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Boca Raton London New York Washington, D.C.CRC Press
SHALLOW FOUNDATIONSBearing Capacity and SettlementBearing Capacity and Settlement
Braja M. Das
PREFACE
Shallow Foundations: Bearing Capacity and Settlement is intended for use asa reference book by university faculty members and graduate students in thearea of geotechnical engineering as well as by consulting engineers.
The text is divided into eight chapters. Chapters 2, 3, and 4 present thevarious theories developed during the past fifty years for estimating the ultimatebearing capacity of shallow foundations under various types of loading andsubsoil conditions.
Chapter 5 discusses the principles for estimating the settlement offoundations—both elastic and consolidation. In order to calculate the founda-tion settlement, it is desirable to know the principles for estimating the stressincrease in a soil mass supporting a foundation which carries the load trans-mitted from the superstructure. These principles are also discussed in thisChapter 5. Recent developments regarding the ultimate bearing capacity ofshallow foundations due to earthquake loading are presented in Chapter 6. Alsoincluded in Chapter 6 are some details regarding the permanent founda-tionsettlement due to cyclic and transient loading derived from experimentalobservations obtained from laboratory and field tests.
During the past fifteen years, steady progress has been made to evaluatethe possibility of using reinforcement in soil to increase the ultimate andallowable bearing capacities of shallow foundations and also to reduce theirsettlement under various types of loading conditions. The reinforcementmaterials include galvanized steel strips, geotextile, and geogrid. Chapter 7presents the state-of-the-art on this subject.
Shallow foundations (such as transmission tower foundations) are, onsome occasions, subjected to uplifting forces. The theories relating to the esti-mation of the ultimate uplift capacity of shallow foundations in granular andclay soils are presented in Chapter 8.
Example problems to illustrate the theories are given in each chapter.I am grateful to my wife, Janice, for typing the manuscript in camera-
ready form and preparing the necessary artwork. It will be satisfying to knowfrom the users of the text if it serves the intended purpose.
Braja M. DasSacramento, California
To Janice and Valerie
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 Failure2.4 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 SoilReferences
THREE ULTIMATE BEARING CAPACITY UNDERINCLINED 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 Load3.6 Ultimate Load on Rectangular Foundations References
FOUR SPECIAL CASES OF SHALLOW FOUNDATIONS 4.1 Introduction4.2 Foundation Supported by a Soil With a Rigid
Rough Base at a Limited Depth 4.3 Foundation on Layered Saturated Anisotropic
Clay 4.4 Foundation on Layered c– Soil—Stronger Soil
Underlain by Weaker Soil 4.5 Foundation on Layered c– Soil—Weaker Soil
Underlain by a Stronger Soil 4.6 Continuous Foundation on Weak Clay With a
Granular Trench4.7 Shallow Foundations Above a Void4.8 Foundations on a Slope4.9 Foundations on Top of a Slope References
FIVE SETTLEMENT AND ALLOWABLE BEARINGCAPACITY 5.1 Introduction5.2 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 Area5.6 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 Clay5.10 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
ParametersReferences
SIX DYNAMIC BEARING CAPACITY AND SETTLEMENT 6.1 Introduction
( ! = 0) !
!
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 SoilDue 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 FoundationReferences
SEVEN SHALLOW FOUNDATIONS ON REINFORCED SOIL7.1 Introduction
FOUNDATIONS ON METALLIC STRIP-REINFORCED
GRANULAR SOIL
7.2 Failure Mode7.3 Force in Reinforcement Ties7.4 Factor of Safety Against Tie Breaking
and Tie Pullout7.5 Design Procedure for a Continuous Foundation
FOUNDATIONS ON GEOTEXTILE-REINFORCED SOIL
7.6 Laboratory Model Test Results 7.7 Comments on Geotextile Reinforcement
FOUNDATIONS ON GEOGRID-REINFORCED SOIL
7.8 General Parameters 7.9 Relationships for Critical Nondimensional
Parameters for Foundations on Geogrid-Reinforced Sand
7.10 Relationship Between BCRu and BCRs in Sand 7.11 Critical Nondimensional Parameters for
Foundations on Geogrid-Reinforced Clay ( = 0 condition)
7.12 Bearing Capacity Theory 7.13 Settlement of Foundations on Geogrid-
Reinforced Soil Due to Cyclic Loading 7.14 Settlement Due to Impact Loading References
EIGHT UPLIFT CAPACITY OF SHALLOW FOUNDATIONS 8.1 Introduction
FOUNDATIONS IN SAND
8.2 Balla’s Theory8.3 Theory of Meyerhof and Adams
!
8.4 Theory of Vesic 58.5 Saeedy’s Theory 8.6 Discussion of Various Theories 8.7 Effect of Backfill on Uplift Capacity
FOUNDATIONS IN SATURATED CLAY
( = 0 CONDITION) 8.8 Ultimate Uplift Capacity—General 8.9 Vesic’s Theory 8.10 Meyerhof’s Theory 8.11 Modifications to Meyerhof’s Theory 8.12 Factor of SafetyReferences
APPENDIX Conversion Factors
!
FIGURE 1.1 Individual footing
CHAPTER 1
INTRODUCTION
1.1 SHALLOW FOUNDATIONS—GENERAL
The lowest part of a structure which transmits its weight to the underlying soilor rock is the foundation. Foundations can be classified into two major cate-gories: that is, shallow foundations and deep foundations. Individual footings(Fig. 1.1) square or rectangular in plan which support columns, and stripfootings which support walls and other similar structures are generally referredto as shallow foundations. Mat foundations, also considered shallow founda-tions, are reinforced concrete slabs of considerable structural rigidity whichsupport a number of columns and wall loads. When the soil located immediate-ly below a given structure is weak, the load of the structure may be transmittedto a greater depth by piles and drilled shafts, which are considered deepfoundations. This book is a compilation of the theoretical and experimentalevaluations presently available in literature as they relate to the load-bearingcapacity and settlement of shallow foundations.
FIGURE 1.2 General shear failure in soil
The shallow foundation shown in Fig. 1.1 has a width B and a length L.The depth of embedment below the ground surface is equal to Df . Theoreti-cally, when B/L is equal to zero (that is, L = !), a plane strain case will exist inthe 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 orequal to the width B (Df /B " 1). However, research studies conducted sincethen 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 belowthe ground surface and supported by a dense sand (or stiff clayey soil). If this
FIGURE 1.3 Local shear failure in soil
foundation is subjected to a load Q which is gradually increased, the load perunit area, q = Q/A ( A = area of the foundation), will increase and the foun-dation will undergo increased settlement. When q becomes equal to qu atfoundation settlement S = Su , the soil supporting the foundation undergoessudden shear failure. The failure surface in the soil is shown in Fig. 1.2a, andthe q versus S plot is shown in Fig. 1.2b. This type of failure is called generalshear failure, and qu is the ultimate bearing capacity. Note that, in this type offailure, 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 sandor clayey soil of medium consistency (Fig. 1.3a), the plot of q versus S will beas shown in Fig. 1.3b. Note that the magnitude of q increases with settlementup to q = qu , which is usually referred to as the first failure load [2]. At this
time, the developed failure surface in the soil will be like that shown by thesolid lines in Fig. 1.3a. If the load on the foundation is further increased, theload-settlement curve becomes steeper and erratic with the gradual outward andupward progress of the failure surface in the soil (shown by the broken line inFig. 1.3b) under the foundation. When q becomes equal to qu (ultimate bearingcapacity), the failure surface reaches the ground surface. Beyond that, the plotof 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 softclayey soil. For this case, the load-settlement curve will be like that shown inFig. 1.4b. A peak value of load per unit area, q, is never observed. The ultimatebearing capacity, qu , is defined as the point where � ∆S/�∆ q becomes the largest
FIGURE 1.4 Punching shear failure in soil
FIGURE 1.5 Nature of failure in soil with relative density of sand (Dr) and Df /R
and almost constant thereafter. This type of failure in soil is called punchingshear failure. In this case, the failure surface never extends up to the groundsurface.
The nature of failure in soil at ultimate load is a function of several factorssuch as the strength and the relative compressibility of soil, the depth of thefoundation (Df ) in relation to the foundation width (B), and the width-to-lengthratio (B/L) of the foundation. This was clearly explained by Vesic [2] whoconducted extensive laboratory model tests in sand. The summary of Vesic’sfindings is shown in a slightly different form in Fig. 1.5. In this figure, Dr is therelative density of sand, and the hydraulic radius, R, of the foundation isdefined as
RAP
= (1.1)
RBL
B L=
+2( )
RB=4
where A = area of the foundation = BLP = perimeter of the foundation = 2(B + L)
Thus
(1.2)
For a square foundation, B = L. So
(1.3)
From Fig. 1.5 it can be seen that, when Df /R # about 18, punching shear failure occurs in all cases, irrespective of the relative density of compaction of
1.3 SETTLEMENT AT ULTIMATE LOAD
The settlement of the foundation at ultimate load, Su , is quite variable anddepends on several factors. A general sense can be derived from the laboratorymodel 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, forany given foundation, a decrease in the relative density of sand results in anincrease in the settlement at ultimate load.
Based on laboratory and field test results, the approximate ranges of valuesof Su in various types of soil are given below.
SoilD
Bf S
Bu (%)
SandSandClayClay
0Large0Large
5 to 1225 to 28 4 to 815 to 20
1.4 ULTIMATE AND ALLOWABLE BEARING CAPACITIES
For a given foundation to perform to its optimum capacity, one must ensurethat 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 bearingcapacity, q Considering the ultimate bearing capacity and the uncertainties
sand.
FSu
all =
FIGURE 1.6 Variation of Su /B for surface foundation (Df /B) on sand (after Vesic [3])
involved in evaluating the shear strength parameters of the soil, the allowablebearing capacity, qall , can be obtained as
(1.4)
A factor of safety of 3 to 4 is generally used. However, based on limitingsettlement conditions, there are other factors which must be taken into accountin deriving the allowable bearing capacity. The total settlement, St , of afoundation will be the sum of the following:
1. Elastic or immediate settlement, S (described in Section 1.3), and2. Primary and secondary consolidation settlement, Sc , of a clay layer
(located below the ground water level) if located at a reasonably smalldepth below the foundation.
Most building codes provide an allowable settlement limit for a foundation
FIGURE 1.7 Settlement of a structure
which may be well below the settlement derived corresponding to qall given byEq. (1.4). Thus, the bearing capacity corresponding to the allowable settlementmust also be taken into consideration.
A given structure with several shallow foundations may undergo uniformsettlement (Fig. 1.7a). This occurs when a structure is built over a very rigidstructural mat. However, depending on the load on various foundationcomponents, a structure may experience differential settlement. A foundationmay undergo uniform tilt (Fig. 1.7b) or nonuniform settlement (Fig. 1.7c). Inthese cases, the angular distortion, ∆, can be defined as
∆ =−
′
S S
Lt t(max) (min) (
1
for nonuniform settlement)
(1.5)
and
(1.6)
Limits for allowable differential settlement of various structures are alsoavailable in building codes. Thus, the final decision on the allowable bearingcapacity of a foundation will depend on (a) the ultimate bearing capacity, (b)the allowable settlement, and (c) the allowable differential settlement for thestructure.
REFERENCES
1. Terzaghi, K., Theoretical Soil Mechanics. Wiley, New York, 1943.2. Vesic, A. S., Analysis of ultimate loads of shallow foundations. J. Soil Mech.
Found. Div., ASCE, 99(1), 45, 1973.3. Vesic, A. S., Bearing capacity of deep foundations in sand. Highway Res. Rec.
39, National Research Council, Washington, D.C.,112, 1963.
∆ =−
′
S S
Lt t(max) (min) (for uniform tilt)
r r e= 0θ φtan
CHAPTER TWO
ULTIMATE BEARING CAPACITY THEORIES –CENTRIC VERTICAL LOADING
2.1 INTRODUCTION
During the last fifty years, several bearing capacity theories were proposed forestimating the ultimate bearing capacity of shallow foundations. This chaptersummarizes some of the important works developed so far. The cases consid-ered in this chapter assume that the soil supporting the foundation extends toa great depth and also that the foundation is subjected to centric verticalloading. The variation of the ultimate bearing capacity in anisotropic soils isalso considered.
2.2 TERZAGHI’S BEARING CAPACITY THEORY
In 1948, Terzaghi [1] proposed a well-conceived theory to determine the ultimatebearing capacity of a shallow rough rigid continuous (strip) foundationsupported by a homogeneous soil layer extending to a great depth. Terzaghidefined a shallow foundation as a foundation where the width, B, is equal to orless than its depth, Df . The failure surface in soil at ultimate load (that is, qu , perunit area of the foundation) assumed by Terzaghi is shown in Fig. 2.1. Referringto Fig. 2.1, the failure area in the soil under the foundation can be divided intothree major zones. They are:
1. Zone abc. This is a triangular elastic zone located immediately belowthe bottom of the foundation. The inclination of sides ac and bc of thewedge 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 alsolocated to the left of the elastic triangular zone abc; however, they are notshown in Fig. 2.1.
Line cf is an arc of a log spiral, defined by the equation
(2.1)
Lines bf and fg are straight lines. Line fg actually extends up to the groundsurface. Terzaghi assumed that the soil located above the bottom of thefoundation could be replaced by a surcharge q = (Df .
FIGURE 2.1 Failure surface in soil at ultimate load for a continuous rough rigid foundation as assumed by Terzaghi
s c= ′ +σ φtan
φ+==
245tan 2
)1( ddppqHHqKP
The shear strength, s, of the soil can be given as
(2.2)
where F´ = effective normal stress c = cohesionThe ultimate bearing capacity, qu , of the foundation can be determined if
we consider faces ac and bc of the triangular wedge abc and obtain the passiveforce on each face required to cause failure. Note that the passive force Pp willbe a function of the surcharge q = (Df , cohesion c, unit weight (, and angle offriction of the soil N. So, referring to Fig. 2.2, the passive force Pp on the facebc 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 , andPp( , the method of superposition can be used; however, it will not be an exactsolution.
Relationship for Ppq (NN …… 0, (( = 0, q …… 0, c = 0)
Consider the free body diagram of the soil wedge bcfj shown in Fig. 2.2 (alsoshown in Fig. 2.3). For this case the center of the log spiral, of which cf is an arc,will be at point b. The forces per unit length of the wedge bcfj due to thesurcharge q only are shown in Fig. 2.3a, and they are:
1. Ppq
2. Surcharge, q3. The Rankine passive force, Pp(1)
4. The frictional resisting force along the arc cf, FThe Rankine passive force, Pp(1) , can be expressed as
(2.4)
where Hd = bbf j Kp = Rankine passive earth pressure coefficient = tan2(45 + N/2)According to the property of a log spiral defined by the equation r =
r0e2tanN
, 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 thelog spiral (as shown in Fig. 2.3a). Taking the moment about point b
FIGURE 2.2 Passive force on the face bc of wedge abc shown in Fig. 2.1
FIGURE 2.3 Determination of Ppq (N… 0, (= 0, q …0, c' 0)
22)(
4 )1(
d
ppq
HP
bjbjq
Bp +
=
φ
== sec20
Brbc
(2.5)
Let
(2.6)
From Eq. (2.1)
φ
−=2
45cos1
rbj
φ−=
245sin
1rH
d
22
45tan2
45sin
22
45cos
4
222
1
22
1
φ+
φ−
+
φ−
=qrqr
BPpq
φ
−=2
45cos4 22
1qr
BP
pq
φ+
=
φ−
φ=
φ
φ−
π
φ
φ−
π
245cos4
245cossec
2
tan24
32
2tan
24
32
2
qBe
eqBPpq
(2.7)
So
(2.8)
and
(2.9)
Combining Eqs. (2.4), (2.5), (2.8), and (2.9)
or
(2.10)
Now, combining Eqs.(2.6), (2.7), and (2.10)
(2.11)
Considering the stability of the elastic wedge abc under the foundation asshown in Fig.2.3b
φ
φ
−π
==tan
24
3
01errbf
q
qN
pq
qqN
eq
B
Pq =
φ+
==φ
φ−
π
444 3444 212
45cos2
2
2
tan24
32
φ
+==2
45tan22)2( ddppcHHKcP
cppcM
r
PB
P +
φ
−=
2
245sin
4
1
)2(
qq (B × 1) = 2Ppq
where qq = load per unit area on the foundation, or
(2.12)
Relationship for Ppc (NN …… 0, (( = 0, q = 0, c …… 0)
Figure 2.4 shows the free body diagram for the 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 atpoint b. The forces on the wedge which are due to cohesion c are also shownin Fig.2.4, and they are
1. Passive force, Ppc
2. Cohesive force, C c bc= ×( )1
3. Rankine passive force due to cohesion,
4. Cohesive force per unit area along arc cf, c.Taking the moment of all the forces about point b
(2.13)
where Mc = moment due to cohesion c along arc cf
= (2.14)c
r r2 1
202
tan( )
φ−
So
FIGURE 2.4 Determination of Ppc (N… 0, ( = 0, q = 0, c … 0)
)(tan2
2
245sin
245tan2
4
2
0
2
1
1
rrc
r
cHB
Pdpc
−
φ
+
φ−
φ+=
(2.15)
The relationships for Hd , r0 , and r1 in terms of B and N are given in Eqs.(2.9), (2.6), and (2.7), respectively. Combining Eqs. (2.6), (2.7), (2.9), and (2.15),and noting that sin2(45 ! N/2) × tan(45 + N/2) = ½cosN
φ
φ
+
φ
φ=
φ
φ−π
φ
φ−π
tan24
32
2
tan24
32
2
sectan2
2
cos)(sec
eBc
eBcPpc
φ+φφ−
φφ+φ=
φ
φ
−π
φ
φ
−π
tantan
sec
tan
secsec
2
tan24
322tan
243
2
cc
ec
ecqc
φ−
φφ
−
φφ
+φ=φ
φ−π
tantan
sec
tan
secsec
22tan24
32
cceqc
φ+
φ=
φφ+φ=
φφ+
φ=
φφ+φ
245cos2
1cot
cos
sin1cot
sincos
1
cos
1
tan
secsec
2
2
2
(2.16)
Considering the equilibrium of the soil wedge abc (Fig. 2.4b)
qc (B × 1) = 2C sinN + 2Ppc
or
qc B = cB secN sinN + 2Ppc (2.17)
where qc = load per unit area of the foundationCombining Eqs. (2.16) and (2.17)
(2.18)
or
(2.19)
However
(2.20)
Also
φ=
φφφ=
φφ−
φφ=
φ−φφ=φ−φφ
cotcos
coscot
cos
sin
cos
1cot
)tan(seccottantan
sec
2
2
2
2
2
222
)1(cot1
245cos2
cot2
tan24
32
−φ==
−
φ+
φ=φ
φ−π
qc
cN
cNccN
ecq
4444 34444 21
φ+γ=
245tan
2
1 22
)3( dpHP
P l Wl P lp p w p Rγ = + ( )3
(2.21)
Substituting Eqs. (2.20) and (2.21) into Eq. (2.19)
(2.22)
Relationship for Pp(( (NN …… 0, (( …… 0, q = 0, c = 0)
Figure 2.5a shows the free body diagram of wedge bcfj. Unlike the free bodydiagrams shown in Figs. 2.3 and 2.4, the center of the log spiral of which bf is anarc is at a point O along line bf and not at b. This is because the minimum valueof Pp( has to be determined by several trials. Point O is only one trial center. Theforces per unit length of the wedge that need to be considered are:
1. Passive force, Pp(
2. The weight of wedge bcfj, W3. The resultant of the frictional resisting force acting along arc cf, F4. The Rankine passive force, Pp (3)
The Rankine passive force Pp (3) can be given by the relation
(2.23)
Also note that the line of action of force F will pass through O. Taking themoment about O
FIGURE 2.5 Determination of Pp( (N… 0, (… 0, q = 0, c = 0)
Pl
Wl P lpp
w p Rγ = +1
3( )
or
(2.24)
If a number of trials of this type are made by changing the location at the centerof the log spiral O along line bf, then the minimum value of Pp( can be deter-mined.
Considering the stability of wedge abc as shown in Fig. 2.5, we can writethat
q( B = 2Pp( !Ww (2.25)
φγ−=
φγ=
γγ tan4
21
tan4
2
2
BP
Bq
BW
p
w
So
φγ=
φγ=γ= γγγγ
22
2
2 tan8
1
2
tan
2
1
2
1pppp
KBKB
KhP
γ
γ
γ
γγ
γ=
φ−φγ=
φγ−φγ=
BNKB
BKB
Bq
N
p
p
2
1
2
tantan
2
1
2
1
tan4
tan4
11
2
222
4444 34444 21
q cN qN BNu c q= + +1
2γ γ
where q( = force per unit area of the foundationWw = weight of wedge abc
However,
(2.26)
(2.27)
The passive force Pp( can be expressed in the form
(2.28)
where Kp( = passive earth pressure coefficientSubstituting Eq. (2.28) into Eq. (2.27)
(2.29)
Ultimate Bearing Capacity
The ultimate load per unit area of the foundation (that is, the ultimate bearingcapacity qu ) for a soil with cohesion, friction, and weight can now be given as
qu = qq + qc + q( (2.30)
Substituting the relationships for qq , qc , and q( given by Eqs. (2.12), (2.22), and(2.29) into Eq. (2.30) yields
(2.31)
where Nc , Nq , and N( = bearing capacity factors, and
φ+
=φ
φ−π
245cos2 2
tan24
32
eN
q
N K pγ γ φ φ= −1
2 22tan
tan
N
N
c
q
= +−
= +−
228 4 3
40
40 5
40
. φφ
φφ
(2.32)
Nc = cot!(Nq!1) (2.33)
(2.34)
Table 2.1 gives the variations of the bearing capacitiy factors with soilfriction angle ! given by Eqs. (2.32), (2.33), and (2.34). The values of N" wereobtained 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"
0123456789
10111213141516
5.70 6.00 6.30 6.62 6.97 7.34 7.73 8.15 8.60 9.09 9.6110.1610.7611.4112.1112.8613.68
1.001.11.221.351.491.641.812.002.212.442.692.983.293.634.024.454.92
0.000.010.040.060.100.140.200.270.350.440.560.690.851.041.261.521.82
1718192021222324252627282930313233
14.6015.1216.5717.6918.9220.2721.7523.3625.1327.0929.2431.6134.2437.1640.4144.0448.09
5.456.046.707.448.269.19
10.2311.4012.7214.2115.9017.8119.9822.4625.2828.5232.23
2.18 2.59 3.07 3.64 4.31 5.09 6.00 7.08 8.34 9.8411.6013.7016.1819.1322.6526.8731.94
3435363738394041424344454647484950
52.6457.7563.5370.0177.5085.9795.66
106.81119.67134.58151.95172.28196.22224.55258.28298.71347.50
36.5041.4447.1653.8061.5570.6181.2793.85
108.75126.50147.74173.28204.19241.80287.85344.63415.14
38.0445.4154.3665.2778.6195.03
115.31140.51171.99211.56261.60325.34407.11512.84650.87831.99
1072.80
Krizek [3] gave simple empirical relations for Terzaghi’s bearing capacityfactors Nc , Nq , and N" with a maximum deviation of 15%. They are as follows:
(2.35a)
(2.35b)
φ−φ
=γ40
6N
qc q B
u = + + + +−
° °( . ) ( )228 4 3 40 5 340
φ φ φγφ
φ(for = 0 to 35 )
(2.35c)where N = soil friction angle, in degrees
Equations (2.35a), (2.35b), and (2.35c) are valid for N = 0 to 35E. Thus,substituting Eqs. (2.35) into (2.31)
(2.36)
For foundations that are rectangular or circular in plan, a plane straincondition in soil at ultimate load does not exist. Therefore, Terzaghi [1]proposed the following relationships for square and circular foundations.
qu = 1.3cNc + qNq + 0.4(BN( (square foundation; plan B × B) (2.37)
and
qu = 1.3cNc + qNq + 0.3(BN( (circular foundation; plan B × B) (2.38)
Since Terzaghi’s founding work, numerous experimental studies to estimatethe ultimate bearing capacity of shallow foundations have been conducted.Based on these studies, it appears that Terzaghi’s assumption of the failuresurface in soil at ultimate load is essentially correct. However, the angle " thatthe sides ac and bc of the wedge (Fig. 2.1) make with the horizontal is closer to45 + N/2, and not N as assumed by Terzaghi. In that case, the nature of the soilfailure surface would be as shown in Fig. 2.6.
The method of superposition was used to obtain the bearing capacityfactors, Nc , Nq , and N( . For derivation of Nc and Nq , the center of the arc of thelog spiral cf is located at the edge of the foundation. However, for derivation ofN( , 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 THEORYFOR LOCAL SHEAR FAILURE
It is obvious from Section 2.2 that Terzaghi’s bearing capacity theory wasobtained by assuming general shear failure in soil. However, the local shearfailure in soil, Terzaghi [1] suggested the following relationships:
FIGURE 2.6 Modified failure surface in soil supporting a shallow foundation at ultimate load
k D D Dr r r= + − ≤ ≤0 67 0 75 2. . (for 0 0.67)
Strip foundation (B/L = 0; L = length of foundation)
qu = c´Nc + q´Nq + –12!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 factorsc´ = 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!
0123456789
10111213141516
5.705.906.106.306.516.746.977.227.477.748.028.328.638.969.319.67
10.06
1.001.071.141.221.301.391.491.591.701.821.942.082.222.382.552.732.92
0.000.0050.020.040.0550.0740.100.1280.160.200.240.300.350.420.480.570.67
1718192021222324252627282930313233
10.4710.9011.3611.8512.3712.9213.5114.1414.8015.5316.0317.1318.0318.9920.0321.1622.39
3.133.363.613.884.174.484.825.205.606.056.547.077.668.319.039.82
10.69
0.760.881.031.121.351.551.741.972.252.592.883.293.764.394.835.516.32
3435363738394041424344454647484950
23.7225.1826.7728.5130.4332.5334.8737.4540.3343.5447.1351.1755.7360.9166.8073.5581.31
11.6712.7513.9715.3216.8518.5620.5022.7025.2128.0631.3435.1139.4844.5450.4657.4165.60
7.228.359.41
10.9012.7514.7117.2219.7522.5026.2530.4036.0041.7049.3059.2571.4585.75
Vesic [4] suggested a better mode to obtain "´ for estimating Nc and Nq forfoundations on sand in the form
"´ = tan!1(k tan") (2.42)
(2.43)
FIGURE 2.7 Slip line fields for a rough continuous foundation
where Dr = relative density
2.4 MEYERHOF’S BEARING CAPACITY THEORY
In 1951, Meyerhof published a bearing capacity theory which could be appliedto rough shallow and deep foundations. The failure surface at ultimate loadunder a continuous shallow foundation assumed by Meyerhof [5] is shown inFig. 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 amixed shear zone in which the shear varies between the limits of radial andplane shear, depending on the depth and roughness of the foundation. Theplane be is called an equivalent free surface. The normal and shear stresses onplane be are po and so , respectively. The superposition method is used todetermine the contribution of cohesion, c; po ; and ! and " on the ultimatebearing capacity, qu , of the continuous foundation and is expressed as
qu = cN c + qN q + –12!BN ! (2.44)
where Nc , Nq , and N! = bearing capacity factorsB = width of the foundation
Rs
= 1
cosφ
s Rs
o = + =+
cos( )cos( )
cos2
21η φη φφ
cos( )cos
tan
( tan ) cos
tan2
1 1
η φφ
φφ φ
φ+ =
+=
++
s
c p
m c p
c po o
Rc p
=+ 1 tan
cos
φφ
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 isassumed that, along be,
so = m(c + po tan") (2.45)
where c = cohesion" = soil friction anglem = degree of mobilization of shear strength (0 # m #1)
Now, consider the linear zone bde (Fig. 2.8a). Plastic equilibrium requiresthat 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 onzone bde. Note that P is the pole. The traces of planes bd and be are also shownin the figure. For the Mohr’s circle
(2.47)
where R = radius of the Mohr’s circleAlso
(2.48)
Combining Eqs. (2.45), (2.46), and (2.48)
(2.49)
Again, referring to the trace of plane de (Fig. 2.8c)
s1 = R cos"
(2.50)
Note that
FIGURE 2.8 Determination of Nq and Nc
FIGURE 2.8 (Continued)
FIGURE 2.8 (Continued)
p R p R
p R p
c pp
o
o
o
1
1
1
2
2
2
+ = + +
= + − +
=+
+ − +
sin sin( )
[sin( ) sin ]
tan
cos[sin( ) sin ]
φ η φ
η φ φ
φφ
η φ φ
022
2
0
2
11
=+
′−
cp
Mr
pr
p
r bc0 =
(2.51)
Figure 2.8d shows the free body diagram of zone bcd. Note that the normaland shear stresses on the face bc are pp and sp , or
sp = c + pp tan"
orpp = (sp ! c)cot" (2.52)
Taking the moment about b
(2.53)
where
r bd r eo1 = = θ φtan
Mc
r rc = −2 1
202
tan( )
φ
Bq
B
s
B
ppp
′=
φ+
φ+
′+
φ+
φ+
′2
45sin
245cos
222
45cos
245cos
22
φ−′+′=′
245cot
ppspq
qoc
qN
o
cN
NpcN
ep
ecq
+=
φ+ηφ−
φ++
φ+ηφ−
φ+φ=′φθφθ
444 3444 21444 3444 21)2sin(sin1
)sin1()2sin(sin1
)sin1(cot
tan2tan2
It can be shown that
(2.55)
Substituting Eqs. (2.54) and (2.55) into Eq. (2.53) yields
pp = p1 e2#tan" + c cot"(e2#tan" ! 1) (2.56)
Combining Eqs. (2.52) and (2.56)
sp = (c + p1 tan")e2#tan" (2.57)
Figure 2.8e shows the free body diagram of wedge abc. Resolving theforces in the vertical direction
where q´ = load per unit area of the foundation, or
(2.58)
Substituting Eqs. (2.51), (2.52), and (2.57) into Eq. (2.58), and furthersimplifying yields
(2.59)
(2.55)
cos( )( tan ) cos
tan2
1
η φφ φ
φ+ =
++
m c p
c po
φ−φ+= φπ
sin1sin1taneN
q
where Nc , Nq = bearing capacity factorsThe bearing capacity factors will depend on the degree of mobilization, m,
of shear strength on the equivalent free surface. This is because m controls η.From Eq. (2.49)
for m = 0, cos(2$ + ") = 0, or
$ = 45 ! –"2
(2.60)
For m = 1, cos(2$ + ") = cos", or
$ = 0 (2.61)
Also, the factors Nc and Nq are influenced by the angle of inclination of theequivalent free surface %. From the geometry of Fig. 2.7
# = 135$ + % ! $ ! –"2
(2.62)
From Eq. (2.60), for m = 0, the value of $ is (45!"/2). So
# = 90 + % (for m = 0) (2.63)
Similarly, for m = 1, since $ = 0 [Eq. (2.61)]
# = 135$ + % ! –"2 (for m = 1) (2.64)
Figures 2.9 and 2.10 show the variation of Nc and Nq with ", %, and m. Itis of interest to note that, if we consider the surface foundation condition (asdone 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 areas follows (surface foundation condition)
(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 (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
FIGURE 2.11 Determination of Ng
q´ = cNc + qNq (2.68) % %Eq. (2.66) Eq. (2.67)
PW l P l
lpw p R R
pγ =
+ ( )
γ
γ
γ=
φ+−
γ
φ+
γ=′′ BNB
PB
qp
2
1
245tan
2
1245sin4
2 2
Note that Ww is the weight of wedge abc in Fig. 2.11b.
Derivation of N!!!! ("""" """" 0, !!!! """" 0, po = 0, c = 0)
N! is determined by trial and error as in the case of the derivation of Terzaghi’s bearing capacity factor N! (Section 2.2). Referring to Fig. 2.11a,following is a step-by-step approach for the derivation of N! .
1. Choose values for " and the angle % (such as +30$, + 40$, !30$, ...).2. Choose a value for m (such as, m = 0 or m = 1).3. Determine the value of # from Eqs. (2.63) or (2.64) for m = 0 or m
= 1, as the case may be.4. With known values of # and %, draw lines bd and be.5. Select a trial center such as O and draw an arc of a log spiral
connecting points c and d. The log spiral follows the equation r =r0e
#tan".6. Draw line de. Note that lines bd and de make angles of 90 ! " due to
the restrictions on slip lines in the linear zone bde. Hence the trialfailure surface is not, in general, continuous at d.
7. Consider the trial wedge bcdf. Determine the following forces perunit length of the wedge at right angles to the cross section shown:(a) weight of wedge bcdf —W, and (b) Rankine passive force on theface df —Pp(R) .
8. Take the moment of the forces about the trial center of the log spiralO, or
(2.69)
where Pp! = passive force due to !Note that the line of action of Pp! acting on the face bc is located ata distance of 2bc/3.
9. For given values of %, ", and m, and by changing the location ofpoint O (that is, the center of the log spiral), repeat Steps 5 through 8to obtain the minimum value of Pp!.
10. Refer to Figure 2.11b. Resolve the forces acting on the triangularwedge abc in the vertical direction, or
(2.70)
where q&& = force per unit area of the foundation N ! = bearing capacity factor
and f only.
FIGURE 2.12 Meyerhof’s bearing capacity factor—variation of with b, f, and m [Eq. (2.70)]
The variation of Ng (determined in the above manner) with b, f, and m is givenin Fig. 2.12.
Combining Eqs. (2.59) and (2.70), the ultimate bearing capacity of a con-tinuous foundation (for the condition c " 0, ! " 0, and " " 0) can be given as
N!
qu = q´ + q&& = cNc + po Nq + –12!BN!
The above equation is the same form as Eq. (2.44). Similarly, for surfacefoundation conditions (that is, % = 0 and m =0), the ultimate bearing capacityof a continuous foundation can be given as
qu = q´ + q! = cNc + qNq + –12!BN! (2.71)
% % % % Eq. (2.68) Eq. (2.70) Eq. (2.67) Eq. (2.66)
For shallow foundation design, the ultimate bearing capacity relationship givenby Eq. (2.71) is presently used. The variation of N! for surface foundationconditions (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)]
\
0123456789
10111213141516
5.145.385.635.906.196.496.817.167.537.928.358.809.289.81
10.3710.9811.63
1 001.091.201.311.431.571.721.882.062.252.472.712.973.263.593.944.34
0.000.0020.010.020.040.070.110.150.210.280.370.470.600.740.92 1.131.38
1718192021222324252627282930313233
12.3413.1013.9314.8315.8216.8818.0519.3220.7222.2523.9425.8027.8630.1432.6735.4938.64
4.775.265.806.407.077.828.669.60
10.6611.8513.2014.7216.4418.4020.6323.1826.09
1.662.002.402.873.424.074.825.726.778.009.46
11.1913.2415.6718.5622.0226.17
3435363738394041424344454647484950
42.1646.1250.5955.6361.3567.8775.3183.8693.71
105.11118.37133.88152.10173.64199.26229.93266.89
29.4433.3037.7542.9248.9355.9664.2073.9085.3899.02
115.31134.88158.51187.21222.31265.51319.07
31.1537.1544.4353.2764.0777.3393.69
113.99139.32171.14211.41262.74328.73414.32526.44674.91873.84
" Nc Nq N ! " Nc Nq N ! " Nc Nq N !
φ−φ+= φπ
sin1
sin1taneNq
2.5 GENERAL DISCUSSION ON THE RELATIONSHIPSOF BEARING CAPACITY FACTORS
At this time, the general trend among geotechnical engineers is to accept themethod of superposition as a proper means to estimate the ultimate bearingcapacity of shallow rough foundations. For rough continuous foundations, thenature 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 , whichare the same as Meyerhof’s [5] solution for surface foundations. Or
(2.66)
and
Nc = (N q ! 1)cot! (2.67)
There has been considerable 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’ssolution [Eq. (2.67)]. Caquot and Kerisel [10] assumed that the elastic tri-angular soil wedge under a rough continuous foundation to be of the shapeshown in Fig. 2.6. Using integration of Boussinesq’s differential equation, theypresented 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] developednumerical methods (using the theory of plasticity) for exact determination ofrupture 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 analysistheorem suggested by Drucker and Prager [13]. Biarez et al. [14] alsorecommended the following relationship for N"
N" = 1.8(Nq ! 1)tan! (2.75)
Soilfrictionangle, !
(deg)
N"
Terzaghi[Eq. (2.34)]
Meyerhof[Eq. (2.72)]
Vesic[Eq. (2.74)]
Hansen[Eq. (2.73)]
0 1 2 3 4 5 6 7 8 9101112131415161718192021222324252627282930313233343536373839404142434445
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 95.03115.31140.51171.99211.56261.60325.34
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 93.69113.99139.32171.14211.41262.74
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 92.25109.41130.22155.55186.54224.64271.76
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 95.05113.95137.10165.58200.81
Table 2.4 gives a comparison of the N" values recommended by Meyerhof[8], Terzaghi [1], Caquot and Kerisel [10], Vesic [4], and Hansen [9].
TABLE 2.4 Comparison of N"""" Values
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 arepresentative value of the soil friction angle ! for computing the bearingcapacity. The parameter ! depends on many factors, such as intermediate prin-cipal stress condition, friction angle anisotropy, and curvature of the Mohr-
tL
B φ
−=φ 1.01.1
φ α φ< < +min 452
Coulomb failure envelope. Ingra and Baecher [15] compared the theoreticalsolutions of N" with the experimental results obtained by several investigatorsfor foundations with B/L = 1 and 6 (B = width and L = length of the founda-tion). It was noticed that, when triaxial friction angles were used to deduceexperimental N" , their values were substantially higher than those obtainedtheoretically. 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 angleIt was previously suggested that the plane strain soil friction angle !p
instead of !t be used to estimate the bearing capacity [9]. To that effect, Vesic[4] raised the issue that this type of assumption might help explain the dif-ferences between the theoretical and experimental results for long rectangularfoundations. However, it does not help to interpret results of tests with squareor circular foundations. Ko and Davidson [16] also concluded that when planestrain angles of internal friction are used in commonly accepted bearing capa-city formulas, the bearing capacity for rough footings could be seriously over-estimated for dense sands. To avoid the controversy, Meyerhof [8] suggestedthe following:
(2.80)
2.6 OTHER BEARING CAPACITY THEORIES
Hu [17] proposed a theory according to which the base angle, #, of thetriangular wedge below the foundation (see Fig. 2.1) is a function of severalparameters, or
# = f (", !, q) (2.81)
The minimum and maximum values of # can be given as follows
FIGURE 2.15 Hu’s bearing capacity factors
α φmax = +45
2
and
The values of Nc , Nq , and N" determined by this procedure are shown in Fig.2.15.
FIGURE 2.16 Nature of failure surface considered for Balla’s [18] bearing capacity theory
Balla [18] proposed a bearing capacity theory which was developed for an as-sumed failure surface in soil (Fig. 2.16). For this failure surface, the curve cdwas assumed to be an arc of a circle having a radius r. The bearing capacitysolution was obtained using Kötter’s equation to determine the distribution ofthe normal and tangential stresses on the slip surface. According to this solutionfor a continuous foundation
qu = cNc + qNq + –12"Bn"
The bearing capacity factors can be determined as follows:1. Obtain the magnitude of c/B" and Df /B.2. With the values obtained in Step 1, go to Fig. 2.17 to obtain the
magnitude of $ = 2r/B.3. With known values of $, go to Figs. 2.18, 2.19, and 2.20 respectively
to determine Nc , Nq , and N" .
2.7 SCALE EFFECTS ON BEARING CAPACITY
The problem of estimating the ultimate bearing capacity becomes complicatedif the scale effect is taken into consideration. The scale effect, which has cometo the limelight more recently, shows that the ultimate bearing capacity de-creases with the increase in the size of the foundation. This condition is morepredominant in granular soils. Figure 2.21 shows the general nature of thedecrease in N" with the increase in foundation width B. The magnitude of N"
initially decreases with B and remains almost constant for larger values of B.The reduction in N" for larger foundations may ultimately result in a sub-stantial decrease in the ultimate bearing capacity which can primarily be attri-buted to the following reasons.
FIGURE 2.17 Variation of $ with soil friction angle for determination of Balla’s bearing capacity factors
FIGURE 2.18 Balla’s bearing capacity factor Nc
1. For larger-sized foundations, the rupture along the slip lines in soil isprogressive, and the average shear strength mobilized (and so !)along a slip line decreases with the increase in B.
2. There are zones of weakness which exist in the soil under thefoundation.
3. The curvature of the Mohr-Coulomb envelope.
2.8 EFFECT OF WATER TABLE
The preceding sections assume that the water table is located below the failuresurface in the soil supporting the foundation. However, if the water table ispresent 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-
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 depthd 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" shouldbe changed to "´.
Case II– 0 < d #### Df
For this case, q will be equal to "d + (Df ! d)"´, and the term " associated withN" should be changed to "´.
FIGURE 2.20 Balla’s bearing capacity factor N"
FIGURE 2.21 Nature of variation of N" with B
FIGURE 2.22 Effect of ground water table on ultimate bearing capacity
)( γ′−γ
−+γ′=γ
B
Ddf
Case III– Df #### d#### B
This condition is one in which the ground water table is located at or below thebottom of the foundation. In such case, q = "Df and the last term " should bereplaced by an average effective unit weight of soil, !", or
(2.82)
Case IV– d > Df + B
For d > Df + B, q = !Df and the last term should remain !. This implies thatthe 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 thepreceding 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 ofthe foundation on the increase in the ultimate bearing capacity. Therefore, ageneral bearing capacity may be written as
qu = cNc "cs "cd + qNq "qs "qd + –12!BN! "!s "!d (2.83)
where "cs , "qs , "!s = shape factors "cd , "qd , "!d = depth factors
Most of the shape and depth factors available in literature are empirical and/orsemi-empirical, and they are given in Table 2.5.
It is recommended that, if Eqs.(2.67), (2.66), and (2.74) are used respec-tively for Nc , Nq , and N! , then DeBeer’s shape factors and Hansen’s depthfactors should be used. However, if Eqs. (2.67), (2.66), and (2.72) are used forbearing capacity factors Nc , Nq , and N! , then Meyerhof’s shape and depthfactors 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 soilsupporting the foundation has the following parameters: # = 25!, c = 48kN/m2, and ! = 18 kN/m3. Determine the ultimate vertical load that thefoundation 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 factorsproposed 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 inTable 2.5.
Solution From Eq. (2.83):
qu = cNc "cs "cd + qNq "qs "qd + –12!BN! "!s "!d
TABLE 2.5 Summary of Shape and Depth Factors
Factor Relationship Reference
Shape For # = 0!:
1
1
2.01
=λ
=λ
+=λ
γs
qs
cs L
B
For # "10!:
φ+
+=λ=λ
φ+
+=λ
γ 245tan1.01
245tan2.01
2
2
L
B
L
B
sqs
cs
Meyerhof [8]
+=λ
L
B
N
N
c
q
cs1
[Note: Use Eq. (2.67) for Nc and Eq. (2.66) for Nq as given inTable 2.3]
−=λ
φ
+=λ
γ L
B
L
B
s
qs
4.01
tan1
DeBeer [19]
Depth For # = 0!:
1
2.01
=λ=λ
+=λ
γdqd
f
cd B
D
For # " 10!:
φ+
+=λ=λ
φ+
+=λ
γ 245tan1.01
245tan2.01
B
D
B
D
f
dqd
f
cd
Meyerhof [8]
Factor Relationship Reference
For Df /B # 1:
1
)sin1(tan21
4.01
2
=λ
φ−φ+=λ
+=λ
γd
f
qd
f
cd
B
D
B
D
For Df /B > 1:
1
tan)sin1(tan21
tan4.01
12
1
=λ
φ−φ+=λ
+=λ
γ
−
−
d
f
qd
f
cd
B
D
B
D
radiansin is tan:Note 1-
B
Df
Hansen [9]
8.02.1
6.0)4.0(14.01
233.125tan2.1
6.01tan1
257.12.1
6.0
72.20
66.1011
=
−=
−=λ
=
+=φ
+=λ
=
+=
+=λ
γ L
B
L
B
L
B
N
N
s
qs
c
q
cs
1
155.16.0
6.0)25sin1)(25(tan21
)sin1(tan21
4.16.0
6.0)4.0(1)4.0(1
2
2
=λ
=
−+=
φ−φ+=λ
=
+=
+=λ
γd
f
qd
f
cd
B
D
B
D
246.1225
45tan2.16.0
2.012
45tan2.01 22 =
+
+=
φ+
+=λ
L
Bcs
Hansen’s depth factors are as follows:
So
qu = (48)(20.72)(1.257)(1.4) + (0.6)(18)(10.66)(1.233)(1.155)
+ –12
(18)(0.6)(10.88)(0.8)(1)
a. From Table 2.3, for # = 25 !, Nc = 20.72 and Nq = 10.66. Also, fromTable 2.4, for # = 25 !, Vesic’s value of N! = 10.88. DeBeer’s shapefactors are as follows:
= 1750.2 + 163.96 + 47 $ 1961 kN/m2
b. From Table 2.3 for # = 25!, Nc = 20.72, Nq = 10.66, and N! = 6.77.Now, referring to Table 2.5, Meyerhof’s shape and depth factors areas follows:
123.1
2
2545tan
2.1
6.01.01
245tan1.01 22
=
+
+=
φ+
+=λ=λ γ L
Bsqs
157.1225
45tan6.06.0
1.01
245tan1.01
314.1225
45tan6.06.0
2.01
245tan2.01
=
+
+=
φ+
+=λ=λ
=
+
+=
φ+
+=λ
γ B
D
B
D
f
dqd
f
cd
So
qu = (48)(20.72)(1.246)(1.314)
+ (0.6)(18)(10.66)(1.123)(1.157)
+ –12
(18)(0.6)(6.77)(1.123)(1.157)
= 1628.3 + 149.6 + 47.7 = 1825.6 kN/m2 !!
2.10 EFFECT OF SOIL COMPRESSIBILITY
In Section 2.3, the ultimate bearing capacity equations proposed by Terzaghi[1] for local shear failure were given [Eqs. (2.39)–(2.41)]. Also, suggestions byVesic [4] shown in Eqs. (2.42) and (2.43) address the problem of soil com-pressibility and its effect on soil bearing capacity. In order to account for soilcompressibility, Vesic [4] proposed the following modifications to Eq. (2.83).Or
qu = cNc "cs "cd "cc + qNq "qs "qd "qd + –12!BN! "!s "!d "!c (2.84)
where "cc , "qd , "!c = soil compressibility factorsThe soil compressibility factors were derived by Vesic [4] from the
analogy of expansion of cavities [20]. According to this theory, in order tocalculate "cc , "qd , and "!c , the following steps should be taken.
IG
c qr = + tanφ
φ−
−=
245cot45.03.3exp
2
1( L
BI
r cr)
φ+φ
+
φ
+−
=λ=λγ
sin1
)2)(logsin07.3(
tan6.04.4
exp
r
qcc
I
L
B
λ λλφcc qc
qc
qN= −
−1
tan
1. Calculate the rigidity index, Ir , of the soil (approximately at a depthof B/2 below the bottom of the foundation, or
(2.85)
where G = shear modulus of the soil ! = soil friction angle q = effective overburden pressure at the level of the foundation
2. The critical rigidity index of the soil, Ir(cr) , can be expressed as
(2.86)
3. If Ir ! Ir(cr) , then use λcc , λqc , and λγc equal to one. However if Ir < Ir(cr),
(2.87)
For ! = 0
"cc = 0.32 + 0.12 –BL
+0.6 logIr (2.88)
For other friction angles
(2.89)
Figures 2.23 and 2.24 show the variations of "#c = "qc [Eq. (2.87)] with ! andIr .
EXAMPLE 2.2
Refer to Example 2.1a. For the soil, the given modulus of elasticity, E = 620kN/m2; Poisson’s ratio, $ = 0.3. Determine the ultimate bearing capacity.
IG
c q
E
c qr =+
=+ +
=+ + ×
=
tan ( )[ tan ]
( . )[( . ) tan ].
φ ν φ2 1
6202 1 0 3 48 18 0 6 25
4 5
46.622
2545cot
2.1
6.045.03.3exp
2
1
245cot45.03.3exp
2
1
=
−
×−=
φ
−
−=
L
BI
r (cr)
Solution
From Eq. (2.86)
FIGURE 2.23 Variation of "#c = "qc with ! and Ir for square foundation(B/L = 1)
FIGURE 2.24 Variation of "#c = "qc with ! and Ir for foundations withL/B > 5
353.0
25sin1
)5.42log()25sin07.3(
25tan2.16.0
6.04.4
exp
sin1
)2)(logsin07.3(
tan6.04.4
exp
=
+×
+
×+−
=
φ+φ
+φ
+−
=λ=λ γ
r
cqc
I
L
B
Since Ir(cr) > Ir , use "cc , "qc , and "#c relationships from Eqs. (2.87) and (2.89)
λ λλφcc qc
qc
qN= −
−= − − =
10 353
1 0 353
10 66 250 228
tan.
.
. tan.
°°φ−φ−φ=φ
90)(
211
i
Also
Equation (2.84):
qu = (48)(20.72)(1.257)(1.4)(0.228)
+ (0.6)(18) (10.66) (1.233) (1.155)(0.353)
+ –12
(18)(0.6)(10.88)(0.8)(1)(0.353)
= 399.05 + 57.88 + 16.59 " 474 kN/m2 !!
2.11 BEARING CAPACITY OF FOUNDATIONS ON ANISOTROPIC SOIL
Foundation on Sand (c = 0)
Most natural deposits of cohesionless soil have an inherent anisotropicstructure due to their nature of deposition in horizontal layers. Initial depositionof the granular soil and subsequent compaction in the vertical direction causesthe soil particles to take a preferred orientation. For a granular soil of this typeMeyerhof suggested that, if the direction of application of deviator stressmakes an angle i with the direction of deposition of soil (Fig. 2.25), then thesoil friction angle ! can be approximated in a form
(2.90)
where!1 = soil friction angle with i = 0#!2 = soil friction angle with i = 90#
Figure 2.26 shows a continuous (strip) rough foundation on an anisotropicsand deposit. The failure zone in the soil at ultimate load is also shown in thefigure. 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 Zone2. 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 continuousfoundation on an anisotropic sand could be calculated by assuming an equiva-lent friction angle ! = !eq , or
FIGURE 2.25 Aniostropy in sand deposit
FIGURE 2.26 Continuous rough foundation on anisotropic sand deposit
φφ φ φ
eq =+
=+( ) ( )2
3
2
31 2 1n
where friction ratio = 2n =φφ1
(2.91)
(2.92)
Once the equivalent friction angle is determined, the ultimate bearing capacityfor vertical loading conditions on the foundation can be expressed as(neglecting the depth factors)
qu = q Nq(eq) "qs + –12#B N#(eq) "#s (2.93)
FIGURE 2.27 Variation of N#(eq) [Eq. (2.93)]
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 2.28present the plots of Nq(eq) and N#(eq) in terms of n and !1 . Note that the soil
−γ+
φ
+= γ L
BBN
L
BqNq
qu4.01
21
tan1(eq)eq(eq)
b
a
c
c c
u i
uV uH
= = °( )
( )( )
45
+=
2)(uHuV
icu
ccNq
friction angle ! = !eq was used in Eqs. (2.66) and (2.72) to prepare the graphs.So combining the relationships for shape factors (Table 2.5) given by DeBeer[19]
(2.94)
Foundations on Saturated Clay (!!!! = 0 concept)
As in the case of sand discussed above, saturated clay deposits also exhibitanisotropic undrained shear strength properties. Figures 2.29a and 2.29b showthe nature of variation of the undrained shear strength of clays, cu , with respectto the direction of principal stress application [22]. Note that the undrainedshear strength plot shown in Fig. 2.29b is elliptical. However, the center of theellipse does not match the origin. The geometry of the ellipse leads to theequation
(2.95)
where cuV = undrained shear strength with i = 0#cuH = undrained shear strength with i = 90#
A continuous foundation on a saturated clay layer (! = 0) whosedirectional strength variation follows Eq. (2.95) is shown in Fig. 2.29c. Thefailure 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 themajor principal stress is horizontal in Zone III; however, it gradually changesfrom vertical to horizontal in Zone II. Using the stress characteristic solution,Davis and Christian [22] determined the bearing capacity factor Nc(i) for thefoundation. For a surface foundation
(2.96)
The variation of Nc(i) with the ratio of a/b (Fig. 2.29b) is shown in Fig. 2.30.Note that, when a = b, Nc(i) becomes equal to Nc = 5.14 [isotropic case; Eq.(2.67)].
In 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 approximationmay be used
FIGURE 2.29 Bearing capacity of continuous foundation on anisotropicsaturated clay
14.5
29.0
=
↑
+≈ uHuV
cu
ccNq
(2.97)
FIGURE 2.30 Variation of Nc(i) with a/b based on the analysis of Davis and Christian
qdqsqcdcs
iuHiuV
icuqN
ccNq λλ+λλ
+= °=°=
2)90()0(
)(
qdqsfcdcsuHuV
icuD
ccNq λλγ+λλ
+=
2)(
The preceding equation, which was suggested by Davis and Christian [22], isbased on the undrained shear strength results of several clays. So, in general,for a rectangular foundation with vertical loading condition
(2.98)
For ! = 0 condition, Nq = 1 and q = #Df . So
(2.99)
The desired relationships for the shape and depth factors can be taken fromTable 2.5 and the magnitude of qu can be estimated.
Foundation on c–!!!! Soil
The ultimate bearing capacity of a continuous shallow foundation supported
FIGURE 2.31 Anisotropic clay soil—assumptions for bearing capacity evaluation
by anisotropic c–! soil was studied by Reddy and Srinivasan [23] using themethod of characteristics. According to this analysis the shear strength of a soilcan be given as
s = %´tan! + c
However, it is assumed that the soil is anisotropic only with respect tocohesion. As mentioned previously in this section, the direction of the majorprincipal stress (with respect to the vertical) along a slip surface located belowthe foundation changes. In anisotropic soils, this will induce a change in theshearing resistance to the bearing capacity failure of the foundation. Reddy andSrinivasan [23] assumed the directional variation of c at a given depth z belowthe foundation as (Fig. 2.31a)
Kc
cV z
H z
= ( )
( )
β α
γ
cV z
V z
l
c
lc
= ′
=
=
=
( )
( )
0
0where characteristic length =
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 anisotropycoefficient K is defined as the ratio of cV(z) to cH(z) .
(2.101)
In overconsolidated soils K is less than one and, for normally consolidatedsoils the magnitude of K is greater than one.
For many consolidated soils, the cohesion increases linearly with depth(Fig. 2.31c). Thus
cV(z) = cV(z=0) + &´s (2.102)
where cV(z) , cV(z=0) = cohesion in the vertical direction (that is, i = 0) at depths of z and z = 0, respectively &´ = the rate of variation with depth z
According to this analysis, the ultimate bearing capacity of a continuous foun-dation may be given as
qu = cV(z=0) Nc(i´) + qNq(i´) + –12#BN#(i´) (2.103)
where Nc(i´) , Nq(i´) , N#(i´) = bearing capacity factors q = #Df
This equation is similar to Terzaghi’s bearing capacity equation for continuousfoundations [Eq. (2.31)].
The bearing capacity factors are functions of the parameters 'c and K. Theterm 'c can be defined as
(2.104)
(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 thefoundation, B´
B´ = –Bl
(2.106)
The variations of the bearing capacity factors with 'c , B´, !, and Kdetermined using the method of analysis by Reddy and Srinivasan [23] areshown in Figs. 2.32 to 2.37. This study shows that the rupture surface in soilat ultimate load extends to a smaller distance below the bottom of the founda-tion for the case where the anisotropic coefficient K is greater than one. Also,when K changes from one to two with &´ = 0, the magnitude of Nc(i´) is reducedby about 30% – 40%.
FIGURE 2.33 Reddy and Srinivasan’s bearing capacity factor Nc(i — influence of K ('c = 0.2)
FIGURE 2.34 Reddy and Srinivasan’s bearing capacity factor, Nc(i%) — influence of K ('c = 0.4)
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 thefollowing: 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)
Characteristic length,
Nondimensional width,
lc
BB
l
V z= = =
′ = = =
=( ) .
..
0 250
1102 27
9
2 27396
γ
Also
FIGURE 2.36 Reddy and Srinivasan’s bearing capacity factors, N#(i%) and Nq(i%) — influence of K ('c = 0)
FIGURE 2.37 Reddy and Srinivasan’s bearing capacity factors, N#(i%) and Nq(i%) — influence of K ('c = 0.2)
β αc
V z
l
c= ′ = =
=( )
( )( . ).
0
25 2 27
2500 227
Now, referring to Figs. 2.33, 2.34, 2.36, and 2.37, for ! = 20#, 'c = 0.227, K= 2, and B´ = 3.96 (by interpolation)
Nc(i´) " 14.5; Nq(i´) " 6, and N#(i´) " 4
FSu
all =
FSu
all(net)(net)=
From Eq. (2.103)
qu = cV(z=0) Nc(i´) + qNq(i´) + –12!BN!(i´)
= (250)(14.5) + (3)(110)(6) + –12
(110)(10)(4)
= 7,805 lb/ft2 !!
2.12 ALLOWABLE BEARING CAPACITY WITH RESPECT TO FAILURE
Allowable bearing capacity for a given foundation may be (a) to protect thefoundation against a bearing capacity failure, or (b) to ensure that the foun-dation does not undergo undesirable settlement. There are three definitions forthe allowable capacity with respect to a bearing capacity failure. They are:
Gross Allowable Bearing Capacity
The gross allowable bearing capacity is defined as
(2.107)
where qall = gross allowable bearing capacity FS = factor of safetyIn 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 areaof the foundation that can be supported by the soil in excess of the pressurecaused by the surrounding soil at the foundation level. If the difference be-tween the unit weight of concrete used in the foundation and the unit weight ofthe surrounding soil is assumed to be negligible, then
qu(net) = qu ! q (2.108)
where q = !Df
qu(net) = net ultimate bearing capacityThe net allowable bearing capacity can now be defined as
(2.109)
A factor of safety of 3 to 4 in the preceding equation is generally considered
φ=φ
=
−
(shear)
(shear)
FS
FS
cc
d
d
tantan 1
FSu
all = = ≈1961
4490 kN / m 2
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) , whichmay be in the range of 1.3 to 1.6 is adopted. In order to evaluate qall(shear) , thefollowing procedure may be used.
1. Determine the developed cohesion, cd , and the developed angle offriction, "d , as
(2.110)
(2.111)
2. The gross and net ultimate allowable bearing capacities with respectto shear failure can now be determined as [Eq. (2.83)]
qall(shear)—gross = cd Nc #cs #cd + qNq #qs #qd + –12!BN! #!s #!d (2.112)
qall(shear)—net = qall(shear)—gross ! q
= cd Nc #cs #cd + q(Nq ! 1)#qs #qd + –12!BN! #!s #!d (2.113)
where Nc , Nq , and N! = bearing capacity factors for friction angle "d
EXAMPLE 2.4
Refer to Example Problem 2.1a and determinea. 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.
8.02.16.0
4.014.01
156.13.17tan2.1
6.01tan1
192.12.1
6.0
5.12
8.411
=
−=
−=λ
=
+=φ
+=λ
=
+=
+=λ
γ L
B
L
B
L
B
N
N
s
dqs
c
q
cs
1
308.16.06.0
)3.17sin1)(3.17)(tan2(1
)sin1(tan21
4.16.0
6.04.014.01
2
=λ
=
−+=
φ−φ+=λ
=
+=
+=λ
γd
f
qd
f
cd
B
D
B
D
b. qq q
FSu
all(net) =−
=−
≈1961 0 6 18
4
( . )( )488 kN / m 2
c.
°=
=
φ=φ
===
−− 3.175.125tan
tantan
tan
325.1
48
11
(shear)
2
(shear)
kN/m
FS
FS
cc
d
d
For " = 17.3", Nc = 12.5, Nq = 4.8 (Table 2.3), and N! = 3.6 (Table2.4)
From Eq. (2.112)
qall(shear)—gross = cd Nc #cs #cd + qNq #qs #qd + –12!BN! #!s #!d
= (32)(12.5)(1.192)(1.4) + (0.6)(18)(4.8)(1.156)×
(1.308) + –12
(18)(0.6)(3.6)(0.8)(1)
= 667.5 + 78.4 + 15.6 = 761.5 kN/m2
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 bearingcapacity of single rough continuous foundations supported by a homogeneoussoil medium extending to a great depth were discussed. However, if founda-tions are placed close to each other with similar soil conditions, the ultimatebearing capacity of each foundation may decrease due to the interference effectof 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 massto be the same as that assumed by Terzaghi (Fig. 2.1). According to Stuart, thefollowing conditions may arise (Fig. 2.38)
1. Case 1 (Fig. 2.38a): If the center-to-center spacing of the twofoundations is x $ x1 , the rupture surface in the soil under eachfoundation will not overlap. So the ultimate bearing capacity of eachcontinuous foundation can be given by Terzaghi’s equation [Eq.(2.31)]. For c = 0
qu = qNq + –12!BN! (2.114)
where Nq , N! = Terzaghi’s bearing capacity factors (Table 2.1)2. Case 2 (Fig. 2.38b): If the center-to-center spacing of the two
foundations (x = x2 < x1) are such that the Rankine passive zones justoverlap, then the magnitude of qu will still be given by Eq. (2.114).However, the foundation settlement at ultimate load will change(compared to the case of an isolated foundation).
3. Case 3 (Fig. 2.38c): This is the case where the center-to-centerspacing of the two continuous foundations is x = x3 < x2 . Note that thetriangular wedges in the soil under the foundation make angles of180" ! 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 ofthe logarithmic spirals d2 g2 and d2 e are tangent to each other at pointd2 . For this case, the ultimate bearing capacity of each foundation canbe given as (c = 0)
qu = qNq $q + –12!BN! $! (2.115)
where $q , $! = efficiency ratios
FIGURE 2.38 Assumptions for the failure surface in granular soil under two closely spacedrough continuous foundations rough continuous foundations (Note: α1 = φ, α2 = 45 − φ/2, α3 = 180 − φ)
FIGURE 2.38 (Continued)
FIGURE 2.39 Stuart’s interference factor %q
FIGURE 2.40 Stuart’s interference factor %!
FIGURE 2.41 Comparison of experimental and theoretical %q
The efficiency ratios are functions of x/B and soil friction angle ".The theoretical variations of $q and $! are given in Figs. 2.39 and2.40.
4. Case 4 (Fig. 2.38d): If the spacing of the foundation is furtherreduced such that x = x4 < x3 , blocking will occur, and the pair offoundations will act as a single foundation. The soil between theindividual units will form an inverted arch which travels down withthe foundation as the load is applied. When the two foundationstouch, the zone of arching disappears and the system behaves as asingle foundation with a width equal to 2B. The ultimate bearingcapacity for this case can be given by Eq. (2.114), with B beingreplaced by 2B in the third term.
Das and Larbi-Cherif [26] conducted laboratory model tests to determinethe interference efficiency ratios ($q and $!) of two rough continuous founda-tions resting on sand extending to a great depth. The sand used in the modeltests was highly angular, and the tests were conducted at a relative density ofabout 60%. The angle of friction " at this relative density of compaction was39". Load-displacement curves obtained from the model tests were of localshear type. The experimental variations of $q and $! obtained from these testsare 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 2.42 Comparison of experimental and theoretical %!
similar to those predicted by theory, there is a large variation in the magnitudesbetween the theory and experimental results. Figure 2.43 shows theexperimental variations of Su /B with x/B (Su = settlement at ultimate load).The elastic settlement of the foundation decreases with the increase in thecenter-to-center spacing of the foundation and remains constant at x > about4B.
REFERENCES
1. Terzaghi, K., Theoretical Soil Mechanics, John Wiley, New York, 1943.2. Kumbhojkar, A. S., Numerical evaluation of Terzaghi’s N! , J. Geotech. Eng.,
ASCE, 119(3), 598, 1993.3. Krizek, R. J., Approximation for Terzaghi’s bearing capacity, J. Soil Mech.
Found. Div., ASCE, 91(2), 146, 1965.4. Vesi!, A. S., Analysis of ultimate loads of shallow foundations, J. Soil Mech.
Found. Div., ASCE, 99(1), 45, 1973.5. Meyerhof, G. G., The ultimate bearing capacity of foundations, Geotechnique, 2,
301, 1951.
FIGURE 2.43 Variation of experimental elastic settlement (Si /B) with center- to-center spacing of two continuous rough foundations
6. Reissner, H., Zum erddruckproblem, in Proc., First Intl. Conf. Appl. Mech.,Delft, The Netherlands, 1924, 295.
7. Prandtl, L., Uber die eindringungs-festigkeit plastisher baustoffe und diefestigkeit von schneiden, Z. Ang. Math. Mech., 1(1), 15, 1921.
8. Meyerhof, G. G., Some recent research on the bearing capacity of foundations,Canadian Geotech. J., 1(1), 16, 1963.
9. Hansen, J. B., A Revised and Extended Formula for Bearing Capacity, BulletinNo. 28, Danish Geotechnical Institute, Copenhagen, 1970.
10. Caquot, A., and Kerisel, J., Sue le terme de surface dans le calcul des fondationsen milieu pulverulent, in Proc., III Intl. Conf. Soil Mech. Found. Eng., Zurich,Switzerland, 1, 1953, 336.
11. Lundgren, H., and Mortensen, K., Determination by the theory of plasticity of thebearing capacity of continuous footings on sand, in Proc., III Intl. Conf. Mech.Found. Eng., Zurich, Switzerland, 1, 1953, 409.
12. Chen, W. F., Limit Analysis and Soil Plasticity, Elsevier Publishing Co., NewYork, 1975.
13. Drucker, D. C., and Prager, W., Soil mechanics and plastic analysis of limitdesign, Q. Appl. Math., 10, 157, 1952.
14. Biarez, J., Burel, M., and Wack, B., Contribution à l’étude de la force portantedes fondations, in Proc., V Intl. Conf. Soil Mech. Found. Eng., Paris, France, 1,1961, 603.
15. Ingra, T. S., and Baecher, G. B., Uncertainty in bearing capacity of sand, J.Geotech. Eng., ASCE, 109(7), 899, 1983.
16. Ko, H. Y., and Davidson, L. W., Bearing capacity of footings in plane strain, J.Soil Mech. Found. Div., ASCE, 99(1), 1, 1973.
17. Hu, G. G. Y., Variable-factors theory of bearing capacity, J. Soil Mech. Found.Div., ASCE, 90(4), 85, 1964.
18. Balla, A., Bearing capacity of foundations, J. Soil Mech. Found. Div., ASCE,88(5), 13, 1962.
19. DeBeer, E. E., Experimental determination of the shape factors of sand,Geotechnique, 20(4), 307, 1970.
20. Vesi!, A., Theoretical Studies of Cratering Mechanisms Affecting the Stabilityof Cratered Slopes, Final Report, Project No. A-655, Engineering ExperimentStation, Georgia Institute of Technology, Atlanta, Ga., 1963.
21. Meyerhof, G. G., Bearing capacity of anisotropic cohesionless soils, CanadianGeotech. J., 15(4), 593, 1978.
22. Davis, E., and Christian, J. T., Bearing capacity of anisotropic cohesive soil, J.Soil Mech. Found. Div., ASCE, 97(5), 753, 1971.
23. Reddy, A. S., and Srinivasan, R. J., Bearing capacity of footings on anisotropicsoils, J.Soil Mech. Found. Div., ASCE, 96(6), 1967, 1970.
24. Casagrande, A., and Carrillo, N., Shear failure in anisotropic materials, inContribution to Soil Mechanics 1941-53, Boston Society of Civil Engineers, 122,1944.
25. Stuart, J. G., Interference between foundations with special reference to surfacefooting on sand, Geotechnique, 12(1), 15, 1962
26. Das, B. M., and Larbi-Cherif, S., Bearing capacity of two closely spaced shallowfoundations on sand, Soils and Foundations, 23(1), 1, 1983.
CHAPTER THREE
ULTIMATE BEARING CAPACITY UNDER INCLINEDAND ECCENTRIC LOADS
3.1 INTRODUCTION
Due to bending moments and horizontal thrusts transferred from the super-structure, shallow foundations are many times subjected to eccentric andinclined loads. Under such circumstances, the ultimate bearing capacity theoriespresented in Chapter 2 will need some modification, and this is the subject ofdiscussion in this chapter. The chapter is divided into two major parts. The firstpart discusses the ultimate bearing capacity of shallow foundations subjectedto centric inclined load, and the second part is devoted to the ultimate bearingcapacity 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 undervertical loading (Section 2.4) to the case with inclined load. Figure 3.1 showsthe plastic zones in the soil near a rough continuous (strip) foundation withsmall 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 pointedout 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 shearstresses on plane ae are po and so , respectively. Also, the unit base adhesion isc a . The solution for the ultimate bearing capacity, qu , can be expressed as
qu(v) = qucos# = cNc+ po Nq+ –12$BN$ (3.2)
where Nc , Nq , N$ = bearing capacity factors for inclined loading condition $ = unit weight of soil
Similar to Eqs. (2.71), (2.59), and (2.70), we can write
FIGURE 3.1 Plastic zones in soil near a foundation with inclined load
φθ
φθ
φ+ηφ−φ−ψφ+=
−
φ+ηφ−φ−ψφ+φ=
tan2
tan2
)2sin(sin1
)2sin(sin1
1)2sin(sin1
)2sin(sin1cot
eN
eN
q
c
qu(v) = qucos# = q u(V) + q!u(V) (3.3)
where
q u(v) = cNc + po Nq (for ""0, $"0, po"0, c"0) (3.4)
and
q!u(v) = –12$BN$ (for ""0, $"0, po"0, c"0) (3.5)
It was shown by Meyerhof [1] in Eq. (3.4) that
(3.6)
(3.7)
Note that the horizontal component of the inclined load per unit area on thefoundation, qh , cannot exceed the shearing resistance at the base, or
qu(h) # ca + q u(v) tan% (3.8)
where ca = unit base adhesion% = unit base friction angle
In order to determine the minimum passive force per unit length of thefoundation, Pp$(min) (see Fig. 2.11 for comparison), to obtain N$ , one can takea numerical step-by-step approach as shown by Caquot and Kerisel [2] or asemi-graphical approach based on the logarithmic spiral method as shown byMeyerhof [3]. Note that the passive force Pp$ acts at an angle " with thenormal drawn to the face bc of the elastic wedge abc (Fig. 3.1). The
)(for δ≤αφ
φ−ψψ−
φ−ψ+
φ−ψψ
γ= γ
γ cos
)cos(sin)cos(
)cos(
sin2 2
2
(min)
B
PN p
FIGURE 3.2 Meyerhof’s [1] bearing capacity factor Ncq for purely cohesive soil (" = 0)
relationship for N$ is
(3.9)
The ultimate bearing capacity expression given by Eq. (3.2) can also bedepicted as
qu(v) = qu cos# = cNcq + –12$BN$q (3.10)
where Ncq , N$q = bearing capacity factors which are functions of the soil friction angle, ", and the depth of the foundation, Df
For a purely cohesive soil (" = 0)
qu(v) = qu cos# = cNcq (3.11)
Figure 3.2 shows the variation of Ncq for a purely cohesive soil (" = 0) forvarious load inclinations (#).
FIGURE 3.3 Meyerhof’s [1] bearing capacity factor N$q for cohesionless soil (c =0, % = ")
For cohesionless soils, c = 0 and, hence, Eq. (3.10) gives
qu(v) = qu cos# = –12$BN$q (3.12)
Figure 3.3 shows the variation of N$q with #.
2
2
1
901
°φ°α−=λ
°°α−=λ=λ
γi
qici
5
5
cotcos
sin7.01
1
1
cotcos
sin5.01
φ+α
α−=λ
↑
−λ−
−λ=λ
φ+α
α−=λ
γ BLcQ
Q
N
BLcQ
Q
u
u
i
q
qi
qici
u
u
qi
2.3 Table
3.3 GENERAL BEARING CAPACITY EQUATION
The general ultimate bearing capacity equation for a rectangular foundationgiven by Eq. (2.82) can be extended to account for inclined load and can beexpressed as
qu = cNc &cs &cd &ci + qNq &qs &qd &qi + –12$BN$ &$s &$d &$i (3.13)
where Nc , Nq , N$ = bearing capacity factors [for Nc and Nq , use Table 2.3; for N$ , see Table 2.4 — Eqs. (2.72), (2.73), (2.74)]
&cs , &qs , &$s = shape factors (Table 2.5)&cd , &qd , &$d = depth factors (Table 2.5)&ci , &qi , &$i = inclination factors
Meyerhof [4] provided the following inclination factor relationships
(3.14)
(3.15)
Hansen [5] also suggested the following relationships for inclination factors
(3.16)
(3.17)
(3.18)
where, in Eqs. (3.14) to (3.18)# = inclination of the load on the foundation with the vertical
Qu = ultimate load on the foundation = qu BLB = width of the foundationL = length of the foundation
3.4 OTHER RESULTS FOR FOUNDATIONS WITHCENTRIC INCLINED LOAD
Based on the results of field tests, Muhs and Weiss [6] concluded that the ratioof the vertical component Qu(v) of the ultimate load with the inclination # with
Q
Qu v
u
( )
( )
( tan )α
α=
= −0
21
Q
BLQ
BL
q
q
u v
u
u v
u
( )
( )
( )
( )
( tan )α α
α= =
= = −0 0
21
qBN
uq=
γαγ
2 cos
D
Bf = =12
121
.
.;
the vertical to the ultimate load Qu when the load is vertical (that is, # = 0) andis approximately equal to (1$tan#)2.
or
(3.19)
Dubrova [7] developed a theoretical solution for the ultimate bearingcapacity of a continuous foundation with centric inclined load and expressedit in the following form
qu = c(Nq* $ 1)cot" +2qNq
* + B$N$* (3.20)
where Nq*, N$
* = bearing capacity factorsq = $Df
The variations of Nq* and N$
* are given in Figs. 3.4. and 3.5.
EXAMPLE 3.1
Consider a continuous foundation in a granular soil with the following: B = 1.2m; Df = 1.2 m; unit weight of soil, $ = 17 kN/m3; soil friction angle, " = 40%;load inclination, # = 20%. Calculate the gross ultimate load bearing capacity qu .
a. Use Eq. (3.12).b. Use Eq. (3.13) and Meyerhof’s bearing capacity factors (Table 2.3),
his shape and depth factors (Table 2.5); and inclination factors [Eqs.(3.14) and (3.15)].
Solution
a. From Eq. (3.12)
" = 40%; and #=20 %. From Fig. 3.3, N$q & 100. So
FIGURE 3.4 Variation of Nq*
qu = =( )( . )( )
cos
17 12 100
2 201085.5 kN / m 2
b. With c = 0 and B/L = 0, Eq. (3.13) becomes
qu = qNq &qd &qi + –12$BN$ &$d &$i
For "= 40 %, from Table 2.3, Nq = 64.2 and N$ = 93.69. From Table2.5,
214.12
4045tan
2.1
2.11.01
245tan1.01
=
+
+=
φ
+
+=λ=λ γ B
Df
dqd
25.04020
11
605.09020
190
1
22
22
=
−=
φα−=λ
=
−=
α−=λ
γi
qi
From Eqs. (3.14) and (3.15)
FIGURE 3.5 Variation of N$*
qBN
uq=
γαγ
2 cos
qu = ≈( )( . )( )
cos
17 12 65
2 20706 kN / m 2
So
qu = (1.2 × 17)(64.2)(1.214)(0.65) + –12
(17)(1.2)(93.69)(1.214)(0.25)
= 1323.5 kN / m2 !!
EXAMPLE 3.2
Consider the continuous foundation described in Example 3.1. Other quan-tities remaining the same, let " = 35%.
a. Calculate qu using Eq. (3.12).b. Calculate qu using Eq. (3.20).
Solution
a. From Eq. (3.12)
From Fig. 3.3, N$q & 65
b. For c = 0, Eq. (3.20) becomes
qu = 2qNq* + B$N$
*
Using Figs. 3.4 and 3.5, for " = 35 % and tan# = tan20=0.36, Nq* & 8.5
and N$* & 6.5 (extrapolation)
qu = (2)(17 × 1.2)(8.5) + (1.2)(17)(6.5) & 480 kN / m2
Note: Eq. (3.20) does not provide depth factors. !!
3.5 CONTINUOUS FOUNDATION WITH ECCENTRIC LOAD
When a shallow foundation is subjected to an eccentric load, it is assumed thatthe contact pressure decreases linearly from the toe to the heel. However, atultimate load, the contact pressure is not linear. This problem was analyzed byMeyerhof [1], who suggested the concept of effective width, B´. The effectivewidth is defined as (Fig. 3.6)
B´ = B $ 2e (3.21)
FIGURE 3.6 Effective width B!
where e = load eccentricityAccording to this concept, the bearing capacity of a continuous foundation
can be determined by assuming that the load acts centrally along the effectivecontact width as shown in Fig. 3.6. Thus, for a continuous foundation [fromEq. (2.83)] with vertical loading
qu = cNc !cd + qNq !qd + –12" B´N" !"d (3.22)
Note that the shape factors for a continuous foundation are equal to one. Theultimate 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 loadedcontinuous foundations using the method of slices proposed by Janbu [9].Based on that analysis, they proposed that
Rq
qku c
u
= −1 ( )
( )
eccentri
centric
k
k B
eaR
=
−=−=
k
ukuu B
eaqRqq 1)1(
)()()( centriccentriceccentric
(3.23)
where Rk = reduction factorqu(eccentric) = ultimate bearing capacity of eccentrically loaded
continuous foundationsqu(centric) = ultimate bearing capacity of centrally loaded continuous
foundationsThe magnitude of Rk can be expressed as
where a and k are functions of the embedment ratio Df /B (Table 3.1).
TABLE 3.1 Variations of a and k [Eq. (3.24)]
Df /B a k
00.250.51.0
1.8621.8111.7541.820
0.730.7850.800.888
Hence, combining Eqs. (3.23) and (3.24)
(3.25)
where
qu(centric) = cNc !dc + qNq !dq + –12" BN" !"d (3.26)
Theory of Prakash and Saran
Prakash and Saran [10] provided a comprehensive mathematical formulationto estimate the ultimate bearing capacity for rough continuous foundationsunder eccentric loading. According to this procedure, Fig. 3.7 shows theassumed failure surface in a c–φ soil under a continuous foundation subjected
FIGURE 3.7 Derivation of the bearing capacity theory of Prakash and Saran for eccentrically loaded rough continuous foundation
BBN D N cNu
ue f q e c e=
×= + +
( ) ( ) ( ) ( )1
1
2γ γγ
32
83.563.263.10.1
+
−
−=
B
e
B
e
B
e
S
S
o
e
to eccentric loading. Let Qu be the ultimate load per unit length of the foun-dation of width B with an eccentricity e. In Fig. 3.7, Zone I is an elastic zonewith wedge angles of #1 and #2 . Zones II and III are similar to those assumedby Terzaghi (that is, Zone II is a radial shear zone and Zone III is a Rankinepassive zone).
The bearing capacity expression can be developed by considering theequilibrium 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 equalto Bx1 . Neglecting the self-weight of the wedge
Qu = Pp cos(#1 " $) + Pm cos(#2 " $m) + Ca sin #1 + Ca sin#2 (3.27)
wherePp , Pm = passive forces per unit length of the wedge along the wedge faces bc and ac, respectively
$ = soil friction angle$m = mobilized soil friction angle (#$)
Ca = adhesion along wedge face bc =cBx1 2
1 2
sin
sin( )
ψψ ψ+
Ca = adhesion along wedge face ac =mcBx1 1
1 2
sin
sin( )
ψψ ψ+
m = mobilization factor (#1)c = unit cohesion
Equation (3.27) can be expressed in the form
(3.28)
where N"(e) , Nq(e) , Nc(e) = bearing capacity factors for an eccentrically loaded continuous foundation
The above-stated bearing capacity factors will be functions of e/B, $, andalso 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 thosegiven by Terzaghi [11] for a centrically loaded foundation.
Prakash [12] also gave the relationships for settlement of a given founda-tion under centric and eccentric loading conditions for a equal factor of safety,FS. They are as follows (Fig.3.11)
(3.29)
FIGURE 3.8 Prakash and Saran’s bearing capacity factors, Nc(e)
32
54.3161.2231.20.1
+
−
−=
B
e
B
e
B
e
S
S
o
m
and
(3.30)
where So = settlement of a foundation under centric loading at
qall(centric) =qu(centric)
FS
Se , Sm = settlements of the same foundation under eccentric loading at
qall(eccentric) = qu(eccentric)
FS
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 thedepth of the foundation Df = 1 m, determine the ultimate load per unit meterlength 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 factormethod.
Solution Since c = 0, B/L = 0. From Eq. (3.26)
qu(centric) = qNq !dq + –12" BN" !"d
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:
107.12
4045tan
2
11.01
245tan1.01
=
+
+=
φ+
+=λ=λ γ B
Df
dqd
−=−=
k
ukuu B
eaqRqq 1)1(
)()()( centriccentriceccentric
So
qu(centric) = (1)(17.5)(64.2)(1.107) + –12
(17.5)(2)(93.69)(1.107)
= 1243.7 + 1815.2 = 3058.9 kN / m2
According to Eq. (3.25)
FIGURE 3.10 Prakash and Saran’s bearing capacity factors, N"(e)
FIGURE 3.11 Notations for Eqs. (3.28) and (3.29)
2kN/m 22092
2.0754.119.3058
8.0
)(≈
−=
eccentricuq
+γ+γ×= γ )()()(2
1)1(
eceqfeucNNDBNBQ
For Df /B = 1/2 = 0.5, from Table 3.1, a = 1.754 and k = 0.80. So
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)
kN/m 3888=+=
+×=
)6.9815.962)(2(
)09.56)(1)(5.17()55)(2)(5.17(21
)12(u
Q
107.12
4045tan
2
11.01
245tan1.01
=
+
+=
φ+
+=λ=λ γ B
Df
dqd
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
!!
EXAMPLE 3.5
Solve Example Problem 3.3 using Eq. (3.22)..
Solution For c = 0, from Eq. (3.22)
qu = qNq !qd + –12" B´N" !"d
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 depthfactors are as follows:
qu = (1 × 17.5)(64.2)(1.107) + –12
(17.5)(1.6)(93.69)(1.107)
= 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 beextended to determine the ultimate load on rectangular foundations. Eccentricloading of shallow foundations occurs when a vertical load Q is applied ata location other than the centroid of the foundation (Fig. 3.12a), or when afoundation 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
FIGURE 3.12 Eccentric load on rectangular foundation
eM
Q
eM
Q
LB
BL
=
=
and
(3.31)
(3.32)
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
FIGURE 3.13 One-way eccentricity of load on foundation
According to Meyerhof [1], the ultimate bearing capacity qu and theultimate load Qu of an eccentrically loaded foundation (vertical load) can begiven as
qu = cNc !cs !cd + qNq !qs !qd + –12"B´N" !"s !"d (3.33)
and
Qu = (qu)A´ (3.34)
where A´ = effective area = B´L´B´ = effective widthL´ = effective length
The effective area A´ is a minimum contact area of the foundation such thatits centroid coincides with that of the load. For one-way eccentricity, that is, ifeL =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 .
FIGURE 3.14 Rectangular foundation with one-way eccentricity
λ+
λγ+λγ==
γγ
)()(
)()()()(2
1
)()(
ecsec
eqseqfese
uu
cN
NDBNBLBLqQ
2
)( 2
343.068.0
20.1
−+
−+=λ γ L
B
B
e
L
B
B
eBB
es
Based on their model test results Prakash and Saran [10] suggested that, forrectangular foundations with one-way eccentricity in the width direction (Fig.3.14), the ultimate load may be expressed as
(3.37)
where !"s(e) ,!qs(e) , !cs(e) = shape factors
The shape factors may be expressed by the following relationships
(3.38)
where L = length of the foundation
!qs(e) = 1 (3.39)
and
+=λ
L
Becs
2.01)(
−=
−=
L
eLL
B
eBB
L
B
35.1
35.1
1
1
and
(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
(3.41)
(3.42)
So, the effective area
A´ = –12
B1 L1 (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
A´ = –12
(L1 + L2)B (3.44)
The effective length L´ is the larger of the two values L1 or L2 . The effectivewidth is equal to
B´ = —A´L´
(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
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])
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 ofeL /L and eB /B, the magnitudes of B1 and B2 can be obtained from Figs. 3.20 and3.21. So the effective area can be obtained as
A´ = –12
(B1 + B2)L (3.46)
In this case, the effective length is equal to
L´ = L (3.47)
The effective width can be given as
B´ = —A´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 inFig. 3.22. For this case, the eL /L curves sloping upward in Fig. 3.23 representthe values of B2 /B on the abscissa. Similarly, in Fig. 3.24 the family of eL /L
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])
FIGURE 3.22 Effective area for the case ofeL /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])
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), theeccentricity 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 foun-dation, once the magnitudes of the effective area and the effective width aredetermined, they can be used in Eqs. (3.33) and (3.34) to determine the ulti-mate 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 bedetermined from those presented in Tables 2.3 and 2.4.
2. The shape and depth factors are determined by using the relationshipsgiven in Table 2.5 by replacing B´ for B and L´ for L whenever theyappear.
FIGURE 3.25 Normalized effective dimension of circularfoundations (after Highter and Anders [13])
3. The depth factors are determined from the relationships given in Table2.5. However, for calculating the depth factor, the term B is notreplaced by B´.
e
B
e
LB L= = = =0 4
401
12
60 2
.. ;
..
′ = ′′
= ′ = =BA
L
A
L1
13 02
5192 51
.
.. ft
339.135tan19.551.2
1tan1 =
+=φ
′′
+=λL
Bqs
806.019.551.2
)4.0(14.01 =
−=
′′
−=λγ L
Bs
EXAMPLE 3.5
A shallow foundation measuring 4 ft × 6 ft in plan is subjected to a centric loadand a moment. If eB = 0.45 ft, eL = 1.2 ft, and the depth of the foundation is 3ft, determine the allowable load the foundation can carry. Use a factor of safetyof 4. For the soil given unit weight, ! = 115 lb / ft3 ; friction angle, " = 35!; andcohesion, c = 0. Use Vesic’s N! (Table 2.4), DeBeer’s shape factors (Table2.5), and Hansen’s depth factors (Table 2.5).
Solution For this case
For this type of condition, Case II as shown in Fig. 3.16 applies. Referring toFigs. 3.17 and 3.18
—L1
L = 0.865, or L1 = (0.865)(6) = 5.19 ft
—L2
L = 0.22, or L2 = (0.22)(6) = 1.32 ft
From Eq. (3.44)
A´ = –12
(L1 + L2)B = –12
(5.19 + 1.32)(4) = 13.02 ft2
So
Since c = 0
qu = qNq #qs #qd + –12!B´N! #!s #!d
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)
1
191.14
3)35sin1)(35)(tan2(1
)sin1(tan21
2
2
=λ
=
−+=
φ−φ+=λ
γd
f
qd B
D
Qq A
FSu=
′= =
( , )( . )23 908 13 02
477,820 lb
The depth factors given by Hansen are as follows:
So
qu = (115)(3)(33.3)(1.339)(1.191) + –12
(115)(2.51)(48.03)(0.806)(1)
= 18,321 + 5,587 = 23,908 lb / ft2
So the allowable load on the foundation is
!!
REFERENCES
1. Meyerhof, G. G., The bearing capacity of foundations under eccentric andinclined loads, in Proc., III Intl. Conf. on Soil Mech. Found. Eng., Zurich,Switzerland, 1, 1953, 440.
2. Caquot, A., and Kerisel, J., Tables for the Calculation of Passive Pressure,Active Pressure, and the Bearing Capacity of Foundations, Gauthier-Villars,Paris, 1949.
3. Meyerhof, G. G., The ultimate bearing capacity of foundations, Geotechnique,2, 301, 1951.
4. Meyerhof, G. G., Some recent research on the bearing capacity of foundations,Canadian Geotech. J., 1(1), 16, 1963.
5. Hansen, J. B., A Revised and Extended Formula for Bearing Capacity, BulletinNo. 28, Danish Geotechnical Institute, Copenhagen, 1970.
6. 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.
7. Dubrova, G. A., Interaction of Soils and Structures, Rechnoy Transport,Moscow, 1973.
8. Purkayastha, R. D., and Char, R. A. N., Stability analysis for eccentricallyloaded footings, J. Geotech. Eng. Div., ASCE, 103(6), 647, 1977.
9. Janbu, N., Earth pressures and bearing capacity calculations by generalizedprocedure of slices, in Proc., IV Int. Conf. Soil Mech. Found. Eng., London, 2,1957, 207.
10. Prakash, S., and Saran, S., Bearing capacity of eccentrically loaded footings, J. Soil Mech. Found. Div., ASCE, 97(1), 95, 1971.
11. Terzaghi, K., Theoretical Soil Mechanics, John Wiley, New York, 1943.12. Prakash, S., Soil Dynamics, McGraw-Hill, New York, 1981. 13. Highter, W. H., and Anders, J. C., Dimensioning footings subjected to
eccentric loads., J. Geotech. Eng., ASCE, 111(5), 659, 1985.
CHAPTER FOUR
SPECIAL CASES OF SHALLOW FOUNDATIONS
4.1 INTRODUCTION
The bearing capacity problems described in Chapters 2 and 3 assume that thesoil supporting the foundation is homogeneous and extends to a great depthbelow the bottom of the foundation. They also assume that the ground surfaceis horizontal; however, this is not true in all cases. It is possible to encounter arigid layer at a shallow depth, or the soil may be layered and have differentshear strength parameters. It may be necessary to construct foundations on ornear a slope. Bearing capacity problems related to these special cases will bedescribed in this chapter.
4.2 FOUNDATION SUPPORTED BY A SOIL WITHA RIGID ROUGH BASE AT A LIMITED DEPTH
Figure 4.1a shows a shallow rigid rough continuous foundation supported bya soil which extends to a great depth. The ultimate bearing capacity of thisfoundation can be expressed (neglecting the depth factors) as (Chapter 2)
qu= cNc + qNq + –12!BN! (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 thefailure zone in soil at ultimate load qu is equal to D. The magnitude of Dobtained 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 theevaluation of N! is given in Fig. 4.3.
Now if a rigid rough base is located at a depth of H < D below the bottomof the foundation, full development of the failure surface in soil will berestricted. In such a case, the soil failure zone and the development of slip linesat 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 integrationusing the theory of plasticity. According to Mandel and Salencon’s theory, theultimate bearing capacity of a rough continuous foundation with a rigid roughbase located at a shallow depth can be given by the relation
qu! cNc* " qNq
* " –12! BN!
* (4.2)
FIGURE 4.1 Failure surface under a rigid rough continuous foundation:(a) Homogeneous soil extending to a great depth; (b) Witha 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 rigidbase located at a shallow depth can be given as
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! )
qu = qNq*#*
qs + –12!BN!
*#*!s (4.3)
where #*qs , #
*!s = modified shape factors
The above-mentioned shape factors are functions of H/B and ". Based onthe work of Meyerhof and Chaplin [5] and with simplifying the assumptionthat, in radial planes, the stresses and shear zones are identical to those intransverse planes, Meyerhof [6] evaluated the approximate values of #*
qs and#*
!s as
FIGURE 4.4 Mandel and Salencon’s bearing capacity factor N*c [Eq. (4.2)]
−=λ
−=λ
γ L
Bm
L
Bm
s
qs
2
*
1
*
1
1
and
(4.4)
(4.5)
where L = length of the foundation
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 thecomparison of the experimental evaluation of N!
* for a rough surfacefoundation (Df = 0) on a sand layer with theory. The angle of friction of thesand used for these tests was 35". From Fig. 4.9 the following conclusions canbe 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.
FIGURE 4.6 Mandel and Salencon’s bearing capacity factor N*! [Eq. (4.2)]
3. 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 substantiallylower 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 athigh normal stress levels.
For saturated clay (that is, " = 0), Eq. (4.2) will simplify to the form
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
FIGURE 4.7 Variation of m1 (Meyerhof’s values) for use in the modified shape factor equation [Eq. (4.4)]
for
square
≥−
+
−++π=
02
2
2
22
22
)(
H
B
qcH
Bq
uu
44444 344444 21*
)(
)( 14.5
707.05.0114.5
square
square
cN
uuqcH
B
q +
−+=
obtaining the ultimate bearing capacity of square foundations, or
(4.7)
where cu = undrained shear strengthEquation 4.7 can be rewritten as
(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)
FIGURE 4.10 Foundation on a weaker clay layer underlain by a stronger clay layer [Note: cu (1) < cu (2) ]
Table 4.1 gives the values of Nc*for continuous and square foundations.
TABLE 4.1 Values of Nc*for Continuous
and Square Foundations (""""!!!!0 Condition)
BH
Nc*
Squarea Continuousb
2 3 4 5 6 810
5.435.936.446.947.438.439.43
5.245.716.226.687.208.179.05
aBuisman’s analysis [10]bMandel and Salencon’s analysis [9]
Equations 4.6 and 4.7 assume the existence of a rough rigid layer at alimited depth. However, if a soft saturated clay layer of limited thickness[undrained shear strength = cu (1) ] is located over another saturated clay with asomewhat larger shear strength cu (2) [Note: cu (1) < cu (2) ; see Fig. 4.10], then thefollowing relationship suggested by Vesic [11] and DeBeer [12] may be usedto estimate the ultimate bearing capacity
qc
L
BH
B
c
c
L
Bq
u
u
u
u+
+
−
−+
+=
)1(
)2(
)1(
12
2114.52.01
=
−
−
1)(
2)(
)(,
vu
vu
Lc c
c
B
HfN
(4.9)
where L = length of the foundation
4.3 FOUNDATION ON LAYERED SATURATEDANISOTROPIC CLAY ("""" = 0)
Figure 4.11 shows a shallow continuous foundation supported by layered satu-rated anisotropic clay. The width of the foundation is B, and the interfacebetween the clay layers is located at a depth H measured from the bottom of thefoundation. It is assumed that the clays are anisotropic with respect to strengthfollowing 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
horizontalcu(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 factorHowever, the bearing capacity factor, Nc(L) , will be a function of H/B andcu(v)#2/cu(v)#1, or
(4.12)
where cu(v)#2 = undrained shear strength of the bottom clay layer when the major principal stress is vertical
FIGURE 4.11 Shallow continuous foundation on layered anisotropic saturated clay
α+α=−θ ∫∫θ
−
θ
θ
− dcrdcrbrbqiuiuu
1
0
2)(
2
1
1)(
2 ][2][2)sin(2
Reddy and Srinivasan [14] developed a procedure to determine the varia-tion of Nc(L) . In developing their theory, they assumed that the failure surfacewas cylindrical as shown in Fig. 4.12 when the center of the trial failure surfacewas 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 asshown in Fig. 4.12b.
In Fig. 4.12a, for equilibrium of the foundation, considering forces per unitlength, and taking the moment about point O
(4.13)
where b = half-width of the foundation = B/2r = 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 thedirection 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(%"$) (4.14)
FIGURE 4.12 Assumptions in deriving Nc (L) for a continuous foundation on anisotropic layered clay
Along arc CE
cu (i)#2 = cu (h)#2 + [cu (v)#2#cu (h)#2]sin2(%"$) (4.15)
Similarly, along arc DB
cu (i)#1 = cu (h)#1 + [cu (v)#1#cu (h)#1]sin2(%#$) (4.16)
and, along arc ED
cu (i)#2 = cu (h)#2 + [cu (v)#2#cu (h)#2]sin2(%#$) (4.17)
Kc
c
c
cu v
u h
u v
u h
= =−
−
−
−
( )
( )
( )
( )
1
1
2
2
111)(
2)(
1)(
2)( −
=−
=
−
−
−
−
hu
hu
vu
vu
c
c
c
cn
αψ−α−+++
αψ+α−+++
αψ−α−++
αψ+α−+=−θ
−−−
θ
−−−
θ
−−−
θ
θ
−−−
θ
θ
∫
∫
∫
∫
dcccnr
dcccnr
dcccr
dcccrbrbq
huvuhu
huvuhu
huvuhu
huvuhuu
)}(sin][){1(
)}(sin][{)1(
)}(sin][{
)}(sin][{)sin(2
2
1)(1)(1)(
1
0
2
2
1)(1)(1)(
1
0
2
2
1)(1)(1)(
1
2
2
1)(1)(1)(
1
2
ψ−θ+
ψ+θ−−
ψ−θ+ψ+θ−−
θ−+θ−+θ+θ
−θ
=−
2
)(2sin
2
)(2sin
2
)1(
2)(2sin
2)(2sin
21
)1()1(22
1sin2
11
11
2
2
1)(
Kn
K
KnKn
b
rK
b
r
c
q
vu
u
Note that for the portion of the arc AE, i = %"$ and, for the portion BE, i =%#$. Let the anisotropy coefficient be defined as
(4.18)
The magnitude of the anisotropy coefficient K is less than one for overcon-solidated clays and K > 1 for normally consolidated clays. Also, let
(4.19)
where n = a factor representing the relative strength of two clay layersCombining Eqs. (4.13), (4.14), (4.15), (4.16), (4.17), and (4.18)
(4.20)
Or, combining Eqs. (4.18) and (4.20)
(4.21)
+θ=θ −
r
Hcoscos 1
1 where
Nq
cc Lu
u v( )
( )
=−1
∂∂θ
∂∂
N
N
r
c L
c L
( )
( )
=
=
0
0
and
From Eq. (4.11) note that, with q = 0 (surface foundation),
(4.22)
In order to obtain the minimum value of Nc(L) = qu /cu (v)-1 , the theorem ofmaxima and minima need to be used, or
(4.23)
(4.24)
Equations (4.21), (4.23), and (4.24) will yield two relationships in termsof the variables ! and r/b. So, for given values of H/b, K, n, and ", the aboverelationships may be solved to obtain values of ! and r/b. These can then beused 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 themajor principal stress for anisotropic soils can be taken to be approximatelyequal to 35!. The variations of the bearing capacity factor Nc (L) obtained in thismanner 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 theform (# = 0 condition)
qu = cu (v)-1 Nc (L) $cs $cd + q $qs $qd (4.25)
where $cs , $qs = shape factors$cd , $qd = depth factors
The proper shape and depth factors can be selected from Table 2.5.
FIGURE 4.13 Bearing capacity factor Nc(L) (After Reddy and Srinivasan [15])
FIGURE 4.13 (Continued)
FIGURE 4.13 (Continued)
67.045
30
4.1;15.0
5.0
2
1)(
2)( ==
===
=
−
−
vu
vu
c
c
KB
H
b
H
1
16.10.18.0
)2.0(12.01
1
125.16.1
1)2.0(12.01
=λ
=
+=
+=λ
=λ
=
+=
+=λ
cd
f
cd
qs
cs
B
D
L
B
FSu
all = = =29314
4
.73.29 kN / m 2
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 loadbearing 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
n = 1 " 0.67 = 0.33
So, from Fig. 4.13(d), the value of Nc (L) = 4.75Using Meyerhof’s shape and depth factors given in Table 2.5
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
!!
FIGURE 4.14 Rough continuous foundation on layered soil — stronger over weaker
q qC P
BHu b
a p= ++
−2
1
( sin )δγ
4.4 FOUNDATION ON LAYERED c – #### SOIL — STRONGER SOIL UNDERLAIN BY WEAKER SOIL
Meyerhof and Hanna [16] developed a theory to estimate the ultimate bearingcapacity of a shallow rough continuous foundation supported by a strong soillayer underlain by a weaker soil layer as shown in Fig. 4.14. According to theirtheory, at ultimate load per unit area, qu , the failure surface in soil will be asshown in Fig. 4.14. If the ratio H/B is relatively small, a punching shear failurewill occur in the top (stronger) soil layer followed by a general shear failure inthe bottom (weaker) layer. Considering the unit length of the continuousfoundation, the ultimate bearing capacity can be given as
(4.26)
where B = width of the foundation%1 = unit weight of the stronger soil layerCa = 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
δ
+γ=
δ
γ+
δ
γ=
cos
21
2
1
cos))((
cos2
1
2
1
1
2
1
pHf
pH
f
pH
p
K
H
DH
KHD
KHP
HB
K
H
DH
B
Hcq
HB
K
H
DH
B
Hcqq
pHfab
pHfabu
1
2
1
1
2
1
tan21
2
sin
cos
21
2
12
2
γ−δ
+γ++=
γ−
δ
δ
+γ++=
HB
K
H
DH
B
Hcqq sfa
bu 112
1
tan21
2γ−
φ
+γ++=
Note that, in Eq. (4.26)
Ca = ca H (4.27)
where ca = unit adhesion
(4.28)
where KpH = horizontal component of the passive earth pressure coefficientAlso
qb = c2 Nc (2) + %1 (Df + H)Nq (2) + –12%2 BN%(2) (4.29)
where 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)
(4.30)
Let
KpH tan& = Ks tan#1 (4.31)
where Ks = punching shear coefficientSo
(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 pres-sure coefficient charts proposed by Caquot and Kerisel [17]. Figure 4.15 givesthe variation of Ks with q2 /q1 and #1 . Note that q1 and q2 are the ultimatebearing capacities of a continuous surface foundation of width B under verticalload on homogenous beds of upper and lower soils, respectively, or
q1 = c1 Nc (1) + –12%1 BN%(1) (4.33)
where Nc (1) , N% (1) = bearing capacity factors corresponding to soil friction angle #1
q2 = c2 Nc (2) + –12%2 BN%(2) (4.34)
FIGURE 4.16 Continuous rough foundation on layered soil — H/B is relatively small
t
sfa
buqH
B
K
H
DH
B
Hcqq ≤γ−
φ
+γ++=
112
1
tan21
2
tssf
aa
bu
qHB
K
H
DH
L
B
B
Hc
L
Bqq
≤γ−λ
φ
+γ
++
λ
++=
112
1
tan211
21
If the height H (Fig. 4.14) is large compared to the width B, then the failuresurface will be completely located in the upper stronger soil layer as shown inFig. 4.16. In such case, the upper limit for qu will be of the following form
qu = qt = c1 Nc (1) + qNq (1) + –12%1 BN%(1) (4.35)
Hence, combining Eqs. (4.32) and (4.35)
(4.36)
For rectangular foundations, the preceding equation can be modified as
(4.37)
where $a , $s = shape factors
qb = c2 Nc(2)$cs(2) + %1(Df + H)Nq(2) $qs(2) + –12%2BN%(2) $%s(2) (4.38)
)40.4(
tan2112.0114.5
1
12
12
tf
s
sf
u
qD
B
K
H
DH
L
B
L
Bcq
≤γ+
λφ
+γ
++
+=
φ
+
+γ+
φ
+
+γ=
γ 245tan1.01
2
1
245tan1.01
12
)1(1
12
)1(1
L
BBN
L
BNDq
qft where
q
q
c N
BN
c
BNc2
1
2 2
1 1
2
1 11
2
514
0 5= =( )
( )( )
.
.γ γγ
γ
qt = c1 Nc(1) !cs(1) + "1Df Nq(1) !qs(1) + –12"1BN"(1) !"s(1) (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)], somespecial 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 asgiven in Table 2.5). So
(4.41)
In Eq. (4.41) the relationships for the shape factors !qs and !"s are thosegiven by Meyerhof [18] as shown in Table 2.5. Note that Ks is a function ofq2/q1 [Eqs. (4.33) and (4.34) ]. For this case
(4.42)
Once q2 /q1 is known, the magnitude of Ks can be obtained from Fig. 4.15which, in turn, can be used in Eq. (4.40) to determine the ultimate bearingcapacity of the foundation qu . The value of the shape factor !s for a stripfoundation can be taken as one. For square or circular foundations, as per the
FIGURE 4.17 Hanna and Meyerhof’s analysis — variation of $/#1 with q2 /q1 and #1 (for stronger sand over weaker clay)
experimental work for Hanna and Meyerhof [19], the magnitude of !s appearsto 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 alterna-tive design charts to determine the punching shear coefficient Ks , and thesecharts are shown in Figs. 4.17 and 4.18. In order to use these charts, theensuing 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.
EXAMPLE 4.2
Figure 4.19 shows a continuous foundation. If H = 1.5 m, determine theultimate 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
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!
FIGURE 4.19
++=
+
++=
γ+φ
+γ+=
HKH
K
HH
DB
K
H
DHcq
s
s
fsf
u
4.21342.72.175
)2.1)(5.17(2
40tan)2.1)(2(1))(5.17()30)(14.5(
tan2114.5
2
2
112
12
q
q
c
BN2
1
2
1
514
0 5=
.
. ( )γ γ
q
q2
1
14 30
0 5 17 5 2 93 70 094= =(5. )( )
( . )( . )( )( . ).
(a)
To determine Ks , we need to obtain q2/q1 . From Eq. (4.42)
From Table 2.4 for #1 = 40!, Meyerhof’s value of N" (1) is equal to 93.7. so
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
tuq
HHq ≤
++= 4.2156.282.175 2
φ
+
+γ+
φ
+
+γ=
γ 245tan1.01
2
1
245tan1.01
12
)1(1
12
)1(1
L
BBN
L
BNDq
qft
2kN/m 3.3425.1
4.21)5.1)(56.28(2.175 2 =
++=
uq
ts
sf
buqH
B
K
H
DH
L
Bqq ≤γ−λ
φ
+γ
++=
112
1
tan211
(b)
From Eq. (4.41)
For a continuous foundation B/L = 0. So
qt = "1 Df Nq (1) + –12"1BN"(1)
For #1 = 40 !, Meyerhof’s value of N" (1) = 93.7 and Nq (1) = 62.4 (Table 2.4).Hence
qt = (17.5)(1.2)(62.4) + –12
(17.5)(2)(93.7)
= 1348.2 + 1639.75 = 2987.95 kN / m2 (c)
If H = 1.5 m is substituted into Eq. (b)
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)
(4.43)
where qb = "1(Df + H)Nq(2) !qs(2) + –12"2BN"(2) !"s(2) (4.44)
φ
+
+=λ=λ
φ
+
+=λ=λ
γ
γ
245tan1.01
245tan1.01
22
)2()2(
12
)1()1(
L
B
L
B
sqs
sqs
and
q
q
BN
BN
N
N2
1
2 2
1 1
2 2
1 1
0 5
0 5= =
.
.( )
( )
( )
( )
γγ
γγ
γ
γ
γ
γ
η =q
q2
1
2
2
1
21
H
q
q
a
φ
−φ
=
2
2
2
1
21
2
1
2 zH
q
q
q
qz
′
φ
−φ
+φ
=δ ′
qt = "1Df Nq(1) !qs(1) + –12"1BN"(1) !"s(1) (4.45)
Using Meyerhof’s shape factors given in Table 2.5
(4.46)
(4.47)
For conservative design, for all B/L ratios, the magnitude of !s can be takenas one. For this case
(4.48)
Once the magnitude of q2/q1 is determined, the value of the punching shearcoefficient Ks can be obtained from Fig. 4.15. Hanna [20] suggested that thefriction angles obtained from direct shear tests should be used.
Hanna [20] also provided an improved design chart for estimating thepunching shear coefficient Ks in Eq. (4.43). In this development he assumedthat the variation of $ for the assumed failure surface in the top stronger sandlayer will be of the nature shown in Fig. 4.20, or
$z" = %#2 + az"2 (4.49)
where
(4.50)
(4.51)
So
(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 thetwo soil layers)
2
1
2 φ
=δ
q
q
2
2
2
1
21
2
1
2 )( zHH
q
q
q
qz
−
φ
−φ
+φ
=δ
∫ +
δ
γ=
H
f
z
zpH
pdzDz
KP
0
)(1 )(cos
(4.53)
and at the level of the foundation, that is z! = H
! = "1 (4.54)
Equation (4.49) can also be rewritten as
(4.55)
where ! is the angle of inclination of the passive pressure with respect to thehorizontal at a depth z measured from the bottom of the foundation. So, thepassive force per unit length of the vertical surface aa! (or bb!) is
(4.56)
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 theexpression !z given in Eq. (4.55), can be determined. In order to determine themagnitude of the punching shear coefficient Ks given in Eq. (4.31), we need toknow an average value of !. In order to achieve that, the following steps aretaken:
1. Assume an average value of ! and obtain KpH as given in the tables byCaquot 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) tocalculate 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 designchart for Ks , and this design chart is shown in Fig. 4.22.
FIGURE 4.21 Hanna’s analysis — variation of !/"1
tfaa
cuqD
B
Hc
L
BNc
L
Bq ≤γ+λ
++
+=
1)2(2
212.01
fctDNc
L
Bq
1)1(12.01 γ+
+= where
q
q
c N
c N
c
c
c
cc
c
2
1
2 2
1 1
2
1
2
1
514
514= = =( )
( )
.
.
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)
(4.57)
(4.58)
For conservative design the magnitude of the shape factor $a may be taken asone. The magnitude of the adhesion ca is a function of q2 /q1 . For this condition
(4.59)
Figure 4.23 shows the theoretical variation of ca with q2 /q1 [16].
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
q
q
c
c2
1
2
1
32
800 4= = = .
2kN/m 32.470)1)(18()14.5)(80(3
5.12.01
2.011)1(11
=+
+=
γ+
+=
fcDNc
L
Bq
+=
+
++=
γ+
++
+=
B
H
B
H
DB
Hc
L
BNc
L
Bq
fa
cu
21693.198
)1)(18()72)(2()5.1()14.5)(32)(1.01(
212.01
1)2(2
+==
B
Hq
t21693.19832.470 2kN/m
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 = 0condition) = 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)
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)
With $s = 1, Eq. (4.57) yields
Note that qu = 198.93 + 96(H/B) gives a straight line variation. In order todetermine the value of H/B at which qu is equal to qt , we set
So
FIGURE 4.24
ttbtuq
D
Hqqqq ≥
−−+=
2
1)(
—HB
= 1.26
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 bya 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 thefoundation B, the failure surface at ultimate load will be fully located in theweaker soil layer (see the right-hand side of Fig. 4.25a). For estimating theultimate bearing capacity of such foundations, Meyerhof [6] and Meyerhof andHanna [16] proposed the following semiempirical relationship.
(4.60)
where D = depth of failure surface beneath the foundation in the thick bed of the upper weaker soil layerqt = ultimate bearing capacity in a thick bed of the upper soil layerqb = ultimate bearing capacity in a thick bed of the lower soil layer
So
qt = c1 Nc (1) $cs (1) + #1 Df Nq (1) $qs (1) + –12#1 N# (1) $#s (1) (4.61)
and
qb = c2 Nc (2) $cs (2) + #2 Df Nq (2) $qs (2) + –12#2 N# (2) $#s (2) (4.62)
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) , $#s (2) = shape factors corresponding to the soil friction angle "2
FIGURE 4.25 (a) Foundation on weaker soil layer underlain by stronger sandlayer; (b) Nature of variation of qu with H/B [Eq. (4.60)]
ttbtuq
B
Hqqqq ≥
−−+=
8.1
21)(
[ ]
[ ])1()2(
)(
)2()(
)1()2(
)(
)2()(
qqqmq
m
NND
HNN
NND
HNN
−
−=
−
−=
γ
γγγ
γγ
Equations (4.60), (4.61), and (4.62) imply that the maximum and minimumlimits of qu are qb and qt , respectively (Fig. 4.25). The magnitude of D/B variesfrom 1 for loose sand and clay to about 2 for dense sands [16]. The authorconducted several laboratory model tests and found a better approximation isobtained by modifying Eq. (4.60) as
(4.63)
Based on several laboratory model tests, Hanna [21] proposed the follow-ing relationship for estimating the ultimate bearing capacity qu for a founda-tionresting on a weak sand layer underlain by a strong sand layer.
qu = –12#1 B$
*#s N# (m) + #1 Df $*
qs Nq (m)
" –12#2 B$#s(2) N# (2) + –
12#2 Df $qs(2) Nq (2) (4.64)
where N# (2) , Nq (2) = Meyerhof’s bearing capacity factors with reference to the soil friction angle "2 (Table 2.4)
$#s(2) , $qs(2) = Meyerhof’s shape factors (Table 2.5) with reference to soil friction angle "2
=
φ
+
+
245tan1.01 22
L
B
N# (m) , Nq (m) = modified bearing capacity factors$*#s , $*
qs = modified shape factorsThe modified bearing capacity factors can be obtained as follows
(4.65)
(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. Therelationships 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 bysubstituting D(q) for H and "1 for ". Similarly, the term m2 [Eq. (4.5)] can bedetermined from Fig. 4.8 by substituting D(#) for H and "1 for ".
FIGURE 4.26
D
B
D
Bq( ) ( ). ; .γ = =10 19
5.121)3.3388.134(8.35.0
88.134
3.206)1.377.262(2
5.07.262
)(
)(
=−
−=
=−
−=γ
mq
m
N
N
EXAMPLE 4.4
A shallow foundation is shown in Fig. 4.26. Use Eq. (4.64) and determine theultimate bearing capacity qu .
Solution H = 0.5 m, !1 = 35 , !2 = 45 . From Figs. 4.2 and 4.3 for !1 =35
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)
From Eq. (4.64)
°=φ== 3525.0
25.0
:1
and (4.5) and (4.4) Eqs. FromB
HNote
28.022
72.011
27.02
273.011
2
*
1
*
=
−≈
−=λ
=
−≈
−=λ
γ L
Bm
L
Bm
s
qs
and
583.12
4545tan
2
21.01
245tan1.01
2
22
)2()2(
=
+
+=
φ
+
+=λ=λ γ L
Bqss
qu = –12"1 B#
*"s N" (m) + "1 Df #
*qs Nq (m)
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
qu = qb = –12"2 B#"s(2) N" (2) + –
12"2 Df #qs(2) Nq (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 CLAYWITH A GRANULAR TRENCH
In practice, there are several techniques to improve the load bearing capacityand settlement of shallow foundations on weak compressible soil layers. One
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. Figure4.27 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 widthof the foundation is B, and the depth of the trench is H. The width of thetrench, W, can be smaller or larger than B. The parameters of the strongertrench material and the weak soil for bearing capacity calculation are asfollow:
Trench material Weak soil
Angle of frictionCohesionUnit weight
!1
c1
"1
!2
c2
"2
Madhav and Vitkar [22] assumed a general shear failure mechanism in the soilunder the foundation to analyze the ultimate bearing capacity of the foundationusing the upper bound limit analysis suggested by Drucker and Prager [23], andthis 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
)(2
)(2)(2 2 TTqfTcuN
BNDNcq γ
γ
+γ+=
where !1 = angle of friction of the trench material.3. 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
(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 andc2 = cu ) determined by Madhav and Vitkar [22] are given in Figs. 4.28, 4.29and 4.30. The values of N" (T) given in Fig. 4.30 are for "1 /"2 = 1. In an actualcase, 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 verifythe above theory. Hamed, Das, and Echelberger [24] conducted severallaboratory model tests to determine the variation of the ultimate bearing capa-city of a strip foundation resting on a granular trench (sand, c1 = 0) made in asaturated soft clay medium (c2 = cu ; !2 =0). For these tests the width of thefoundation 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 experi-mental qu with H/B. In both cases the experimental ultimate bearing capacityincreased with H/B up to a maximum value and remained practically constantthereafter. 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 ofqu when compared to the experimental maximum value of qu .
2. The minimum height H of the granular trench necessary to obtain themaximum 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 forestimating the maximum qu (for c1 = 0, c2 = cu , !2 = 0, and B = W). This pro-cedure can be explained by referring to Fig. 4.33, in which A and B are two soilelements. For soil element A the major principal stress is 'I , and the minorprincipal stress is 'III . So
FIGURE 4.28 Madhav and Vitkar’s bearing capacity factor, Nc (T)
FIGURE 4.29 Madhav and Vitkar’s bearing capacity factor, Nq (T)
φ
+
=
245tan 12
)1(
=
tcoefficien pressureearth passive Rankine wherep
K
σ σ1 3 2 22= +K c Kp u p( ) ( )
'I = qu = Kp (1) 'III (4.68)
For soil element B the major principal stress is '1 and the minor principalstress is '3 . However,
'3 = "2 Df
and
φ+
=
245tan 22
)2(
=
tcoefficien pressureearth passive Rankine wherep
K
Since 2 = 0, the value of Kp (2) = 1. Hence
Again referring to Figure 4.33, it can be seen that, at failure (which is of thebulging type)
FIGURE 4.30 Madhav and Vitkar’s bearing capacity factor, N " (T)
'1 = '3 + 2cu = "2 Df + 2cu (4.69)
!
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)
FIGURE 4.33 Derivation of Eq. (4.71)
φ
++γ=+γ=2
45tan)2()2( 12
22)1( ufufpucDcDKq
'1 = 'III (4.70)
Now, combining Eq.s (4.68), (4.69), and (4.70)
(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 thetheoretical qu [Eq. (4.71)] is equal to or somewhat less than the maximumexperimental 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 ofthe soil and the bedrock leaving void spaces. Estimating the ultimate bearingcapacity of shallow foundations constructed over these voids, as well as thestability of the foundations, is gradually becoming an important issue. Only afew studies have been reported in published literature so far. Baus and Wang[25] reported some experimental results for the ultimate bearing capacity of ashallow 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 Hbelow the bottom of the foundation. The void is continuous and has cross-
FIGURE 4.34 Shallow continuous rough foundation over a void
f J J k= + = ′α 1 2
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 = 0.28Unit weight of compacted soil, " = 18.42 kN / m 3
The results of Baus and Wang [25] are shown in a nondimensional form inFig. 4.35. Note that the results of the tests which constitute Fig. 4.35 are for thecase 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 in-crease in the void width, W!.
2. For any given W!/B, there is a critical H/B ratio beyond which thevoid 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 oftheir experimental findings. In the finite element analysis, the soil was treatedas an elastic, perfectly plastic material. They also assumed that Hooke’s lawis void in the elastic range, and the soil follows the von Mises yield criterion inthe perfectly plastic range, or
(4.72)
FIGURE 4.35 Experimental bearing capacity of a continuous foundation as a function of void size and location [1]
&f = 0
where yield function
=tan
(9 +12tan )
(9 +12tan )
0.5
0.5
f
kc
=
′ =
αφ
φ
φ3
(4.73)
(4.74)
(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 ofDrucker and Prager [23]. The results of the finite element analysis have showngood agreement with experiments.
4.8 FOUNDATION ON A SLOPE
In 1957, Meyerhof [26] proposed a theoretical solution to determine theultimate bearing capacity of a shallow foundation located on the face of aslope. Figure 4.36 shows the nature of the plastic zone developed in the soil
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 amixed 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. Theshear 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
qu = cNc + po Nc + –12"BN" (4.76)
The preceding relationship can also be expressed as
qu = cNcq + –12"BN"q (4.77)
where Ncq , N"q = bearing capacity factorsFor 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
Ns = –—"Hc
(4.79)
where H = height of the slopeIn a similar manner, for a granular soil (c = 0)
qu = –12"BN"q (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 ofa slope of height H. It is located at a distance b from the edge of the slope. Theultimate bearing capacity of the foundation can be expressed by Eq. (4.77), or
qu = cNcq + –12"BN"q (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
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 cohesivesoil (! = 0) and N"q for a granular soil (c = 0), and these variations are shownin Figs. 4.40 and 4.41. Note that, for purely cohesive soil (Fig. 4.40)
qu = cNcq
and, for granular soil (Fig. 4.41)
qu = –12"BN"q
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 betaken as zero when B < H. If B # H the curve for the actual stability numbershould be used.
Typical plots of load per unit area q versus settlement s obtained by theauthor 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 seenthat, for similar foundation conditions, the settlement at ultimate load decreaseswith 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
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)
λ λ ββ γβq = = −( tan )1 2
located at the edge of the slope), Hansen [27] proposed the followingrelationship for the ultimate bearing capacity of a continuous foundation
qu = cNc $c# + qNq $q# + –12"BN" $"# (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 factorsq = "Df
According to Hansen [27]
(4.83)
λλ
φ
λ βπ
φ
ββ
β
cq q
q
c
N
N=
−−
>
= −+
=
1
1
12
2
(for 0)
(for 0)
22 )tan1(sin)tan1(14.5
21)14.5( β−βγ−β−γ+
β−= BDcq
fu
(4.84)
(4.85)
For ! = 0 condition, Vesic [11] pointed out that, with the absence ofweight due to the slope, the bearing capacity factor N" has a negative value andcan be given as
N" = $2sin# (4.86)
Thus for ! = 0 condition, with Nc = 5.14 and Nq = 1, Eq. (4.82) takes the form
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 ultimatebearing capacity of shallow continuous foundations on the top of a slope (Fig.4.39) using the limit equilibrium and limit analysis approach. According tothis theory, for a strip foundation
qu = cNc + qNc + –12"BN" (4.88)
where Nc , Nq , N" = bearing capacity factorsq = "Df
Referring to the notations used in Fig. 4.39, the numerical values of Nc , Nq , andN" are given in Table 4.2.
Stress Characteristics SolutionAs shown in Eq. (4.81), for granular soils (that is, c = 0)
qu = –12"BN"q (4.89)
TABLE 4.2 Bearing Capacity Factors Based on Saran, Seed, and Handa’s Analysis
Factor#
(deg)
Df
B
bB
Soil friction angle, ! (deg)
40 35 30 25 20 15 10
N" 30 20 10 0
0 0 25.3753.48
101.74165.39
12.4124.5443.3566.59
6.1411.6219.6528.98
3.205.619.19
13.12
1.264.274.356.05
0.701.791.962.74
0.100.450.771.14
30 20 10 0
0 1111 60.0685.98
125.32165.39
34.0342.4955.1566.59
18.9521.9325.8628.89
10.3311.4212.2613.12
5.455.896.056.05
0.001.352.742.74
30 25 20%15
1111 0 91.87115.65143.77165.39
49.4359.1266.0066.59
26.3928.8028.8928.89
30 25%20
111 111 131.34151.37166.39
64.3766.5966.59
28.8928.8928.89
Nq 30 20%10
111 0 12.1312.6781.30
16.4219.4841.40
8.9816.8022.50
7.0412.7012.70
5.007.407.40
3.604.404.40
30 20%10
111 111 28.3142.2581.30
24.1441.441.4
22.522.522.5
Nc 50 40 30 20%10
0 0 21.6831.8044.8063.2088.96
16.5222.4428.7241.2055.36
12.6016.6422.0028.3236.50
10.0012.8016.2020.6024.72
8.6010.0412.2015.0017.36
7.108.008.60
11.3012.61
5.506.256.708.769.44
50 40 30 20%10
0 11111 38.8048.0059.6475.1295.20
30.4035.4041.0750.0057.25
24.2027.4230.9235.1636.69
19.7021.5223.6027.7224.72
16.4217.2817.3617.3617.36
50 40 30 20%10
11111 0 35.9751.1670.5993.7995.20
28.1137.9550.3757.2057.20
22.3829.4236.2036.2036.20
18.3822.7524.7224.7224.72
15.6617.3217.3617.3617.36
10.0012.1612.1612.1612.16
50 40 30%20
1111 1111 53.6567.9885.3895.20
42.4751.6157.2557.25
35.0036.6936.6936.69
24.7224.7224.7224.72
Graham, Andrews, and Shields [29] provided a solution for the bearing capa-city factor, N"q , for a shallow continuous foundation on the top of a slope in
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 showsthe 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 methodare 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 con-tinuous foundations located at the top of a slope of cohesionless sand. Basedon these 215 tests, Gemperline proposed that the ultimate bearing capacity ofa continuous foundation can be expressed as
FIGURE 4.44 Graham et al.’s theoretical values of N!q (Df /B = 0)
FIGURE 4.45 Graham et al.’s theoretical values of N!q (Df /B = 0.5)
FIGURE 4.46 Graham et al.’s theoretical values of N!q (Df /B = 1)
( )[ ]
)90.4(
tan2
2tan33.01
tan2
2tan118.0165.01
2
2
2
β
+
β
−×
β
+
β−−−
+=
γ
γ
B
bB
D
B
bB
D
N
N
f
f
qR
q
Given Since use D
B
b
B
H
BN
fs= = = = > =15
151
0
150 1 0
.
.,
.. , .
qu = –12!BN!q
Shields, Chandler, and Garnier [31] normalized the N!q values proposed byGemperline in the following form
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
FIGURE 4.47
From Fig. 4.40, for and the
value of is about 5.85. So
D
B
b
BN
N
fs
cq
= = = ° =1 0 30 0, , , ,β
2kN/m 203=
−−
−+
×π−=
β−βγ−β−γ+β−=
2
2
22
)30tan1)(30)(sin5.1)(5.18(
)30tan1)(5.1)(5.18()49(30180
)2(14.5
)tan1(sin)tan1()214.5( BDcqfu
qu = (49)(5.85) = 286.7 kN / m 2
b. From Eq. (4.87)
!!
EXAMPLE 4.6
Figure 4.48 shows a continuous foundation on a slope of a granular soil. Esti-mate 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)]
FIGURE 4.48
Given and = 30 .b
B
D
Bf= = = = = ° °15
151
15
151 40
.
.,
.
., ,φ β
For and = 30 , the value of
and the value of (Table 4.2)
b
B
D
BN
N
f
q
= = = ° ° =
=
1 1 40
13134 28 31
, , ,
. .
φ β γ
Solution
a. For granular soil (c = 0) from Eq. (4.81)
qu = –12!BN!q
From Fig. 4.41, N!q # 120. So
qu = –12
(16.8)(1.5)(120) = 1512 kN / m 2
b. For c = 0, from Eq. (4.88)
qu = qNq + –12!BN!
qu = (16.8)(1.5)(28.31) + –12
(16.8)(1.5)(131.34) # 2368 kN / m 2
c. From Eq. (4.89)
qu = –12!Bn!q
From Fig. 4.46b, N!q # 110
qu = –12
(16.8)(1.5)(110) = 1386 kN / m 2
!!
REFERENCES
1. Prandtl, L., Uber die eindringungsfestigkeit plastisher baustoffe und diefestigkeit 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 ofthe bearing capacity of continuous footings on sand, in Proc., III Int. Conf. SoilMech. Found. Engg., Zurich, Switzerland, 1, 409, 1953.
4. Mandel, J., and Salencon, J., Force portante d’un sol sur une assise rigide (étudetheorizue), Geotechnique, 22(1), 79, 1972.
5. Meyerhof, G. G., and Chaplin, T. K., The compression and bearing capacity ofcohesive soils, Br. J. Appl. Phys., 4, 20, 1953.
6. Meyerhof, G. G., Ultimate bearing capacity of footings on sand layer overlyingclay. Canadian Geotech. J., 11(2), 224, 1974.
7. Milovic, D. M., and Tournier, J. P., Comportement de foundations reposant surune coche compressible d´épaisseur limitée, in Proc., Conf. Comportement desSols Avant la Rupture, Paris, France, 1971, 303.
8. Pfeifle, T. W., and Das, B. M., Bearing capacity of surface footings on sand layerresting 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, inProc., 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 andApplications, Sayed, S. M., Ed., Gulf Publishing Co., 1975, 212.
13. Casagrande, A., and Carrillo, N., Shear failure in anisotropic materials, inContribution 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 layeredclays, 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 onlayered soils under inclined load, Canadian Geotech. J., 15(4), 565,1978.
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 capacityfor 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 overlyinga strong deposit, Canadian Geotech. J., 19(3), 392, 1982.
22. Madhav, M. R., and Vitkar, P. P., Strip footing on weak clay stabilized with agranular trench or pile, Canadian Geotech. J., 15(4), 605, 1978.
23. Drucker, D. C., and Prager, W., Soil mechanics and plastic analysis of limitdesign, Q. Appl. Math., 10, 157, 1952.
24. Hamed, J. T., Das, B. M., and Echelberger, W. F., Bearing capacity of a stripfoundation on granular trench in soft clay. Civil Engineering for Practicing andDesign 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, inProc., 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 toslopes, J. Geotech. Eng., ASCE, 115(4), 553, 1989.
29. Graham, J., Andrews, M., and Shields, D. H., Stress characteristics for shallowfootings 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 inslopes, J. Geotech. Eng., ASCE, 116(3), 528, 1990.
17. Caquot, A., and Kerisel, J., Tables for the Calculation of Passive Pressure,
FIGURE 5.1 Load-settlement curve for shallow foundation
CHAPTER FIVE
SETTLEMENT AND ALLOWABLE BEARING CAPACITY
5.1 INTRODUCTION
Various theories relating to the ultimate bearing capacity of shallow founda-tions were presented in Chapters 2, 3, and 4. In Section 2.12, a number ofdefinitions for the allowable bearing capacity were discussed. In the design ofany foundation, one must consider the safety against bearing capacity failureas well as against excessive settlement of the foundation. In the design of mostfoundations, there are specifications for allowable levels of settlement. Referto Fig. 5.1 which is a plot of load per unit area q versus settlement S for a foun-dation. The ultimate bearing capacity is realized at a settlement level of Su . LetSall be the allowable level of settlement for the foundation and qall(S) be thecorresponding allowable bearing capacity. If FS is the factor of safety againstbearing capacity failure, then the allowable bearing capacity is qall (b) = qu /FS.
The settlement corresponding to qall (b) is S´. For foundations with smallerwidths 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, forlarger foundation widths, the allowable settlement controls. This chapterdescribes the procedures for estimating the settlement of foundations underload and thus estimating the allowable bearing capacity.
The settlement of a foundation can have three components: (a) elasticsettlement, Se ; (b) primary consolidation settlement, Sc ; (c) secondary consoli-dation 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 ornegligible.
Elastic settlement is caused by deformation of dry soil, as well as moist andsaturated soils, without any change in moisture content. Primary consoli-dationsettlement is a time-dependent process which occurs in clayey soils locatedbelow the ground water table as a result of the volume change in soil becauseof the expulsion of water that occupies the void spaces. Secondary consolida-tion settlement follows the primary consolidation process in saturated clayeysoils and is a result of the plastic adjustment of soil fabrics. The pro-cedures forestimating the above three types of settlement are discussed in this chapter.
Any type of settlement is a function of the additional stress imposed on thesoil by the foundation. Hence, it is desirable to know the relationships forcalculating the stress increase in the soil caused by application of load to thefoundation. These relationships are given in Section 5.2 and are derivedassuming 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 dueto a point load Q acting on the surface of a semi-infinite mass. In Fig. 5.2 thestress increase at a point A is shown in the Cartesian coordinate system, and thestress increase in the cylindrical coordinate system is shown in Fig. 5.3. Thecomponents of the stress increase can be given by the following relationships.
Cartesian coordinate system (Fig. 5.2)
FIGURE 5.2 Boussinesq’s problem—Stress increase at a point in the Cartesiancoordinate system due to a point load on the surface
−
++−
+ν−+
π=σ
π=σ
323
2
5
2
5
3
)(
)2(
)(
1
3
21
2
3
2
3
R
z
zRR
xzR
zRRR
zxQ
R
Qz
x
z
+
+ν−−
π=τ
−
++
−+
ν−+
π=σ
235
323
2
5
2
)(
)2(
3
21
2
3
)(
)2(
)(
1
3
21
2
3
zRR
xyzR
R
xyzQ
R
z
zRR
yzR
zRRR
zyQ
xy
y
τπxz
Q xz
R=
3
2
2
5
(5.1)
(5.2)
(5.3)
(5.4)
(5.5)
FIGURE 5.3 Boussinesq’s problem—Stress increase at a point in the cylindricalcoordinate system due to a point load on the surface
τπyz
Q yz
R= 3
2
2
5
R z r
r x y
= +
= +
2 2
2 2
(5.6)
where ! = normal stress" = shear stress
# = Poisson’s ratio
Cylindrical coordinate system (Fig. 5.3):
+ν−−
π=σ
π=σ
)(213
2
23
5
2
5
3
zRRR
zrQ
R
Qz
r
z
5
2
3
2
3
)(1
)21(2
R
Qrz
R
z
zRR
Q
rz π=τ
−
+ν−
π=σθ
dz qrd dr
r zzσ θ
π=
+3
2
3
2 2 5/2( )
∫ ∫ ∫
+−=
+πθ=σ=σ
=
π
=θ
2/322
3
0
2
0
2/322
3
)(1
)(2
3
zR
zq
zr
drqrdzd
R
r
zz
+
++
ν+−ν+=σ=σ θ 2/322
3
2/122 )()(
)1(221
2 zR
z
zR
zqr
(5.7)
(5.8)
(5.9)
(5.10)
Uniformly Loaded Flexible Circular Area
Boussinesq’s solution for a point load can be extended to determine the stressincrease due to a uniformly loaded flexible circular area on the surface of asemi-infinite mass (Fig. 5.4). In Fig. 5.4 the circular area has a radius R, and theuniformly distributed load per unit area is q. If the components of stressincrease at a point A below the center are to be determined, then we consideran 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 for R in Eq.r z2 2+5.7. Thus
The vertical stress increase, !z , due to the entire loaded area is then
(5.11)
Similarly, the magnitudes of !$ and !r below the center can be obtained as
(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
dqz dxdy
x y zzσ
π=
+ + +3
2
3
2 2 2 5 2( ) /
∫ ∫ ∫ =++π
=σ=σ= =
qIzyx
dxdyqzd
B
y
L
x
zz
0 0
2/5222
3
)(2
3
Table 5.1 gives the variation of !z /q at any point A below a circularly loadedflexible 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 byAhlvin and Ulery [2].
TABLE 5.1 Variation of !!!!z /q at a point A (Figure 5.5)
r/R !
"z/R
!z /q
0 0.2 0.4 0.6 0.8 1.0
0 0.20.40.60.81.01.52.02.53.04.05.0
1.0000.9920.9790.8640.7560.6460.4240.2840.2000.1460.0870.057
1.0000.9910.9430.8520.7420.6330.4160.2810.1970.1450.0860.057
1.0000.9870.9200.8140.6990.5910.3920.2680.1960.1410.0850.056
1.0000.9700.8600.7320.6190.5250.3550.2480.1880.1350.0820.054
1.0000.8900.7130.5910.5040.4340.3080.2240.1670.1270.0800.053
0.5000.4680.4350.4000.3660.3320.2880.1960.1510.1180.0750.052
Uniformly Loaded Flexible Rectangular Area
Figure 5.6 shows a flexible rectangular area of length L and width B subjectedto a uniform vertical load of q per unit area. The load on the elemental area dAis equal to dQ = qdxdy. This can be treated as an elemental point load. Thevertical stress increase d!z due to this at A, which is located at a depth z belowthe corner of the rectangular area, can be obtained by using Eq. (5.7), or
(5.13)
Hence, the vertical stress increase at A due to the entire loaded area is
(5.14)
FIGURE 5.6 Stress increase below the corner of a uniformly-loaded flexiblerectangular area
z
Ln
z
Bm
nmnm
nmmn
nm
nm
nmnm
nmmn
I
=
=
+−+++
+
++++
×+++
++
π=
−
1)1(2
tan
12
1)1(2
41
2222
5.0221
22
22
2222
5.022
where (5.15)
Table 5.2 shows the variation of I with m and n. The stress below any otherpoint C below the rectangular area (Fig. 5.7) can be obtained by dividing it intofour 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.
TABLE 5.2 Variation of I with m and n
n
m
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
TABLE 5.2 Continued
n
m
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 fourrectangles can be determined. The total stress increase below point C at depthz 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 loadedrectangular area is of primary importance. The vertical stress increase below thecenter of a uniformly-loaded flexible rectangular area can be calculated as
=
=
+++
++++
++
π=σ
−
2
1sin
))(1(
21
12
1
1
2
1
2
1
2
1
11
2
1
2
1
2
1
2
1
2
1
2
1
2
1
11
)(
B
zn
B
Lm
nnm
m
nmn
nm
nm
nm
qcz
where
(5.17)
(5.18)
(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)]
z/B
L/B
1 2 3 4 5 6 7 8 9 10
0.1 0.994 0.997 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
FIGURE 5.8
L
B
z
B= = = =6
415
10
42 5. ; .
EXAMPLE 5.1
Figure 5.8 shows the plan of a flexible loaded area located at the groundsurface. The uniformly distributed load q on the area is 150 kN / m2. Deter-mine the stress increase !z below points A and C at a depth of 10 m below theground surface. Note C is at the center of the area.
Solution
Stress increase below point A: The following table can now be prepared.
AreaNo.
B(m)
L(m)
z(m) m=B/z n=L/z
I(Table 5.2)
1234
2222
2442
10101010
0.20.20.20.2
0.20.40.40.2
0.01790.03280.03280.0179
!0.1014
From Eq. (5.14)
!z = qI = (150)(0.1014) = 15.21 kN / m2
Stress increase below point C:
σ
σ
z
z
q≈
= =
0104
0104 150
.
( . )( ) 15.6 kN / m 2
)foundation s(continuou 20
21
2
−π
=σ =
B
x
qz
)foundation(circular 20
212
−
=σ =
B
x
qz
From Table 5.3
!!
ELASTIC FOUNDATIONS
5.3 FLEXIBLE AND RIGID FOUNDATIONS
Before discussing the relationships for elastic settlement of shallow founda-tions, it is important to understand the fundamental concepts and the dif-ferences between a flexible foundation and a rigid foundation. When a flexiblefoundation 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 alsoshows the settlement profile of the foundation. If a similar foundation is placedon a granular soil it will undergo larger elastic settlement at the edges ratherthan 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 distri-bution will be as shown in Fig. 5.10a. If this rigid foundation is placed on agranular 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 perfectlyelastic material, the contact pressure can be expressed as (Fig. 5.11)
(5.20)
(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
FIGURE 5.9 Contact pressure and settlement for flexible foundation: (a) Elastic material; (b) Granular soil
3
2
2
21
1
6
1
ν−ν−
=
Bt
E
E
K
s
f
f
s=
factor stiffness relative where
!z=0 = f (K) (5.22)
(5.23)
"s = Poisson’s ratio of the elastic material"f = Poisson’s ratio of the foundation materialt = 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 ofelasticity may be used to estimate the settlement of shallow foundations atallowable loads. Judicious use of these results have done well in the design,construction, and maintenance of structures.
[ ])(1
θσ+σν−σ=∈rz
s
z E
[ ]dzE
dzrz
z sz
z)(
1θ
∞∞
σ+σν−σ= ∫∫∈
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) onthe surface of an elastic material (in this case soil which is assumed to beelastic). The radius of the circular area is R. The vertical strain #z at a pointA due to the loading can be given as
(5.24)
where Es = modulus of elasticity of soil" = Poisson’s ratio of soil
The elastic settlement at A can be evaluated as
(5.25)
FIGURE 5.11 Contact pressure distribution under an infinitely rigidfoundation supported by a perfectly elastic material
FIGURE 5.12 Settlement under a uniformly loaded flexible circular area
ν−+ν+=
21)1(
1II
R
zR
EqS
s
e
=
R
r
R
zfII and and where
21
S qE
RIes
= −1 2
2ν
Sq R
E
qB
E
SqB
E
qB
E
es s
es s
( )( ) ( )
( ). ( ) . ( )
center
edge
= − = −
= − = −
2 1 1
1273 1
2
0 636 1
2 2
2 2
ν ν
ν ν
By substituting the proper relations for !z , !$ , and !r into Eq. (5.25), weobtain
(5.26)
The variations of I1 and I2 at r/R = 0 and 1 (that is, below the center andedge of the loaded area) are given in Table 5.4 and 5.5.
At the surface, that is, z/R = 0
(5.27)
Using Tables 5.4 and 5.5 and Eq. (5.27), at the surface
(5.28)
(5.29)
where B = 2R = diameter of the loaded area
TABLE 5.4 Variations TABLE 5.5 Variations of I1 and I2 for r/R = 0 of I1 and I2 for r/R =1
z/R I1 I2 z/R I1 I2
00.20.40.60.81.01.52.02.53.04.05.0
1.0000.8040.6290.4860.3750.2930.1680.1060.0720.0510.0300.019
2.0001.6401.3501.1300.9610.8280.6060.4720.3850.3250.2460.198
00.20.40.60.81.01.52.02.53.04.05.0
0.5000.3830.3100.2560.2130.1790.1190.0830.0600.0450.0270.018
1.2731.1000.9620.8490.7560.6780.5310.4320.3620.3100.2390.195
FIGURE 5.13 Average settlement under a uniformly loaded flexible circular area (Note: Diameter = B = 2R)
B
dxS
S
B
B
e
e
∫−=
2/
2/)(average
SqB
Ees
( ). ( )
average = −0 85 1 2ν
S SqB
Ee es
( ) .. ( )
rigid (average flexible)≈ ≈−
0 930 79 1 2ν
The average settlement at the surface can now be determined as (Fig.5.13)
For most practical purposes, Se (average) ! 0.85Se (center). Hence, at thesurface
(5.30)
For rigid foundations, the surface settlement (z/R = 0) is
(5.31)
EXAMPLE 5.2
A flexible circularly loaded area located on the ground surface has a radius of1.5 m. The uniformly distributed load on the area is 250 kN / m2. Determinethe settlement at the ground surface for the following conditions:
a. Below the centerb. At the edgec. Average
Assume Poisson’s ratio to be 0.3 and Es = 9500 kN / m2.
SqB
Ees
= − = × − =( ) ( )( . )( . ).
1 250 2 15 1 0 3
95000 0718
2 2ν m = 71.8 mm
SqB
Ees
= − = =0 636 10 636 718
2. ( )( . )( . )
ν45.66 mm
SqB
Ees
=−
= =0 85 1
0 85 7182. ( )
( . )( . )ν
61.03 mm
SqB
Ees
= − = × −
=
0 79 1 0 79 250 2 15 1 0 3
9500
0 0568
2 2. ( ) ( . )( )( . )( . )
.
ν
m = 56.8 mm
ν−ν−−ν−=
43
2
1
21)1(
2)( II
E
qBS
s
ecorner
Solution
a. From Eq. (5.28)
b. From Eq. (5.29)
c. From Eq. (5.30)
!!
EXAMPLE 5.3
Refer to Example 5.2. Assume the loaded area to be rigid and determine theelastic surface settlement.
Solution From Eq. (5.31)
!!
5.5 SETTLEMENT UNDER A RECTANGULAR AREA
The elastic settlement at any depth below the corner of a flexible rectangulararea of dimension L × B (Fig. 5.6) can be obtained by proper integration of theexpression for vertical strain as [4]
(5.32)
−′+′++′+′+′+
′−′+′+′+′+′+
π=
11
11ln
1
1ln
122
22
22
22
3nm
nmm
mnm
mnmI where
B
zn
B
Lm
nmn
mnI
=′
=′
′+′+′
′π′
= −
22
1
41
tan
SqB
EIe ( ) ( )corner = −
21 2
5ν
−′++′+′+
′−′+′+′+
π=
11
11ln
1
1ln
12
2
2
2
5m
mm
mm
mmI
SqB
EIe
s
( ) ( )center = −1 25ν
SqB
EI
qB
EIe
s s
( ).
( ) ( )average = − = −0 851 12
52
6ν ν
(5.33)
(5.34)
(5.35)
(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 thesurface
(5.37)
where
(5.38)
Using the method of superposition it can be shown that, at the surface,
(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
(5.40)
SqB
EI
qB
EIe
s s
( ).
( ) ( )rigid = − = −0 791 12
52
7ν ν
TABLE 5.6 Variation of I3
n!
m!
1 2 3 4 5 6 7 8 9 10
0.000.250.500.751.001.251.501.752.002.252.502.753.003.253.503.754.004.254.504.755.005.255.505.756.006.256.506.757.007.257.507.758.008.258.508.759.009.259.509.75
10.00
1.1221.0951.0250.9330.8380.7510.6740.6080.5520.5040.4630.4270.3960.3690.3460.3250.3060.2890.2740.2600.2480.2370.2270.2170.2080.2000.1930.1860.1790.1730.1680.1620.1580.1530.1480.1440.1400.1370.1330.1300.126
1.5321.5101.4521.3711.2821.1921.1061.0260.9540.8880.8290.7760.7280.6860.6470.6120.5800.5510.5250.5010.4790.4580.4400.4220.4060.3910.3770.3640.3520.3410.3300.3200.3100.3010.2930.2850.2770.2700.2630.2570.251
1.7831.7631.7081.6321.5471.4611.3781.2991.2261.1581.0951.0370.9840.9350.8890.8480.8090.7740.7410.7100.6820.6550.6310.6080.5860.5660.5470.5290.5130.4970.4820.4680.4550.4420.4300.4190.4080.3980.3880.3790.370
1.9641.9441.891.8161.7341.6501.5701.4931.4211.3541.2911.2331.1791.1281.0811.0370.9950.9570.9210.8870.8550.8250.7970.7700.7450.7220.6990.6780.6580.6390.6210.6040.5880.5730.5580.5440.5310.5180.5060.4940.483
2.1052.0852.0321.9591.8781.7961.7171.6411.5711.5051.4441.3861.3321.2811.2341.1891.1471.1071.0701.0341.0010.9690.9390.9110.8840.8580.8340.8100.7880.7670.7470.7280.7100.6920.6760.6600.6440.6300.6160.6020.589
2.2202.2002.1482.0761.9951.9141.8361.7621.6921.6271.5671.5101.4571.4061.3591.3151.2731.2331.1951.1591.1251.0931.0621.0321.0040.9770.9520.9270.9040.8810.8600.8390.8200.8010.7830.7650.7480.7320.7170.7020.688
2.3182.2982.2462.1742.0942.0131.9361.8621.7941.7301.6701.6131.5611.5111.4651.4211.3791.3391.3011.2651.2311.1991.1671.1371.1091.0821.0551.0301.0060.9830.9600.9390.9180.8990.8790.7610.8430.8260.8100.7940.778
2.4032.3832.3312.2592.1792.0992.0221.9491.8811.8171.7581.7021.6501.6011.5551.5111.4701.4311.3931.3581.3231.2911.2601.2301.2011.1731.1471.1211.0971.0731.0501.0281.0070.9870.9670.9480.9300.9120.8950.8780.862
2.4772.4582.4062.3342.2552.1752.0982.0251.9581.8941.8361.7801.7291.6801.6341.5911.5501.5111.4761.4381.4041.3721.3411.3111.2821.2551.2281.2031.1781.1541.1311.1091.0871.0661.0461.0271.0080.9900.9720.9550.938
2.5442.5252.4732.4012.3222.2422.1662.0932.0261.9631.9041.8501.7981.7501.7051.6621.6211.5821.5451.5101.4771.4441.4131.3841.3551.3281.3011.2751.2511.2271.2041.8111.1601.1391.1181.0991.0801.0611.0431.0261.009
(5.41)
The variations of I6 and I7 with L/B are given in Table 5.9.
TABLE 5.7 Variation of I4
n!
m!
1 2 3 4 5 6 7 8 9 10
0.250.500.751.001.251.501.752.002.252.502.753.003.253.503.754.004.254.504.755.005.255.505.756.006.256.506.757.007.257.507.758.008.258.508.759.009.259.509.75
10.00
0.0980.2480.1660.1670.1600.1490.1390.1280.1190.1100.1020.0960.0900.0840.0790.0750.0710.0670.0640.0610.0590.0560.0540.0520.0500.0480.0460.0450.0430.0420.0400.0390.0380.0370.0360.0350.0340.0330.0320.032
0.1030.1670.2020.2180.2220.2200.2130.2050.1960.1860.1770.1680.1600.1520.1450.1380.1320.1260.1210.1160.1110.1070.1030.0990.0960.0930.0890.0870.0840.0810.0790.0770.0740.0720.0700.0690.0670.0650.0640.062
0.1040.1720.2120.2340.2450.2480.2470.2430.2370.2300.2230.2150.2080.2000.1930.1860.1790.1730.1670.1610.1550.1500.1450.1410.1360.1320.1280.1240.1210.1170.1140.1110.1080.1050.1030.1000.0980.0950.0930.091
0.1050.1740.2160.2410.2540.2610.2630.2620.2590.2550.2500.2440.2380.2320.2260.2190.2130.2070.2010.1950.1900.1850.1790.1740.1700.1650.1610.1560.1520.1490.1450.1410.1380.1350.1320.1290.1260.1230.1200.118
0.1050.1750.2180.2440.2590.2670.2710.2730.2720.2690.2660.2620.2580.2530.2480.2430.2370.2320.2270.2210.2160.2110.2060.2010.1970.1920.1880.1830.1790.1750.1710.1680.1640.1600.1570.1540.1510.1470.1450.142
0.1050.1750.2190.2460.2620.2710.2770.2790.2790.2780.2770.2740271
0.2670.2630.2590.2540.2500.2450.2410.2360.2320.2270.2230.2180.2140.2100.2050.2010.1970.1930.1900.1860.1820.1790.1760.1720.1690.1660.163
0.1050.1750.2200.2470.2640.2740.2800.2830.2840.2840.2830.2820.2790.2770.2730.2700.2670.6230.2590.2550.2510.2470.2430.2390.2350.2310.2270.2230.2190.2160.2120.2080.2050.2010.1980.1940.1910.1880.1850.182
0.1050.1760.2200.2480.2650.2750.2820.2860.2880.2880.2880.2870.2850.2830.2810.2780.2760.2720.2690.2660.2630.2590.2550.2520.2480.2450.2410.2380.2340.2310.2270.2240.2200.2170.2140.2100.2070.2040.2010.198
0.1050.1760.2200.2480.2650.2760.2830.2880.2900.2910.2910.2910.2900.2880.2870.2850.2820.2800.2770.2740.2710.2680.2650.2620.2590.2560.2520.2490.2460.2430.2400.2360.2330.2300.2270.2240.2210.2180.2150.212
0.1050.1760.2200.2480.2660.2770.2840.2890.2920.2930.2940.2940.2930.2920.2910.2890.2870.2850.2830.2810.2780.2760.2730.2700.2670.2650.2620.2590.2560.2530.2500.2470.2440.2410.2380.2350.2330.2300.2270.224
FIGURE 5.14
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
1.1221.5321.7831.9642.105
6 7 8 910
2.2202.3182.4032.4472.544
15 20 30 50100
2.8022.9853.2433.5684.010
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
0.9541.3021.5161.6991.789
0.8861.2101.4091.5521.663
6 7 8 9 10
1.8871.9702.0432.1052.162
1.7541.8311.8981.9572.010
15 20 30 50100
2.3822.5372.7573.0333.409
2.2142.3582.5622.8193.168
EXAMPLE 5.4
Figure 5.14 shows a flexible rectangular area measuring 10 ft × 5 ft in plan onthe ground surface. The area is subjected to a uniformly distributed load of4000 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 pointb. The surface settlement below the center of the loaded area
Solution
a. From Eq. (5.32)
ν−ν−−ν−=
43
2
1
21)1(
2II
E
qBS
s
e
15
5
25
10
1
21)1(
2 43
2
===′
===′
ν−ν−−ν−=
B
zn
B
Lm
IIE
qBS
s
e
in. 0.346=ft 0288.0
218.035.01
35.021282.1)35.01(
)1442500)(2(
)5)(4000( 2
=
−×−−−
×=
eS
SqB
EI
L
B I
S
es
e
= −
= = =
=×
− =
( )
. .
( )(5)( . )( . ) .
1
10
52 1532
4000
2500 1441 0 35 1532 0 075
25
2
ν
From Table 5.8,
ft =
5
0.9 in.
From Tables 5.6 and 5.7 for m ! = 2 and n! = 1, I3 = 1.282 and I4 =0.218. Hence,
b. From Eq. (5.39)
!!
5.6 EFFECT OF A RIGID BASEAT A LIMITED DEPTH
The relationships for elastic settlement given in the preceding section assumethat the elastic soil layer extends to an infinite depth. However, if the soil layeris underlain by a rigid base at a limited depth h as shown in Fig. 5.15, theelastic settlement of the compressible soil layer can be calculated as
FIGURE 5.15 Effect of a rigid base located at a limited depth on elastic settlement of a foundation
∫∫∞∞
∈∈ −=h
zzedzdzS
0
SRq
Ees
(center, flexible) = ( )1 21− ν α
SRq
Ees
(center, rigid) = −( )1 22ν α
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 uni-formly loaded circular area of radius R and a rectangular area with dimensionsof L × B. He assumed that no friction exists between the soil layer and therigid base. According to this method [4]
Circular area:
(5.43)
(5.44)
where R = radius of the circular area
Rectangular Area (B × L)
SBq
E
SBq
E
es
es
(center, flexible) =
(center, rigid)
( )
( )
1
1
23
24
−
= −
ν α
ν α
(5.45)
(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
00.51.02.03.05.010.0
00.521.001.441.621.781.88
00.450.791.161.321.481.64
Table 5.11 Variations of """"3 and """"4
h/B
L/B
1 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 conditionswhere the load is at the surface of the soil layer (Fig. 5.16a). In actuality, foun-dations 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
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 theoreticalstudies available on this subject. Fox [6] evaluated the effect of embedment onthe average settlement of a uniformly loaded flexible rectangular area (L × B),and this was presented in a graphical form. The interpolated values from thisgraph are given in Table 5.12. Note that Table 5.12 is valid only for ! = 0.5.
TABLE 5.12 Variation ofSe (average, Df )/Se (average, Df =0)
Df
LB
L/B
1 5 10
0 0.10.20.30.40.50.60.81.0
10.9810.9590.9280.8940.8630.8250.7630.732
10.9750.9470.9060.8750.8440.8190.7750.737
10.9690.9310.8930.8560.8310.8130.7810.738
φφ
φreltc
relrelative friction angle = (0=− °
°− °≤ ≤
25
45 251)
TABLE 5.13 Suggested Values for Poisson’s Ratio
Soil type Poisson’s ratio, !
Coarse sandMedium loose sandFine sandSandy silt and siltSaturated clay (undrained)Saturated clay--lightly overconsolidated (drained)
0.15 – 0.200.20 – 0.250.25 – 0.300.30 – 0.35
0.500.2 – 0.4
5.8 ELASTIC PARAMETERS
Parameters such as the modulus of elasticity Es and Poisson’s ratio ! for agiven soil must be known to calculate the elastic settlement of a foundation. Inmost cases, if laboratory test results are not available, they are estimated fromempirical correlations. Table 5.13 provides some suggested values forPoisson’s ratio.
Trautman and Kulhawy [7] used the following relationship for Poisson’sratio (drained state)
! = 0.1 + 0.3"rel (5.47)
(5.48)
where "tc = friction angle from drained triaxial compression testA 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 Es (kN / m2)
Coarse and medium coarse sand Loose Medium dense DenseFine sand Loose Medium dense DenseSandy silt Loose Medium dense Dense
25,000 – 35,00030,000 – 40,00040,000 – 45,000
20,000 – 25,00025,000 – 35,00035,000 – 40,000
8,000 – 12,00010,000 – 12,00012,000 – 15,000
= plasticityhigh tomoderatefrom clays inorganiclean for 1500 to1000
u
s
c
E
SqB
Ees
= µ µ1 2
A number of correlations for the modulus of elasticity of sand with thefield standard penetration resistance N and cone penetration resistance qc weremade 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 forestimating 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
(5.52)
Duncan and Buchignani [11] correlated Es /cu with the overconsolidationratio OCR and plasticity index PI of several clay soils. This broadly generalizedcorrelation is shown in Fig. 5.17.
5.9 SETTLEMENT OF FOUNDATIONSON SATURATED CLAYS
Method of Janbu et al.
Janbu et al. [12] proposed a generalized equation for estimating the averageelastic settlement of a uniformly loaded flexible foundation located on a satu-rated 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) thepossible existence of a rigid layer at a shallow depth under the foundation asshown in Fig. 5.18, or
(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
=µ
=µ
B
L
B
h
B
Df f
,2
1 where
L = foundation lengthB = foundation width
Christian and Carrier [13] made a critical evaluation of the factors µ1 andµ2 , and the results were presented in a graphical form. The interpolated valuesof µ1 and µ2 from these graphs are given in Tables 5.15 and 5.16.
TABLE 5.15 Variation of µ1 with Df /0 [Eq. (5.53)]
Df /B µ1
0 2 4 6 8101214161820
1.00.90.880.8750.870.8650.8630.8600.8560.8540.850
TABLE 5.16 Variation of µ2 with h/B
h/B Circle
L/B
1 2 5 10 "
1 2 4 6 8102030
0.360.470.580.610.620.630.640.66
0.360.530.630.670.680.700.710.73
0.360.630.820.880.900.920.930.95
0.360.640.941.081.131.181.261.29
0.360.640.941.141.221.301.471.54
0.360.640.941.161.261.421.741.84
Method of D’Appolonia et al.
D’Appolonia et al. [10] proposed an improved method for estimating the
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 purposeof this theory was to account for the limitation of the elastic theory and to ac-count for the stress redistribution and the strains occurring after local yielding.
fK
co
u
v
=−
′
12
σ
ρ
ρ
=
=
S
S
SS
e
i
ie
So
This hypothesis can be explained by referring to Fig. 5.19, which shows anidealized plot of applied load per unit area on the foundation q versussettlement Se plot. In this figure OA is a linear plot. The first local yield occursat A. It is followed by nonlinear segment AB. When elastic theories are usedto calculate settlement the nonlinearity of segment AB is not taken intoaccount.
According to this approach, the steps to estimate the initial settlementfollow (refer to Fig. 5.20).
1. Using any procedure, estimate the ultimate bearing capacity, qu , of thefoundation.
2. Determine the allowable load per unit area, q, and then the appliedstress ratio, q/qu .
3. Calculate the initial shear stress ratio, f, as
(5.54)
where Ko = coefficient of lateral at-rest earth pressurecu = undrained shear strength##v = initial vertical effective stress
The initial shear stress ratio f can be estimated from Fig. 5.21, whichis the average of tests conducted on several clays [10].
4. Estimate the magnitude of the modulus of elasticity, Es (Section 5.8).5. Calculate the elastic settlement, Se , by using one of the applicable
equations given in Sections 5.4 and 5.5.6. For a given applied stress ratio, q/qu , f, and h/B (h = thickness of
compressible layer; B = width of the rectangular foundation ordiameter of the circular foundation), determine the settlement ratio,$, using Fig. 5.22
(5.55)
(5.56)
From Fig. 5.22 it is clear than h/B has minor influence on the settlementratio, $. The result for h/B = 1.5 will apply for all values of h/B > 1.5.
FIGURE 5.21 Relationship between f and OCR (after D’Appolonia et al. [10])
ft) 4>(for =)(kip/ft
and
ft) 4(for =)(kip/ft
2
net(all)
2
net(all)
BB
BNq
BN
q
21
6
4
+′
≤′
5.10 SETTLEMENT OF FOUNDATIONS ON SAND
Correlation of Standard Penetration Resistanceand Cone Penetration Resistance
Geotechnical engineers have used empirical approaches based on a largenumber of case studies to estimate the elastic settlement of foundationsconstructed on sand. The two most widely used empirical approaches utilizethe corrected standard penetration resistance N´ and the cone penetration resis-tance number qc . Meyerhof [14] proposed a correlation for the net allowablebearing capacity for foundations for one inch (25.4 mm) of estimatedmaximum settlement with the corrected standard penetration resistance N´ as
(5.57)
(5.58)
FIGURE 5.22 Relationship between $ and q/qu for continuousfoundation (after D’Appolonia et al. [10])
where B = width of the foundationN# = corrected standard penetration resistance
qnet(all) = qall $ %Df (5.59)
ft) 4>(for =)(kip/ft
ft) 4(for =)(kip/ft
2
2
net(all)
2
net(all)
BSFB
BNq
BSFN
q
ed
ed
+′
≤′
1
4
5.2
+=
B
DF f
d33.01=factordepth
m) 1(for =)(kN/m 2
net(all)22.
4.2516.19 ≤
′ BS
FNq e
d
m) 1.22>(for 11.98=)(kN/m 2
net(all)B
SF
B
BNq e
d
+′
4.2528.3128.3
2
qqc
net(all)2
2
(kip / ft )kip / ft
(for B 4 ft and settlement of 1 in. )
=
≤
( )
15
The N# value to be used in the above equation should be the average N# valuein the influence zone of the foundation, that is, 0.5Df above the foundation and2B 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 safelyincreased by 50%. Thus, these two equations can be modified as
(5.60a)
(5.60b)
where Se = estimate maximum elastic settlement, in inchesB = width, in ft
(5.61)
Df = depth of foundationIn SI units, Eqs. (5.60a) and (5.60b) can be expressed as
(5.62)
(5.63)
where B = width, in metersSe = estimated maximum elastic settlement, in mm
Meyerhof [14] also developed empirical relationships for the net allowablebearing capacity of foundations based on cone penetration resistance, qc , or
(5.64)
and
)settlement of in. 1 andft 4>for
kip/ft(kip/ft
22
net(all)
B
B
Bqq c
(
1
25
)()
2
+=
+=
≤=
mm 25.4 of settlement
and m 1.22>for
and
mm 25.4 of settlement
and m 1.22for
kN/mkN/m
net(all
net(all
B
B
Bqq
Bqq
c
c
2
)
22
)
28.3128.3
25
15
)()(
B
B
D
B
BN
fqS
fe
+
′
′=
4.01
1
71.11
87.0
all
Sq B
N
Sq B
N
e
e
=′
=′
2 (for sand and gravel)
(for silty sand)
(5.65)
In SI units, the preceding two equations can be written as
(5.66)
(5.67)
Based on the observation of the settlement of 48 buildings all constructed onsand, Schultze and Sherif [16] gave the following expression for predictingelastic settlement
(5.68)
where qall is in kip/cm2
f´ = influence factor of the settlementB1 = 1 cmDf = depth of foundation
Based on the study of Schultze and Sherif [16], Meyerhof [17] suggested thefollowing relations for estimating elastic settlement
(5.69)
(5.70)
where Se = settlement, in inches, q is in U.S. ton / ft2, and B is in inches.
[ ])(1
θσ+σν−σ=∈rz
s
z E
[ ]BAE
q
s
z′+′ν−ν+=∈ )21(
)1(
])21)[(1( BAq
EI sz
z′+′ν−ν+==
∈
Use of Strain Influence Factor
The equation for vertical strain, !z , below the center of a flexible circular loadof radius R was in given in Eq. (5.24) as
(5.24)
After proper substitution for "z , "r , and "# in the preceding equation, oneobtains
(5.71)
where A!, B! = nondimensional factors and functions of z/RThe 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!!!! (Belowthe Center of a Flexible Loaded Area)
z/R A! B! z/R A! B!
00.20.40.60.81.01.52.0
1.00.8040.6290.4860.3750.2930.1680.106
00.1890.3200.3780.3810.3540.2560.179
2.53.04.05.06.07.08.09.0
0.0720.0510.0300.0190.1040.0100.0080.006
0.1280.0950.0570.0380.0270.0200.0150.012
From Eq. (5.71) we can write
(5.72)
Figure 5.23 shows plots of Iz versus z/R obtained from the experimentalresults of Eggstad [18] along with the theoretical values calculated from Eq.(5.72). Based on Fig. 5.23, Schmertmann [8] proposed a practical variation ofIz and z/B (B = foundation width) for calculating the elastic settlement of foun-dations. This model was later modified by Schmertmann and Hartman [9], andthe variation is shown in Fig. 5.24 for L/B = 1 and L/B " 10. Interpolations canbe used to obtain the Iz –z/B variation for other L/B values. Using the simplifiedstrain influence factor, elastic settlement can be calculated as
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)
∑ ∆
−= z
E
IqqccS
s
z
e)(
21
=
−
=
0.1yearsin time
0.2log+1=
soilin creepfor factor correction a
0.5-1=
foundation ofdepth for factor correction a
2
1
c
q
c
(5.73)
FIGURE 5.24 Variation of Iz versus z/B
q = $Df
!q = stress at the level of the foundationThe use of Eq. (5.73) can be explained by the following example. Figure 5.25ashows a continuous foundation for which B = 8 ft; Df = 4 ft; unit weight ofsand, $ = 110 lb / ft3; !q = 25 lb / in.2.
For this case, L/B is greater than 10. Accordingly, the plot of Iz with depthis shown in Fig. 5.25a. Note that
Iz = 0.2 at z = 0Iz = 0.5 at z = 8 ft (= B)Iz = 0 at z = 32 ft (= 4B)
FIGURE 5.25
Based on the results of the standard penetration test or cone penetration test, thevariation of Es can be calculated by using Eq. (5.49) or (5.51) (or similarrelationships). The variation is shown by the broken line in Fig. 5.25b. The
93.094.21
06.305.15.01
1=
−=
−
qc
4.11.0
10log2.01
2=
+=c
.in 2.53 ==
∆
−= ∑
)0886.0)(94.21)(4.1)(93.0(
)(21
zE
IqqccS z
e
actual variation of Es can be approximated by several linear plots, and this isalso shown in Fig. 5.25b (solid lines). For elastic settlement, Table 5.18 cannow be prepared.
TABLE 5.18 Elastic Settlement Calculation (Fig. 5.25)
LayerNo. %z
(in.)Es
(lb / in.2)
z to themiddle ofthe layer
(in.)
Iz at themiddle ofthe layer
Iz
Es
Δz
(in.3 / lb)
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
#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
Assume the time for creep is 10 years. Hence,
Thus,
5.11 FIELD PLATE LOAD TESTS
In some cases it is desirable to conduct field load tests to determine theallowable and ultimate bearing capacities of foundations and associated elastic
q qu B u B( ) ( )1 2=
where ultimate bearing capacity of foundations withwidths and , respectively
q qB B
u B u B( ) ( ),1 2
1 2
=
≈
1
2)2()1( B
BSS
BeuBeu
where elastic settlement of foundations with widths and respectively, at ultimate load2
S SB B
eu B eu B( ) ( ),,
1 1
1
=
≈
1
2)1()2( B
BSS
BeBe
≈
1
2)1()2( B
Bqq
BuBu
2
1
2
2
1
2)1()2( 128.3
128.3
++
≈
B
B
B
BSS
BeuBeu
settlement. The standard method for field load tests is given by the AmericanSociety for Testing and Materials under Designation D-1194. Circular steelbearing 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 thebearing capacity and associated elastic settlement of full-scale foundations canbe 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 ina clay soil (& = 0 condition). In this case B2 > B1 . Similarly, in Fig. 5.27 arethe plots for two plates supported by a sand extending to a great depth. Fromthese two figures we can see that, in clay
(5.74)
However
(5.75)
In a similar manner, for a given load intensity q (< qu ) on the foundation
(5.76)
For foundations on sand
(5.77)
(5.78)
where B is in metersAlso, for a given load intensity q (< qu ) on the foundation
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
++
≈
128.3
128.3
1
2
2
1
2)1()2( B
B
B
BSS
BeBe
)
log1
1
2
21c(for σ′≥σ′
σ′σ′
−=
eeC
c
)
log4
3
4
43c(for σ′≤σ′
σ′σ′
−=
eeC
s
(5.79)
where B is in metersThe 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 in-creases 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 controlledby settlement criteria.
CONSOLIDATION SETTLEMENT
5.12 GENERAL PRINCIPLES OF CONSOLIDATION SETTLEMENT
As explained in Section 5.1, consolidation settlement is a time-dependentprocess which occurs due to the expulsion of excess pore water pressure insaturated clayey soils below the ground water table and is created by the in-crease in stress created by the foundation load. For a normally consolidatedclay, 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 alsoshown in Fig. 5.28b. In this figure the preconsolidation pressure is "!c . Theslope of the e vs. log "! plot for the normally consolidated portion of the soilis referred to as compression index Cc , or
(5.80)
Similarly, the slope of the e vs. log "! plot for the overconsolidated portion ofthe clay is called the swell index Cs , or
(5.81)
For normally consolidated clays, Terzaghi and Peck [19] gave a correlation forthe compression index as
Cc = 0.009(LL $ 10) (5.82a)
FIGURE 5.28 Nature of variation of void ratio with effectivestress: (a) Normally consolidated clay; (b) Over-consolidated clay
where LL = liquid limitThe preceding relation has a reliability in the range of ±30% and should not beused for clays with sensitivity ratios greater than 4.
∫=
=
σ∆−
=σ∆2
112
)(1
Hz
Hz
dzHHav
σ′
σ∆+σ′+
=+∆=
o
o
o
cc
o
c e
HC
e
eS avlog
11
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 indexproperties 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 beforethe construction of the foundation be eo , and let the average effective verticalstress on the clay layer be "!o . The foundation located at a depth Df is subjectedto a net average pressure increase of q. This will result in an increase in thevertical stress in the soil. If the vertical stress increase at any point below thecenter line of the foundation is %", the average vertical stress increase %"av
in the clay layer can thus be given as
(5.88)
The consolidation settlement Sc, due to this average stress increase can becalculated as follows
(5.89)
(for normally consolidated clay, that is, "!o = "!c )
FIGURE 5.29 Primary consolidation settlement calculation
σ′
σ∆+σ′+
=+∆=
o
o
o
cs
o
c e
HC
e
eS avlog
11
σ′
σ∆+σ′+
+
σ′σ′
+=
+∆=
c
o
o
cc
o
c
o
cs
o
c e
HC
e
HC
e
eS avlog
1log
11
(5.90)
(for overconsolidated clay and !!o + "!av " !!c )
(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 calculatethe primary consolidation settlement. They are as follows:
FIGURE 5.30 Average stress increase, "!av
Method A
According to this method, !!o is the in situ average effective stress (that is, theeffective stress at the middle of the clay layer). The magnitude of "!av can becalculated as (Fig. 5.29)
"!av = –16
("!t + 4"!m + "!b ) (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 previouslyin this chapter.
The average stress increase, "!av , from z = 0 to z = H below the center ofa uniformly loaded flexible rectangular area (Fig. 5.30) was obtained using theintegration method by Griffiths [21], or
FIGURE 5.31 Variation of Iav with a/h and b/h
=
h
b
h
afI ,
av where
Ih I h I
h hh hh av h
avav
( / )( ) ( )
1 2
2 12 1
2 1
=−−
"!av = qIav (5.93)
(5.94)
a, b = half-length and half-width of the foundationThe 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 ofaverage stress increase from z = 0 to z = h (Fig. 5.30). For calculating theaverage stress increase in a clay layer as shown in Fig. 5.32
FIGURE 5.32 Average stress increase in a clay layer
c
h
h
Hhh
h
b
h
afI
h
b
h
afI
=−
=
=
12
11
)1(
22
)2(
,
,
av
av where
−=σ∆
c
hh
H
IhIhq )1(1)2(2 avav
av
So
(5.95)
Method B
In this method, a given clay layer can be divided into several thin layershaving thicknesses of Hc(1) , Hc(2) , . . . , Hc(n) (Fig. 5.33). The in situ effectivestresses at the middle of each layer are !!o(1) , !!o(2) , . . ., !!o(n) . The averagestress increase for each layer can be approximated to be equal to the vertical
FIGURE 5.33 Consolidation settlement calculation using Method B
Se
eHc
i
o ii
i n
i=+=
=∑
∆11 ( )
σ′
σ∆+σ′+
=o
o
o
cc
c e
HCS avlog
1
aL
bB
= = =
= = =
2
3
215
2
15
20 75
.
..
m
m
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 entirelayer can be calculated as
(5.96)
EXAMPLE 5.5
Refer to Fig. 5.34. Using Method A, determine the primary consolidationsettlement of a foundation measuring 1.5 m × 3 m (B × L) in plan.
Solution From Eq. (5.89)
Given Cc = 0.27, Hc = 3 m, eo = 0.92
!!o = (1 + 1.5)(16.5) + (1.5)(17.8 $ 9.81) + –32
(18.2 $ 9.81) = 65.82 kN / m2
FIGURE 5.34
a
h
b
h1 1
15
30 5
0 75
30 25= = = =.
. ;.
.
From Fig. 5.31, av(I h10 54) .=
Similarly, a
h
b
h2 2
15
60 25
0 75
60125= = = =.
. ;.
.
From Fig. 5.31, av(I h20 34) .=
2
avav
av
kN/m 8.233
)54.0)(3()34.0)(6(170
)1(1)2(2
=
−=
−=σ∆
c
hh
H
IhIhq
mm 57=m 057.082.65
8.2382.65log
9.01)3)(27.0( =
+
+=
cS
h1 = 1.5 + 1.5 = 3 m
h2 = 1.5 + 1.5 + 3 = 6 m
From Eq. (5.95)
!!
EXAMPLE 5.6
Solve Example 5.5 by Method B. (Note: Divide the clay layer into threelayers, each 1 m thick).
Solution The following tables can now be prepared.
Calculation of !!o :
LayerNo.
Layerthickness, Hi
(m)
Depth to the middle of clay layer(m)
!!o (kN / m2)
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
mm 47=m
av
047.0)065.0096.0168.0)(142.0(
21.74
9.1121.74log
82.65
15.1682.65log
43.57
2.2743.57log
9.01
)1)(27.0(
log1
)(
)()(
=++=
++
++
+
+=
σ′σ∆+σ′
+= ∑
io
iio
o
icc e
HCS
Calculation of !"av :
LayerNo.
LayerthicknessHi
(m)
Depth to middle oflayer from bottomof foundation, z(m)
L/Ba z/B
b
Δσ(av)
q !"avc
1 1 3.5 2 2.33 0.16 27.2
2 1 4.5 2 3.0 0.095 16.15
3 1 5.5 2 3.67 0.07 11.9
aB = 1.5 m; L = 3 m; b Table 5.3; c q = 170 kN / m2
!!
5.14 THREE-DIMENSIONAL EFFECT ON PRIMARY CONSOLIDATION SETTLEMENT
The procedure described in the preceding section is for one-dimensional con-solidation and will provide a good estimation for a field case where the widthof the foundation is large relative to the thickness of the compressible stratum,Hc , and also when the compressible material lies between two stiffer soillayers. This is because the magnitude of horizontal strains is relatively less inthe above cases.
In order to account for the three-dimensional effect, Skempton andBjerrum [22] proposed a correction to the one-dimensional consolidation set-tlement for normally consolidated clays. This can be explained by referring toFig. 5.35, which shows a circularly loaded area (diameter = B) on a layer ofnormally consolidated clay of thickness Hc . Let the stress increases at a depthz 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 begiven as
!u = !"3 + A(!"1 ! !"3) (5.97)
FIGURE 5.35 Three-dimensional effect on primary consolidationsettlement (circular foundation of diameter B)
))(()1(
1
dzue
edzumdS
o
vc∆
σ∆+
∆=⋅∆⋅=
[ ]
dzAAmS
dzAe
edSS
cH
vc
cH
o
cc
∫
∫ ∫
−
σ∆σ∆
+σ∆=
σ∆−σ∆+σ∆
σ∆+
∆==
0 1
3
1
313
0 1
)1(
)()1(
or
where A = pore water pressure parameterThe consolidation settlement dSc of an elemental soil layer of thickness dz is
(5.98)
where mv = volume coefficient of compressibility !e = change in void ratio eo = initial void ratio
Hence,
(5.99)
For conventional one-dimensional consolidation (Section 5.12)
∫∫∫ σ∆=σ∆+σ∆
∆=+∆=
cH
v
cH
o
cH
o
oedcdzmdz
e
edz
e
eS
0
1
0
1
10
)( )1(1
1
0
1
0
3
0
1
1
3
0
1
)(
)(
)1()1(
)1(
MAA
dz
dz
AA
dzm
dzAAm
S
S
cH
cH
cH
v
cH
v
oedc
cNCc
−+=
σ∆
σ∆−+=
σ∆
−
σ∆σ∆
+σ∆==µ
∫
∫
∫
∫
dz
dz
McH
cH
∫
∫σ∆
σ∆=
0
1
0
3
1 where
)5.0)(2
1
3
1
2
3313
=νσ∆−σ∆
+
−+σ∆=∆ (for Au
zNNmScH
vc∆
σ∆σ∆
−+σ∆= ∫1
3
0
1)1(
21
31
23 +
−= AN where
(5.100)
From Eqs. (5.99) and (5.100)
(5.101)
(5.102)
The variation of µc(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 thecenter line of the foundation. For this condition it can be shown that
(5.103)
In a similar manner as in Eq. (5.99)
(5.104)
(5.105)
Thus
FIGURE 5.36 Variation of µc (NC) with A and Hc /B [Eq. (5.101)]
FIGURE 5.37 Three-dimensional effect on primary consolidationsettlement (continuous foundation of width B)
2
0
1
1
3
0
1
)(
)(
)1(
)1(
MNN
dzm
dzNNm
S
ScH
v
cH
v
oedc
cNCs
−+=
σ∆
σ∆σ∆
−+σ∆
==µ
∫
∫
dz
dz
McH
cH
∫
∫σ∆
σ∆
=
0
1
0
3
2 where
FIGURE 5.38 Variation of µs (NC) with A and Hc /B [Eq. (5.105)]
(5.106)
(5.107)
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-dimen-
sional consolidation effect in the field for a circular foundation located overoverconsolidated clays. Referring to Fig. 5.39
Sc = µc(OC) Sc(oed) (5.108)
FIGURE 5.39 Three-dimensional effect on primary consolidationsettlement of overconsolidated clay (circular foundation)
o
c
c
OCc
OCR
H
BOCRf
σ′σ′
=
=µ ,
)( where (5.109)
(5.110)
""c = preconsolidation pressure ""o = present effective consolidation pressure
The interpolated values of µc(OC) from the work of Leonard [24] are given inTable 5.19.
EXAMPLE 5.7
Refer to Example 5.5. Assume that the pore water pressure parameter A for theclay is 0.6. Considering the three-dimensional effect, estimate the consoli-dation 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 weassume that the load from the foundation spreads out along planes havingslopes of 2V:1H, then the dimension of the loaded area on the top of the claylayer is
FIGURE 5.40
TABLE 5.19 Variation of µc(OC) with OCRand B/Hc
OCR
µ c(OC)
B/Hc = 4.0 B/Hc = 1.0 B/Hc = 0.2
1 2 3 4 5 6 7 8 910111213141516
1 0.9860.9720.9640.9500.9430.9290.9140.9000.8860.8710.8640.8570.8500.8430.843
1 0.9570.9140.8710.8290.8000.7570.7290.7000.6710.6430.6290.6140.6070.6000.600
1 0.9290.8420.7710.7070.6430.5860.5290.4930.4570.4290.4140.4000.3860.3710.357
B" = 1.5 + –12
(3) = 3 m
L" = 3 + –12
(3) = 4.5 m
The diameter of an equivalent circular area, Beq , can be given as
m
or
eq
eq
15.4)5.4)(3(44
42
=
π=′′π=
′′=π
LBB
LBB
H
Bc = =3
4 150 723
..
∆=α
1
2logt
t
eC =indexn compressiosecondary
+
= α
1
2log1 t
t
e
HCS
p
c
s
From Fig. 5.36, for A = 0.6 and Hc /B = 0.723, the magnitude of µc(NC) ! 0.76. So
Sc = Sc(oed) µc(NC) = (57)(0.76) = 43.3 mm !!
5.15 SECONDARY CONSOLIDATION SETTLEMENT
Secondary consolidation follows the primary consolidation process and takesplace under essentially constant effective stress as shown in Fig. 5.41. Theslope of the void ratio versus log-of-time plot is equal to C! , or
(5.111)
The secondary consolidation settlement, Ss , can be calculated as
(5.112)
where ep = void ratio at the end of primary consolidation t2 , t1 = time
The magnitude of the secondary compression index can vary widely, and somegeneral ranges are as follow:
Overconsolidated clays (OCR > 2 to 3) . . . . . . . . >0.001Organic soils . . . . . . . . . . . . . . . . . . . . . . 0.025 or moreNormally consolidated clays . . . . . . . . . . . . 0.004–0.025
FIGURE 5.41 Secondary consolidation settlement
In a majority of cases, secondary consolidation is small compared to primaryconsolidation settlement. It can, however, be substantial for highly plasticclays and organic soils.
DIFFERENTIAL SETTLEMENT
5.16 GENERAL CONCEPT OF DIFFERENTIAL SETTLEMENT
In most instances the subsoil is not homogeneous and the load carried byvarious 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 agiven building. The differential settlement of various parts of a building canlead to damage of the superstructure. Hence, it is important to define certainparameters to quantify differential settlement and develop limiting values forthese parameters for desired safe performance of structures. Burland andWorth [24] summarized the important parameters relating to differential settle-ment. 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
FIGURE 5.42 Definition of parameters for differential settlement
β = =
=
angular distortion
( distance between points and
∆S
l
Note l i j
T ij
ij
ij
( )
: )
∆L
= deflection ratio
is BB", . . . . Based on this figure the definitions of the various parametersfollow:
ST = total settlement of a given point"ST = difference between total settlement between any two parts! = gradient between two successive points
# = tilt" = relative deflection (that is, movement from a straight line joining
two reference points
Since the 1950's, attempts have been made by various researchers andbuilding codes to recommend allowable values for the above parameters. Asummary of some of these recommendations is given in the following section.
5.17 LIMITING VALUE OF DIFFERENTIAL SETTLEMENT PARAMETERS
In 1956, Skempton and MacDonald [25] proposed the following limitingvalues for maximum settlement, maximum differential settlement, and maxi-mum angular distortion to be used for building purposes
Maximum settlement, ST(max)
In sand . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 mmIn clay . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 mm
Maximum differential settlement, " ST(max)
Isolated foundations in sand . . . . . . . . . . . . . . . . 51 mmIsolated foundations in clay . . . . . . . . . . . . . . . . . 76 mmRaft in sand . . . . . . . . . . . . . . . . . . . . . . . . . . 51–76 mmRaft in clay . . . . . . . . . . . . . . . . . . . . . . . . . . 76–127 mm
Maximum angular distortion, $max . . . . . . . . . . . . . . . . 1/300
Based on experience, Polschin and Tokar [26] provided the allowable de-flection ratios for buildings as a function of L/H (L = length; H = height ofbuilding), 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 L/H "/L
Multistory buildings and civil dwellings
#3 0.0003 (for sand)0.0004 (for clay)
$5 0.0005 ( for sand)0.0007 (for clay)
One-story mills 0.001 (for sand and clay
Bjerrum [27] recommended the following limiting angular distortion ($max)for various structures
Category of potential damage $max
Safe limit for flexible brick wall (L/H > 4)Danger of structural damage to most buildingsCracking of panel and brick wallsVisible tilting of high rigid buildingsFirst cracking of panel wallsSafe limit for no cracking of buildingDanger to frames with diagonals
1/1501/1501/1501/2501/3001/5001/600
Grant et al.[28] correlated ST(max) and $max for several buildings with thefollowing results.
Soil type Foundation type Correlation
ClayClaySandSand
Isolated shallow foundationRaftIsolated shallow foundationRaft
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 isknown, the magnitude of the allowable ST(mzx) can be calculated.
The European Committee for Standardization more recently providedvalues for limiting values for serviceability limit states [29] and the maximumaccepted 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 forserviceability [29]
ST
"ST
$
25 mm50 mm5 mm10 mm20 mm1/500
Isolated shallow foundationRaft foundationFrames with rigid claddingFrames with flexible claddingOpen frames—
Maximum acceptablefoundation movement [30]
ST
"ST
$
5020!1/500
Isolated shallow foundationIsolated shallow foundation—
REFERENCES
1. Boussinesq, J., Application des Potentials a L’Etude de L’Equilibre et dueMouvement des Solides Elastiques, Gauthier-Villars, Paris, 1883.
2. Ahlvin, R. G., and Ulery, H. H., Tabulated values for determining thecomplete pattern of stresses, strains, and deflections beneath a uniform load ona homogeneous half space, High-way Res. Rec., Bulletin 342, 1, 1962.
3. Borowicka, H., Influence of rigidity of a circular foundation slab on thedistribution of pressures over the contact surface, In. Proc. I Int. Conf. SoilMech. Found. Eng., 2, 1936, 144.
4. Harr, M. E., Fundamentals of Theoretical Soil Mechanics, McGraw-Hill, NewYork, 1966.
5. Egorov, K. E., Concerning the question of the deformation of bases of finitethickness, Mekhanika Gruntov, Sb. Tr. 34, Gosstroiizdat, Moscow, 1958.
6. Fox, E. N., The mean elastic settlement of a uniformly loaded area at a depthbelow the ground surface, in Proc., II Int. Conf. Soil Mech. Found. Eng., 1,1948, 129.
7. Trautmann, C. H., and Kulhawy, F. H., CUFAD—A Computer Program forCompression and Uplift Foundation Analysis and Design, Report EL-4540-CCM, Electric Power Research Institute, 16, Palo Alto, 1987.
8. Schmertmann, J. H., Static cone to compute settlement over sand, J. SoilMech. Found. Div., ASCE, 96(8), 1011, 1970.
9. Schmertmann, J. H., and Hartman, J. P., Improved strain influence factordiagrams, J. Geotech. Eng. Div., ASCE, 104(8), 1131, 1978.
10. D’Appolonia, D. T., Poulos, H. G., and Ladd, C. C., Initial settlement ofstructures on clay, J. Soil Mech. Found. Div., ASCE, 97(10), 1359, 1971.
11. Duncan, J. M., and Buchignani, A. L., An Engineering Manual for SettlementStudies, Department of Civil Engineering, University of California, Berkeley,1976.
12. Janbu, N., Bjerrum, L., and Kjaernsli, B., Veiledning ved losning avFundamenteringsoppgaver, Norwegian Geotechnical Institute Publication 16,Oslo, 1956.
13. Christian, J. T., and Carrier III, W. D., Janbu, Bjerrum and Kjaernsli’s chartreinterpreted, Canadian Geotech. J., 15(1), 124, 1978.
14. Meyerhof, G. G., Penetration tests and bearing capacity of cohesionless soils,J. Soil Mech. Found.. Div., ASCE, 82(1), 1, 1956.
15. Meyerhof, G. G., Shallow foundations, J. Soil Mech. Found. Div., ASCE,91(2), 21, 1965.
16. Schultze, E., and Sherif, G., Prediction of settlement from evaluation ofsettlement observations for sand, In Proc., VIII Int. Conf. Soil Mech. Found.Eng., Moscow, U.S.S.R., 1(3), 1973, 225.
17. Meyerhof, G. G., General report: state-of-the-art of penetration testing incountries outside Europe, in Proc. Eur. Symp. Pen. Test., 1974.
18. Eggstad, A., Deformation measurements below a model footing on the surfaceof dry sand, in Proc. Eur Conf. Soil Mech. Found. Eng., Weisbaden, W.Germany, 1, 1963, 223.
19. Terzaghi, K., and Peck, R. B., Soil mechanics in engineering practice, 2nd Ed.,Wiley, New York, 1967.
20. Azzouz, A. S., Krizek, R. T., and Corotis, R. B., Regression analysis of soilcompressibility, Soils and Foundations, 16(2), 19, 1976.
21. Griffiths, D. V., A chart for estimating the average vertical stress increase inan elastic foundation below a uniformly loaded rectangular area, CanadianGeotech. J., 21(4), 710, 1984.
22. Skempton, A. W., and Bjerrum, L., A contribution to settlement analysis offoundations in clay, Geotechnique, 7, 168, 1957.
23. Leonards, G. A., Estimating Consolidation Settlement of Shallow Foundationson Over-consolidated Clay, Transportation Research Board, Special Report163, Washington, D.C., 13, 1976.
24. Burland, J. B., and Worth, C. P., Allowable and differential settlement ofstructures, including damage and soil-structure interaction, in Proc. Conf. OnSettlement of Structures, Cambridge University, U.K., 1970, 611.
25. Skempton, A. W., and MacDonald, D. H., The allowable settlement ofbuildings, in Proc. of Institute of Civil Engineers, 5, Part III, 1956, 727.
26. Polshin, D. E., and Tokar, R. A., Maximum allowable non-uniform settlementof structures, in Proc., IV Int. Conf. Soil Mech. Found. Eng., London, 1, 1957,402.
27. Bjerrum, L., Allowable settlement of structures, in Proc. European Conf. SoilMech. Found. Eng., Weisbaden Germany, 3, 1963, 135.
28. Grant, R., Christian, J. T., and Vanmarcke, E. H., Differential settlement ofbuildings, J. Geotech. Eng. Div., ASCE, 100(9), 1974, 973.
29. European Committee for Standardization, Basis of design and actions onstructures, Eurocode 1, Brussels, Belgium, 1994.
30. European Committee for Standardization, Geotechnical design, generalrules—Part I, Eurocode 7, Brussels, Belgium, 1994.
Nq
Bs
uγ γλ
γ=
12
CHAPTER 6
DYNAMIC BEARING CAPACITY AND SETTLEMENT
6.1 INTRODUCTION
Depending on the type of superstructure and the type of loading, a shallowfoundation may be subjected to dynamic loading. The dynamic loading may beof various types, such as (a) monotonic loading with varying velocities; (b)earthquake loading; (c) cyclic loading; and (d) transient loading. The ultimatebearing capacity and settlement of shallow foundations subjected to dynamicloading is the subject of discussion of this chapter.
6.2 EFFECT OF LOAD VELOCITY ONULTIMATE BEARING CAPACITY
The static ultimate bearing capacity of shallow foundations was discussed inChapters 2, 3, and 4. Vesic, Banks, and Woodward [1] conducted laboratorymodel tests to study the effect of the velocity of loading on the ultimate bearingcapacity. These tests were conducted on a rigid rough circular model foun-dation having a diameter of 101.6 mm. The model foundation was placed on thesurface of a dense sand layer. The velocity of loading to cause failure wasvaried from 10-5 in. / sec to 10 in. / sec. The tests were conducted in dry andsubmerged sand. From Eq. (2.83), for a surface foundation in sand subjectedto vertical loading
qu = –12 !BN! "!s
or
(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 ofVesic et al. [1] is shown in Fig. 6.1. It can be seen from this figure that, whenthe loading velocity is between 10-3 in. / sec to 10-2 in. / sec, the ultimate bearing
FIGURE 6.1 Variation of N! λ!s with loading velocity (after Vesic et al. [1])
∆∆
=∈•B
S
te
2
1
capacity reaches a minimum value. Vesic [2] suggested that the minimumvalue 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 con-ventionally 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 isvery high is due to the fact that the soil particles in the failure zone do notalways follow the path of least resistance, resulting in high shear strength ofsoil and thus ultimate bearing capacity.
Unlike in the case of sand, the undrained shear strength of saturated clayincreases with the increase in the strain rate of loading. An excellent examplecan be obtained from the unconsolidated undrained triaxial tests conducted byCarroll [4] on Buckshot clay. The tests were conducted with a chamberconfining pressure of 2000 lb / ft2 (#96 kN / m2), and the moisture contentsof 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. Fora given foundation the strain rate, $!, can be approximated as (Fig. 6.2)
(6.2)
where t = time Se = settlement
So, if the undrained cohesion cu (# = 0 condition) for a given soil at a givenstrain rate is known, this value can be used in Eq. (2.83) to calculate theultimate bearing capacity.
FIGURE 6.2 Strain rate definition under a foundation
[ ]
φ+φδ+φ−φδ+φ+φφ+φ=α −
)cot(tantan1
tancottan1)(cot(tantantan
5.0
1
A
6.3 ULTIMATE BEARING CAPACITYUNDER EARTHQUAKE LOADING
Richards et al. [5] more recently proposed a bearing capacity theory for acontinuous foundation supported by a granular soil under earthquake loading.This theory assumes a simplified failure surface in soil at ultimate load. Figure6.3a shows this failure surface under static conditions based on Coulomb’sactive and passive pressure wedges. Note that, in Zone I, %A is the angle thatCoulomb’s active wedge makes with the horizontal at failure
(6.3)
Similarly, in Zone II, %P is the angle that Coulomb’s passive wedge makes withthe horizontal at failure, or
FIGURE 6.3 Bearing capacity of a continuous foundation on sand--static condition
[ ]
φ+φδ+φ+φδ+φ+φφ+φ−=α −
)cot(tantan1
tancottan1)(cot(tantantan
5.0
1
P(6.4)
where # = soil friction angle& = wall friction angle (AB in Fig. 6.3a)
Considering a unit length of the foundation, Fig. 6.3b shows the equilib-rium analysis of wedges I and II. In this figure the following notations areused.
PA = Coulomb’s active pressurePP = Coulomb’s passive pressure
2
2
cos
sin)sin(1cos
cos
δ
φδ+φ+δ
φ=A
K
2
2
cos
sin)sin(1cos
cos
δ
φδ−φ−δ
φ=P
K
′ = =q qK
KqNu
P
Aq
RA = resultant of shear and normal force along ACRP = resultant of shear and normal force along CD
WI , WII = weight of wedges ABC and BCD, respectivelyNow, if # $ 0, ! = 0, and q $ 0, then
qu = q%u
and
PA cos& = PP cos& (6.5)
However
PA cos& =q%u KA H (6.6)
where H = !!ABKA = horizontal component of Coulomb’s active earth pressure coefficient, or
(6.7)
Similarly
PP cos& = qKP H (6.8)
where KP = horizontal component of Coulomb’s passive earth pressure coeffi- cient, or
(6.9)
Combining Eqs. (6.5), (6.6), and (6.8)
(6.10)
−γ=′′
−γ=′′
121
1)(
2
1 2
A
P
u
A
APu
K
KHq
HKKKHq
or
γγ=
−αγ=′′ BN
K
KBq
A
P
Au 21
1tan21
−α=γ 1tan
A
P
A K
KN =factor capacity bearing where
where Nq = bearing capacity factorAgain, if # $ 0, ! $ 0, and q = 0, then qu = q&u .
PA cos& = q&u HKA + –12 !H 2KA (6.11)
Also
PP cos& = –12 !H 2KP (6.12)
Equating the right-hand sides of Eqs. (6.11) and (6.12)
q&u HKA + –12 !H 2KA = –
12 !H 2KP
(6.13)
However,
H = Btan%A (6.14)
Combining Eqs. (6.13) and (6.14)
(6.15)
(6.16)
If # $ 0, ! $ 0, and q$ 0, using the superposition we can write
qu = q%u + q&u = qNq + –12 !BN! (6.17)
Richards et al. [5] suggested that, in calculating the bearing capacity factorsNq and N! which are functions of # and &, we may assume & = #/2. With thisassumption, the variations of Nq and N! are given in Table 6.1.
FIGURE 6.4 Variation of Nc with soil friction angle #
TABLE 6.1 Variation of Nq , N!!!! , and Nc (Assumption: &&&& = ####/2)
Soil frictionangle, #
(deg)&
(deg)Nq N! Nc
10203040 0 5101520
1 2.37
5.916.5159.04
0 1.38 6.0623.76111.9
6 7.7713.4626.8658.43
It can also be shown that for # = 0 condition if Coulomb’s wedge analysisis performed, it will give a value of 6 for the bearing capacity factor, Nc . Forbrevity, 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 becalculated, and these values are shown in Table 6.1. Figures 6.4, 6.5, and 6.6show the variations of the bearing capacity factors with soil friction angle, #.
FIGURE 6.5 Variation of Nq with soil friction angle #
+θ+δ+−θ+δ++
+=α −
)cot)(tantan(1
tan]cot)tan(1)[tan1(tan
21
aa
aaaa
AE
Thus, the ultimate bearing capacity qu for a continuous foundation supportedby a c–# soil can be given as
qu = cNc + qNq + –12 !BN! (6.18)
The ultimate bearing capacity of a continuous foundation under earth-quake loading can be evaluated in a manner similar to that for the static condi-tion shown above. Figure 6.7 shows the wedge analysis for this condition fora 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 foractive and passive conditions, or
(6.19)
and
FIGURE 6.6 Variation of N! with soil friction angle #
FIGURE 6.7 Bearing capacity of a continuous foundation on sand--earthquake
loading
+θ+δ++θ−δ++
+−=α −
)cot)(tantan(1
tan]cot)tan(1)[tan1(tan
21
aa
aaaa
PE
θ =−
−tan 1
1
k
kh
v
(6.20)
where a = !! " (6.21)
(6.22)
kh = horizontal coefficient of accelerationkv = vertical coefficient of acceleration
NK
KqEPE
AE
=
−α=γ 1tan
AE
PE
AEE K
KN
2
2
)cos(
)sin()sin(1)cos(cos
)(cos
θ+δ
θ−φδ+φ+θ+δθ
θ−φ=AE
K
2
2
)cos(
)sin()sin(1)cos(cos
)(cos
θ+δ
θ−φδ+φ−θ+δθ
θ−φ=PE
K
Figure 6.7b shows the equilibrium analysis of wedges I and II as shownin Fig. 6.7a. As in the static analysis [similar to Eq. (6.17)]
quE = qNqE + –12 #BN#E (6.23)
where quE = ultimate bearing capacity,NqE , N#E = bearing capacity factors
Similar to Eqs. (6.10) and (6.16)
(6.24)
(6.25)
where KAE , KPE = horizontal coefficients of active and passive earth pressure (under earthquake conditions), respectively, or
(6.26)
and
(6.27)
Using $ = !/2 as before, the variations of KAE and KPE for various valuesof " can be calculated. They can then be used to calculate the bearing capacityfactors NqE and N#E . Again, for a continuous foundation supported by a c–!soil
quE = cNcE + qNqE + –12
#BN#E (6.28)
where NcE = bearing capacity factorThe magnitude of NcE can be approximated as
NcE " (NqE !1)cot! (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 beused 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 ultimatebearing capacity quE .
Solution From Eq. (6.28)
quE = cNcE + qNqE + –12
#BN#E
FIGURE 6.9 Variation of Nqe /Nq with tan" and ! (after Richards et al. [5])
N
NN
N
NN
N
NN
cE
ccE
qE
qqE
EE
= = =
= = =
= = =
0 44 0 44 20 8 8
0 38 0 38 10 38
013 013 14 182
. ; ( . )( ) .
. ; ( . )( ) .
. ; ( . )( ) .γ
γγ
For ! = 25 #, from Figs. 6.4, 6.5, and 6.6, Nc " 20, Nq " 10, N# " 14. FromFigs. 6.8, 6.9, and 6.10, for tan" = kh /(1 ! kv) = 0.25/(1 ! 0) = 0.25,
So
FIGURE 6.10 Variation of Nce /Nc with tan" and ! (after Richards et al. [5])
*
011 h
h
v
h kk
k
k≈
−
≈
−
crcr
quE = (30)(8.8) + (1 × 17)(3.8) + –12
(17)(1.5)(1.82) = 351.8 kN / m2
!!
6.4 SETTLEMENT OF FOUNDATION ON GRANULARSOIL 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 ratiokh/(1!kv) reaches a certain critical value. Thus, if kv " 0, then
(6.30)
FIGURE 6.11 Critical acceleration kh*for incipient foundation settlement (after Richards et al. [5])
FIGURE 6.11 (Continued)
FIGURE 6.12 Variation of tan %AE with kh and ! (after Richards et al. [5])
SV
Ag
k
Aeh
AE=−
01742 4
. tan*
α
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 frictionangle (!). Richards et al. [5] developed this relationship, and it is shown in agraphical form in Fig. 6.11. According to Richards et al. [5], the settlementof a foundation during an earthquake can be given as
(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 mayundergo.
FSq
qu= = =
all
1015
3402 98.
SV
Ag
k
Aeh
AE=−
01742 4
. tan*
α
Se = =−
( . )( .
( . )( . )
.
.( . ) .0174
0 35
0 32 9 81
0 28
0 320 95 0 011
4 m / s)
m / s m =
2
211 mm
Solution From Eq. (6.17)
qu = qNq + –12 #Bn#
From Figs. 6.5 and 6.6, for ! = 35#, Nq " 30; N# " 42. So
qu = (1 × 16.5)(30) + –12
(16.5)(1.5)(42) " 1015 kN / m2
Given qall = 340 kN / m2, so
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)
From Fig. 6.12 for ! = 35# and k*h = 0.28, tan%AE " 0.95. So
!!
6.5 FOUNDATION SETTLEMENT DUE TO CYCLICLOADING—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 low-frequency (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 wereconducted for foundation widths (B) of 75 mm and 228 mm. The unit weightof sand was 16.97 kN / m3. Since the settlement of the foundation Se after thefirst cycle of load application was primarily due to the placement of the foun-dation 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
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 numberof load cycles, the settlement increased with the increase in &d/qu and, (b) fora given &d/qu , Se increased with N. These load-settlement curves can be ap-proximated by the relation (for N = 2 to 105)
FIGURE 6.15 Variation of Se (after first load cycle) with &d /qu and N — B = 228 mm (after Raymond and Komos [6])
Sa
Nb
e =−1
log
−
σ+=
+
σ+−=
1.230000363.0153579.0
09.60000693.015125.0
821.0
18.1
u
d
u
d
qBb
qBa where
(6.32)
(6.33)
(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 designingrailroad ties.
Settlement of Machine Foundations
Machine foundations subjected to sinusoidal vertical vibration (Fig. 6.18) mayundergo permanent settlement (Se). In Fig. 6.18, the weight of the machine andthe foundation is W, and the diameter of the foundation is B. The impressedcyclic force Q is given by the relationship
FIGURE 6.16 Contours of variation of Se with σd and N --- B = 75 mm (after Raymond and Komos [6])
FIGURE 6.17 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 = timeMany investigators believe that the peak acceleration is the primary con-
trolling parameter for the settlement. Depending on the degree of compactionof the granular soil, the solid particles come to an equilibrium condition for agiven peak acceleration resulting in a settlement Se (max) as shown in Fig. 6.19.This threshold acceleration must be acceded before additional settlement cantake place.
Brumund and Leonards [7] evaluated the settlement of circular founda-tions subjected to vertical sinusoidal loading by laboratory model tests. For thisstudy, the model foundation had a diameter of 4 in., and 20-30 Ottawa sand
FIGURE 6.20 Variation of Se (max) with peak acceleration and weight of foundation (after Brumund and Leonards [7])
kGB
s
=−
21 ν
compacted at a relative density of 70% was used. Based on their study, itappears 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 forweights of foundation, W = 48.8 lb, 73.4 lb, and 98 lb. The frequency of vibra-tion was kept constant at 20 Hz for all tests. It is obvious that, for a given valueof W, the magnitude of Se increases linearly with the peak acceleration level.
The maximum energy transmitted to the foundation per cycle of vibrationcan be theorized as follows. Figure 6.21 shows the schematic diagram of alumped-parameter one degree-of-freedom vibrating system for the machinefoundation. The soil supporting the foundation has been taken to be equivalentto a spring and a dashpot. Let the spring constant be equal to k and the viscousdamping constant of the dashpot be c. The spring constant k and the viscousdamping constant c can be given by the following relationships (for furtherdetails see any soil dynamics text, for example, Das [8]).
(6.36)
FIGURE 6.21 Lumped–parameter one degree –of–freedom vibrating system
c BG
gs
=−0 85
12.
νγ
F kz cdz
dtdynamic = +
(6.37)
where G = shear modulus of the soil"s = Poisson’s ratio of the soilB = diameter of the foundation# = unit weight of soilg = 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 forceThe dynamic force transmitted by the foundation can be given as
(6.39)
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 + $)
where magnitude of maximum dyanmic force = dynamic(max)A F
A A Z k c
=
= + = +( cos ) ( sin ) ( )β β ω2 2 2 2
or
Fdynamic = Acos(!t + $ + %) (6.40)
(6.41)
The energy transmitted to the soil per cycle of vibration, Etr , is
Etr = "F dz = Fav Z (6.42)
where F = total contact force on soil Fav = average contact force on the soil
However
Fav = –12 (Fmax + Fmin ) (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, andc. a maximum downward dynamic force of 0.3W to W.
The results show that Se (max) increases linearly with Etr . Figure 6.23 shows aplot of the experimental results of Se (max) against peak acceleration for differentranges of Etr . This clearly demonstrates that, if the value of the transmittedenergy is constant, the magnitude of Se (max) remains constant irrespective of thelevel of peak acceleration.
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 TOCYCLIC LOADING IN SATURATED CLAY
Das and Shin[9] provided small-scale model test results for the settlement ofa continuous surface foundation (Df = 0) supported by a saturated clay andsubjected 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 11.9kN / m2. The load to the foundation was applied in two stages (Fig. 6.24):
FIGURE 6.24 Load application sequence to observe foundation settlement insaturated 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, andStage II— application of a cyclic load, the intensity of which has an amplitudeof &d as shown in Fig. 6.24b. The frequency of the cyclic load was 1 Hz. Figure6.24c shows the variation of the total load intensity on the foundation. Typicalexperimental plots obtained from these laboratory tests are shown by the
FIGURE 6.25 Typical plot of Se/B versus N for FS 3.33 and &d /qu 4.38%, 9.38%, and 18.75%
FIGURE 6.26 General nature of plot of Se versus N for given valuesof FS and &d /qu
broken lines in Fig. 6.25 (FS = 3.33; &d /qu = 4.38%, 9.38%, and 18.75%). Itis important to note that Se in this figure refers to the settlement obtained dueto cyclic load only (that is, after application of Stage II load; Fig 6.24b). Thegeneral nature of these plots is shown in Fig. 6.26. They consist ofapproximately 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 settle-
ment 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 con-clusions were drawn:
1. The initial rapid settlement is completed within the first 10 cycles ofloading.
2. The magnitude of Ncr varied between 15,000 to 20,000 cycles. This isindependent 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 inFS.
Figure 6.27 shows a plot of Se(max) /Se(u) versus &d /qu for various values ofFS. Note that Se(u) is the settlement of the foundation corresponding to static
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
and 1.
6.27) Fig.
,
(
1
1
)(
(max)
!"!#$
n
u
d
ue
e
qm
S
S
σ=
FIGURE 6.29 Nature of transient load in the laboratory tests of Cunny and Sloan [10]
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
2. for any FS and &d /qu (Fig.6.28), the limiting value of Se(r) may beabout 0.8Se(max) .
6.7 SETTLEMENT DUE TO TRANSIENT LOADON FOUNDATION
A limited number of test results are available in the literature which relate tothe evaluation of settlement of shallow foundations (supported by sand andclay) subjected to transient loading. The findings of these tests will be dis-cussed in this section.
Cunny and Sloan [10] conducted several model tests on square surfacefoundations (Df = 0) to observe the settlement when the foundations weresubjected to transient loading. The nature of variation of the transient loadwith time used for this study in shown in Fig. 6.29. Tables 6.2 and 6.3 showthe results of these tests conducted in sand and clay, respectively. Other detailsof the tests are as follows:
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
Width of model foundation, B (in.) 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 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
Width of model foundation, B (in.) 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 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
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 threecorners by linear potentiometers. Based on the results of these tests, the fol-lowing general conclusions can be drawn.
1. The settlement of the foundation under transient loading is generallyuniform.
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 Table6.2, the settlement at ultimate load, Qu (static bearing capacity test)was about 2.620 in. However, when subjected to a transient load withQd (max) = 1.35Qu , the observed settlement was about 0.410 in. Simi-larly for Test No. 2 in Table 6.3, the settlement at ultimate load wasabout 2 in. Under transient load with Qd (max) = 1.26Qu , the observedsettlement was only about 0.7 in.
Jackson and Hadala [11] reported several laboratory model test results onsquare surface foundation with width B varying from 4.5 in. to 8 in. whichwere supported by saturated Buckshot clay. For these tests, the nature of thetransient 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 onthese tests, it showed that there is a unique relationship between Qd(max)/(B
2cu)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, determinethe relationship between load Q and Se /B.
2. Plot a graph of Q/B2cu versus Se /B as shown by the broken lines inFig. 6.31.
3. Since the strain-rate factor in clays is about 1.5 (Section 6.2), deter-mine 1.5Q/B2cu and develop a plot of 1.5Q/B2cu versus Se /B as shownby the solid lines in Fig. 6.31. This will be the relationship betweenQd(max)/(B
2cu) versus Se /B.
FIGURE 6.30 Nature of transient load in the laboratory tests of Johnson andHadala [11]
FIGURE 6.31 Relationship of Qd (max) /B 2cu versus Se /B from plate
load tests (plate size B × B)
REFERENCES
1. Vesic, A. S., Banks, D. C., and Woodward, J. M., An experimental study ofdynamic bearing capacity of footings on sand, in Proceedings, VI Int. Conf. Soil Mech. Found. Eng., Montreal, Canada, 2, 1965, 209.
2. Vesic, A. S., Analysis of ultimate loads of shallow foundations, J. Soil Mech.Found. Engg. Div., ASCE, 99(1), 45, 1973.
3. Whitman, R. V., and Healy, K. A., Shear strength of sands during rapidloading, J. Soil Mech. Found. Engg. Div., ASCE, 88(2), 99, 1962.
4. Carroll, W. F., Dynamic Bearing Capacity of Soils: Vertical Displacement ofSpread Footing on Clay: Static and Impulsive Loadings, Technical Report 3-599, Report 5, U.S. Army Corps of Engineers, Waterways Experiment Station,Mississippi, 1963.
5. Richards, R., Jr., Elms, D. G., and Budhu, M., Seismic bearing capacity andsettlement of foundations, J. Geotech. Eng., ASCE, 119(4), 622, 1993.
6. Raymond, G. P., and Komos, F. E., Repeated load testing of a model planestrain footing, Canadian Geotech. J., 15(2), 190, 1978.
7. Brumund, W. F., and Leonards, G. A., Subsidence of sand due to surfacevibration, J. Soil Mech. Found. Eng. Div., ASCE, 98(1), 27, 1972.
8. Das, B. M., Principles of Soil Dynamics, PWS Publishers, Boston,Massachusetts, 1993.
9. Das, B. M., and Shin, E. C., Laboratory model tests for cyclic load-inducedsettlement of a strip foundation on a clayey soil, Geotech. Geol. Eng., London,14, 213, 1996.
10. Cunny, R. W., and Sloan, R. C., Dynamic Loading Machine and Results ofPreliminary Small-Scale Footing Tests, Spec. Tech. Pub. 305, ASTM, 65,1961.
11. 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.
CHAPTER SEVEN
SHALLOW FOUNDATIONS ON REINFORCED SOIL
7.1 INTRODUCTION
Reinforced soil, or mechanically stabilized soil, is a construction technique thatconsists of soil that has been strengthened by tensile elements such as metalstrips, geotextiles, or geogrids. In the 1960's the French Road ResearchLaboratory conducted extensive research to evaluate the beneficial effects ofusing reinforced soil as a construction technique. Results of the early workwere well documented by Vidal [1]. During the last thirty years many retainingwalls and embankments were constructed all over the world using reinforcedsoil, and they have performed very well.
The metallic strips that are used for reinforced soil are usually galvanizedsteel strips. However, the galvanized steel strips are subject to corrosion at therate of about 0.025 to 0.05 mm/year. Hence, depending on the projected servicelife of a given structure, allowances must be made for the rate of corrosion.
Geotextiles and geogrids are non-biodegradable materials. They are madefrom 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 protectfine-grained soil from being washed into the coarse-grained soil.
Geogrids are made by tensile drawing of polymer materials such as poly-ethylene and polypropylene. They are relatively stiff material compared to geo-textiles. They have large apertures which allow interlocking with surroundingsoil to perform the functions of reinforcement and/or segregation. Commer-cially available geogrids are generally of two types—uniaxial and biaxial.Figures 7.1a and 7.1b show these two types of geogrids. Geogrids are manu-factured so that the open areas of the grids are greater than 50% of the totalarea. They develop reinforcing strength at low strain levels such as 2%.
FOUNDATIONS ON METALLIC STRIP-REINFORCEDGRANULAR SOIL
7.2 FAILURE MODE
Binquet and Lee [2,3] conducted several laboratory tests and proposed a theoryfor 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
FIGURE 7.1 Geogrids: (a) Uniaxial; (b) Biaxial
FIGURE 7.2 Foundation on metallic strip reinforced granular soil
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. Thedistance between each layer of reinforcement is h. It was experimentallyshown [2, 3] that the most beneficial effect of reinforced earth is obtainedwhen 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 tiesbreak. This phenomenon is shown in Fig. 7.3.
Figure 7.4 shows an idealized condition for the development of a failure
FIGURE 7.5 Variation of x"/B with z/B
surface in reinforced earth which consists of two zones. Zone I is immediatelybelow the foundation which settles with the foundation during the applicationof 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 pointsat 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 maximumshear stress occurs is a function of z/B. This is shown in a nondimensionalform 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 rein-forcing 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 rightangle 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 thefoundation supported by reinforced earth qR to the load per unit areaon 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.
FIGURE 7.6 Assumptions to calculate the force in reinforcement ties
( )
β−α
−= hB
q
NT
o
Ro
11
With the above assumptions it can be seen that
(7.1)
where T = tie force per unit length of the foundation at a depth z (lb / ft or kN / m)
FIGURE 7.7 Variation of " with z/B
FIGURE 7.8 Variation of # with z/B
N = number of reinforcement layersqo = load per unit area of the foundation on unreinforced soil for a foun-
dation settlement level of Se = S"e
qR = load per unit area of the foundation on reinforced soil for a founda- tion settlement level of Se = S"e
", # = parameters which are functions of z/BThe variations of " and # with z/B are shown in Figs. 7.7 and 7.8 respectively.
FSwtnf
T
tf LDR
TBy y= =
( )
+′−γ+σφ= ∫
=
′=
µ
Xx
xx
fPDzxXwndxwnF ))((tan2
FIGURE 7.9 Frictional resistance against tie pullout
7.4 FACTOR OF SAFETY AGAINSTTIE BREAKING AND TIE PULLOUT
In designing a foundation it is essential to determine if the reinforcement tieswill fail either by breaking or by pullout. Let the width of a single tie (at rightangles to the cross section shown in Fig. 7.2) be w and its thickness be t. If thenumber of ties per unit length of the foundation placed at any depth z is equalto n, then the factor of safety against the possibility of tie break (FSB) is
(7.2)
where 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. Thefrictional resistance against tie pullout at that depth can be calculated as
(7.4)
+′−γ+
δφ= µ ))(()(tan2
f
o
RoP
DzxXq
qBqLDRF
FSF
TPP=
FIGURE 7.10 Variation of $ with z/B
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 foundationX # 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 factthat frictional resistance is derived from the top and bottom of the ties. Thus,from Eq. (7.4)
(7.5)
The term $ is a function of z/B and is shown in Fig. 7.10. Figure 7.11 shows aplot of X/B versus z/B. Hence, at any given depth z, the factor of safety againsttie pullout, FSP , can be given as
(7.6)
FIGURE 7.11 Variation of X/B with z/B
7.5 DESIGN PROCEDURE FOR ACONTINUOUS FOUNDATION
Following is a step-by-step procedure for designing a continuous foundationon a granular soil reinforced with metallic strips.
Step 1. Establish the following parametersa. Foundation
$ Net load per unit length, Q$ Depth, Df
$ Factor of safety, FS, against bearing capacity failure on unrein-forced soil
$ Allowable settlement, Se
b. Soil$ Unit weight, '$ Friction angle, %$ Modulus of elasticity, Es
$ Poisson’s ratio, µs
c. Reinforcement ties$ Width, w
′ ≈ =+
FS
qN BN
FSu
q
all
12
γ γ
′′ =−
qE S
B Is e
sall
( )1 27µ
BR =
$ 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 thefoundation
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
(7.8)
where qu = ultimate bearing capacity on unreinforced soilq = 'Df
Nq , N' = bearing capacity factors (Table 2.3)
Step 5. Determine the allowable bearing capacity, q'all , based on allowablesettlement. From Eq. (5.41),
(7.9)
The magnitude of I7 for continuous foundations can be taken to beapproximately 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
(7.10)
Step 8. Calculate T for all layers of reinforcement using Eq. (7.1).
Number of strips, mnLDR
w= = =
0 6
0 0078 57
.
.. /
′ =+
qqN BN
FS
q
all
12
γ γ
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 in-creased by substituting X´ (>X) in Eq. (7.5) so that FP /T is equal toFSP .
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 / mDf = 1.2 mFactor of safety against bearing capacity failure in unreinforced soil, Fs = 3.5Tolerable 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 = 3FSP = 2fy = 2.5 × 105 kN / m2
Solution Let B = 1.2 m, u = 0.5 m, h = 0.5 m, N = 4, and LDR = 60%. WithLDR = 60%
From Eq. (7.8)
From Table 2.3, for " = 36 " the magnitudes of Nq and N! are 37.75 and 44.43respectively. So
′ = × + =qall2 kN / m
( . . )( . ) ( . )( . )( . )( . )
..
12 16 5 37 75 0 5 16 5 12 44 43
35339 23
′′ =−
= ×−
=qE S
B Is e
sall
2 kN / m( )
( . )( . )
( . )[ ( . ) ]( ).
1
3 4 10 0 025
12 1 0 3 2389 2
27
4
2µ
BR = = × =15 101250
3. kN
1.2 kN / m2
)(1 hBq
q
N
qT
o
Ro β−α
−
=
+′−γ+
δ
φ= µ ))((
)(tan2f
o
R
o
P DzxXq
qBq
T
LDR
T
F
From Eq. (7.9)
Since q!all < qall , qo = qall = 339.23 kN / m2. Thus
Now the tie forces can be calculated using Eq. (7.1)
LayerNo.
−
1
o
Ro
q
q
N
qz
(m)z/B #B#$h T
(kN/m)
1234
227.7227.7227.7227.7
0.51.01.52.0
0.470.831.251.67
0.2850.3000.3250.330
64.8968.3174.0075.14
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)
+′−′γ+
δ
φ= µ ))((
)(tan2f
o
R
o
P DzxXq
qBq
T
LDR
T
F
FStf LDR
T
tFS T
f LDR
TT
By
B
y
=
= =×
= × −
( )
( )( )
( )( )
( )( )
( . )( . )
3
2 5 10 0 62 10
55
Parameter
Layer
1 2 3 4
0.0086 0.0082 0.0076 0.00752 tan ( )φµ LDR
T (m / kN)
z/B 0.47 0.83 1.25 1.67
% 0.12 0.14 0.15 0.16
(kN/m)
δ
o
Ro q
qBq 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
(kN/m)))((f
DzxX +′−γ 23.56 65.34 117.6 145.7
FP /T 1.75 2.26 2.6 2.89
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 /Tis equal to 2. So for Layer 1
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)
FIGURE 7.12
The following table can now be prepared.
LayerNo.
T(kN/m)
t(mm)
1234
64.8968.3174.0075.14
$1.3$1.4$1.5$1.503
.
A tie thickness of 1.6 mm will be sufficient for all layers. Figure 7.12 showsa diagram of the foundation with the ties.
!!
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'sto determine the bearing capacity of surface foundations (that is, Df =0) restingon geotextile-reinforced soils can be found in the literature. Guido et al. [4] re-ported results of several laboratory model tests for a square surface foundationmeasuring 0.31 m × 0.31 m (B × B) and supported by a loose to medium sandreinforced with multiple layers of nonwoven melt-bonded geotextile (size b ×b). Figure 7.13 shows the geometric parameters of the problem under con-sideration, and Fig. 7.14 shows some of the results of these tests. For the testsreported in Fig. 7.14, the following parameters apply: relative density of sand,
FIGURE 7.13 Foundation on geotextile-reinforced soil
FIGURE 7.14 Model test results of Guido et al. [4] on geotextile-reinforced sand for a square surface foundation
FIGURE 7.15 Model test results of Sakti and Das [5] on geotextile-reinforcedsaturated clay for a continuous surface foundation
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 bya saturated clay (! = 0 condition) reinforced by heat-bonded nonwoven geo-textile were reported by Sakti and Das [5]. The load-settlement curves for someof these tests are given in Fig. 7.15. For these tests the following parametersapply: width, B = 76.2 mm; undrained cohesion of clay, cu = 22.5 kN/m2, andu/B = h/B = 0.33. The tests clearly show that the ultimate bearing capacity offoundations increases when geotextile reinforcement is used.
7.7 COMMENTS ON GEOTEXTILE REINFORCEMENT
Figures 7.14 and 7.15 show that geotextile reinforcement contributes to theincrease in ultimate bearing capacity of foundations on sand and saturated clay.However, at low settlement level of the foundation, geotextile reinforcementhardly contributes to the load bearing capacity. This is primarily because geo-textiles are made of flexible material. Sufficient settlement of the foundationwill be necessary to give the layers of geotextile a catenary shape and to developtension to resist the stress transmitted from the foundation. The design of mostfoundations with a width B greater than about 1 m will be controlled by theallowable level of settlement rather than the ultimate bearing capacity. For thatreason, geotextiles may not be a suitable material for soil reinforcement in theimprovement of bearing capacity.
FIGURE 7.16 Geometric parameters of a rectangular foundationsupported by geogrid-reinforced soil
FOUNDATIONS ONGEOGRID-REINFORCED SOIL
7.8 GENERAL PARAMETERS
Geogrids are stiffer material than geotextiles. Since the mid 1980's, a numberof laboratory model studies have been reported relating to the evaluation of theultimate and allowable bearing capacities of shallow foundations supported bysoil reinforced with multiple layers of geogrids. The results obtained so far seempromising. In this section the general parameters of the problem are defined.
BCRq
quu R
u
= ( )
FIGURE 7.17 General nature of the load-settlement curvesfor unreinforced and geogrid-reinforced soilsupporting a foundation
Figure 7.16 shows the general parameters of a rectangular surfacefoundation on a soil layer reinforced with several layers of geogrid. The sizeof the foundation is B × L (width × length) and the size of the geogrid layersis b × l (width × length). The first layer of geogrid is located at a depth u belowthe foundation, and the vertical distance between consecutive layers of geogridis h. The total depth of reinforcement d can be given as
d = u + (N ! 1)h (7.11)
where N = number of reinforcement layersThe beneficial effects of reinforcement to increase the bearing capacity can
be expressed in terms of a nondimensional parameter called the bearingcapacity ratio (BCR). The bearing capacity ratio can be expressed with respectto the ultimate bearing capacity or the allowable bearing capacity (at a givensettlement level of the foundation). Figure 7.17 shows the general nature of theload-settlement curve of a foundation both with and without geogrid reinforce-ment. Based on this concept the bearing capacity ratio can be defined as
(7.12)
BCRq
qsR=
FIGURE 7.18 Definition of critical nondimensional parameters — (d/B)cr , (b/B)cr , and (l /B)cr
and
(7.13)
whereBCRu = bearing capacity ratio with respect to the ultimate loadBCRs = bearing capacity ratio at a given settlement level Se for the
foundationFor 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 thecritical-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 . Thisconcept is schematically illustrated in Fig. 7.18. As an example, Fig. 7.19shows the variation of BCRu with d/B for four model foundations (B/L = 0, 1/3,
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 laboratorymodel tests [6,7] that for a given foundation, if b/B, l/B, d/B, and h/B are keptconstant, 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 . Foru/B > (u/B)cr the magnitude of BCRu decreases (Zone 2). For u/B > (u/B)max theplot of BCRu versus u/B generally flattens out (Zone 3). The present under-standing (in general) among investigators is that, in Zones 1 and 2, the natureof the failure surface in soil will be as shown in Fig. 7.21a. In Zone 1 the initialincrease in BCRu with u/B is due to the increase in confining pressure on thegeogrid layers. In Zone 3 [that is, u/B " (u/B)max ] the failure surface in soilbelow the foundation is located fully above the first layer of geogrid, whichacts as a semi-rigid rough base (Fig. 7.21b).
7.9 RELATIONSHIPS FOR CRITICAL NONDIMENSIONAL PARAMETERS FOR FOUNDATIONS ON GEOGRID-REINFORCED SAND
Based on the results of their model tests and other existing results, Omar et al.[6] developed the following empirical relationships for the nondimensionalparameters (d/B)cr , (b/B)cr , and (l/B)cr described in the preceding section.
FIGURE 7.20 Nature of variation of BCRu with u/B
FIGURE 7.21 Failure surface in geogrid-reinforced soil under afoundation (a) u/B < (u/B)max ; (b) u/B = (u/B)max
≤≤
−=
≤≤
−=
126.043.1
5.04.12
L
B
L
B
b
d
L
B
L
B
b
d
0.5for
0for
cr
cr
51.0
5.38
−=
L
B
B
b
cr
B
L
L
B
B
l +
=
5.3cr
Critical Reinforcement-Depth Ratio
(7.14)
(7.15)
The preceding relationships suggest that the bearing capacity increase isrealized only when the reinforcement is located within a depth of 2B for acontinuous foundation and a depth of 1.2B for a square foundation.
Critical Reinforcement-Width Ratio
(7.16)
According to Eq. (7.16), (b/B)cr is about 8 for a continuous foundation andabout 4.5 for a square foundation. It needs to be realized that generally, withother parameters remaining constant, about 70% or more of BCRu is realizedwith b/B # 2. The remaining 30% of BCRu is realized when b/B increased fromabout 2 to (b/B)cr .
Critical Reinforcement-Length Ratio
(7.17)
The magnitude of (u/B)max was recommended by Binquet and Lee [2,3] to beabout 0.67. However, model test results of Guido et al. [8] on a square founda-tion and Omar et al. [9] on a continuous foundation show that (u/B)max is about0.9 to 1. Based on the existing studies it appears that (u/B)cr is approximatelyequal 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 Bgreater than about 1 m will be generally controlled by the allowable level ofsettlement Se , not by the ultimate bearing capacity. For that reason, it is
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)
important to determine BCRs at various levels of settlement [see definition inEq. (7.13)] and its relationship with BCRu .
Omar et al. [9] conducted several laboratory model tests on square andcontinuous surface foundations (that is, Df = 0) supported by unreinforced andgeogrid-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 mmRelative density of sand, Dr = 70%Soil friction angle, ! = 40.3$Size of square foundation = 76.2 mm × 76.2 mmWidth of continuous foundation, B = 76.2 mmGeogrid 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 variationof BCRs shown in Fig. 7.22 is the average plot for Se /Se (u) = 0.25, 0.50, and0.75 [Se (u) = settlement at ultimate load for unreinforced sand] since the scatter
≤==
≤==
75.014.1
75.007.1
)(
)(
ue
eus
ue
esu
S
S
L
BBCRBCR
S
S
L
BBCRBCR
and for 1.45 to
and
and for 1.8 to
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)
(7.18)
(7.19)
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 thedetermination of bearing capacity of foundations supported by saturated clayare relatively scarce. Shin et al. [10] reported some laboratory model test resultswhich were intended to determine the nondimensional parameters as definedby Section 7.8. These tests were conducted with a continuous surface founda-tion (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 forreinforcement. 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/Bvalues, the magnitude of BCR increases from u/B = 0.25 to u/B = 0.4and 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 fromFig. 7.23.
5. Critical reinforcement-depth ratio: (d/B)cr # 1.8. This is illustrated inFig. 7.24. Note that (d/B)cr is independent of the undrained cohesionof the clay, cu .
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 Shinet al. [4]) ( Note: b/B = 4, u/B = 0.4, h/B = 1/3)
7.12 BEARING CAPACITY THEORY
Most of the existing data available relating to the ultimate and allowable bear-ing capacities for shallow foundations supported by geogrid-reinforced soil isbased on small-scale laboratory model tests. More recently, Adams and Collin[11] provided large-scale model test results in sand which could be used toverify laboratory test results. Also Huang and Meng [12] recently proposed anultimate bearing capacity theory for foundations on geogrid-reinforced sandbased on the wide-slab effect. However, more study is necessary to develop abearing capacity theory that will take into account the failure mode in soil, thestrength and stiffness of geogrid reinforcement, and scale effects.
7.13 SETTLEMENT OF FOUNDATIONS ON GEOGRID- REINFORCED SOIL DUE TO CYCLIC LOADING
In many cases shallow foundations supported by geogrid-reinforced soil maybe subjected to cyclic loading. This problem will primarily be encountered byvibratory machine foundations. Das [13] and Das and Shin [14] reportedlaboratory model test results on settlement caused by cyclic loading on surfacefoundations supported, respectively, by reinforced sand and saturated clay.
The model tests of Das [13] were conducted with a square model founda-tion on unreinforced and geogrid-reinforced sand. Details of the sand and geo-grid parameters were:
Model foundation: Square; B = 76.2 mmSand: Relative density of compaction, Dr = 76%
Angle of friction, ! = 42$Reinforcement: Geogrid; TENSAR BX1000
Reinforcement-width ratio:
33.0
33.0
=
=
≈
≈
B
h
B
u
B
u
B
b
B
b
cr
cr
(7.16)] Eq. [see
Reinforcement-depth ratio: (7.15)] Eq. [see cr
33.1=
≈
B
d
B
d
Number of layers of reinforcement: N = 4
FIGURE 7.25 Nature of load application—cyclic load test
FIGURE 7.26 Plot of Sec /B versus n (after Das [13]) ( Note: Forreinforced 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 ofintensity qs (= qu (R) /FS; FS = factor of safety) followed by a cyclic load of lowfrequency (1 cps). The amplitude of the intensity of cyclic load was qdc (max) . Thenature of load application described is shown in Fig. 7.25. Figure 7.26 showsthe nature of variation of foundation settlement due to cyclic load application
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 amaximum value Sec (max) at n = ncr . Based on these tests the followingconclusions can be drawn.
1. For given values of FS and n, the magnitude of Sec /B increases withthe 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 between1.5 × 105 and 2.0 × 105 cycles.
The variations of Sec (max) /B obtained from these tests for various values ofqdc (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)
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)
ρ =−
−S
Sec
ec
(max)
(max)
reinforced
unreinforced
the reduction of the level of permanent settlement caused by geogrid reinforce-ment due to cyclic loading. Using the results of Sec (max) given in Fig. 7.27, thevariation of settlement ratio ! for various combinations of qdc (max) /qu (R) and FSare plotted in Fig. 7.28. The settlement ratio is defined as
(7.21)
From Fig. 7.28 it can be seen that, although some scattering exists, the settle-ment 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 similarloading conditions as those described above (Fig. 7.25) in reinforced saturatedclay were provided by Das and Shin [14]. General parameters of the testprogram were as follows:
73.1
5
5
33.0
4.0
=
=
=
=
≈
=
=
=
cr
cr
cr
B
d
B
d
N
B
b
B
b
B
h
B
u
B
u
FIGURE 7.29 Nature of variation of foundation settlement in clay due to cyclic load application
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
Range of test: q
q
FSq
q
d
u R
u R
s
(max) to 14.6%
to 6.85
( )
( )
.
.
=
= =
3 4
315
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 andqdc (max) /qu (R) ] obtained is shown in Fig. 7.29, which can be divided into threemajor zones. Zone 1 (for n = 1 to n = nr ) is a rapid settlement zone duringwhich about 70% of the maximum settlement [Sec (max) ] takes place. The magni-tude of nr is about 10. Zone 2 (n = nr to n = ncr ) is a zone in which the settle-ment 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 magni-tude 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 ofSec(max)/B for various combinations of qdc (max) /qu (R) and FS. It is important tonote 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 thatare likely to be subjected to impact loading. This is shown in the results of
FIGURE 7.31 Nature of transient load
laboratory model tests in sand reported by Das [13]. The tests were conductedwith a square surface foundation (Df = 0; B = 76.2 mm). TENSAR BX1000geogrid was used as reinforcement. Following are the physical parameters ofthe soil and reinforcement:
Sand: Relative density of compaction = 76%Angle of friction, " = 42"
Reinforcement: u
B
b
B
h
B= = =0 33 4 0 33. ; ; .
Number of reinforcement layers, = 0,1, 2, 3, and 4N
The idealized shape of the impact load applied to the model foundation isshown in Fig. 7.31, in which tr and td are rise and decay times, and qt (max) is themaximum 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 settle-ments observed due to the impact loading Set (max) are shown in a nondimen-sional form in Fig. 7.32. In this figure, qu and Se (u) , respectively, are theultimate bearing capacity and the corresponding foundation settlement onunreinforced 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 factorR or
FIGURE 7.32 Variation of Set(max) /Se(u) with qt(max) /qu and d/B (after Das [13])
RS
Set d
et d
==
(max)-
(max)- 0
where Set (max)-d = maximum settlement due to impact load with reinforce-ment 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 andd/dcr is shown in Fig. 7.33. From the plot it is obvious that the geogridreinforcement acts as an excellent settlement retardant under impact loading.
FIGURE 7.33 Plot of settlement reductionfactor 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 desTravaus 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 ageotextile 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 reinforcedwith geotextile layers, Trans. Res. Rec. No. 1153, National Academy ofSciences, Washington, DC, 40, 1987.
6. Omar, M. T., Das, B. M., Yen, S. C., Puri, V. K., and Cook, E. E., Ultimatebearing 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 rectangularfootings 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 ongeogrid-reinforced earth slabs, in Proc., Geosynthetics ‘87, 1987, 216.
9. Omar, M. T., Das, B. M., Yen, S. C., Puri, V. K., and Cook, E. E., Shallowfoundations on geogrid-reinforced sand, Trans. Res. Rec. No. 1414, NationalAcademy of Sciences, Washington, DC, 59, 1993.
10. Shin, E. C., Das, B. M., Puri, V. K., Yen, S. C., and Cook, E. E., Bearingcapacity 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 ongeosynthetic 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 onreinforced sandy ground, J. Geotech. Geoenviron. Eng., ASCE, 123(1), 30,1997.
13. Das, B. M., Dynamic loading on foundation on reinforced soil, in Geosyntheticsin Foundation Reinforcement and Erosion Control Systems, Bowders, J. J.,Scranton, H. B., and Broderick, G. P., Eds., Geotechnical Special PublicationNo. 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.
FIGURE 8.1 Shallow foundation subjected to uplift
CHAPTER EIGHT
UPLIFT CAPACITY OF SHALLOW FOUNDATIONS
8.1 INTRODUCTION
Foundations and other structures may be subjected to uplift forces underspecial circumstances. For those foundations, during the design process it isdesirable to apply sufficient factor of safety against failure by uplift. Duringthe last thirty or so years, several theories have been developed to estimate theultimate uplift capacity of foundations embedded in sand and clay soils, andsome of those theories are detailed in this chapter. The chapter is divided intotwo major parts: foundations in granular soil and foundations in saturated claysoil (! = 0).
Figure 8.1 shows a shallow foundation of width B. The depth of embed-ment is Df . The ultimate uplift capacity of the foundation Qu can be expressedas
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 inthe 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 relativedensity of compaction in the case of sand and with the consistency in the caseof clay soils.
FIGURE 8.2 Balla’s theory for shallow circular foundations
When the failure surface in soil extends up to the ground surface atultimate load, it is defined as a shallow foundation under uplift. For largervalues of Df /B, failure takes place around the foundation and the failure sur-face does not extend to the ground surface. These are called deep foundationsunder uplift. The embedment ratio, Df /B, at which a foundation changes fromshallow 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, insaturated 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 thesetheories 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 soilwill be as shown in Fig. 8.2. Note from the figure that aa! and bb! are arcs ofa circle. The angle " is equal to 45! !/2. The radius of the circle, of which aa!and bb! are arcs, is equal to
φ+
=
245sin
fD
r
φ+
φγ=
B
DF
B
DFDQ ff
fu,,
31
3
FQ
ADqu
f
=γ
(8.2)
As mentioned before, the ultimate uplift capacity of the foundation is the sumof two components: (a) the weight of the soil and the foundation in the failurezone and (b) the shearing resistance developed along the failure surface. Thus,assuming that the unit weight of soil and the foundation material are approxi-mately the same
(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 theembedment ratio, Df /B.
In general, Balla’s theory is in good agreement with the uplift capacity ofshallow foundations embedded in dense sand at an embedment ratio of Df /B" 5. However for foundations located in loose and medium sand, the theoryoverestimates the ultimate uplift capacity. The main reason Balla’s theoryoverestimates the ultimate uplift capacity for Df /B > about 5 even in densesand is because it is essentially deep foundation condition, and the failuresurface does not extend to the ground surface.
The simplest procedure to determine the embedment ratio at which deepfoundation condition is reached may be determined by plotting the nondimen-sional breakout factor Fq against Df /B as shown in Fig. 8.4. The breakoutfactor is derived as
(8.4)
where A = area of the foundation.
The breakout factor increases with Df /B up to a maximum value of Fq" F*q at
Df /B" (Df /B)cr . For Df /B > (Df /B)cr the breakout factor remains practicallyconstant (that is, F*
q).
FIGURE 8.3 Variation of F1 + F3 [Eq. (8.3)]
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 ofa shallow foundation was proposed by Meyerhof and Adams [2], and it is des-cribed in detail in this section. Figure 8.5 shows a continuous foundation ofwidth B subjected to an uplifting force. The ultimate uplift capacity per unitlength of the foundation is equal to Qu . At ultimate load the failure surface in
)(cos
121
cos2
fphh
pDK
PP γ
δ
=
δ′
=′
δγ+=δ
γ+= tantan
2
12 22
fphfphuDKWDKWQ
soil makes an angle " with the horizontal. The magnitude of " depends onseveral factors, such as the relative density of compaction and the angle offriction of the soil, and it varies between 90#! a! to 90#! b!. Let usconsider the free body diagram of the zone abcd. For stability consideration, thefollowing forces per unit length of the foundation need to be considered.
a. the weight of the soil and concrete, W, andb. the passive force Pp per unit length along the faces ad and bc. The
force Pp is inclined at an angle $ to the horizontal. For an averagevalue of "" 90! !/2, the magnitude of $ is about b!.
If we assume that the unit weights of soil and concrete are approximatelythe same, then
W = #Df B
(8.5)
where Ph = horizontal component of the passive forceKph = 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
(8.6)
The passive earth pressure coefficient based on the curved failure surfacefor $ = b! can be obtained from Caquot and Kerisel [3]. Furthermore, it isconvenient to express Kphtan$ in the form
Ku tan! = Kphtan$ (8.7)
Combining Eqs. (8.6) and (8.7)
Qu = W + Ku #Df² tan! (8.8)
FIGURE 8.6 Variation of Ku
FQ
BDsu
f
=γ
φ
+=
φγ+= tan1
tan2
B
DK
W
DKWF f
u
fu
q
where Ku = nominal uplift coefficientThe 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 takenas equal to 0.95 for all values of ! varying from 30# to about 48#. The ulti-mate 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 continuousfoundation, the breakout factor per unit length is
or
(8.9)
For circular foundations, Eq. (8.8) can be modified to the form
Qu = W + –%2
SF #BDf²Ku tan! (8.10)
W% –%4
B2Df # (8.11)
+=
B
DmS f
F1
φγ
+π+γπ= tan1
2422
uf
f
fuKBD
B
DmDBQ
φ
++=
πγ
φγ
+π+γπ
=γ
=
tan121
4
tan124
2
22
u
ff
f
uf
f
f
f
uq
KB
D
B
Dm
DB
KBDB
DmDB
AD
QF
+=
B
DmS f
F1
where SF = shape factor B = diameter of the foundation
The shape factor can be expressed as
(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
(8.13)
The breakout factor Fq can be given as
(8.14)
For rectangular foundations having dimensions of B × L, the ultimatecapacity 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 por-tions of length B/2 are governed by the shape factor SF , while the passivepressure along the central portion of length L &B is the same as the continuousfoundation. In Eq. (8.15)
W % #BLDf (8.16)
and
(8.17)
φ
−+
+γ+γ= tan122
u
f
ffuKBLB
B
DmDBLDQ
FQ
BLDqu
f
=γ
φ
+
++= tan1211
u
ff
qK
B
D
L
B
B
DmF
Thus
(8.18)
The breakout factor Fq can now be determined as
(8.19)
Combining Eqs. (8.18) and (8.19), we obtain [4]
(8.20)
The coefficient m given in Eq. (8.12) was determined from experimentalobservations [2] and its values are given in Table 8.1. In Fig. 8.7, m is alsoplotted as a function of the soil friction angle !.
TABLE 8.1 Variationof m [Eq. (8.12)]
Soil frictionangle, ! m
20253035404548
0.050.10.150.250.350.50.6
As shown in Fig. 8.4, the breakout factor Fq increases with Df /B to amaximum value of Fq
* at (Df /B)cr and remains constant thereafter. Based onexperimental observations, Meyerhof and Adams [2] recommended thevariation 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
FIGURE 8.7 Variation of m
FIGURE 8.8 Variation of (Df /B)cr for square and circularfoundations
FIGURE 8.9 Plot of Fq [Eqs. (8.14) and (8.20)] for square and circular foundations
calculate the breakout factor (Fq) variation with embedment ratio (Df /B). Themaximum 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 circularfoundations with the soil friction angle ! is shown in Fig. 8.10.
Laboratory experimental observations have shown that the critical embed-ment ratio (for a given soil friction angle !) increases with the L/B ratio.Meyerhof [5] indicated that, for a given value of !,
FIGURE 8.10 Fq* for deep square and circular foundations
5.1≈
square-cr
continuous-cr
B
D
B
D
f
f
S-crS-crR-cr
≤
+
=
B
D
B
L
B
D
B
Dfff 4.1867.0133.0
(8.21)
Based on laboratory model test results, Das and Jones [6] gave an empiricalrelationship for the critical embedment ratio of rectangular foundations in theform
(8.22)
BB
B
D
BL
B
D
f
f
×
=
×
=
of dimensions
having foundation square a of ratioembedment critical
of dimensions having
foundationr rectangula a of ratioembedment critical where
S-cr
R-cr
FIGURE 8.11 Vesic’s theory of expansion of cavities
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 aspherical cavity close to the surface of a semi-infinite, homogeneous, isotropicsolid (in this case, the soil). Now, referring to Fig. 8.11, it can be seen that, ifthe distance Df is small enough, there will be an ultimate pressure po that willshear away the soil located above the cavity. At that time, the diameter of thespherical cavity is equal to B. The slip surfaces ab and cd will be tangent tothe 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 verticaldirection we can determine the ultimate pressure po in the cavity. Forces thatwill be involved are:
1. Vertical component of the force inside the cavity, PV ;2. Effective self-weight of the soil, W = W1 + W2 ; and3. Vertical component of the resultant of internal forces, FV .
+
=
+
+
−=
22
22
23
20.1
43
2
21
B
DA
B
DAF
B
DA
B
DA
D
B
F
ff
c
ff
f
q where
FIGURE 8.12 Cavity expansion theory applied to circular foundation uplift
For a c–! soil, we can thus determine that
po = c!Fc + #Df !Fq (8.23)
(8.24)
(8.25)
where A1 , A2 , A3 , A4 = functions of the soil friction !For granular soils c = 0, so
po = #Df !Fq (8.26)
Vesic [8] applied the preceding concept to determine the ultimate upliftcapacity of shallow circular foundations. In Fig. 8.12 consider that the circularfoundation ab having a diameter B is located at a depth Df below the ground
γ
π=
3
3 23
2 BW
γ=
π
γ
π
=
π
=23
2
2
23
2
2
2
3
23
1
B
B
B
B
Wp
+γ=
γ+γ=+=
π==
f
qf
qfouu
u
D
B
FD
BFDpp
B
Q
A
232
23
2
)(2
12
factorBreakout ↑
γ=
+
+γ==qf
ff
fu
uFD
B
DA
B
DAD
A
2
21
22
1
surface. Assuming that the unit weight of the soil and the unit weight of thefoundation are approximately the same, if the hemispherical cavity above thefoundation (that is, ab) is filled with soil, it will have a weight of
(8.27)
This weight of soil will increase the pressure by p1 , or
If the foundation is embedded in a cohesionless soil (c = 0), then the pressurep1 should be added to Eq. (8.26) to obtain the force per unit area of the anchor,qu , needed for complete pullout. Thus
(8.28)
or
(8.29)
The variation of the breakout factor Fq for shallow circular foundationsis given in Table 8.2 and Fig. 8.13. In a similar manner, using the analogy of
FIGURE 8.13 Vesic’s breakout factor, Fq , for shallow circular foundations
TABLE 8.2 Vesic’s Breakout Factor, Fq , forCircular Foundations
Soil frictionangle, !
(deg)
Df /B
0.5 1.0 1.5 2.5 5.0
01020304050
1.01.181.361.521.651.73
1.01.371.752.112.412.61
1.01.592.202.793.303.56
1.02.083.254.415.436.27
1.0 3.67 6.71 9.8913.015.7
FIGURE 8.14 Vesic’s breakout factor, Fq , for shallow continuous foundation
expansion of long cylindrical cavities, Vesic determined the variation of thebreakout factor Fq for shallow continuous foundations. These values are givenin Table 8.3 and are also plotted in Fig. 8.14.
TABLE 8.3 Vesic’s Breakout Factor, Fq , for Continuous Foundations
Soil frictionangle, !
(deg)
Df /B
0.5 1.0 1.5 2.5 5.0
01020304050
1.01.091.171.241.301.32
1.01.161.331.471.581.64
1.01.251.491.711.87 2.04
1.01.421.832.192.462.60
1.01.832.653.383.914.20
8.5 SAEEDY’S THEORY
A theory for the ultimate uplift capacity of circular foundations embedded insand was proposed by Saeedy [9] in which the trace of the failure surface wasassumed to be an arc of a logarithmic spiral. According to this solution, forshallow foundations the failure surface extends to the ground surface. How-ever, for deep foundations [that is, Df > Df (cr)] the failure surface extends onlyto 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 ) forvarious values of " and the Df /B ratio. The author converted the solution intoa plot of breakout factor Fq = Qu /!ADf (A = area of the foundation) versus thesoil friction angle " as shown in Fig. 8.15. According to Saeedy, during thefoundation uplift the soil located above the anchor gradually becomes com-pacted, in turn increasing the shear strength of the soil and, hence, the ultimateuplift capacity. For that reason, he introduced an empirical compaction factorµ, which is given in the form
µ = 1.044Dr + 0.44 (8.30)
where Dr = relative density of sandThus, 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 makesome general observations:
1. The only theory that addresses the problem of rectangular foundationsis 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 embedmentratio (Df /B)cr for square and circular foundations as a function of thesoil friction angle.
3. Experimental observations generally tend to show that, for shallowfoundation in loose sand, Balla’s theory [1] overestimates the ultimateuplift capacity. Better agreement, however, is obtained for foundationsin dense soil.
4. Vesic’s theory [8] is, in general, fairly accurate in estimating the ulti-mate uplift capacity of shallow foundations in loose sand. However,laboratory experimental observations have shown that, for shallowfoundations in dense sand, this theory can underestimate the actualuplift capacity by as much as 100% or more.
FIGURE 8.15 Plot of Fq based on Saeedy’s theory
Q D F Fu f= +31 3
3
γ ( ).Fig. 8
! "# $#
Figure 8.16 shows a comparison of some published laboratory experi-mental results for the ultimate uplift capacity of circular foundations with thetheories of Balla, Vesic, and Meyerhof and Adams. Table 8.4 gives the refer-ences to the laboratory experimental curves shown in Fig. 8.16. In developingthe theoretical plots for " = 30" (loose sand condition) and " = 45" (densesand condition) the following procedures were used.
1. According to Balla’s theory [1], from Eq. (8.3) for circular founda-tions
FIGURE 8.16 Comparison of theories with laboratory experimental results for circular foundations
AD
QD
B
BD
QB
D
QFF
f
u
f
f
u
f
u
γ
π
=
πγ
π
=γ
=+
2
23
2
331
4
4
4
231
4
π
+=
γ=
f
f
uq
D
B
FF
AD
QF
TABLE 8.4 References to Laboratory Experimental Curves Shown in Fig. 8.16
Curve
Reference
Circularfoundationdiameter, B Soil properties
123456789
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]
25.438.150.876.238.1–152.438.1–152.476.276.261–119.4
" = 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
or
(8.32)
So, for a given soil friction angle the sum of F1 + F3 was obtainedfrom Fig. 8.3, and the breakout factor was calculated for various valuesof Df /B. These values are plotted in Fig. 8.16.
2. For Vesic’s theory [8] the variations of Fq versus Df /B for circularfoundations are given in Table 8.2. These values of Fq are also plottedin Fig. 8.16.
3. The breakout factor relationship for circular foundations based onMeyerhof 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 alsoplotted in Fig. 8.16.
kN 101.5=
π= )2.2)(5.1)(4.17()5.1(
42
uQ
Based on the comparison between the theories and the laboratory experi-mental 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 anestimate as any for the ultimate uplift capacity. So this theory is recommendedfor use. However, it needs to be kept in mind that the majority of the experi-mental results presently available in the literature for comparison with thetheory are from laboratory model tests. When applying these results to thedesign of an actual foundation, the scale effect needs to be taken intoconsideration. For that reason, a judicious choice is necessary in selecting thevalue 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 theultimate 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
!!
φ
++= tan121
u
ff
qK
B
D
B
DmF
kN 122.7=
π=γ= )5.1()5.1(
4)4.17)(66.2( 2
fquADFQ
EXAMPLE 8.3
Redo Example Problem 8.1 using Meyerhof and Adams’ theory.
Solution From Eq. (8.14)
For " = 35", m = 0.25 (Table 8.1). So
Fq = 1+2[1 + (0.25)(1)](1)(0.95)(tan35) = 2.66
So
!!
8.7 EFFECT OF BACKFILL ON UPLIFT CAPACITY
Spread foundations constructed for electric transmission towers are subjectedto uplifting force. The uplift capacity of such foundations can be estimated byusing the same relationship described in the preceding sections. During theconstruction of such foundations, the embedment ratio Df /B is usually kept at3 or less. For foundation construction, the native soil is first excavated. Oncethe foundation construction is finished, the excavation is backfilled and com-pacted. The degree of compaction of the backfill material plays an importantrole in the actual ultimate uplift capacity of the foundation. Kulhawy,Trautman, and Nicolaides [13] conducted several laboratory model tests to ob-serve the effect of the degree of compaction of the backfill compared to thenative soil. According to their observations, failure in soil in most cases takesplace by side shear as shown in Fig. 8.17. However, wedge or combined shearfailure occurs for foundations with Df /B < about 2 in medium to dense nativesoil 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 breakoutfactor Fq when the native soil is loose. Similarly, Fig. 8.20 is for the casewhere 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 influenceon the breakout factor of the foundation, and the ultimate uplift capacitygreatly increases with the degree of backfill compaction.
FIGURE 8.17 Failure by side shear
FIGURE 8.18 Wedge or combined shear failure
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])
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 deter-mining the ultimate uplift capacity of foundations embedded in saturated claysoil are rather limited. In the following sections, the results of some of theexisting studies are reviewed.
Figure 8.21 shows a shallow foundation in a saturated clay. The depth ofthe foundation is Df , and the width of the foundation is B. The undrainedshear strength and the unit weight of the soil are cu and γ, respectively. If weassume that the unit weight of the foundation material and the clay are approx-imately the same, then the ultimate uplift capacity can be expressed as [8]
Qu = A("Df + cu Fc) (8.33)
where A = area of the foundationFc = breakout factor" = saturated unit weight of the soil
8.9 VESIC’S THEORY
Using the analogy of the expansion of cavities, Vesic [8] presented thetheoretical variation of the breakout factor Fc (for ! = 0 condition) with
FIGURE 8.22 Vesic’s breakout factor Fc
TABLE 8.5 Variation of Fc (!!!! = 0 Condition)
Foundation type
Df /B
0.5 1.0 1.5 2.5 5.0
Circular (diameter = B)Continuous (width = B)
1.760.81
3.801.61
6.122.42
11.64.04
30.38.07
the embedment ratio Df /B, and these values are given in Table 8.5. A plot ofthese same values of Fc against Df /B is also shown in Fig. 8.22. Based on thelaboratory model test results available at the present time, it appears thatVesic’s theory gives a closer estimate only for shallow foundations embeddedin softer clay.
In general, the breakout factor increases with embedment ratio up to amaximum value and remains constant thereafter as shown in Fig. 8.23. Themaximum value of Fc = Fc
* is reached at Df /B = (Df /B)cr . Foundations locatedat Df /B > (Df /B)cr are referred to as deep foundations for uplift capacity consid-eration. For these foundations at ultimate uplift load, local shear failure in soil
FIGURE 8.23 Nature of variation of Fc with Df /B
92.1 ≤
=
B
DF f
c
86.0 ≤
=
B
DF f
c
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 followingrelationship
Qu = A("Df + Fc cu)
For circular and square foundations
(8.34)
and, for strip foundations
(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 stripfoundations.
98 to≤
≈
B
DnF f
c
75.2107.0 ≤+=
−
u
f cB
D
Scr
2
Scr
kN/min cohesion, undrained
)foundation (or sfoundation of ratioembedment critical where
=
=
−
u
f
c
B
D
circularsquare
ScrScrRcr −−−
≤
+
=
B
D
B
L
B
D
B
Dfff 55.127.073.0
foundation oflength
sfoundation of ratioembedment critical whereRcr
=
=
−
L
B
Df rrectangula
8.11 MODIFICATIONS TO MEYERHOF’S THEORY
Das [14] compiled a number of laboratory model test results on circularfoundations in saturated clay with cu varying from 5.18 kN / m2 to about 172.5kN / m2. Figure 8.24 shows the average plots of Fc versus Df /B obtained fromthese studies, along with the critical embedment ratios.
From Fig. 8.24 it can be seen that, for shallow foundations
(8.36)
where n = a constant
The magnitude of n varies between 5.9 to 2.0 and is a function of the un-drained cohesion. Since n is a function of cu and Fc = Fc
* is about 8 to 9 in allcases, it is obvious that the critical embedment ratio (Df /B)cr will be a functionof cu .
Das [14] also reported some model test results with square and rectangularfoundations. Based on these tests, it was proposed that
(8.37)
It was also observed by Das [19] that
(8.38)
FIGURE 8.24 Variation of Fc with Df/B from various experimental observations—circular foundations; diameter = B
cr
=α′
B
DB
D
f
f
′ =βF
Fc
c*
+=− L
BF
c44.156.7*
R
+==
L
BFF
cc44.156.7*
γ+
+=
fuuDc
L
BAQ 44.156.7
Based on the above findings, Das [19] proposed an empirical procedureto obtain the breakout factors for shallow and deep foundations. According tothis procedure, #" and $" are two nondimensional factors defined as
(8.39)
and
(8.40)
For a given foundation, the critical embedment ratio can be calculated by usingEqs. (8.37) and (8.38). The magnitude of Fc
* can be given by the followingempirical relationship
(8.41)
where F*c-R = breakout factor for deep rectangular foundations
Figure 8.25 shows the experimentally derived plots (upper limit, lowerlimit, and average of $" and #". Following is a step-by-step procedure to esti-mate 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
Thus
(8.42)
where A = area of the foundation
FIGURE 8.25 Plot of $" versus #"
γ+
+β′=γ+β′=
fufucuDc
L
BADcFAQ 44.156.7)( *
06.85.2)52)(107.0(5.2107.0 =+=+=
−
u
f cB
D
Scr
6. For Df /B! (Df /B)cr
(8.43)
The value of $" can be obtained from the average curve of Fig. 8.25. Theprocedure outlined above gives fairly good results in estimating the netultimate 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 upliftcapacity.
Solution From Eq. (8.37)
89.85.1
327.073.07
27.073.0
=
+=
+
=
−−B
L
B
D
B
Dff
ScrRcr
85.10)7)(55.1( ==
−Scr
1.55 :CheckB
Df
13.089.82.1 ==
=α′
crB
D
B
D
f
f
kN 540.6=
+
+=
γ+
+β′=
)8.1)(9.18()52(3
5.144.156.7)2.0()3)(5.1(
44.156.7fuu
DcL
BAQ
So use (Df /B)cr-S = 7. Again, from Eq. (8.38)
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.
Referring to the average curve of Fig. 8.25, for #" = 0.13 the magnitude of $"= 0.2. From Eq. (8.43)
!!
8.12 FACTOR OF SAFETY
In most cases of foundation design, it is recommended that a minimum factorof 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.
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, Gauthier-Villars, Paris, 1949.
4. Das, B. M., and Seeley, G. R., Breakout resistance of horizontalanchors, J. Geotech. Eng. Div., ASCE, 101(9), 999, 1975.
5. Meyerhof, G. G., Uplift resistance of inclined anchors and piles, inProc., VIII Int. Conf. Soil Mech. Found. Eng., Moscow, USSR, 2.1,1973, 167.
6. Das, B. M., and Jones, A. D., Uplift capacity of rectangular founda-tions in sand, Trans. Res. Rec. 884, National Research Council,Washington, DC, 54, 1982.
7. Vesic, A. S., Cratering by explosives as an earth pressure problem, inProc., VI Int. Conf. Soil Mech. Found. Eng., Montreal, Canada, 2,1965, 427.
8. Vesic, A. S., Breakout resistance of objects embedded in ocean bottom,J. Soil Mech. Found. Div., ASCE, 97(9), 1183, 1971.
9. Saeedy, H. S., Stability of circular vertical earth anchors, CanadianGeotech. J., 24(3), 452, 1987.
10. Baker, W. H., and Kondner, R. L., Pullout load capacity of a circularearth anchor buried in sand, Highway Res. Rec.108, National ResearchCouncil, Washington, DC, 1, 1966.
11. Sutherland, H. B., Model studies for shaft raising through cohesionlesssoils, in Proc., VI Int. Conf. Soil Mech. Found. Eng., Montreal Canada,2, 1965, 410
12. Esquivel-Diaz, R. F., Pullout Resistance of Deeply Buried Anchors inSand, M.S. Thesis, Duke University, Durham, NC, USA, 1967.
13. Kulhawy, F. H., Trautman, C. H., and Nicolaides, C. N., Spreadfoundations in uplift: experimental study, Found. for TransmissionTowers, Geotech. Spec. Pub. 8, ASCE, 110, 1987.
14. Das, B. M., Model tests for uplift capacity of foundations in clay, Soilsand Foundations, Japan, 18(2), 17, 1978.
15. Ali, M., Pullout Resistance of Anchor Plates in Soft Bentonite Clay,M.S. Thesis, Duke University, Durham, NC, USA, 1968.
16. Kupferman, M., The Vertical Holding Capacity of Marine Anchors inClay Subjected to Static and Dynamic Loading, M.S. Thesis, Univer-sity of Massachusetts, Amherst, , USA, 1971.
17. Adams, J. K., and Hayes, D. C., The uplift capacity of shallowfoundations, Ontario Hydro. Res. Quarterly, 19(1), 1, 1967.
18. Bhatnagar, R. S., Pullout Resistance of Anchors in Silty Clay, M.S.Thesis, Duke University, Durham, NC, USA, 1969.
19. Das, B. M., A procedure for estimation of ultimate uplift capacity offoundations in clay, Soils and Foundations, Japan, 20(1), 77, 1980.
APPENDIX
CONVERSION FACTORS
CONVERSION FACTORS FROM SITO ENGLISH UNITS
Length: 1 m = 3.281 ft1 cm = 3.281 × 10-2 ft1 mm = 3.281 × 10-3 ft1 m = 39.37 in.1 cm = 0.3937 in.1 mm = 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 = 35.32 ft3
1 cm3 = 35.32 × 10-4 ft3
1 m3 = 61,023.4 in.3
1 cm3 = 0.061023 in.3
Force: 1 N = 0.2248 lb1 kN = 224.8 lb1 kgf = 2.2046 lb1 kN = 0.2248 kip1 kN = 0.1124 U.S. ton1 metric ton = 2204.6 lb1 N / 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
Unit weight: 1 kN / m3 = 6.361 lb / ft3
1 kN / m3 = 0.003682 lb / in.3
Moment: 1 N · m = 0.7375 lb-ft1 N · m = 8.851 lb-in.
CONVERSION FACTORS FROM ENGLISH TO SI UNITS
Length: 1 ft = 0.3048 m1 ft = 30.48 cm1 ft = 304.8 mm1 in. = 0.0254 m1 in. = 2.54 cm1 in. = 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 = 4.448 N1 lb = 4.448 × 10-3 kN1 lb = 0.4536 kgf1 kip = 4.448 kN1 U.S. ton = 8.896 kN1 lb = 0.4536 × 10-3 metric ton1 lb / ft = 14.593 N / m
Stress: 1 lb / ft2 = 47.88 N / m2
1 lb / ft2 = 0.04788 kN / m2
1 U.S. ton / ft1 = 95.76 kN / m2
1 kip/ft2 = 47.88 kN / m2
1 lb/in.2 = 6.895 kN / m2
Unit weight: 1 lb / ft3 = 0.1572 kN/m3
1 lb / in.3 = 271.43 kN/m3
Moment 1 lb-ft = 1.3558 N · m1 lb-in. = 0.11298 N · m
