation.
5. For Df /B > (Df /B)cr
B
Fc = Fc* = 7.56 + 1.44
L
Thus
B
Qu = A 7.56 + 1.44 cu + γD f
L
where A = area of the foundation
© 1999 by CRC Press LLC
(8.42)
FIGURE 8.25 Plot of $" versus #"
6. For Df /B! (Df /B)cr
B
Qu = A(β′ Fc* cu + γD f ) = A β′ 7.56 + 1.44 cu + γD f
L
(8.43)
The value of $" can be obtained from the average curve of Fig. 8.25. The
procedure outlined above gives fairly good results in estimating the net
ultimate capacity of foundations.
EXAMPLE 8.4
A rectangular foundation in a saturated clay measures 1.5 m × 3 m. Given: Df
= 1.8 m; cu = 52 kN / m2; " = 18.9 kN / m3. Estimate the ultimate uplift
capacity.
Solution From Eq. (8.37)
Df
B = 0.107cu + 2.5 = ( 0.107)(52) + 2.5 = 8.06
cr −S
© 1999 by CRC Press LLC
So use (Df /B)cr-S = 7. Again, from Eq. (8.38)
Df
D
L
= f 0 .73 + 0.27
B
B
cr − R B cr − S
3
= 7 0 .73 + 0.27
= 8.89
1 .5
D
Check : 1.55 f = (1.55)( 7) = 10.85
B cr −S
So use (Df /B)cr-R = 8.89. The actual embedment ratio is Df /B = 1.8/1.5 = 1.2.
Hence this is a shallow foundation.
Df
B
= 1.2 = 0.13
α′ =
8.89
Df
B
cr
Referring to the average curve of Fig. 8.25, for #" = 0.13 the magnitude of $"
= 0.2. From Eq. (8.43)
B
Qu = Aβ′7.56 + 1.44 cu + γD f
L
1.5
= (1.5)(3)(0.2)7.56 + 1.44 (52) + (18.9)(1.8) = 540.6 kN
3
!!
8.12 FACTOR OF SAFETY
In most cases of foundation design, it is recommended that a minimum factor
of safety of 2 to 2.5 be used to arrive at the allowable ultimate uplift capacity.
REFERENCES
1.
Balla, A., The resistance to breaking out of mushroom foundations for pylons,
in Proc., V Int. Conf. Soil Mech. Found. Eng., Paris, France, 1, 1961,569.
© 1999 by CRC Press LLC
2.
Meyerhof, G. G., and Adams, J. I., The ultimate uplift capacity of foundations,
Canadian Geotech. J., 5(4)225, 1968.
3.
Caquot, A., and Kerisel, J., Tables for Calculation of Passive Pressure,
Active Pressure, and Bearing Capacity of Foundations, GauthierVillars, Paris, 1949.
Das, B. M., and Seeley, G. R., Breakout resistance of horizontal
anchors, J. Geotech. Eng. Div., ASCE, 101(9), 999, 1975.
Meyerhof, G. G., Uplift resistance of inclined anchors and piles, in
Proc., VIII Int. Conf. Soil Mech. Found. Eng., Moscow, USSR, 2.1,
1973, 167.
Das, B. M., and Jones, A. D., Uplift capacity of rectangular foundations in sand, Trans. Res. Rec. 884, National Research Council,
Washington, DC, 54, 1982.
Vesic, A. S., Cratering by explosives as an earth pressure problem, in
Proc., VI Int. Conf. Soil Mech. Found. Eng., Montreal, Canada, 2,
1965, 427.
Vesic, A. S., Breakout resistance of objects embedded in ocean bottom,
J. Soil Mech. Found. Div., ASCE, 97(9), 1183, 1971.
Saeedy, H. S., Stability of circular vertical earth anchors, Canadian
Geotech. J., 24(3), 452, 1987.
Baker, W. H., and Kondner, R. L., Pullout load capacity of a circular
earth anchor buried in sand, Highway Res. Rec.108, National Research
Council, Washington, DC, 1, 1966.
Sutherland, H. B., Model studies for shaft raising through cohesionless
soils, in Proc., VI Int. Conf. Soil Mech. Found. Eng., Montreal Canada,
2, 1965, 410
Esquivel-Diaz, R. F., Pullout Resistance of Deeply Buried Anchors in
Sand, M.S. Thesis, Duke University, Durham, NC, USA, 1967.
Kulhawy, F. H., Trautman, C. H., and Nicolaides, C. N., Spread
foundations in uplift: experimental study, Found. for Transmission
Towers, Geotech. Spec. Pub. 8, ASCE, 110, 1987.
Das, B. M., Model tests for uplift capacity of foundations in clay, Soils
and Foundations, Japan, 18(2), 17, 1978.
Ali, M., Pullout Resistance of Anchor Plates in Soft Bentonite Clay,
M.S. Thesis, Duke University, Durham, NC, USA, 1968.
Kupferman, M., The Vertical Holding Capacity of Marine Anchors in
Clay Subjected to Static and Dynamic Loading, M.S. Thesis, University of Massachusetts, Amherst, , USA, 1971.
Adams, J. K., and Hayes, D. C., The uplift capacity of shallow
foundations, Ontario Hydro. Res. Quarterly, 19(1), 1, 1967.
Bhatnagar, R. S., Pullout Resistance of Anchors in Silty Clay, M.S.
Thesis, Duke University, Durham, NC, USA, 1969.
Das, B. M., A procedure for estimation of ultimate uplift capacity of
foundations in clay, Soils and Foundations, Japan, 20(1), 77, 1980.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
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APPENDIX
CONVERSION FACTORS
CONVERSION FACTORS FROM SI
TO ENGLISH UNITS
Length:
1m
1 cm
1 mm
1m
1 cm
1 mm
= 3.281 ft
= 3.281 × 10-2 ft
= 3.281 × 10-3 ft
= 39.37 in.
= 0.3937 in.
= 0.03937 in.
Area:
1 m2 = 10.764 ft2
1 cm2 = 10.764 × 10-4 ft2
1 mm2 = 10.764 × 10-6 ft2
1 m2 = 1550 in.2
1 cm2 = 0.155 in.2
1 mm2 = 0.155 × 10-2 in.2
Volume:
1 m3
1 cm3
1 m3
1 cm3
Force:
1N
= 0.2248 lb
1 kN
= 224.8 lb
1 kgf
= 2.2046 lb
1 kN
= 0.2248 kip
1 kN
= 0.1124 U.S. ton
1 metric ton = 2204.6 lb
1N/m
= 0.0685 lb / ft
Stress:
1 N / m2 = 20.885 × 10-3 lb / ft2
1 kN / m2 = 20.885 lb / ft2
1 kN / m2 = 0.01044 U.S. ton / ft2
1 kN / m2 = 20.885 × 10-3 kip / ft2
1 kN / m2 = 0.145 lb / in.2
= 35.32 ft3
= 35.32 × 10-4 ft3
= 61,023.4 in.3
= 0.061023 in.3
Unit weight: 1 kN / m3 = 6.361 lb / ft3
1 kN / m3 = 0.003682 lb / in.3
Moment:
1 N · m = 0.7375 lb-ft
1 N · m = 8.851 lb-in.
© 1999 by CRC Press LLC
CONVERSION FACTORS FROM
ENGLISH TO SI UNITS
Length:
1 ft
1 ft
1 ft
1 in.
1 in.
1 in.
= 0.3048 m
= 30.48 cm
= 304.8 mm
= 0.0254 m
= 2.54 cm
= 25.4 mm
Area
1 ft2 = 929.03 × 10-4 m2
1 ft2 = 929.03 cm2
1 ft2 = 929.03 × 102 mm2
1 in.2 = 6.452 × 10-4 m2
1 in.2 = 6.452 cm2
1 in.2 = 645.16 mm2
Volume:
1 ft3 = 28.317 × 10-3 m3
1 ft3 = 28.317 × 103 cm3
1 in.3 = 16.387 × 10-6 m3
1 in.3 = 16.387 cm3
Force:
1 lb
1 lb
1 lb
1 kip
1 U.S. ton
1 lb
1 lb / ft
Stress:
1 lb / ft2
= 47.88 N / m2
2
1 lb / ft
= 0.04788 kN / m2
1
1 U.S. ton / ft = 95.76 kN / m2
1 kip/ft2
= 47.88 kN / m2
2
1 lb/in.
= 6.895 kN / m2
= 4.448 N
= 4.448 × 10-3 kN
= 0.4536 kgf
= 4.448 kN
= 8.896 kN
= 0.4536 × 10-3 metric ton
= 14.593 N / m
Unit weight: 1 lb / ft3 = 0.1572 kN/m3
1 lb / in.3 = 271.43 kN/m3
Moment
1 lb-ft = 1.3558 N · m
1 lb-in. = 0.11298 N · m
© 1999 by CRC Press LLC
