research article retaining structure force-deformation

Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2013 Article ID 549491 18 pageshttpdxdoiorg1011552013549491

Research ArticleRetaining Structure Force-Deformation Analysis Model foran Ultradeep Foundation Pit

Qi Hu

College of Civil Engineering and Architecture Zhejiang University of Technology Hangzhou 310014 China

Correspondence should be addressed to Qi Hu huqizjueducn

Received 5 April 2013 Revised 17 September 2013 Accepted 3 October 2013

Academic Editor Asier Ibeas

Copyright copy 2013 Qi HuThis is an open access article distributed under theCreative CommonsAttribution License which permitsunrestricted use distribution and reproduction in any medium provided the original work is properly cited

An ultradeep foundation pit is a complex system composed of a retaining structure foundation soil and groundwater Design andconstruction of foundation pits for use at greater depths than in the past require continual improvement in the design methodsand analysis methods applied In this paper a load-deformation analysis model of a retaining structure based on a bearing-modeanalysis of an ultra-deep foundation pit is proposed A calculation method was theoretically derived for the horizontal foundationstiffness coefficient for this model and the influences of factors such as space size stress path load level and seepage were analyzedA numerical example and a case study of an ultra-deep foundation pit in the Hangzhou Metro Line number 1 test section arepresented The calculated results for deformation of the structure and for earth pressure were found to be similar to the resultsobtained from elastic-plastic finite element analysis and similar to the measured results The results of this study indicate that theproposed analysis model adequately reflects the force-deformation characteristics of an ultra-deep foundation pit and show thatthe proposed analysis model appropriately considers the influences of various factors

1 Introduction

Numerous achievements have been reported in the studyof deep foundation pits Codes for the construction offoundation pits have been developed in numerous countries[1 2] and technical specifications have been prepared bylocal governments Valuable experience in the design andconstruction of ultradeep foundation pits has been reportedDue to increasing depths of excavation new supportmethodsand construction technologies have been applied and morestringent safetymeasures have been requiredThese advancesrequire continuous improvement in the design methods andanalysis theories for retaining structures

The principle underlying current codes and technicalspecifications for the design of a foundation pit which is astatic design problem is illustrated in Figure 1Themain stepsin the design process can be summarized as follows First aninitial state is considered in which the external earth pressureis equivalent to the active earth pressure 119875

119886 and this state is

constant Second part of the initial internal earth pressureis offset by the external earth pressure and the change inthe internal earth pressure is represented by the soil spring

force 119865119894= 119870119894sdot 120575119894 The influence of seepage on the water-earth

pressure and the horizontal foundation stiffness coefficient isnot considered This basic design approach has been provento be applicable to shallow foundation pit excavation

An ultradeep foundation pit is a complex system thatis composed of a retaining structure foundation soil andgroundwater The following characteristics of an ultradeepfoundation pit are illustrated in Figure 2 (1) The water-earthpressure on a retaining structure is large and the earthpressure on a flexible retaining structure is closely related tothe deformation of the retaining structure which cannot bedescribed by a single model (2) Because the unloading of thesoil inside the pit is large and the stress path is complex theinfluences of the stress level and stress path on soil parametersmust be considered (3)The deformation mode of a retainingstructure consists of the lateral deformation of the entire pilegroup The magnitude of the lateral support that the soilinside the pit provides to the retaining structure is related notonly to the properties of the soil but also to the space sizethat is the excavation width 119897 and excavation depth ℎ (4) Inareas with high groundwater levels the influence of seepageon the water-earth pressure and soil parameters is significant

2 Mathematical Problems in Engineering

Dh

l

Initial earth pressure inside pit

Active earth pressure

Ground surface

Excavationsurface

120575

Figure 1 Schematic of current design approach for a foundation pit

Seepage

Groundwater

Supporting structure

Soil pressure outside pit

Excavation surface

Soil spring inside pit

Retaining structure

Ground surface

Figure 2 Schematic of an ultradeep foundation pit system

E

Moving Moving backwardforward

E0

Ea

Ep

Figure 3 Relationship between earth pressure and deformation

due to the large difference between the water level inside thepit and the water level outside the pit Under the influence ofseepage the effective soil stress inside the pit is reduced andthe deformation of the retaining structure is increased

Many researchers have studied these problems AsFigure 3 shows the displacement-dependent earth pressuretheory assumes that the earth pressure consists of activeearth pressure and passive earth pressure Xu [3] usedtrigonometric functions to describe the relationship between

earth pressure and deformation Chen et al [4] andZhao et al[5] used exponential functions to describe the relationshipbetween earth pressure and deformation Bei and Zhao [6]analyzed the relationship between active earth pressure andthe deformation of a retaining structure Many researcherssuch as Lade andDuncan [7] Yuan et al [8] Liu andHou [9]Liu [10] and Charles and Qun [11] have conducted soil stresspath experiments on foundation pits The current methodsfor determining horizontal foundation stiffness coefficientvalues can be classified into three categoriesThefirst categoryencompasses empirical methods The value of the horizontalfoundation stiffness coefficient is selected on the basis ofanalyses of soil geological conditions as well as experiencewith similar projects and codes [12 13] These methods aresometimes arbitrary The second category encompasses fieldtest methods including horizontal static load tests pressuremeter tests and flat dilatometer tests [14] The third categoryencompasses laboratory test methods According to certaintheories the relationships between the horizontal foundationstiffness coefficient 119870 and the soil modulus 119864

119904and shear

strength 119862119906can be expressed by the equations 119870 = 120572119864

119904[15]

and 119870 = 120573 sdot 119862119906[16] respectively

Although these three types of methods have yieldeduseful results none of them reflects the force-deformationbehavior of an ultradeep foundation pit or considers

Mathematical Problems in Engineering 3

Static earth pressure

(a) Loading system

Soil spring outside Soil spring inside

(b) Supporting system

Figure 4 Analysis model for an ultradeep foundation pit

the influence of various factors on that behavior This paperproposes an analysis model for an ultradeep foundationpit based on the force-deformation behavior of a retainingstructure and soil A calculation method was theoreticallyderived for the horizontal foundation stiffness coefficient inthis model The influences of the space size stress path loadlevel and seepage on the force-deformation behavior of aretaining structure were examined The proposed methodwas verified using a numerical example and a case study ofan ultradeep foundation pit in the Hangzhou Metro Linenumber 1 test section

2 Analysis Model for an UltradeepFoundation Pit

An analysis model framework for an ultradeep foundationpit which can be decomposed into the processes describedbelow is shown in Figure 4 If deformation of a retainingstructure does not occur after excavation the earth pressureoutside the pitmaintains a static state Unbalanced earth pres-sure is sustained by the support system which is composed ofsoil springs inside and outside the pit Due to the effect of theunbalanced earth pressure between the inside and the outsideof the pit the soil springs inside the pit are compressed theearth pressure inside the pit increases the soil springs outsidethe pit are stretched and the earth pressure outside the pitdecreases until a new balanced state is achieved and formedThree significant differences between this model and existinganalysis models exist (1) The initial loading state consists ofstatic earth pressure without active earth pressure (2) Thechange in earth pressure outside the pit is considered usingthe force of the soil springs outside the pit (3)The soil springstiffness or horizontal foundation stiffness coefficient 119870 isrelated not only to the soil properties but also to the loadingmodes of the retaining structure and the soil

h

oSurface

x

uixΔhi

qi

Figure 5 Illustration of the concept of the horizontal foundationstiffness coefficient

As Figure 5 shows according to the definition of a Win-kler elastic foundation the force on a soil spring is definedby 119902119894sdot 119908 sdot Δℎ and the soil spring stiffness is defined by 119870

119894sdot

119908 sdot Δℎ The relationships among the force the stiffness andthe horizontal displacement of the soil spring are describedby the following equations

(119902119894sdot 119908 sdot Δℎ)

(119870119894sdot 119908 sdot Δℎ)

=119902119894

119870119894

= 119906119894119909

or 119870119894=

119902119894

119906119894119909

(1)

where 119902119894denotes the horizontal strip load (in units of

pressure) 119908 denotes the calculated horizontal widthand Δℎ denotes the calculated thickness

As Figure 6 shows the soil applies lateral pressures 119875119900

and 119875119894to the retaining structure and the retaining structure

applies lateral pressures 1198751015840119900and 119875

1015840

119894to the soil If the relation-

ship between the change in the horizontal strip load Δ119902119894and

the horizontal displacement 119906119894119909is established the value of the

horizontal foundation stiffness coefficient 119870119894= Δ119902119894119906119894119909for

any depth can be determined


Outside pit

Deformation

Inside pit

Po

Pi

(a)

Inside pit

Outside pit

h

uix

Δqi

hi

P998400o

P998400i

Δh

Δh

(b)

Figure 6 Force-deformation behavior of the retaining structure and the soil of an ultradeep foundation pit

The solutions for the components of stress at any pointin a semi-infinite elastic space due to a linear horizontalload 119902 applied at a depth 119889 (as illustrated in Figure 7) wereproposed by Melan [17] The solutions for stress at any pointin a semi-infinite elastic space solution due to a uniformhorizontal strip load 119902 (as illustrated in Figure 8) as wellas the displacement solution and the horizontal foundationstiffness coefficient can be obtained by integrating Melanrsquossolutions These solutions for a uniform horizontal strip loadreflect the actual force-deformation pattern of an ultradeepfoundation pit

As mentioned above the solutions for the componentsof stress at any point in a semi-infinite elastic space due toa linear horizontal load 119902 applied at a depth 119889 (as illustratedin Figure 7) were proposed by Melan [17]

120590119909=

119902119909

2120587 (1 minus 120583)1199092

1199034

1

+1199092

+ 8119889119911 + 61198892

1199034

2

+8119889119911(119889 + 119911)

2

1199036

2

+1 minus 2120583

2[1

1199032

1

+3

1199032

2

minus4119911 (119889 + 119911)

1199034

2

]

120590119911=

119902119909

2120587 (1 minus 120583)(119911 minus 119889)

2

1199034

1

minus1198892

minus 1199112

+ 6119889119911

1199034

2

+81198891199111199092

1199036

2

minus1 minus 2120583

2[1

1199032

1

minus1

1199032

2

minus4119911 (119889 + 119911)

1199034

2

]

120590119909+ 120590119911=

119902119909

2120587 (1 minus 120583)

times 1

1199032

1

+5

1199032

2

minus

4 (1199112

+ 1199092

)

1199034

2

+2 (1 minus 2120583)

1199032

2

(2)

where 119902 denotes the linear load 120583 denotes Poissonrsquos ratio 119889denotes the depth of the linear load 119909 denotes the horizontalcoordinate of a point in the semi-infinite space and 119911 denotesthe vertical coordinate of a point in the semi-infinite space

z

xSurface

o

qz

d

d

r2 = [x2 + (z + d)2]12

r1 = [x2 minus(z+ d)2]12

r = x

Figure 7 Schematic diagram of a linear horizontal load

r1 = [x2 minus(z+ d)2]12

z

xSurface

o

z

r2 = [x2 + (z + d)2]12

r = x

q

d2

d2

d1

d1

Figure 8 Schematic diagram of a horizontal uniform strip load


Retaining structure

Unilateral

Inside pit

Inside pit

Inside pit

Inside pit

RightOutside pit Outside pit

Outside pit

Outside pitMapped areaMapped area

(b3)

(b2)(b1)

(c1) (c2)

(a)

Left

infin infin infin infin

infin

infin

infininfin

infininfin

Pi Pi

2Pi

Po Po

2Po

+ +

Figure 9 Assumption 1 for the analysis model

Surface

P

ub

um

ut

(a) Flexible load

P ua

Surface

(b) Rigid base


The solutions for the components of stress at any pointin a semi-infinite elastic space due to a uniform horizontalstrip load 119902 (as illustrated in Figure 8) can be obtained byintegrating Melanrsquos solutions

120590119911= int

1198892

1198891

119902119909

2120587 (1 minus 120583)

times (119911 minus 119889)

2

1199034

1

minus1198892

minus 1199112

+ 6119889119911

1199034

2

+81198891199111199092

1199036

2


2[1

1199032

1

minus1

1199032

2

minus4119911 (119889 + 119911)

1199034

2

] d119889

=119902

2120587 (1 minus 120583)

times 1

2arctan 119889 minus 119911

119909minus

119909 (119889 minus 119911)

21199032

1

minus1

2arctan 119889 + 119911

119909+

2119909119911119889 (119889 + 119911)

1199034

2

+119909 (119889 + 119911)

21199032

2


2( arctan 119889 minus 119911

119909

minus arctan 119889 + 119911

119909+

2119909119911

1199032

2

)

100381610038161003816100381610038161003816100381610038161003816

119889=1198892

119889=1198891

120590119909+ 120590119911= int

1198892

1198891

119902119909

2120587 (1 minus 120583)

1

1199032

1

+5

1199032

2

minus

4 (1199112

+ 1199092

)

1199034

2

+2 (1 minus 2120583)

1199032

2

d119889

=119902

2120587 (1 minus 120583)arctan 119889 minus 119911

119909+ 5 arctan 119889 + 119911

119909

minus 2 (1199112

+ 1199092

)

times (119889 + 119911

1199091199032

2

+1

1199092arctan 119889 + 119911

119909)

+2 (1 minus 2120583) arctan 119889 + 119911

119909

10038161003816100381610038161003816100381610038161003816

119889=1198892

119889=1198891

(3)


l

h

L L

o

i

z

z

Pileft Piright

PoPo

l2 l2

infin infin

Figure 11 Boundary conditions for a strip foundation pit

40

35

30

25

20

15

10

5

000 02 04 06 08 10 12

120572i (mminus1)

l = 10ml = 20m

l = 40ml = 100m

Foundation pitwidth l

Soil

dept

hz i

(m)

Figure 12 The coefficient 120572119894of the horizontal foundation stiffness

coefficient inside the pit

where 119902 denotes the uniform strip load 1198891denotes the depth

of the top of the uniform strip load and1198892denotes the depth

of the bottom of the uniform strip load

21 Basic Assumptions of the Analysis Model A comparisonof Figures 6 and 8 reveals certain differences between thephysical model of an ultradeep foundation pit and a semi-infinite space Therefore certain assumptions can be made

Assumption 1 The physical model of a strip foundation pitis usually described as a plane-strain elastic problem in asemi-infinite space as shown in Figure 9 The foundation pitis evenly divided into the left side and right side regardlessof the interaction between the left side and the right side(Figure 9(a)) Using a retaining structure as a border thespace inside and outside of the foundation pit is dividedinto two independent regions (Figure 9(b)) The two regions

are asymmetric when subject to a lateral load and canbe expanded into two separate semi-infinite elastic spaces(Figure 9(c))

Assumption 2 Based on the provisions of settlement factorsfor a rigid base and a flexible load [18] if the width ofthe horizontal strip load is sufficiently small the horizontaldeformation of a rigid base is equivalent to the averagedeformation value of a flexible load 119906

119886= (119906119905+ 2119906119898+ 119906119887)4

as shown in Figure 10

Assumption 3 For the purpose of calculating stresses thesoil is considered to be a single-phase homogeneous andisotropic material with a constant modulus

22 Horizontal Foundation Stiffness Coefficient for the Anal-ysis Model According to the definition of the horizontalfoundation stiffness coefficient 119870 = 119902119906

119909 the horizontal

deformation of the isotropic plane-strain problem can becalculated from the following equations

119906119909= int 120576119909d119909

120576119909=

1 minus 1205832

119864119904

(120590119909minus

120583

1 minus 120583120590119911) or

119906119909=

1 minus 1205832

119864119904

(int120590119909d119909 minus

120583

1 minus 120583int120590119911d119909)

(4)

Using the integrals of the stress solutions in (3) the dis-placement in semi-infinite space due to a uniform horizontalstrip load 119902 can be determined as follows

int120590119911d119909 =

119902

2120587 (1 minus 120583)

times minus119911119889 (119889 + 119911)

1199032

2

+ 120583119909 arctan 119889 minus 119911

119909

minus 120583119909 arctan 119889 + 119911

119909+

(1 minus 2120583) (119911 minus 119889)

4


times ln(1199031

119889 minus 119911)

2

+(1 minus 2120583) (119889 minus 119911)

4

times ln(1199032

119889 + 119911)

2

100381610038161003816100381610038161003816100381610038161003816

119889=1198892

119889=1198891

int 120590119909d119909 =

119902

2120587 (1 minus 120583)

times 119911119889 (119889 + 119911)

1199032

2

+ (1 minus 120583) 119909 arctan 119889 minus 119911

119909

+ (5119909 minus 3120583119909 +21199112

119909) arctan 119889 + 119911

119909

+(3 minus 2120583) (119889 minus 119911)

4ln(

1199031

119889 minus 119911)

2

+119889 (5 minus 6120583) + 119911 (7 minus 10120583)

4

times ln(1199032

119889 + 119911)

2

100381610038161003816100381610038161003816100381610038161003816

119889=1198892

119889=1198891

(5)

The average displacement within the loading area isexpressed as follows

119906119909=

(119906119909119911=119889

1

+ 2119906119909119911=(119889

1+1198892)2

+ 119906119909119911=119889

2

)

4 (6)

According to Assumption 1 the real load in a quarter-infinite space is 119902 = 1199022 due to the operation of asymmetricmapping When (4) and (5) are used to solve the horizontaldeformation equation the integral range shown in Figure 11must be determined In a symmetric excavation the horizon-tal integral range is finite inside the pit

119906119909= int

1198972

0

120576119909119889119909 minus int

119897

1198972

120576119909119889119909 (7)

The horizontal foundation coefficient 119870119894of the soil

springs inside the pit can be obtained from the averagedisplacement within the loading area

119870119894=

119902

119906119909

=119864119904

1 minus 1205832

119902

2[(int

1198972

0

120590119909d119909 minus int

119897

1198972

120590119909d119909) minus

120583

1 minus 120583(int

1198972

0

120590119911d119909 minus int

119897

1198972

120590119911d119909)]

minus1

(8)

Determination of the theoretical influence zone whichis infinite outside the pit is similar to the problem ofdetermining the thickness of the underlying layer below astrip foundation Based on the provisions of compressiondepth in calculating the foundation settlement [18] whenthe additional stress decreases to 10 of the gravity stressthe depth is defined as the compression depth The influencezone of lateral soil deformation is assumed to satisfy thecalculation when the lateral stress decreases to 10 of thehorizontal load According to the results obtained using thestress solutions in (3) when the additional horizontal stressis 10 of the horizontal load 119902 the corresponding distance isapproximately ten times the loading width Thus ten times

the foundation pit depth was defined as the influence zoneoutside the pit

119906119909= int

10ℎ

0

120576119909119889119909 (9)


springs outside the pit can be obtained from the averagedisplacement within the loading area

119870119900=

119902

119906119909

=119864119904

1 minus 1205832

119902

2(int

10ℎ

0

120590119909d119909 minus

120583

1 minus 120583int

10ℎ

0

120590119911d119909)minus1

(10)

We define

120572119894=

1

1 minus 1205832

119902

2[(int

1198972

0

120590119909d119909 minus int

119897

1198972

120590119909d119909) minus

120583

1 minus 120583(int

1198972

0

120590119911d119909 minus int

119897

1198972

120590119911d119909)]

minus1

120572119900=

1

1 minus 1205832

119902

2(int

10ℎ

0

120590119909d119909 minus

120583

1 minus 120583int

10ℎ

0

120590119911d119909)minus1

(11)


01 02 03 04

40

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)

L = 100mL = 200m

Influence scope L

00120572o (mminus1)


coefficient outside the pit

Then

119870119894= 120572119894119864119904

119870119900= 120572119900119864119904

(12)

where 120572 denotes the coefficient of the horizontal foundationstiffness coefficient 119870 which is related to the foundationpit space size and Poissonrsquos ratio (mminus1) and 119864

119904denotes the

elastic modulus of the soil

3 Parametric Analysis

As shown in (8) and (10) the horizontal foundation stiffnesscoefficient can be described as 119870 = 120572119864

119904 The factors that

influence 119870 include the space size of the foundation pit (afunction of the pit width the pit depth and the influence zoneoutside the pit) and the soil parameters (elastic modulus andPoissonrsquos ratio)

31 Influence of Space Size of the Foundation Pit Thecoefficient 120572

119894of the horizontal foundation stiffness coeffi-

cient for various foundation pit widths and depths is shownin Figures 12 and 13 for a Poissonrsquos ratio of soil of 03

As Figures 12 and 13 show the value of the coefficient 120572decreases when the foundation pit width 119897 or the influencezone 119871 increases Due to the low level of restraint on thesurface soil the value of the coefficient 120572 is also small Whenthe soil depth 119911

119894is half of the foundation pit width 119897 the

coefficient 120572 approaches a constant value

32 Influence of Poissonrsquos Ratio As shown in Figure 14 thecoefficient 120572

119894decreases when Poissonrsquos ratio 120583 increases

When the foundation pit is 20 meters wide 120572119894120583=01

asymp

11120572119894120583=03

and 120572119894120583=05

asymp 09120572119894120583=03

Poissonrsquos ratio 120583

120583 = 01120583 = 02

120583 = 03

120583 = 04120583 = 05

40

35

30

25

20

15

10

5

000 02 04 06 08 10 12

120572i (mminus1)

Soil

dept

hz i

(m)

Figure 14 Influence of Poissonrsquos ratio on the coefficient 120572119894(founda-

tion pit width 119897 = 20m)

0 1000 2000 3000 4000 5000

40

35

30

25

20

15

10

5

0

l = 10ml = 20m

l = 40ml = 100m


Soil

dept

hz i

(m)

mi (kNmiddotmminus4)

Figure 15 Proportional coefficient 119898119894of sandy silt inside the pit

33 Influence of Stress Path The relationship between the soilmodulus the stress path and the consolidation pressure is[19]

119864119904= 120582 sdot 120590

1015840

= 120582 sdot 1205741015840

sdot 119911 (13)

where the coefficient 120582 denotes the influence of the stresspath 1205741015840 denotes the soil effective gravity and 119911 denotes thesoil depth

For the soil outside the pit the lateral modulus coefficient120582 means the lateral unloading stress path For the soil inside


Table 1 Stress path coefficient 120582 of initial tangent modulus (Wei2006 [20] Hu 2008 [19])

Soil layer Stress path Initial tangentmodulus 119864

119894

Stress pathcoefficient 120582

2-3 Verticalcompression

Vertical tangentmodulus 158

2-3 Lateralunloading Lateral tangent

modulus

83

2-3 Verticalunloading 469


the pit the lateral modulus coefficient 120582 means the verticalunloading stress path

Taking the influence of the stress path into considerationthe horizontal foundation coefficient can be expressed asfollows

119870 = 120572 sdot 119864119904= 120572 sdot 120582 sdot 120574

1015840

sdot 119911 = 119898 sdot 119911 (14)

The proportional coefficient 119898 of the horizontal foun-dation stiffness coefficient can be expressed as 120572 sdot 120582 sdot 120574

1015840The influence of the stress path on the coefficient 120582 of theinitial tangent modulus 119864

119894 according to stress path tests on

Hangzhou sandy silt is indicated in the results shown inTable 1

The average effective gravity 1205741015840 of Hangzhou sandy silt

is 90 kNm3 [19] According to the test results shown inTable 1 the lateral unloading stress path coefficient 120582 isapproximately 80 and the vertical unloading stress pathcoefficient 120582 is approximately 470 The results for the pro-portional coefficient 119898 of sandy silt inside and outside thefoundation pit when Poissonrsquos ratio 120583 is 03 are shown inFigures 15 and 16 and Table 2

As the values in Table 2 show the variation in the propor-tional coefficient 119898 in the homogeneous foundation is simi-lar to that of coefficient 120572 The greater the width of the foun-dation pit is the smaller the proportional coefficient 119898 isThe value of the proportional coefficient 119898 is smallest at theground surface When the soil depth is half of the foundationpit width the proportional coefficient 119898 approaches a con-stant value In general the proportional coefficient 119898 of thesoil inside the pit is considerably larger than the proportionalcoefficient 119898 of the soil outside the pit

34 Influence of Load Level The stress-strain behavior of soilis nonlinear As the load level increases the rate of strainthe soil modulus and the horizontal foundation stiffnesscoefficient decrease Thus the effect of load level shouldbe considered The secant modulus for the Duncan andChang [21] hyperbolic equations shown in Figure 17 can beexpressed as follows

119864119902= 119864119894sdot (1 minus

1205901minus 1205903

(1205901minus 1205903)ult

) = 119864119894sdot (1 minus (120590

1minus 1205903) sdot 119887) (15)

where 119887 = (1205901minus 1205903)minus1

ult

0 50 100 150 200mo (kNmiddotmminus4)

40

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)

L = 100mL = 200m

Influence scope L

Figure 16 Proportional coefficient 119898119900of sandy silt outside the pit

Asymptote (1205901 minus 1205903)ult = 1b

1205901 minus 1205903 =120576

120576

a + b120576

EqEi = 1a

(1205901 minus 1205903)

Figure 17 Nonlinear stress-strain relationship according to theDuncan-Chang model

Taking the influence of load level into consideration thehorizontal foundation stiffness coefficient can be expressed asfollows

119870 =119902

119906119909

=119902

(int 120576119909d119909)

120576119909=

1 minus 1205832

119864119902

(120590119909minus

120583

1 minus 120583120590119911)

=1 minus 1205832

119864119894sdot [1 minus (120590

119909minus 120590119911) sdot 119887]

(120590119909minus

120583

1 minus 120583120590119911)

(16)

As in the linear elastic model the initial tangent modulusin (16) is unrelated to the stress levels 120590

119909and 120590119911 and the effect

of load level is only related to the coefficient 120572 Equation (16)is relatively complicated and must be solved by numericalintegrationThe influence of the load level 119902 and the strengthparameter 119887 on the coefficient 120572 when the foundation pit


Table 2 Proportional coefficient119898 of Hangzhou sandy silt

Location Soil inside pit Soil outside pitFoundation pit width Lm 10 20 40 100 Horizontal influence scope Lm 100 200

Ground surface119898119894kNmminus4 1689 1248 990 778

119898119900kNmminus4 100 89

Deep inside 4230 3170 2598 1879 186 150

0 200 400 600 800 10000

1

2

3

4

5

Test resultsFitting curve

b(M

Paminus1) b = 46 minus 00043120590998400z

Vertical consolidation pressure 120590998400z (kPa)

Figure 18 Relationship between 119887 and vertical consolidation pres-sure [19]

width is 20m is shown in Figures 19 and 20 When theload level 119902 increases the secant modulus of the soil andthe coefficient 120572 decrease When the strength parameter119887 increases the soil secant modulus and the coefficient 120572

decreaseAccording to the stress path test results for Hangzhou

sandy silt the strength parameter 119887 in the vertical unloadingstress-strain curve which is illustrated in Figure 18 can bedetermined from the following equation

119887 = 46 minus 000431205901015840

119911 (17)

where 119887 has units of MPaminus1 and 1205901015840

119911denotes the vertical

consolidation pressure (kPa)

35 Influence of Seepage As shown in (13) and (14) thehorizontal foundation stiffness coefficient and themodulus ofsoil are related to the soil stress state At a site with abundantgroundwater if seepage occurs in the foundation pit the soilstress statewill changeThewater-soil pressure andhorizontalfoundation stiffness coefficient will also be affected

Taking the influence of seepage into consideration thehorizontal foundation stiffness coefficient of the soil insidethe pit can be expressed as follows

119870119894= 120572119894sdot 119864119904= 120572119894sdot 120582 sdot 120590

1015840

119911= 120572119894sdot 120582 sdot (120574

1015840

119904minus 120574119908sdot 119894119894) sdot 119911119894 (18)

where 119894119894denotes the average hydraulic gradient inside the pit

02 03 04 05 06 07 08

q = 0kPaq = 2kPaq = 20kPaq = 50kPa

q = 100 kPaq = 150kPaq = 200 kPa

Load level q40

35

30

25

20

15

10

5

0

120572i (mminus1)

Soil

dept

hz i

(m)

Figure 19 Influence of the load level 119902 on the coefficient 120572119894(119887 =

3MPaminus1)

01 02 03 04 05 06 07 08

b = 0

b = 2

b = 4b = 5

b = 6

b = 8

b = 10

b (MPaminus1)40

35

30

25

20

15

10

5

0

120572i (mminus1)

Soil

dept

hz i

(m)

Figure 20 Influence of the strength parameter 119887 on thecoefficient 120572

119894(119902 = 100 kPa)


Supporting structureRetaining structure

Influence scope outsideFoundation pitInfluence scope outside

15 m

5 m

l = 10m and 40m L = 100mL = 100m

empty6095000 steel pipesempty10001500 bored piles

empty1m

(a) Plane graph

Supporting structureGroundwater Groundwater

Ground surface Ground surface

Excavationsurface

l = 10m and 40m L = 100mL = 100m

empty6095000 steel pipes

Retaining structureempty10001500 bored piles

h=20

m20

m

4 m

(b) Sectional graph

Figure 21 Foundation pit layout for the analysis example

Taking the influence of seepage into consideration thehorizontal foundation stiffness coefficient of the soil outsidethe pit can be expressed as follows

119870119900= 120572119900sdot 119864119904= 120572119900sdot 120582 sdot 120590

1015840

119911= 120572119900sdot 120582 sdot (120574

1015840

119904+ 120574119908sdot 119894119900) sdot 119911119900

(19)

where 119894119900denotes the average hydraulic gradient outside the

pitTaking the influence of seepage into consideration the

static earth pressure outside the pit thewater pressure outsidethe pit and the lateral pressure outside the pit can beexpressed as follows

119875119900119904

= (1205741015840

119904+ 120574119908119894119900) sdot 1198700sdot 119911119900

earth pressure

119875119900119908

= 120574119908(1 minus 119894119900) sdot 119911119900

water pressure

119875119900= (1205741015840

119904+ 120574119908119894119900) sdot 1198700sdot 119911119900+ 120574119908(1 minus 119894119900) sdot 119911119900

lateral pressure(20)

where1198700denotes the static earth pressure coefficient

Taking the influence of seepage into consideration thestatic earth pressure inside the pit the water pressure inside

Soil spring inside

pit

Active pressure

Figure 22 Code analysis model

the pit and the lateral pressure inside the pit can be expressedas follows

119875119894119904

= (1205741015840

119904minus 120574119908119894119894) sdot 1198700sdot 119911119894

earth pressure

119875119894119908

= 120574119908(1 + 119894119894) sdot 119911119894

water pressure

119875119894= (1205741015840

119904minus 120574119908119894119894) sdot 1198700sdot 119911119894+ 120574119908(1 + 119894119894) sdot 119911119894



Static pressure Soil spring inside

Soil spring outside

Figure 23 Analysis model proposed in this study

Lateral unloading stress path

Vertical unloading stress path

Horizontal supporting

Groundwater outside

Closed seepage boundaryGroundwater

inside

5lowast4=20

m20

m

5sim20m 100 m

10 m

Figure 24 Continuum elastic-plastic-medium finite element model

4 Analysis Example

The parameters for the analysis example illustrated inFigure 21 are as follows an excavation depth of 20m fiveexcavation steps horizontal supporting structures consistingof 5 layers of Oslash609 times 165000 steel pipes a stiffnessof 119864119860119863119897 = 1400119897MNm2 retaining structures consistingof Oslash10001500 bored piles with embedded depths of 20ma concrete modulus 119864

119888= 30GPa excavation widths of 10m

and 40m a horizontal zone of influence of 100m outside thepit a water surface elevation equal to the ground surface ele-vation and a water pressure unrelated to the earth pressure

The foundation soil consists of sandy silt with the fol-lowing characteristics saturated gravity 120574

119904= 190 kNm3

effective gravity 1205741015840

= 90 kNm3 Poissonrsquos ratio 120583 = 03shear strength parameters 119888

1015840

= 5 kPa and 1205931015840

= 30∘ lateral

unloading stress path coefficient 120582 of the initial tangentmod-ulus of 80 and vertical unloading stress path coefficient 120582 ofthe initial tangent modulus of 470 The calculation methodsand models are shown in Figures 22 23 and 24 and theparameters are listed in Table 3

41 Analysis Results for the Retaining Structures Figures 25to 28 illustrate the following points (1) The influence of the

foundation pit space size and seepage cannot be consideredand the earth pressure outside the pit and the proportionalcoefficient 119898 of the horizontal foundation stiffness coeffi-cient are constant Thus the horizontal displacement andthe bending moment determined by the code method aresmaller than the horizontal displacement and the bendingmoment determined using the other two methods Thedifference increases when the excavation width increases (2)The method proposed in this study considers the influencesof space size stress path load level and seepage The resultsobtained using this method are similar to those obtained forthe continuum elastic-plastic-medium finite element model

Equations (18) to (21) illustrate the following pointsFirst due to the influence of seepage the water pressureoutside the pit 119875

119900119908decreased which caused a decrease in

the horizontal displacement and bending moment of theretaining structures Second due to the influence of seepagethe horizontal foundation stiffness coefficient inside thepit 119870119894decreased which increased the horizontal displace-

ment and bending moment of the retaining structures AsFigures 25 and 26 show when the width of the foundationpit is small the hydraulic gradient 119894

119894inside the pit is large

The horizontal foundation stiffness coefficient 119870119894inside the

pit decreases rapidly under the influence of seepage and


Table 3 Parameters according to different methods

Methods Code method [1] Method proposed in this study Continuum medium finiteelement method

Parameters

Horizontal foundationstiffness coefficient Earth pressure

Horizontalfoundationcoefficient

Earth pressure Soil modulus Model

Proportional coefficientof horizontal foundationstiffness coefficient119898 =

4000

Active earthpressurecoefficient119870119886= 033

Considering theinfluence of spacesize stress pathload level and

seepage

Static earthpressure coefficientconsidering theinfluence of

seepage 119870119900= 05

Initial tangentmodulus 119864

119894

considering theinfluence of stress

path

Mohr-Coulombelastic-plastic

model

Table 4 Physical and mechanical parameters of the soil

Layer number Soil name 120574119904kNmminus3 Void ratio 119890

Shear strength parameters Permeability coefficientlowast10minus4 m sminus1

119888kPa 120593∘ 119870119881

119870119867

2-1 Sandy silt 189 0853 76 285 339 2412-3 Sandy silt 192 0788 56 312 239 2042-4 Sandy silt 190 0858 61 308 242 1822-5 Sandy silt with sand 193 0772 47 315 265 3002-6 Sandy silt 187 0916 79 293 0132-7 Sandy silt with sand 193 0775 55 312 120 2985 Muddy silty clay 183 1067 193 117 (10minus5 to 10minus6 cms)6-2 Silty clay 192 0866 506 1576-3 Silty clay with silt 200 069 349 1968-1 Sand 187 0829 16 329

the horizontal displacement and bending moment of theretaining structures increase significantly As Figures 27 and28 show when the width of the foundation pit is largethe hydraulic gradient 119894

119894inside the pit is similar to the

hydraulic gradient 119894119900outside the pit and the horizontal foun-

dation stiffness coefficient 119870119894inside the pit and the water

pressure 119875119900119908

outside the pit decrease similarly under theinfluence of seepage Thus the horizontal displacement andthe bending moment of the retaining structures vary slightly

42 Analysis Results for Earth Pressure In the modeldescribed in this paper the initial state of the load consistsof static earth pressure and the earth pressure outside the pitchanges when the soil springs are tensed Figures 29 30 31and 32 highlight the results obtained for earth pressure usingthe method proposed in this paper which are similar to theresults obtained for the continuum elastic-plastic-mediumfinite element model A comparison of the results indicatesthat the method proposed in this paper accurately simulatesthe distribution patterns and the changes in earth pressureboth inside and outside the pit

5 Case Study

51 Overview As Figure 33 shows the parameters of theultradeep foundation pit in the Qiutao Road station ofthe Hangzhou Metro Line number 1 test section are asfollows a strip foundation pit a 20m excavation width

a 168m excavation depth a zone of influence of 200moutside the pit 6 excavation steps horizontal supportingstructures consisting of 5 layers of Oslash609 times 164000 steelpipes with stiffnesses of 119864119860119863119897 = 1750119897MNm2 retainingstructures consisting of 30m-long Oslash10001500 bored pilesthe concrete modulus 119864

119888= 30GPa and the water surface

2m below the ground surface The water inside the pit waspumped and the water outside the pit was not pumpedAs shown in Table 4 and Figure 34 the main soil layers arecomposed of permeable sandy silt The soil layer 24m belowthe ground surface is composed of impermeable muddysilty clay Seepage cannot occur in muddy silty clay so theinfluence of seepage can be disregarded

The average effective gravity of sandy silt is 1205741015840

=

91 kNm3 According to the measured results of the stresspath tests shown in Table 1 and (14) the initial tangentmodulus 119864

119894119894for the lateral unloading soil outside the pit

is 073119911119894MPa and the initial tangent modulus 119864

119894119900for the

lateral unloading soil outside the pit is 428119911119900MPa The

values of the proportional coefficient 119898 for the horizontalfoundation stiffness coefficients disregarding the influence ofload level are shown in Figure 35

52 Application of the Calculation Method and Model Thecalculation method and model which are equivalent tothe calculation methods and models used in the exam-ple analysis are shown in Table 3 and Figures 21 through24 The parameters are also similar to the parameters in


140 120 100 80 60 40 20 0

40

35

30

25

20

15

10

5

0

No seepage

Considering seepage

Code methodMethod in this paperContinuum elastoplastic medium FEM

Horizontal displacement (mm)

Method in this paperContinuum elastoplastic medium FEM

Dep

th (m

)

Figure 25 Retaining structure deformation for 10m wide pit

0 800 1600minus3200 minus2400 minus1600 minus800

40

35

30

25

20

15

10

5

0

Dep

th (m

)

Bending moment (kNmiddotmmiddotmminus1)

No seepage

Considering seepage



Figure 26 Retaining structure bending moment for 10m wide pit

the analysis example with the exception that the proportionalcoefficient 119898 in the code method is 3000 kNm4

53 Analysis Results As Figures 36 and 37 show the resultsfor the retaining structure deformation and earth pressureoutside the pit are as follows

140 120 100 80 60 40 20 0

40

35

30

25

20

15

10

5

0

No seepage

Considering seepage




Dep

th (m

)



40

35

30

25

20

15

10

5

0

Dep

th (m

)


No seepage

Considering seepage




(1) The results obtained using the method proposed inthis paper and using the continuum elastic-plastic-medium finite element method are similar to themeasured results which indicates that the methodand the model proposed in this paper can be usedto accurately calculate the forces and deformations of


0 30 60 90 120 150 180 210

40

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)

Earth pressure outside po (kPa)

Active pressureCalculated by method of this paper

Calculated by continuum elastoplastic medium FEMStatic pressure

Figure 29 External earth pressure for 10m wide pit

0 30 60 90 120 150 180 210

40

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)





the retaining structures of an ultradeep foundationpit

(2) The influences of the stress path the size of thefoundation pit and the stress level on the horizontalfoundation stiffness coefficient are not consideredin the code method The maximum horizontal dis-placements calculated using the code method rangedfrom 32mm to 35mm The maximum horizontal

350 300 250 200 150 100 50 0

40

35

30

25

20

15

10

5

0

Soil

dept

hz i

(m)

Earth pressure outside pi (kPa)

Calculated by method of this paperCalculated by continuum elastoplastic medium FEM

Figure 31 Earth pressure inside 10m wide pit

350 300 250 200 150 100 50 0

40

35

30

25

20

15

10

5

0

Soil

dept

hz i

(m)




displacement calculated using the method proposedin this paper was 48mm which was similar to themaximum horizontal displacement of 49mm cal-culated using the continuum elastic-plastic-mediumfinite element method The measured values rangedfrom 46mm to 51mm

(3) The results obtained for earth pressure using themethod proposed in this paper were similar to


Figure 33 The ultradeep foundation pit in the Qiutao Road station of Hangzhou Metro Line number 1

Groundwater

36800sand8 -1

6 -3 silty clay with silt3080029300

silty clay6 -2

5 muddy silty clay25900

23400sandy silt with sand2 -7

2 -6 sandy silt 21500

1


2 -4 sandy silt11000

9000sandy silt2 -3

2 -1 sandy silt5000

2700urban mixed fill

Retaining structures

Excavation surface

Ground surface

16800

Figure 34 Soil profile

the results obtained with the continuum elastic-plastic-medium finite element method and were sim-ilar to the measured results

6 Conclusions

New load-deformation model and method for analysis ofretaining structures in ultradeep foundation pits are pro-posed in this paper The horizontal foundation stiffnesscoefficient for this model can be expressed as 119870 = 120572119864

119904

The coefficient 120572 is related to the size of the foundation pitPoissonrsquos ratio the stress path and the stress level The soilmodulus 119864

119904is also related to the stress path and the stress

level

(1) The value of the coefficient 120572 decreases as the foun-dation pit width or zone of influence increasesThe restraint applied to the surface of the soilis the smallest restraint thus coefficient 120572 has thesmallest value at the surface When the soil depthis half of the depth of the foundation pit widththe coefficient 120572 approaches a constant value Thevalue of the coefficient 120572 decreases as Poissonrsquos ratioincreases

(2) The proportional coefficient 119898 of the horizontalfoundation stiffness coefficient reflects the effect ofthe stress path which can be expressed as 120572 sdot 120582 sdot 120574

1015840The proportional coefficient 119898 at the ground surface


35

30

25

20

15

10

5

0 0 500 1000 1500 2000 2500 3000 3500

Soil insideSoil outside

m (kNmiddotmminus4)

Soil

dept

hz

(m)

Excavated depth h = 2mExcavated depth h = 5mExcavated depth h = 8mExcavated depth h = 11mExcavated depth h = 14mExcavated depth h = 17m

Figure 35 Horizontal foundation stiffness coefficients for each ofsix excavation steps

0 10 20 30 40 50 60

Measured result (1)Measured result (2)Code method


35

30

25

20

15

10

5

0


Dep

th (m

)

Figure 36 Deformation in retaining structures

exhibits the smallest influence When the soil depthis half of the foundation pit width the proportionalcoefficient 119898 approaches a constant value In generalthe value of 119898 of the soil inside the pit is significantlylarger than the value of 119898 of the soil outside the pit

0 30 60 90 120 150

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)

Static pressureActive pressureMethod in this paper

Continuum elastoplasticmedium FEMMeasured result


Figure 37 Earth pressure outside the pit

(3) When the load level 119902 increases the secant mod-ulus of the soil and the value of the coefficient 120572

decrease When the strength parameter 119887 increasesthe soil secant modulus and the value of thecoefficient 120572 decrease

(4) Taking the influence of seepage into considerationthe horizontal foundation stiffness coefficient 119870

119894of

the soil inside the pit can be expressed as 120572 sdot 120582 sdot

(1205741015840

119904minus 120574119908

sdot 119894119894) sdot 119911119894 and the horizontal foundation

stiffness coefficient 119870119900of the soil outside the pit can

be expressed as 120572 sdot 120582 sdot (1205741015840

119904+ 120574119908sdot 119894119900) sdot 119911119900 Seepage will

cause the value of the horizontal foundation stiffnesscoefficient 119870

119894of the soil inside the pit to decrease

and the coefficient 119870119900of the soil outside the pit to

increase

(5) The results obtained for the example analysis andcase study presented indicate that the model andmethod proposed in this paper yield results similar tomeasured results and similar to results obtained usinga continuum elastic-plastic-medium finite elementmodel The good agreement among the three types ofresults indicates that themethod andmodel proposedin this paper are capable of accurately calculating theforces and deformations of retaining structures in anultradeep foundation pit

Acknowledgment

The author would like to acknowledge the financial supportfrom the National Natural Science Foundation of China(NSFC Grant no 51108417)


References

[1] P R Ministry of Metallurgical Industry China ldquoYB9258-97Code for technique of building foundation pit engineeringrdquoBeijing China Metallurgical Industry Press 1997

[2] P R Ministry of Construction China ldquoJGJ120-99 Technicalspecification for retaining and protection of building founda-tion excavationsrdquo BeijingChina China Building Industry Press1999

[3] R-Q Xu ldquoMethods of earth pressure calculation for excava-tionrdquo Journal of Zhejiang University vol 34 no 4 pp 370ndash3752000

[4] Y-K Chen R-Q Xu X-J Yang and X-N Gong ldquoA newmethod calculating earth pressure on flexible structures forexcavation workrdquo Industrial Construction vol 31 no 3 pp 1ndash4 2001

[5] J-P Zhao M Guo-xiong and J-M Zai ldquoEarth pressure modelconsidering settlement and time effectrdquo Journal of YanchengInstitute of Technology vol 4 pp 55ndash63 2003

[6] L Bei and X-H Zhao ldquoA nonlinear earth pressure methodfor deep excavation considering deformation of retaining wallrdquoRock and Soil Mechanics vol 25 no 2 pp 453ndash458 2004

[7] P V Lade and JMDuncan ldquoStress-path dependent behavior ofcohesionless soilrdquoASCE Journal of the Geotechnical EngineeringDivision vol 102 no 1 pp 51ndash68 1976

[8] J Yuan X-W Liu andD-Q Yi ldquoDetermination ofm coefficientduring excavationrdquo Industrial Construction vol 30 no 9 pp46ndash51 2000

[9] G B Liu and X-Y Hou ldquoUnloading modulus of the shanghaisoft clayrdquoChinese Jounal of Geotechnical Engineering vol 18 no6 pp 18ndash23 1996

[10] X-Y Liu ldquoExperimental and numerical simulation of excava-tion process and microstructure studyrdquo Tianjin China TianjinUniversiy 2003

[11] N W W Charles and S Qun ldquoChanges of stress path causedby stress relief during excavationsrdquo China Civil EngineeringJournal vol 32 no 6 pp 53ndash58 1999

[12] Construction department of zhejiang province ldquoDB33T1008-2000 Code for technique of building foundation excavationengineeringrdquo Zhejiang China 2000

[13] Shanghai Uran Construction Committee ldquoDBJ-61-97 Code fordesign of excavation engineeringrdquo Shanghai China 1997

[14] G-M Chen J-D Zhong and Z-Q Tang ldquoThe discussionon calculation of soil lateral subgrade coefficient with flatdialatometer testrdquo Shanghai Geology vol 2 pp 40ndash42 2002

[15] ZHU Bi-tang ldquoLimiting force profile and response of laterallyloaded pilesrdquo Shanghai China Tongji University 2005

[16] M T Davission and H L Gill ldquoLaterally loaded piles in alayered soil systemrdquo ASCE Journal of the Soil Mechanics andFoundations Engineering vol 89 no 3 pp 63ndash94 1963

[17] HG Poulous and EHDavisElastic Solutions for Soil and RockMechanics John Wiley amp Sons New York NY USA 1974

[18] China Architecture amp Building Press ldquoFoundation and Basisrdquo1991

[19] Q Hu ldquoStudy on design method and water-soil-retainingstructure interaction of ultra-deep foundation pitrdquo HangzhouChina Zhejiang University 2008

[20] J-D Wei ldquo Sudy on the earth pressure of silty soil and thebehaviour of retaining structures for pit excavationrdquo HangzhouChina Zhejiang University 2006

[21] J M Duncan and C Y Chang ldquoNonlinear analysis of stressand strain in soilsrdquo ASCE Journal of the Soil Mechanics andFoundations Division vol 96 no 5 pp 1629ndash1653 1970

Submit your manuscripts athttpwwwhindawicom

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Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

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Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Algebra

Discrete Dynamics in Nature and Society



Decision SciencesAdvances in

Discrete MathematicsJournal of


Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of


Dh

l

Initial earth pressure inside pit

Active earth pressure

Ground surface

Excavationsurface

120575

Figure 1 Schematic of current design approach for a foundation pit

Seepage

Groundwater

Supporting structure

Soil pressure outside pit

Excavation surface

Soil spring inside pit

Retaining structure

Ground surface

Figure 2 Schematic of an ultradeep foundation pit system

E

Moving Moving backwardforward

E0

Ea

Ep

Figure 3 Relationship between earth pressure and deformation

due to the large difference between the water level inside thepit and the water level outside the pit Under the influence ofseepage the effective soil stress inside the pit is reduced andthe deformation of the retaining structure is increased

Many researchers have studied these problems AsFigure 3 shows the displacement-dependent earth pressuretheory assumes that the earth pressure consists of activeearth pressure and passive earth pressure Xu [3] usedtrigonometric functions to describe the relationship between

earth pressure and deformation Chen et al [4] andZhao et al[5] used exponential functions to describe the relationshipbetween earth pressure and deformation Bei and Zhao [6]analyzed the relationship between active earth pressure andthe deformation of a retaining structure Many researcherssuch as Lade andDuncan [7] Yuan et al [8] Liu andHou [9]Liu [10] and Charles and Qun [11] have conducted soil stresspath experiments on foundation pits The current methodsfor determining horizontal foundation stiffness coefficientvalues can be classified into three categoriesThefirst categoryencompasses empirical methods The value of the horizontalfoundation stiffness coefficient is selected on the basis ofanalyses of soil geological conditions as well as experiencewith similar projects and codes [12 13] These methods aresometimes arbitrary The second category encompasses fieldtest methods including horizontal static load tests pressuremeter tests and flat dilatometer tests [14] The third categoryencompasses laboratory test methods According to certaintheories the relationships between the horizontal foundationstiffness coefficient 119870 and the soil modulus 119864

119904and shear

strength 119862119906can be expressed by the equations 119870 = 120572119864

119904[15]

and 119870 = 120573 sdot 119862119906[16] respectively

Although these three types of methods have yieldeduseful results none of them reflects the force-deformationbehavior of an ultradeep foundation pit or considers



(a) Loading system







h

oSurface

x

uixΔhi

qi



119894sdot


(119902119894sdot 119908 sdot Δℎ)

(119870119894sdot 119908 sdot Δℎ)

=119902119894

119870119894

= 119906119894119909

or 119870119894=

119902119894

119906119894119909

(1)






1015840







Outside pit

Deformation

Inside pit

Po

Pi

(a)

Inside pit

Outside pit

h

uix

Δqi

hi

P998400o

P998400i

Δh

Δh

(b)




120590119909=

119902119909

2120587 (1 minus 120583)1199092

1199034

1

+1199092

+ 8119889119911 + 61198892

1199034

2

+8119889119911(119889 + 119911)

2

1199036

2

+1 minus 2120583

2[1

1199032

1

+3

1199032

2

minus4119911 (119889 + 119911)

1199034

2

]

120590119911=

119902119909

2120587 (1 minus 120583)(119911 minus 119889)

2

1199034

1

minus1198892

minus 1199112

+ 6119889119911

1199034

2

+81198891199111199092

1199036

2


2[1

1199032

1

minus1

1199032

2

minus4119911 (119889 + 119911)

1199034

2

]

120590119909+ 120590119911=

119902119909

2120587 (1 minus 120583)

times 1

1199032

1

+5

1199032

2

minus

4 (1199112

+ 1199092

)

1199034

2

+2 (1 minus 2120583)

1199032

2

(2)


z

xSurface

o

qz

d

d

r2 = [x2 + (z + d)2]12

r1 = [x2 minus(z+ d)2]12

r = x


r1 = [x2 minus(z+ d)2]12

z

xSurface

o

z

r2 = [x2 + (z + d)2]12

r = x

q

d2

d2

d1

d1



Retaining structure

Unilateral

Inside pit

Inside pit

Inside pit

Inside pit


Outside pit


(b3)

(b2)(b1)

(c1) (c2)

(a)

Left


infin

infin

infininfin

infininfin

Pi Pi

2Pi

Po Po

2Po

+ +


Surface

P

ub

um

ut

(a) Flexible load

P ua

Surface

(b) Rigid base



120590119911= int

1198892

1198891

119902119909

2120587 (1 minus 120583)

times (119911 minus 119889)

2

1199034

1

minus1198892

minus 1199112

+ 6119889119911

1199034

2

+81198891199111199092

1199036

2


2[1

1199032

1

minus1

1199032

2

minus4119911 (119889 + 119911)

1199034

2

] d119889

=119902

2120587 (1 minus 120583)

times 1


119909minus

119909 (119889 minus 119911)

21199032

1

minus1

2arctan 119889 + 119911

119909+

2119909119911119889 (119889 + 119911)

1199034

2

+119909 (119889 + 119911)

21199032

2


2( arctan 119889 minus 119911

119909

minus arctan 119889 + 119911

119909+

2119909119911

1199032

2

)

100381610038161003816100381610038161003816100381610038161003816

119889=1198892

119889=1198891

120590119909+ 120590119911= int

1198892

1198891

119902119909

2120587 (1 minus 120583)

1

1199032

1

+5

1199032

2

minus

4 (1199112

+ 1199092

)

1199034

2

+2 (1 minus 2120583)

1199032

2

d119889

=119902


119909+ 5 arctan 119889 + 119911

119909

minus 2 (1199112

+ 1199092

)

times (119889 + 119911

1199091199032

2

+1

1199092arctan 119889 + 119911

119909)

+2 (1 minus 2120583) arctan 119889 + 119911

119909

10038161003816100381610038161003816100381610038161003816

119889=1198892

119889=1198891

(3)


l

h

L L

o

i

z

z

Pileft Piright

PoPo

l2 l2

infin infin


40

35

30

25

20

15

10

5

000 02 04 06 08 10 12

120572i (mminus1)

l = 10ml = 20m

l = 40ml = 100m


Soil

dept

hz i

(m)










119886= (119906119905+ 2119906119898+ 119906119887)4






119906119909= int 120576119909d119909

120576119909=

1 minus 1205832

119864119904

(120590119909minus

120583

1 minus 120583120590119911) or

119906119909=

1 minus 1205832

119864119904

(int120590119909d119909 minus

120583

1 minus 120583int120590119911d119909)

(4)


int120590119911d119909 =

119902

2120587 (1 minus 120583)

times minus119911119889 (119889 + 119911)

1199032

2

+ 120583119909 arctan 119889 minus 119911

119909

minus 120583119909 arctan 119889 + 119911

119909+

(1 minus 2120583) (119911 minus 119889)

4


times ln(1199031

119889 minus 119911)

2

+(1 minus 2120583) (119889 minus 119911)

4

times ln(1199032

119889 + 119911)

2

100381610038161003816100381610038161003816100381610038161003816

119889=1198892

119889=1198891

int 120590119909d119909 =

119902

2120587 (1 minus 120583)

times 119911119889 (119889 + 119911)

1199032

2


119909

+ (5119909 minus 3120583119909 +21199112

119909) arctan 119889 + 119911

119909

+(3 minus 2120583) (119889 minus 119911)

4ln(

1199031

119889 minus 119911)

2

+119889 (5 minus 6120583) + 119911 (7 minus 10120583)

4

times ln(1199032

119889 + 119911)

2

100381610038161003816100381610038161003816100381610038161003816

119889=1198892

119889=1198891

(5)


119906119909=

(119906119909119911=119889

1

+ 2119906119909119911=(119889

1+1198892)2

+ 119906119909119911=119889

2

)

4 (6)


119906119909= int

1198972

0

120576119909119889119909 minus int

119897

1198972

120576119909119889119909 (7)



119870119894=

119902

119906119909

=119864119904

1 minus 1205832

119902

2[(int

1198972

0

120590119909d119909 minus int

119897

1198972

120590119909d119909) minus

120583

1 minus 120583(int

1198972

0

120590119911d119909 minus int

119897

1198972

120590119911d119909)]

minus1

(8)



119906119909= int

10ℎ

0

120576119909119889119909 (9)



119870119900=

119902

119906119909

=119864119904

1 minus 1205832

119902

2(int

10ℎ

0

120590119909d119909 minus

120583

1 minus 120583int

10ℎ

0

120590119911d119909)minus1

(10)

We define

120572119894=

1

1 minus 1205832

119902

2[(int

1198972

0

120590119909d119909 minus int

119897

1198972

120590119909d119909) minus

120583

1 minus 120583(int

1198972

0

120590119911d119909 minus int

119897

1198972

120590119911d119909)]

minus1

120572119900=

1

1 minus 1205832

119902

2(int

10ℎ

0

120590119909d119909 minus

120583

1 minus 120583int

10ℎ

0

120590119911d119909)minus1

(11)


01 02 03 04

40

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)

L = 100mL = 200m

Influence scope L

00120572o (mminus1)



Then

119870119894= 120572119894119864119904

119870119900= 120572119900119864119904

(12)


119904denotes the















asymp

11120572119894120583=03

and 120572119894120583=05

asymp 09120572119894120583=03


120583 = 01120583 = 02

120583 = 03

120583 = 04120583 = 05

40

35

30

25

20

15

10

5

000 02 04 06 08 10 12

120572i (mminus1)

Soil

dept

hz i

(m)



0 1000 2000 3000 4000 5000

40

35

30

25

20

15

10

5

0

l = 10ml = 20m

l = 40ml = 100m


Soil

dept

hz i

(m)




119864119904= 120582 sdot 120590

1015840

= 120582 sdot 1205741015840

sdot 119911 (13)






119894





modulus

83





119870 = 120572 sdot 119864119904= 120572 sdot 120582 sdot 120574

1015840

sdot 119911 = 119898 sdot 119911 (14)









119864119902= 119864119894sdot (1 minus

1205901minus 1205903

(1205901minus 1205903)ult

) = 119864119894sdot (1 minus (120590

1minus 1205903) sdot 119887) (15)


ult


40

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)

L = 100mL = 200m

Influence scope L



1205901 minus 1205903 =120576

120576

a + b120576

EqEi = 1a

(1205901 minus 1205903)



119870 =119902

119906119909

=119902

(int 120576119909d119909)

120576119909=

1 minus 1205832

119864119902

(120590119909minus

120583

1 minus 120583120590119911)

=1 minus 1205832

119864119894sdot [1 minus (120590

119909minus 120590119911) sdot 119887]

(120590119909minus

120583

1 minus 120583120590119911)

(16)








119898119900kNmminus4 100 89

Deep inside 4230 3170 2598 1879 186 150

0 200 400 600 800 10000

1

2

3

4

5


b(M

Paminus1) b = 46 minus 00043120590998400z






119887 = 46 minus 000431205901015840

119911 (17)






119870119894= 120572119894sdot 119864119904= 120572119894sdot 120582 sdot 120590

1015840

119911= 120572119894sdot 120582 sdot (120574

1015840

119904minus 120574119908sdot 119894119894) sdot 119911119894 (18)


02 03 04 05 06 07 08


q = 100 kPaq = 150kPaq = 200 kPa

Load level q40

35

30

25

20

15

10

5

0

120572i (mminus1)

Soil

dept

hz i

(m)


3MPaminus1)

01 02 03 04 05 06 07 08

b = 0

b = 2

b = 4b = 5

b = 6

b = 8

b = 10

b (MPaminus1)40

35

30

25

20

15

10

5

0

120572i (mminus1)

Soil

dept

hz i

(m)


119894(119902 = 100 kPa)




15 m

5 m

l = 10m and 40m L = 100mL = 100m


empty1m

(a) Plane graph



Excavationsurface

l = 10m and 40m L = 100mL = 100m



h=20

m20

m

4 m

(b) Sectional graph



119870119900= 120572119900sdot 119864119904= 120572119900sdot 120582 sdot 120590

1015840

119911= 120572119900sdot 120582 sdot (120574

1015840

119904+ 120574119908sdot 119894119900) sdot 119911119900

(19)




119875119900119904

= (1205741015840

119904+ 120574119908119894119900) sdot 1198700sdot 119911119900

earth pressure

119875119900119908

= 120574119908(1 minus 119894119900) sdot 119911119900

water pressure

119875119900= (1205741015840





Soil spring inside

pit

Active pressure



119875119894119904

= (1205741015840

119904minus 120574119908119894119894) sdot 1198700sdot 119911119894

earth pressure

119875119894119908

= 120574119908(1 + 119894119894) sdot 119911119894

water pressure

119875119894= (1205741015840





Soil spring outside





Groundwater outside


inside

5lowast4=20

m20

m

5sim20m 100 m

10 m


4 Analysis Example





119904= 190 kNm3



1015840

= 5 kPa and 1205931015840

= 30∘ lateral














Parameters





4000



seepage


seepage 119870119900= 05


119894


path


model




119888kPa 120593∘ 119870119881

119870119867






pressure 119875119900119908



5 Case Study






=




119894119900for the





140 120 100 80 60 40 20 0

40

35

30

25

20

15

10

5

0

No seepage

Considering seepage




Dep

th (m

)



40

35

30

25

20

15

10

5

0

Dep

th (m

)


No seepage

Considering seepage






140 120 100 80 60 40 20 0

40

35

30

25

20

15

10

5

0

No seepage

Considering seepage




Dep

th (m

)



40

35

30

25

20

15

10

5

0

Dep

th (m

)


No seepage

Considering seepage






0 30 60 90 120 150 180 210

40

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)





0 30 60 90 120 150 180 210

40

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)







350 300 250 200 150 100 50 0

40

35

30

25

20

15

10

5

0

Soil

dept

hz i

(m)




350 300 250 200 150 100 50 0

40

35

30

25

20

15

10

5

0

Soil

dept

hz i

(m)








Groundwater

36800sand8 -1


silty clay6 -2




1



9000sandy silt2 -3

2 -1 sandy silt5000



Excavation surface

Ground surface

16800



6 Conclusions


119904



level





35

30

25

20

15

10

5

0 0 500 1000 1500 2000 2500 3000 3500


m (kNmiddotmminus4)

Soil

dept

hz

(m)



0 10 20 30 40 50 60



35

30

25

20

15

10

5

0


Dep

th (m

)



0 30 60 90 120 150

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)








119894of


(1205741015840

119904minus 120574119908








increase


Acknowledgment



References





























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











(a) Loading system







h

oSurface

x

uixΔhi

qi



119894sdot


(119902119894sdot 119908 sdot Δℎ)

(119870119894sdot 119908 sdot Δℎ)

=119902119894

119870119894

= 119906119894119909

or 119870119894=

119902119894

119906119894119909

(1)






1015840







Outside pit

Deformation

Inside pit

Po

Pi

(a)

Inside pit

Outside pit

h

uix

Δqi

hi

P998400o

P998400i

Δh

Δh

(b)




120590119909=

119902119909

2120587 (1 minus 120583)1199092

1199034

1

+1199092

+ 8119889119911 + 61198892

1199034

2

+8119889119911(119889 + 119911)

2

1199036

2

+1 minus 2120583

2[1

1199032

1

+3

1199032

2

minus4119911 (119889 + 119911)

1199034

2

]

120590119911=

119902119909

2120587 (1 minus 120583)(119911 minus 119889)

2

1199034

1

minus1198892

minus 1199112

+ 6119889119911

1199034

2

+81198891199111199092

1199036

2


2[1

1199032

1

minus1

1199032

2

minus4119911 (119889 + 119911)

1199034

2

]

120590119909+ 120590119911=

119902119909

2120587 (1 minus 120583)

times 1

1199032

1

+5

1199032

2

minus

4 (1199112

+ 1199092

)

1199034

2

+2 (1 minus 2120583)

1199032

2

(2)


z

xSurface

o

qz

d

d

r2 = [x2 + (z + d)2]12

r1 = [x2 minus(z+ d)2]12

r = x


r1 = [x2 minus(z+ d)2]12

z

xSurface

o

z

r2 = [x2 + (z + d)2]12

r = x

q

d2

d2

d1

d1



Retaining structure

Unilateral

Inside pit

Inside pit

Inside pit

Inside pit


Outside pit


(b3)

(b2)(b1)

(c1) (c2)

(a)

Left


infin

infin

infininfin

infininfin

Pi Pi

2Pi

Po Po

2Po

+ +


Surface

P

ub

um

ut

(a) Flexible load

P ua

Surface

(b) Rigid base



120590119911= int

1198892

1198891

119902119909

2120587 (1 minus 120583)

times (119911 minus 119889)

2

1199034

1

minus1198892

minus 1199112

+ 6119889119911

1199034

2

+81198891199111199092

1199036

2


2[1

1199032

1

minus1

1199032

2

minus4119911 (119889 + 119911)

1199034

2

] d119889

=119902

2120587 (1 minus 120583)

times 1


119909minus

119909 (119889 minus 119911)

21199032

1

minus1

2arctan 119889 + 119911

119909+

2119909119911119889 (119889 + 119911)

1199034

2

+119909 (119889 + 119911)

21199032

2


2( arctan 119889 minus 119911

119909

minus arctan 119889 + 119911

119909+

2119909119911

1199032

2

)

100381610038161003816100381610038161003816100381610038161003816

119889=1198892

119889=1198891

120590119909+ 120590119911= int

1198892

1198891

119902119909

2120587 (1 minus 120583)

1

1199032

1

+5

1199032

2

minus

4 (1199112

+ 1199092

)

1199034

2

+2 (1 minus 2120583)

1199032

2

d119889

=119902


119909+ 5 arctan 119889 + 119911

119909

minus 2 (1199112

+ 1199092

)

times (119889 + 119911

1199091199032

2

+1

1199092arctan 119889 + 119911

119909)

+2 (1 minus 2120583) arctan 119889 + 119911

119909

10038161003816100381610038161003816100381610038161003816

119889=1198892

119889=1198891

(3)


l

h

L L

o

i

z

z

Pileft Piright

PoPo

l2 l2

infin infin


40

35

30

25

20

15

10

5

000 02 04 06 08 10 12

120572i (mminus1)

l = 10ml = 20m

l = 40ml = 100m


Soil

dept

hz i

(m)










119886= (119906119905+ 2119906119898+ 119906119887)4






119906119909= int 120576119909d119909

120576119909=

1 minus 1205832

119864119904

(120590119909minus

120583

1 minus 120583120590119911) or

119906119909=

1 minus 1205832

119864119904

(int120590119909d119909 minus

120583

1 minus 120583int120590119911d119909)

(4)


int120590119911d119909 =

119902

2120587 (1 minus 120583)

times minus119911119889 (119889 + 119911)

1199032

2

+ 120583119909 arctan 119889 minus 119911

119909

minus 120583119909 arctan 119889 + 119911

119909+

(1 minus 2120583) (119911 minus 119889)

4


times ln(1199031

119889 minus 119911)

2

+(1 minus 2120583) (119889 minus 119911)

4

times ln(1199032

119889 + 119911)

2

100381610038161003816100381610038161003816100381610038161003816

119889=1198892

119889=1198891

int 120590119909d119909 =

119902

2120587 (1 minus 120583)

times 119911119889 (119889 + 119911)

1199032

2


119909

+ (5119909 minus 3120583119909 +21199112

119909) arctan 119889 + 119911

119909

+(3 minus 2120583) (119889 minus 119911)

4ln(

1199031

119889 minus 119911)

2

+119889 (5 minus 6120583) + 119911 (7 minus 10120583)

4

times ln(1199032

119889 + 119911)

2

100381610038161003816100381610038161003816100381610038161003816

119889=1198892

119889=1198891

(5)


119906119909=

(119906119909119911=119889

1

+ 2119906119909119911=(119889

1+1198892)2

+ 119906119909119911=119889

2

)

4 (6)


119906119909= int

1198972

0

120576119909119889119909 minus int

119897

1198972

120576119909119889119909 (7)



119870119894=

119902

119906119909

=119864119904

1 minus 1205832

119902

2[(int

1198972

0

120590119909d119909 minus int

119897

1198972

120590119909d119909) minus

120583

1 minus 120583(int

1198972

0

120590119911d119909 minus int

119897

1198972

120590119911d119909)]

minus1

(8)



119906119909= int

10ℎ

0

120576119909119889119909 (9)



119870119900=

119902

119906119909

=119864119904

1 minus 1205832

119902

2(int

10ℎ

0

120590119909d119909 minus

120583

1 minus 120583int

10ℎ

0

120590119911d119909)minus1

(10)

We define

120572119894=

1

1 minus 1205832

119902

2[(int

1198972

0

120590119909d119909 minus int

119897

1198972

120590119909d119909) minus

120583

1 minus 120583(int

1198972

0

120590119911d119909 minus int

119897

1198972

120590119911d119909)]

minus1

120572119900=

1

1 minus 1205832

119902

2(int

10ℎ

0

120590119909d119909 minus

120583

1 minus 120583int

10ℎ

0

120590119911d119909)minus1

(11)


01 02 03 04

40

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)

L = 100mL = 200m

Influence scope L

00120572o (mminus1)



Then

119870119894= 120572119894119864119904

119870119900= 120572119900119864119904

(12)


119904denotes the















asymp

11120572119894120583=03

and 120572119894120583=05

asymp 09120572119894120583=03


120583 = 01120583 = 02

120583 = 03

120583 = 04120583 = 05

40

35

30

25

20

15

10

5

000 02 04 06 08 10 12

120572i (mminus1)

Soil

dept

hz i

(m)



0 1000 2000 3000 4000 5000

40

35

30

25

20

15

10

5

0

l = 10ml = 20m

l = 40ml = 100m


Soil

dept

hz i

(m)




119864119904= 120582 sdot 120590

1015840

= 120582 sdot 1205741015840

sdot 119911 (13)






119894





modulus

83





119870 = 120572 sdot 119864119904= 120572 sdot 120582 sdot 120574

1015840

sdot 119911 = 119898 sdot 119911 (14)









119864119902= 119864119894sdot (1 minus

1205901minus 1205903

(1205901minus 1205903)ult

) = 119864119894sdot (1 minus (120590

1minus 1205903) sdot 119887) (15)


ult


40

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)

L = 100mL = 200m

Influence scope L



1205901 minus 1205903 =120576

120576

a + b120576

EqEi = 1a

(1205901 minus 1205903)



119870 =119902

119906119909

=119902

(int 120576119909d119909)

120576119909=

1 minus 1205832

119864119902

(120590119909minus

120583

1 minus 120583120590119911)

=1 minus 1205832

119864119894sdot [1 minus (120590

119909minus 120590119911) sdot 119887]

(120590119909minus

120583

1 minus 120583120590119911)

(16)








119898119900kNmminus4 100 89

Deep inside 4230 3170 2598 1879 186 150

0 200 400 600 800 10000

1

2

3

4

5


b(M

Paminus1) b = 46 minus 00043120590998400z






119887 = 46 minus 000431205901015840

119911 (17)






119870119894= 120572119894sdot 119864119904= 120572119894sdot 120582 sdot 120590

1015840

119911= 120572119894sdot 120582 sdot (120574

1015840

119904minus 120574119908sdot 119894119894) sdot 119911119894 (18)


02 03 04 05 06 07 08


q = 100 kPaq = 150kPaq = 200 kPa

Load level q40

35

30

25

20

15

10

5

0

120572i (mminus1)

Soil

dept

hz i

(m)


3MPaminus1)

01 02 03 04 05 06 07 08

b = 0

b = 2

b = 4b = 5

b = 6

b = 8

b = 10

b (MPaminus1)40

35

30

25

20

15

10

5

0

120572i (mminus1)

Soil

dept

hz i

(m)


119894(119902 = 100 kPa)




15 m

5 m

l = 10m and 40m L = 100mL = 100m


empty1m

(a) Plane graph



Excavationsurface

l = 10m and 40m L = 100mL = 100m



h=20

m20

m

4 m

(b) Sectional graph



119870119900= 120572119900sdot 119864119904= 120572119900sdot 120582 sdot 120590

1015840

119911= 120572119900sdot 120582 sdot (120574

1015840

119904+ 120574119908sdot 119894119900) sdot 119911119900

(19)




119875119900119904

= (1205741015840

119904+ 120574119908119894119900) sdot 1198700sdot 119911119900

earth pressure

119875119900119908

= 120574119908(1 minus 119894119900) sdot 119911119900

water pressure

119875119900= (1205741015840





Soil spring inside

pit

Active pressure



119875119894119904

= (1205741015840

119904minus 120574119908119894119894) sdot 1198700sdot 119911119894

earth pressure

119875119894119908

= 120574119908(1 + 119894119894) sdot 119911119894

water pressure

119875119894= (1205741015840





Soil spring outside





Groundwater outside


inside

5lowast4=20

m20

m

5sim20m 100 m

10 m


4 Analysis Example





119904= 190 kNm3



1015840

= 5 kPa and 1205931015840

= 30∘ lateral














Parameters





4000



seepage


seepage 119870119900= 05


119894


path


model




119888kPa 120593∘ 119870119881

119870119867






pressure 119875119900119908



5 Case Study






=




119894119900for the





140 120 100 80 60 40 20 0

40

35

30

25

20

15

10

5

0

No seepage

Considering seepage




Dep

th (m

)



40

35

30

25

20

15

10

5

0

Dep

th (m

)


No seepage

Considering seepage






140 120 100 80 60 40 20 0

40

35

30

25

20

15

10

5

0

No seepage

Considering seepage




Dep

th (m

)



40

35

30

25

20

15

10

5

0

Dep

th (m

)


No seepage

Considering seepage






0 30 60 90 120 150 180 210

40

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)





0 30 60 90 120 150 180 210

40

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)







350 300 250 200 150 100 50 0

40

35

30

25

20

15

10

5

0

Soil

dept

hz i

(m)




350 300 250 200 150 100 50 0

40

35

30

25

20

15

10

5

0

Soil

dept

hz i

(m)








Groundwater

36800sand8 -1


silty clay6 -2




1



9000sandy silt2 -3

2 -1 sandy silt5000



Excavation surface

Ground surface

16800



6 Conclusions


119904



level





35

30

25

20

15

10

5

0 0 500 1000 1500 2000 2500 3000 3500


m (kNmiddotmminus4)

Soil

dept

hz

(m)



0 10 20 30 40 50 60



35

30

25

20

15

10

5

0


Dep

th (m

)



0 30 60 90 120 150

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)








119894of


(1205741015840

119904minus 120574119908








increase


Acknowledgment



References





























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










Outside pit

Deformation

Inside pit

Po

Pi

(a)

Inside pit

Outside pit

h

uix

Δqi

hi

P998400o

P998400i

Δh

Δh

(b)




120590119909=

119902119909

2120587 (1 minus 120583)1199092

1199034

1

+1199092

+ 8119889119911 + 61198892

1199034

2

+8119889119911(119889 + 119911)

2

1199036

2

+1 minus 2120583

2[1

1199032

1

+3

1199032

2

minus4119911 (119889 + 119911)

1199034

2

]

120590119911=

119902119909

2120587 (1 minus 120583)(119911 minus 119889)

2

1199034

1

minus1198892

minus 1199112

+ 6119889119911

1199034

2

+81198891199111199092

1199036

2


2[1

1199032

1

minus1

1199032

2

minus4119911 (119889 + 119911)

1199034

2

]

120590119909+ 120590119911=

119902119909

2120587 (1 minus 120583)

times 1

1199032

1

+5

1199032

2

minus

4 (1199112

+ 1199092

)

1199034

2

+2 (1 minus 2120583)

1199032

2

(2)


z

xSurface

o

qz

d

d

r2 = [x2 + (z + d)2]12

r1 = [x2 minus(z+ d)2]12

r = x


r1 = [x2 minus(z+ d)2]12

z

xSurface

o

z

r2 = [x2 + (z + d)2]12

r = x

q

d2

d2

d1

d1



Retaining structure

Unilateral

Inside pit

Inside pit

Inside pit

Inside pit


Outside pit


(b3)

(b2)(b1)

(c1) (c2)

(a)

Left


infin

infin

infininfin

infininfin

Pi Pi

2Pi

Po Po

2Po

+ +


Surface

P

ub

um

ut

(a) Flexible load

P ua

Surface

(b) Rigid base



120590119911= int

1198892

1198891

119902119909

2120587 (1 minus 120583)

times (119911 minus 119889)

2

1199034

1

minus1198892

minus 1199112

+ 6119889119911

1199034

2

+81198891199111199092

1199036

2


2[1

1199032

1

minus1

1199032

2

minus4119911 (119889 + 119911)

1199034

2

] d119889

=119902

2120587 (1 minus 120583)

times 1


119909minus

119909 (119889 minus 119911)

21199032

1

minus1

2arctan 119889 + 119911

119909+

2119909119911119889 (119889 + 119911)

1199034

2

+119909 (119889 + 119911)

21199032

2


2( arctan 119889 minus 119911

119909

minus arctan 119889 + 119911

119909+

2119909119911

1199032

2

)

100381610038161003816100381610038161003816100381610038161003816

119889=1198892

119889=1198891

120590119909+ 120590119911= int

1198892

1198891

119902119909

2120587 (1 minus 120583)

1

1199032

1

+5

1199032

2

minus

4 (1199112

+ 1199092

)

1199034

2

+2 (1 minus 2120583)

1199032

2

d119889

=119902


119909+ 5 arctan 119889 + 119911

119909

minus 2 (1199112

+ 1199092

)

times (119889 + 119911

1199091199032

2

+1

1199092arctan 119889 + 119911

119909)

+2 (1 minus 2120583) arctan 119889 + 119911

119909

10038161003816100381610038161003816100381610038161003816

119889=1198892

119889=1198891

(3)


l

h

L L

o

i

z

z

Pileft Piright

PoPo

l2 l2

infin infin


40

35

30

25

20

15

10

5

000 02 04 06 08 10 12

120572i (mminus1)

l = 10ml = 20m

l = 40ml = 100m


Soil

dept

hz i

(m)










119886= (119906119905+ 2119906119898+ 119906119887)4






119906119909= int 120576119909d119909

120576119909=

1 minus 1205832

119864119904

(120590119909minus

120583

1 minus 120583120590119911) or

119906119909=

1 minus 1205832

119864119904

(int120590119909d119909 minus

120583

1 minus 120583int120590119911d119909)

(4)


int120590119911d119909 =

119902

2120587 (1 minus 120583)

times minus119911119889 (119889 + 119911)

1199032

2

+ 120583119909 arctan 119889 minus 119911

119909

minus 120583119909 arctan 119889 + 119911

119909+

(1 minus 2120583) (119911 minus 119889)

4


times ln(1199031

119889 minus 119911)

2

+(1 minus 2120583) (119889 minus 119911)

4

times ln(1199032

119889 + 119911)

2

100381610038161003816100381610038161003816100381610038161003816

119889=1198892

119889=1198891

int 120590119909d119909 =

119902

2120587 (1 minus 120583)

times 119911119889 (119889 + 119911)

1199032

2


119909

+ (5119909 minus 3120583119909 +21199112

119909) arctan 119889 + 119911

119909

+(3 minus 2120583) (119889 minus 119911)

4ln(

1199031

119889 minus 119911)

2

+119889 (5 minus 6120583) + 119911 (7 minus 10120583)

4

times ln(1199032

119889 + 119911)

2

100381610038161003816100381610038161003816100381610038161003816

119889=1198892

119889=1198891

(5)


119906119909=

(119906119909119911=119889

1

+ 2119906119909119911=(119889

1+1198892)2

+ 119906119909119911=119889

2

)

4 (6)


119906119909= int

1198972

0

120576119909119889119909 minus int

119897

1198972

120576119909119889119909 (7)



119870119894=

119902

119906119909

=119864119904

1 minus 1205832

119902

2[(int

1198972

0

120590119909d119909 minus int

119897

1198972

120590119909d119909) minus

120583

1 minus 120583(int

1198972

0

120590119911d119909 minus int

119897

1198972

120590119911d119909)]

minus1

(8)



119906119909= int

10ℎ

0

120576119909119889119909 (9)



119870119900=

119902

119906119909

=119864119904

1 minus 1205832

119902

2(int

10ℎ

0

120590119909d119909 minus

120583

1 minus 120583int

10ℎ

0

120590119911d119909)minus1

(10)

We define

120572119894=

1

1 minus 1205832

119902

2[(int

1198972

0

120590119909d119909 minus int

119897

1198972

120590119909d119909) minus

120583

1 minus 120583(int

1198972

0

120590119911d119909 minus int

119897

1198972

120590119911d119909)]

minus1

120572119900=

1

1 minus 1205832

119902

2(int

10ℎ

0

120590119909d119909 minus

120583

1 minus 120583int

10ℎ

0

120590119911d119909)minus1

(11)


01 02 03 04

40

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)

L = 100mL = 200m

Influence scope L

00120572o (mminus1)



Then

119870119894= 120572119894119864119904

119870119900= 120572119900119864119904

(12)


119904denotes the















asymp

11120572119894120583=03

and 120572119894120583=05

asymp 09120572119894120583=03


120583 = 01120583 = 02

120583 = 03

120583 = 04120583 = 05

40

35

30

25

20

15

10

5

000 02 04 06 08 10 12

120572i (mminus1)

Soil

dept

hz i

(m)



0 1000 2000 3000 4000 5000

40

35

30

25

20

15

10

5

0

l = 10ml = 20m

l = 40ml = 100m


Soil

dept

hz i

(m)




119864119904= 120582 sdot 120590

1015840

= 120582 sdot 1205741015840

sdot 119911 (13)






119894





modulus

83





119870 = 120572 sdot 119864119904= 120572 sdot 120582 sdot 120574

1015840

sdot 119911 = 119898 sdot 119911 (14)









119864119902= 119864119894sdot (1 minus

1205901minus 1205903

(1205901minus 1205903)ult

) = 119864119894sdot (1 minus (120590

1minus 1205903) sdot 119887) (15)


ult


40

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)

L = 100mL = 200m

Influence scope L



1205901 minus 1205903 =120576

120576

a + b120576

EqEi = 1a

(1205901 minus 1205903)



119870 =119902

119906119909

=119902

(int 120576119909d119909)

120576119909=

1 minus 1205832

119864119902

(120590119909minus

120583

1 minus 120583120590119911)

=1 minus 1205832

119864119894sdot [1 minus (120590

119909minus 120590119911) sdot 119887]

(120590119909minus

120583

1 minus 120583120590119911)

(16)








119898119900kNmminus4 100 89

Deep inside 4230 3170 2598 1879 186 150

0 200 400 600 800 10000

1

2

3

4

5


b(M

Paminus1) b = 46 minus 00043120590998400z






119887 = 46 minus 000431205901015840

119911 (17)






119870119894= 120572119894sdot 119864119904= 120572119894sdot 120582 sdot 120590

1015840

119911= 120572119894sdot 120582 sdot (120574

1015840

119904minus 120574119908sdot 119894119894) sdot 119911119894 (18)


02 03 04 05 06 07 08


q = 100 kPaq = 150kPaq = 200 kPa

Load level q40

35

30

25

20

15

10

5

0

120572i (mminus1)

Soil

dept

hz i

(m)


3MPaminus1)

01 02 03 04 05 06 07 08

b = 0

b = 2

b = 4b = 5

b = 6

b = 8

b = 10

b (MPaminus1)40

35

30

25

20

15

10

5

0

120572i (mminus1)

Soil

dept

hz i

(m)


119894(119902 = 100 kPa)




15 m

5 m

l = 10m and 40m L = 100mL = 100m


empty1m

(a) Plane graph



Excavationsurface

l = 10m and 40m L = 100mL = 100m



h=20

m20

m

4 m

(b) Sectional graph



119870119900= 120572119900sdot 119864119904= 120572119900sdot 120582 sdot 120590

1015840

119911= 120572119900sdot 120582 sdot (120574

1015840

119904+ 120574119908sdot 119894119900) sdot 119911119900

(19)




119875119900119904

= (1205741015840

119904+ 120574119908119894119900) sdot 1198700sdot 119911119900

earth pressure

119875119900119908

= 120574119908(1 minus 119894119900) sdot 119911119900

water pressure

119875119900= (1205741015840





Soil spring inside

pit

Active pressure



119875119894119904

= (1205741015840

119904minus 120574119908119894119894) sdot 1198700sdot 119911119894

earth pressure

119875119894119908

= 120574119908(1 + 119894119894) sdot 119911119894

water pressure

119875119894= (1205741015840





Soil spring outside





Groundwater outside


inside

5lowast4=20

m20

m

5sim20m 100 m

10 m


4 Analysis Example





119904= 190 kNm3



1015840

= 5 kPa and 1205931015840

= 30∘ lateral














Parameters





4000



seepage


seepage 119870119900= 05


119894


path


model




119888kPa 120593∘ 119870119881

119870119867






pressure 119875119900119908



5 Case Study






=




119894119900for the





140 120 100 80 60 40 20 0

40

35

30

25

20

15

10

5

0

No seepage

Considering seepage




Dep

th (m

)



40

35

30

25

20

15

10

5

0

Dep

th (m

)


No seepage

Considering seepage






140 120 100 80 60 40 20 0

40

35

30

25

20

15

10

5

0

No seepage

Considering seepage




Dep

th (m

)



40

35

30

25

20

15

10

5

0

Dep

th (m

)


No seepage

Considering seepage






0 30 60 90 120 150 180 210

40

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)





0 30 60 90 120 150 180 210

40

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)







350 300 250 200 150 100 50 0

40

35

30

25

20

15

10

5

0

Soil

dept

hz i

(m)




350 300 250 200 150 100 50 0

40

35

30

25

20

15

10

5

0

Soil

dept

hz i

(m)








Groundwater

36800sand8 -1


silty clay6 -2




1



9000sandy silt2 -3

2 -1 sandy silt5000



Excavation surface

Ground surface

16800



6 Conclusions


119904



level





35

30

25

20

15

10

5

0 0 500 1000 1500 2000 2500 3000 3500


m (kNmiddotmminus4)

Soil

dept

hz

(m)



0 10 20 30 40 50 60



35

30

25

20

15

10

5

0


Dep

th (m

)



0 30 60 90 120 150

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)








119894of


(1205741015840

119904minus 120574119908








increase


Acknowledgment



References





























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










Retaining structure

Unilateral

Inside pit

Inside pit

Inside pit

Inside pit


Outside pit


(b3)

(b2)(b1)

(c1) (c2)

(a)

Left


infin

infin

infininfin

infininfin

Pi Pi

2Pi

Po Po

2Po

+ +


Surface

P

ub

um

ut

(a) Flexible load

P ua

Surface

(b) Rigid base



120590119911= int

1198892

1198891

119902119909

2120587 (1 minus 120583)

times (119911 minus 119889)

2

1199034

1

minus1198892

minus 1199112

+ 6119889119911

1199034

2

+81198891199111199092

1199036

2


2[1

1199032

1

minus1

1199032

2

minus4119911 (119889 + 119911)

1199034

2

] d119889

=119902

2120587 (1 minus 120583)

times 1


119909minus

119909 (119889 minus 119911)

21199032

1

minus1

2arctan 119889 + 119911

119909+

2119909119911119889 (119889 + 119911)

1199034

2

+119909 (119889 + 119911)

21199032

2


2( arctan 119889 minus 119911

119909

minus arctan 119889 + 119911

119909+

2119909119911

1199032

2

)

100381610038161003816100381610038161003816100381610038161003816

119889=1198892

119889=1198891

120590119909+ 120590119911= int

1198892

1198891

119902119909

2120587 (1 minus 120583)

1

1199032

1

+5

1199032

2

minus

4 (1199112

+ 1199092

)

1199034

2

+2 (1 minus 2120583)

1199032

2

d119889

=119902


119909+ 5 arctan 119889 + 119911

119909

minus 2 (1199112

+ 1199092

)

times (119889 + 119911

1199091199032

2

+1

1199092arctan 119889 + 119911

119909)

+2 (1 minus 2120583) arctan 119889 + 119911

119909

10038161003816100381610038161003816100381610038161003816

119889=1198892

119889=1198891

(3)


l

h

L L

o

i

z

z

Pileft Piright

PoPo

l2 l2

infin infin


40

35

30

25

20

15

10

5

000 02 04 06 08 10 12

120572i (mminus1)

l = 10ml = 20m

l = 40ml = 100m


Soil

dept

hz i

(m)










119886= (119906119905+ 2119906119898+ 119906119887)4






119906119909= int 120576119909d119909

120576119909=

1 minus 1205832

119864119904

(120590119909minus

120583

1 minus 120583120590119911) or

119906119909=

1 minus 1205832

119864119904

(int120590119909d119909 minus

120583

1 minus 120583int120590119911d119909)

(4)


int120590119911d119909 =

119902

2120587 (1 minus 120583)

times minus119911119889 (119889 + 119911)

1199032

2

+ 120583119909 arctan 119889 minus 119911

119909

minus 120583119909 arctan 119889 + 119911

119909+

(1 minus 2120583) (119911 minus 119889)

4


times ln(1199031

119889 minus 119911)

2

+(1 minus 2120583) (119889 minus 119911)

4

times ln(1199032

119889 + 119911)

2

100381610038161003816100381610038161003816100381610038161003816

119889=1198892

119889=1198891

int 120590119909d119909 =

119902

2120587 (1 minus 120583)

times 119911119889 (119889 + 119911)

1199032

2


119909

+ (5119909 minus 3120583119909 +21199112

119909) arctan 119889 + 119911

119909

+(3 minus 2120583) (119889 minus 119911)

4ln(

1199031

119889 minus 119911)

2

+119889 (5 minus 6120583) + 119911 (7 minus 10120583)

4

times ln(1199032

119889 + 119911)

2

100381610038161003816100381610038161003816100381610038161003816

119889=1198892

119889=1198891

(5)


119906119909=

(119906119909119911=119889

1

+ 2119906119909119911=(119889

1+1198892)2

+ 119906119909119911=119889

2

)

4 (6)


119906119909= int

1198972

0

120576119909119889119909 minus int

119897

1198972

120576119909119889119909 (7)



119870119894=

119902

119906119909

=119864119904

1 minus 1205832

119902

2[(int

1198972

0

120590119909d119909 minus int

119897

1198972

120590119909d119909) minus

120583

1 minus 120583(int

1198972

0

120590119911d119909 minus int

119897

1198972

120590119911d119909)]

minus1

(8)



119906119909= int

10ℎ

0

120576119909119889119909 (9)



119870119900=

119902

119906119909

=119864119904

1 minus 1205832

119902

2(int

10ℎ

0

120590119909d119909 minus

120583

1 minus 120583int

10ℎ

0

120590119911d119909)minus1

(10)

We define

120572119894=

1

1 minus 1205832

119902

2[(int

1198972

0

120590119909d119909 minus int

119897

1198972

120590119909d119909) minus

120583

1 minus 120583(int

1198972

0

120590119911d119909 minus int

119897

1198972

120590119911d119909)]

minus1

120572119900=

1

1 minus 1205832

119902

2(int

10ℎ

0

120590119909d119909 minus

120583

1 minus 120583int

10ℎ

0

120590119911d119909)minus1

(11)


01 02 03 04

40

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)

L = 100mL = 200m

Influence scope L

00120572o (mminus1)



Then

119870119894= 120572119894119864119904

119870119900= 120572119900119864119904

(12)


119904denotes the















asymp

11120572119894120583=03

and 120572119894120583=05

asymp 09120572119894120583=03


120583 = 01120583 = 02

120583 = 03

120583 = 04120583 = 05

40

35

30

25

20

15

10

5

000 02 04 06 08 10 12

120572i (mminus1)

Soil

dept

hz i

(m)



0 1000 2000 3000 4000 5000

40

35

30

25

20

15

10

5

0

l = 10ml = 20m

l = 40ml = 100m


Soil

dept

hz i

(m)




119864119904= 120582 sdot 120590

1015840

= 120582 sdot 1205741015840

sdot 119911 (13)






119894





modulus

83





119870 = 120572 sdot 119864119904= 120572 sdot 120582 sdot 120574

1015840

sdot 119911 = 119898 sdot 119911 (14)









119864119902= 119864119894sdot (1 minus

1205901minus 1205903

(1205901minus 1205903)ult

) = 119864119894sdot (1 minus (120590

1minus 1205903) sdot 119887) (15)


ult


40

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)

L = 100mL = 200m

Influence scope L



1205901 minus 1205903 =120576

120576

a + b120576

EqEi = 1a

(1205901 minus 1205903)



119870 =119902

119906119909

=119902

(int 120576119909d119909)

120576119909=

1 minus 1205832

119864119902

(120590119909minus

120583

1 minus 120583120590119911)

=1 minus 1205832

119864119894sdot [1 minus (120590

119909minus 120590119911) sdot 119887]

(120590119909minus

120583

1 minus 120583120590119911)

(16)








119898119900kNmminus4 100 89

Deep inside 4230 3170 2598 1879 186 150

0 200 400 600 800 10000

1

2

3

4

5


b(M

Paminus1) b = 46 minus 00043120590998400z






119887 = 46 minus 000431205901015840

119911 (17)






119870119894= 120572119894sdot 119864119904= 120572119894sdot 120582 sdot 120590

1015840

119911= 120572119894sdot 120582 sdot (120574

1015840

119904minus 120574119908sdot 119894119894) sdot 119911119894 (18)


02 03 04 05 06 07 08


q = 100 kPaq = 150kPaq = 200 kPa

Load level q40

35

30

25

20

15

10

5

0

120572i (mminus1)

Soil

dept

hz i

(m)


3MPaminus1)

01 02 03 04 05 06 07 08

b = 0

b = 2

b = 4b = 5

b = 6

b = 8

b = 10

b (MPaminus1)40

35

30

25

20

15

10

5

0

120572i (mminus1)

Soil

dept

hz i

(m)


119894(119902 = 100 kPa)




15 m

5 m

l = 10m and 40m L = 100mL = 100m


empty1m

(a) Plane graph



Excavationsurface

l = 10m and 40m L = 100mL = 100m



h=20

m20

m

4 m

(b) Sectional graph



119870119900= 120572119900sdot 119864119904= 120572119900sdot 120582 sdot 120590

1015840

119911= 120572119900sdot 120582 sdot (120574

1015840

119904+ 120574119908sdot 119894119900) sdot 119911119900

(19)




119875119900119904

= (1205741015840

119904+ 120574119908119894119900) sdot 1198700sdot 119911119900

earth pressure

119875119900119908

= 120574119908(1 minus 119894119900) sdot 119911119900

water pressure

119875119900= (1205741015840





Soil spring inside

pit

Active pressure



119875119894119904

= (1205741015840

119904minus 120574119908119894119894) sdot 1198700sdot 119911119894

earth pressure

119875119894119908

= 120574119908(1 + 119894119894) sdot 119911119894

water pressure

119875119894= (1205741015840





Soil spring outside





Groundwater outside


inside

5lowast4=20

m20

m

5sim20m 100 m

10 m


4 Analysis Example





119904= 190 kNm3



1015840

= 5 kPa and 1205931015840

= 30∘ lateral














Parameters





4000



seepage


seepage 119870119900= 05


119894


path


model




119888kPa 120593∘ 119870119881

119870119867






pressure 119875119900119908



5 Case Study






=




119894119900for the





140 120 100 80 60 40 20 0

40

35

30

25

20

15

10

5

0

No seepage

Considering seepage




Dep

th (m

)



40

35

30

25

20

15

10

5

0

Dep

th (m

)


No seepage

Considering seepage






140 120 100 80 60 40 20 0

40

35

30

25

20

15

10

5

0

No seepage

Considering seepage




Dep

th (m

)



40

35

30

25

20

15

10

5

0

Dep

th (m

)


No seepage

Considering seepage






0 30 60 90 120 150 180 210

40

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)





0 30 60 90 120 150 180 210

40

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)







350 300 250 200 150 100 50 0

40

35

30

25

20

15

10

5

0

Soil

dept

hz i

(m)




350 300 250 200 150 100 50 0

40

35

30

25

20

15

10

5

0

Soil

dept

hz i

(m)








Groundwater

36800sand8 -1


silty clay6 -2




1



9000sandy silt2 -3

2 -1 sandy silt5000



Excavation surface

Ground surface

16800



6 Conclusions


119904



level





35

30

25

20

15

10

5

0 0 500 1000 1500 2000 2500 3000 3500


m (kNmiddotmminus4)

Soil

dept

hz

(m)



0 10 20 30 40 50 60



35

30

25

20

15

10

5

0


Dep

th (m

)



0 30 60 90 120 150

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)








119894of


(1205741015840

119904minus 120574119908








increase


Acknowledgment



References





























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










l

h

L L

o

i

z

z

Pileft Piright

PoPo

l2 l2

infin infin


40

35

30

25

20

15

10

5

000 02 04 06 08 10 12

120572i (mminus1)

l = 10ml = 20m

l = 40ml = 100m


Soil

dept

hz i

(m)










119886= (119906119905+ 2119906119898+ 119906119887)4






119906119909= int 120576119909d119909

120576119909=

1 minus 1205832

119864119904

(120590119909minus

120583

1 minus 120583120590119911) or

119906119909=

1 minus 1205832

119864119904

(int120590119909d119909 minus

120583

1 minus 120583int120590119911d119909)

(4)


int120590119911d119909 =

119902

2120587 (1 minus 120583)

times minus119911119889 (119889 + 119911)

1199032

2

+ 120583119909 arctan 119889 minus 119911

119909

minus 120583119909 arctan 119889 + 119911

119909+

(1 minus 2120583) (119911 minus 119889)

4


times ln(1199031

119889 minus 119911)

2

+(1 minus 2120583) (119889 minus 119911)

4

times ln(1199032

119889 + 119911)

2

100381610038161003816100381610038161003816100381610038161003816

119889=1198892

119889=1198891

int 120590119909d119909 =

119902

2120587 (1 minus 120583)

times 119911119889 (119889 + 119911)

1199032

2


119909

+ (5119909 minus 3120583119909 +21199112

119909) arctan 119889 + 119911

119909

+(3 minus 2120583) (119889 minus 119911)

4ln(

1199031

119889 minus 119911)

2

+119889 (5 minus 6120583) + 119911 (7 minus 10120583)

4

times ln(1199032

119889 + 119911)

2

100381610038161003816100381610038161003816100381610038161003816

119889=1198892

119889=1198891

(5)


119906119909=

(119906119909119911=119889

1

+ 2119906119909119911=(119889

1+1198892)2

+ 119906119909119911=119889

2

)

4 (6)


119906119909= int

1198972

0

120576119909119889119909 minus int

119897

1198972

120576119909119889119909 (7)



119870119894=

119902

119906119909

=119864119904

1 minus 1205832

119902

2[(int

1198972

0

120590119909d119909 minus int

119897

1198972

120590119909d119909) minus

120583

1 minus 120583(int

1198972

0

120590119911d119909 minus int

119897

1198972

120590119911d119909)]

minus1

(8)



119906119909= int

10ℎ

0

120576119909119889119909 (9)



119870119900=

119902

119906119909

=119864119904

1 minus 1205832

119902

2(int

10ℎ

0

120590119909d119909 minus

120583

1 minus 120583int

10ℎ

0

120590119911d119909)minus1

(10)

We define

120572119894=

1

1 minus 1205832

119902

2[(int

1198972

0

120590119909d119909 minus int

119897

1198972

120590119909d119909) minus

120583

1 minus 120583(int

1198972

0

120590119911d119909 minus int

119897

1198972

120590119911d119909)]

minus1

120572119900=

1

1 minus 1205832

119902

2(int

10ℎ

0

120590119909d119909 minus

120583

1 minus 120583int

10ℎ

0

120590119911d119909)minus1

(11)


01 02 03 04

40

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)

L = 100mL = 200m

Influence scope L

00120572o (mminus1)



Then

119870119894= 120572119894119864119904

119870119900= 120572119900119864119904

(12)


119904denotes the















asymp

11120572119894120583=03

and 120572119894120583=05

asymp 09120572119894120583=03


120583 = 01120583 = 02

120583 = 03

120583 = 04120583 = 05

40

35

30

25

20

15

10

5

000 02 04 06 08 10 12

120572i (mminus1)

Soil

dept

hz i

(m)



0 1000 2000 3000 4000 5000

40

35

30

25

20

15

10

5

0

l = 10ml = 20m

l = 40ml = 100m


Soil

dept

hz i

(m)




119864119904= 120582 sdot 120590

1015840

= 120582 sdot 1205741015840

sdot 119911 (13)






119894





modulus

83





119870 = 120572 sdot 119864119904= 120572 sdot 120582 sdot 120574

1015840

sdot 119911 = 119898 sdot 119911 (14)









119864119902= 119864119894sdot (1 minus

1205901minus 1205903

(1205901minus 1205903)ult

) = 119864119894sdot (1 minus (120590

1minus 1205903) sdot 119887) (15)


ult


40

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)

L = 100mL = 200m

Influence scope L



1205901 minus 1205903 =120576

120576

a + b120576

EqEi = 1a

(1205901 minus 1205903)



119870 =119902

119906119909

=119902

(int 120576119909d119909)

120576119909=

1 minus 1205832

119864119902

(120590119909minus

120583

1 minus 120583120590119911)

=1 minus 1205832

119864119894sdot [1 minus (120590

119909minus 120590119911) sdot 119887]

(120590119909minus

120583

1 minus 120583120590119911)

(16)








119898119900kNmminus4 100 89

Deep inside 4230 3170 2598 1879 186 150

0 200 400 600 800 10000

1

2

3

4

5


b(M

Paminus1) b = 46 minus 00043120590998400z






119887 = 46 minus 000431205901015840

119911 (17)






119870119894= 120572119894sdot 119864119904= 120572119894sdot 120582 sdot 120590

1015840

119911= 120572119894sdot 120582 sdot (120574

1015840

119904minus 120574119908sdot 119894119894) sdot 119911119894 (18)


02 03 04 05 06 07 08


q = 100 kPaq = 150kPaq = 200 kPa

Load level q40

35

30

25

20

15

10

5

0

120572i (mminus1)

Soil

dept

hz i

(m)


3MPaminus1)

01 02 03 04 05 06 07 08

b = 0

b = 2

b = 4b = 5

b = 6

b = 8

b = 10

b (MPaminus1)40

35

30

25

20

15

10

5

0

120572i (mminus1)

Soil

dept

hz i

(m)


119894(119902 = 100 kPa)




15 m

5 m

l = 10m and 40m L = 100mL = 100m


empty1m

(a) Plane graph



Excavationsurface

l = 10m and 40m L = 100mL = 100m



h=20

m20

m

4 m

(b) Sectional graph



119870119900= 120572119900sdot 119864119904= 120572119900sdot 120582 sdot 120590

1015840

119911= 120572119900sdot 120582 sdot (120574

1015840

119904+ 120574119908sdot 119894119900) sdot 119911119900

(19)




119875119900119904

= (1205741015840

119904+ 120574119908119894119900) sdot 1198700sdot 119911119900

earth pressure

119875119900119908

= 120574119908(1 minus 119894119900) sdot 119911119900

water pressure

119875119900= (1205741015840





Soil spring inside

pit

Active pressure



119875119894119904

= (1205741015840

119904minus 120574119908119894119894) sdot 1198700sdot 119911119894

earth pressure

119875119894119908

= 120574119908(1 + 119894119894) sdot 119911119894

water pressure

119875119894= (1205741015840





Soil spring outside





Groundwater outside


inside

5lowast4=20

m20

m

5sim20m 100 m

10 m


4 Analysis Example





119904= 190 kNm3



1015840

= 5 kPa and 1205931015840

= 30∘ lateral














Parameters





4000



seepage


seepage 119870119900= 05


119894


path


model




119888kPa 120593∘ 119870119881

119870119867






pressure 119875119900119908



5 Case Study






=




119894119900for the





140 120 100 80 60 40 20 0

40

35

30

25

20

15

10

5

0

No seepage

Considering seepage




Dep

th (m

)



40

35

30

25

20

15

10

5

0

Dep

th (m

)


No seepage

Considering seepage






140 120 100 80 60 40 20 0

40

35

30

25

20

15

10

5

0

No seepage

Considering seepage




Dep

th (m

)



40

35

30

25

20

15

10

5

0

Dep

th (m

)


No seepage

Considering seepage






0 30 60 90 120 150 180 210

40

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)





0 30 60 90 120 150 180 210

40

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)







350 300 250 200 150 100 50 0

40

35

30

25

20

15

10

5

0

Soil

dept

hz i

(m)




350 300 250 200 150 100 50 0

40

35

30

25

20

15

10

5

0

Soil

dept

hz i

(m)








Groundwater

36800sand8 -1


silty clay6 -2




1



9000sandy silt2 -3

2 -1 sandy silt5000



Excavation surface

Ground surface

16800



6 Conclusions


119904



level





35

30

25

20

15

10

5

0 0 500 1000 1500 2000 2500 3000 3500


m (kNmiddotmminus4)

Soil

dept

hz

(m)



0 10 20 30 40 50 60



35

30

25

20

15

10

5

0


Dep

th (m

)



0 30 60 90 120 150

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)








119894of


(1205741015840

119904minus 120574119908








increase


Acknowledgment



References





























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










times ln(1199031

119889 minus 119911)

2

+(1 minus 2120583) (119889 minus 119911)

4

times ln(1199032

119889 + 119911)

2

100381610038161003816100381610038161003816100381610038161003816

119889=1198892

119889=1198891

int 120590119909d119909 =

119902

2120587 (1 minus 120583)

times 119911119889 (119889 + 119911)

1199032

2


119909

+ (5119909 minus 3120583119909 +21199112

119909) arctan 119889 + 119911

119909

+(3 minus 2120583) (119889 minus 119911)

4ln(

1199031

119889 minus 119911)

2

+119889 (5 minus 6120583) + 119911 (7 minus 10120583)

4

times ln(1199032

119889 + 119911)

2

100381610038161003816100381610038161003816100381610038161003816

119889=1198892

119889=1198891

(5)


119906119909=

(119906119909119911=119889

1

+ 2119906119909119911=(119889

1+1198892)2

+ 119906119909119911=119889

2

)

4 (6)


119906119909= int

1198972

0

120576119909119889119909 minus int

119897

1198972

120576119909119889119909 (7)



119870119894=

119902

119906119909

=119864119904

1 minus 1205832

119902

2[(int

1198972

0

120590119909d119909 minus int

119897

1198972

120590119909d119909) minus

120583

1 minus 120583(int

1198972

0

120590119911d119909 minus int

119897

1198972

120590119911d119909)]

minus1

(8)



119906119909= int

10ℎ

0

120576119909119889119909 (9)



119870119900=

119902

119906119909

=119864119904

1 minus 1205832

119902

2(int

10ℎ

0

120590119909d119909 minus

120583

1 minus 120583int

10ℎ

0

120590119911d119909)minus1

(10)

We define

120572119894=

1

1 minus 1205832

119902

2[(int

1198972

0

120590119909d119909 minus int

119897

1198972

120590119909d119909) minus

120583

1 minus 120583(int

1198972

0

120590119911d119909 minus int

119897

1198972

120590119911d119909)]

minus1

120572119900=

1

1 minus 1205832

119902

2(int

10ℎ

0

120590119909d119909 minus

120583

1 minus 120583int

10ℎ

0

120590119911d119909)minus1

(11)


01 02 03 04

40

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)

L = 100mL = 200m

Influence scope L

00120572o (mminus1)



Then

119870119894= 120572119894119864119904

119870119900= 120572119900119864119904

(12)


119904denotes the















asymp

11120572119894120583=03

and 120572119894120583=05

asymp 09120572119894120583=03


120583 = 01120583 = 02

120583 = 03

120583 = 04120583 = 05

40

35

30

25

20

15

10

5

000 02 04 06 08 10 12

120572i (mminus1)

Soil

dept

hz i

(m)



0 1000 2000 3000 4000 5000

40

35

30

25

20

15

10

5

0

l = 10ml = 20m

l = 40ml = 100m


Soil

dept

hz i

(m)




119864119904= 120582 sdot 120590

1015840

= 120582 sdot 1205741015840

sdot 119911 (13)






119894





modulus

83





119870 = 120572 sdot 119864119904= 120572 sdot 120582 sdot 120574

1015840

sdot 119911 = 119898 sdot 119911 (14)









119864119902= 119864119894sdot (1 minus

1205901minus 1205903

(1205901minus 1205903)ult

) = 119864119894sdot (1 minus (120590

1minus 1205903) sdot 119887) (15)


ult


40

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)

L = 100mL = 200m

Influence scope L



1205901 minus 1205903 =120576

120576

a + b120576

EqEi = 1a

(1205901 minus 1205903)



119870 =119902

119906119909

=119902

(int 120576119909d119909)

120576119909=

1 minus 1205832

119864119902

(120590119909minus

120583

1 minus 120583120590119911)

=1 minus 1205832

119864119894sdot [1 minus (120590

119909minus 120590119911) sdot 119887]

(120590119909minus

120583

1 minus 120583120590119911)

(16)








119898119900kNmminus4 100 89

Deep inside 4230 3170 2598 1879 186 150

0 200 400 600 800 10000

1

2

3

4

5


b(M

Paminus1) b = 46 minus 00043120590998400z






119887 = 46 minus 000431205901015840

119911 (17)






119870119894= 120572119894sdot 119864119904= 120572119894sdot 120582 sdot 120590

1015840

119911= 120572119894sdot 120582 sdot (120574

1015840

119904minus 120574119908sdot 119894119894) sdot 119911119894 (18)


02 03 04 05 06 07 08


q = 100 kPaq = 150kPaq = 200 kPa

Load level q40

35

30

25

20

15

10

5

0

120572i (mminus1)

Soil

dept

hz i

(m)


3MPaminus1)

01 02 03 04 05 06 07 08

b = 0

b = 2

b = 4b = 5

b = 6

b = 8

b = 10

b (MPaminus1)40

35

30

25

20

15

10

5

0

120572i (mminus1)

Soil

dept

hz i

(m)


119894(119902 = 100 kPa)




15 m

5 m

l = 10m and 40m L = 100mL = 100m


empty1m

(a) Plane graph



Excavationsurface

l = 10m and 40m L = 100mL = 100m



h=20

m20

m

4 m

(b) Sectional graph



119870119900= 120572119900sdot 119864119904= 120572119900sdot 120582 sdot 120590

1015840

119911= 120572119900sdot 120582 sdot (120574

1015840

119904+ 120574119908sdot 119894119900) sdot 119911119900

(19)




119875119900119904

= (1205741015840

119904+ 120574119908119894119900) sdot 1198700sdot 119911119900

earth pressure

119875119900119908

= 120574119908(1 minus 119894119900) sdot 119911119900

water pressure

119875119900= (1205741015840





Soil spring inside

pit

Active pressure



119875119894119904

= (1205741015840

119904minus 120574119908119894119894) sdot 1198700sdot 119911119894

earth pressure

119875119894119908

= 120574119908(1 + 119894119894) sdot 119911119894

water pressure

119875119894= (1205741015840





Soil spring outside





Groundwater outside


inside

5lowast4=20

m20

m

5sim20m 100 m

10 m


4 Analysis Example





119904= 190 kNm3



1015840

= 5 kPa and 1205931015840

= 30∘ lateral














Parameters





4000



seepage


seepage 119870119900= 05


119894


path


model




119888kPa 120593∘ 119870119881

119870119867






pressure 119875119900119908



5 Case Study






=




119894119900for the





140 120 100 80 60 40 20 0

40

35

30

25

20

15

10

5

0

No seepage

Considering seepage




Dep

th (m

)



40

35

30

25

20

15

10

5

0

Dep

th (m

)


No seepage

Considering seepage






140 120 100 80 60 40 20 0

40

35

30

25

20

15

10

5

0

No seepage

Considering seepage




Dep

th (m

)



40

35

30

25

20

15

10

5

0

Dep

th (m

)


No seepage

Considering seepage






0 30 60 90 120 150 180 210

40

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)





0 30 60 90 120 150 180 210

40

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)







350 300 250 200 150 100 50 0

40

35

30

25

20

15

10

5

0

Soil

dept

hz i

(m)




350 300 250 200 150 100 50 0

40

35

30

25

20

15

10

5

0

Soil

dept

hz i

(m)








Groundwater

36800sand8 -1


silty clay6 -2




1



9000sandy silt2 -3

2 -1 sandy silt5000



Excavation surface

Ground surface

16800



6 Conclusions


119904



level





35

30

25

20

15

10

5

0 0 500 1000 1500 2000 2500 3000 3500


m (kNmiddotmminus4)

Soil

dept

hz

(m)



0 10 20 30 40 50 60



35

30

25

20

15

10

5

0


Dep

th (m

)



0 30 60 90 120 150

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)








119894of


(1205741015840

119904minus 120574119908








increase


Acknowledgment



References





























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










01 02 03 04

40

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)

L = 100mL = 200m

Influence scope L

00120572o (mminus1)



Then

119870119894= 120572119894119864119904

119870119900= 120572119900119864119904

(12)


119904denotes the















asymp

11120572119894120583=03

and 120572119894120583=05

asymp 09120572119894120583=03


120583 = 01120583 = 02

120583 = 03

120583 = 04120583 = 05

40

35

30

25

20

15

10

5

000 02 04 06 08 10 12

120572i (mminus1)

Soil

dept

hz i

(m)



0 1000 2000 3000 4000 5000

40

35

30

25

20

15

10

5

0

l = 10ml = 20m

l = 40ml = 100m


Soil

dept

hz i

(m)




119864119904= 120582 sdot 120590

1015840

= 120582 sdot 1205741015840

sdot 119911 (13)






119894





modulus

83





119870 = 120572 sdot 119864119904= 120572 sdot 120582 sdot 120574

1015840

sdot 119911 = 119898 sdot 119911 (14)









119864119902= 119864119894sdot (1 minus

1205901minus 1205903

(1205901minus 1205903)ult

) = 119864119894sdot (1 minus (120590

1minus 1205903) sdot 119887) (15)


ult


40

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)

L = 100mL = 200m

Influence scope L



1205901 minus 1205903 =120576

120576

a + b120576

EqEi = 1a

(1205901 minus 1205903)



119870 =119902

119906119909

=119902

(int 120576119909d119909)

120576119909=

1 minus 1205832

119864119902

(120590119909minus

120583

1 minus 120583120590119911)

=1 minus 1205832

119864119894sdot [1 minus (120590

119909minus 120590119911) sdot 119887]

(120590119909minus

120583

1 minus 120583120590119911)

(16)








119898119900kNmminus4 100 89

Deep inside 4230 3170 2598 1879 186 150

0 200 400 600 800 10000

1

2

3

4

5


b(M

Paminus1) b = 46 minus 00043120590998400z






119887 = 46 minus 000431205901015840

119911 (17)






119870119894= 120572119894sdot 119864119904= 120572119894sdot 120582 sdot 120590

1015840

119911= 120572119894sdot 120582 sdot (120574

1015840

119904minus 120574119908sdot 119894119894) sdot 119911119894 (18)


02 03 04 05 06 07 08


q = 100 kPaq = 150kPaq = 200 kPa

Load level q40

35

30

25

20

15

10

5

0

120572i (mminus1)

Soil

dept

hz i

(m)


3MPaminus1)

01 02 03 04 05 06 07 08

b = 0

b = 2

b = 4b = 5

b = 6

b = 8

b = 10

b (MPaminus1)40

35

30

25

20

15

10

5

0

120572i (mminus1)

Soil

dept

hz i

(m)


119894(119902 = 100 kPa)




15 m

5 m

l = 10m and 40m L = 100mL = 100m


empty1m

(a) Plane graph



Excavationsurface

l = 10m and 40m L = 100mL = 100m



h=20

m20

m

4 m

(b) Sectional graph



119870119900= 120572119900sdot 119864119904= 120572119900sdot 120582 sdot 120590

1015840

119911= 120572119900sdot 120582 sdot (120574

1015840

119904+ 120574119908sdot 119894119900) sdot 119911119900

(19)




119875119900119904

= (1205741015840

119904+ 120574119908119894119900) sdot 1198700sdot 119911119900

earth pressure

119875119900119908

= 120574119908(1 minus 119894119900) sdot 119911119900

water pressure

119875119900= (1205741015840





Soil spring inside

pit

Active pressure



119875119894119904

= (1205741015840

119904minus 120574119908119894119894) sdot 1198700sdot 119911119894

earth pressure

119875119894119908

= 120574119908(1 + 119894119894) sdot 119911119894

water pressure

119875119894= (1205741015840





Soil spring outside





Groundwater outside


inside

5lowast4=20

m20

m

5sim20m 100 m

10 m


4 Analysis Example





119904= 190 kNm3



1015840

= 5 kPa and 1205931015840

= 30∘ lateral














Parameters





4000



seepage


seepage 119870119900= 05


119894


path


model




119888kPa 120593∘ 119870119881

119870119867






pressure 119875119900119908



5 Case Study






=




119894119900for the





140 120 100 80 60 40 20 0

40

35

30

25

20

15

10

5

0

No seepage

Considering seepage




Dep

th (m

)



40

35

30

25

20

15

10

5

0

Dep

th (m

)


No seepage

Considering seepage






140 120 100 80 60 40 20 0

40

35

30

25

20

15

10

5

0

No seepage

Considering seepage




Dep

th (m

)



40

35

30

25

20

15

10

5

0

Dep

th (m

)


No seepage

Considering seepage






0 30 60 90 120 150 180 210

40

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)





0 30 60 90 120 150 180 210

40

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)







350 300 250 200 150 100 50 0

40

35

30

25

20

15

10

5

0

Soil

dept

hz i

(m)




350 300 250 200 150 100 50 0

40

35

30

25

20

15

10

5

0

Soil

dept

hz i

(m)








Groundwater

36800sand8 -1


silty clay6 -2




1



9000sandy silt2 -3

2 -1 sandy silt5000



Excavation surface

Ground surface

16800



6 Conclusions


119904



level





35

30

25

20

15

10

5

0 0 500 1000 1500 2000 2500 3000 3500


m (kNmiddotmminus4)

Soil

dept

hz

(m)



0 10 20 30 40 50 60



35

30

25

20

15

10

5

0


Dep

th (m

)



0 30 60 90 120 150

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)








119894of


(1205741015840

119904minus 120574119908








increase


Acknowledgment



References





























Volume 2014




Journal of











Journal of


Function Spaces






Algebra












119894





modulus

83





119870 = 120572 sdot 119864119904= 120572 sdot 120582 sdot 120574

1015840

sdot 119911 = 119898 sdot 119911 (14)









119864119902= 119864119894sdot (1 minus

1205901minus 1205903

(1205901minus 1205903)ult

) = 119864119894sdot (1 minus (120590

1minus 1205903) sdot 119887) (15)


ult


40

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)

L = 100mL = 200m

Influence scope L



1205901 minus 1205903 =120576

120576

a + b120576

EqEi = 1a

(1205901 minus 1205903)



119870 =119902

119906119909

=119902

(int 120576119909d119909)

120576119909=

1 minus 1205832

119864119902

(120590119909minus

120583

1 minus 120583120590119911)

=1 minus 1205832

119864119894sdot [1 minus (120590

119909minus 120590119911) sdot 119887]

(120590119909minus

120583

1 minus 120583120590119911)

(16)








119898119900kNmminus4 100 89

Deep inside 4230 3170 2598 1879 186 150

0 200 400 600 800 10000

1

2

3

4

5


b(M

Paminus1) b = 46 minus 00043120590998400z






119887 = 46 minus 000431205901015840

119911 (17)






119870119894= 120572119894sdot 119864119904= 120572119894sdot 120582 sdot 120590

1015840

119911= 120572119894sdot 120582 sdot (120574

1015840

119904minus 120574119908sdot 119894119894) sdot 119911119894 (18)


02 03 04 05 06 07 08


q = 100 kPaq = 150kPaq = 200 kPa

Load level q40

35

30

25

20

15

10

5

0

120572i (mminus1)

Soil

dept

hz i

(m)


3MPaminus1)

01 02 03 04 05 06 07 08

b = 0

b = 2

b = 4b = 5

b = 6

b = 8

b = 10

b (MPaminus1)40

35

30

25

20

15

10

5

0

120572i (mminus1)

Soil

dept

hz i

(m)


119894(119902 = 100 kPa)




15 m

5 m

l = 10m and 40m L = 100mL = 100m


empty1m

(a) Plane graph



Excavationsurface

l = 10m and 40m L = 100mL = 100m



h=20

m20

m

4 m

(b) Sectional graph



119870119900= 120572119900sdot 119864119904= 120572119900sdot 120582 sdot 120590

1015840

119911= 120572119900sdot 120582 sdot (120574

1015840

119904+ 120574119908sdot 119894119900) sdot 119911119900

(19)




119875119900119904

= (1205741015840

119904+ 120574119908119894119900) sdot 1198700sdot 119911119900

earth pressure

119875119900119908

= 120574119908(1 minus 119894119900) sdot 119911119900

water pressure

119875119900= (1205741015840





Soil spring inside

pit

Active pressure



119875119894119904

= (1205741015840

119904minus 120574119908119894119894) sdot 1198700sdot 119911119894

earth pressure

119875119894119908

= 120574119908(1 + 119894119894) sdot 119911119894

water pressure

119875119894= (1205741015840





Soil spring outside





Groundwater outside


inside

5lowast4=20

m20

m

5sim20m 100 m

10 m


4 Analysis Example





119904= 190 kNm3



1015840

= 5 kPa and 1205931015840

= 30∘ lateral














Parameters





4000



seepage


seepage 119870119900= 05


119894


path


model




119888kPa 120593∘ 119870119881

119870119867






pressure 119875119900119908



5 Case Study






=




119894119900for the





140 120 100 80 60 40 20 0

40

35

30

25

20

15

10

5

0

No seepage

Considering seepage




Dep

th (m

)



40

35

30

25

20

15

10

5

0

Dep

th (m

)


No seepage

Considering seepage






140 120 100 80 60 40 20 0

40

35

30

25

20

15

10

5

0

No seepage

Considering seepage




Dep

th (m

)



40

35

30

25

20

15

10

5

0

Dep

th (m

)


No seepage

Considering seepage






0 30 60 90 120 150 180 210

40

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)





0 30 60 90 120 150 180 210

40

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)







350 300 250 200 150 100 50 0

40

35

30

25

20

15

10

5

0

Soil

dept

hz i

(m)




350 300 250 200 150 100 50 0

40

35

30

25

20

15

10

5

0

Soil

dept

hz i

(m)








Groundwater

36800sand8 -1


silty clay6 -2




1



9000sandy silt2 -3

2 -1 sandy silt5000



Excavation surface

Ground surface

16800



6 Conclusions


119904



level





35

30

25

20

15

10

5

0 0 500 1000 1500 2000 2500 3000 3500


m (kNmiddotmminus4)

Soil

dept

hz

(m)



0 10 20 30 40 50 60



35

30

25

20

15

10

5

0


Dep

th (m

)



0 30 60 90 120 150

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)








119894of


(1205741015840

119904minus 120574119908








increase


Acknowledgment



References





























Volume 2014




Journal of











Journal of


Function Spaces






Algebra













119898119900kNmminus4 100 89

Deep inside 4230 3170 2598 1879 186 150

0 200 400 600 800 10000

1

2

3

4

5


b(M

Paminus1) b = 46 minus 00043120590998400z






119887 = 46 minus 000431205901015840

119911 (17)






119870119894= 120572119894sdot 119864119904= 120572119894sdot 120582 sdot 120590

1015840

119911= 120572119894sdot 120582 sdot (120574

1015840

119904minus 120574119908sdot 119894119894) sdot 119911119894 (18)


02 03 04 05 06 07 08


q = 100 kPaq = 150kPaq = 200 kPa

Load level q40

35

30

25

20

15

10

5

0

120572i (mminus1)

Soil

dept

hz i

(m)


3MPaminus1)

01 02 03 04 05 06 07 08

b = 0

b = 2

b = 4b = 5

b = 6

b = 8

b = 10

b (MPaminus1)40

35

30

25

20

15

10

5

0

120572i (mminus1)

Soil

dept

hz i

(m)


119894(119902 = 100 kPa)




15 m

5 m

l = 10m and 40m L = 100mL = 100m


empty1m

(a) Plane graph



Excavationsurface

l = 10m and 40m L = 100mL = 100m



h=20

m20

m

4 m

(b) Sectional graph



119870119900= 120572119900sdot 119864119904= 120572119900sdot 120582 sdot 120590

1015840

119911= 120572119900sdot 120582 sdot (120574

1015840

119904+ 120574119908sdot 119894119900) sdot 119911119900

(19)




119875119900119904

= (1205741015840

119904+ 120574119908119894119900) sdot 1198700sdot 119911119900

earth pressure

119875119900119908

= 120574119908(1 minus 119894119900) sdot 119911119900

water pressure

119875119900= (1205741015840





Soil spring inside

pit

Active pressure



119875119894119904

= (1205741015840

119904minus 120574119908119894119894) sdot 1198700sdot 119911119894

earth pressure

119875119894119908

= 120574119908(1 + 119894119894) sdot 119911119894

water pressure

119875119894= (1205741015840





Soil spring outside





Groundwater outside


inside

5lowast4=20

m20

m

5sim20m 100 m

10 m


4 Analysis Example





119904= 190 kNm3



1015840

= 5 kPa and 1205931015840

= 30∘ lateral














Parameters





4000



seepage


seepage 119870119900= 05


119894


path


model




119888kPa 120593∘ 119870119881

119870119867






pressure 119875119900119908



5 Case Study






=




119894119900for the





140 120 100 80 60 40 20 0

40

35

30

25

20

15

10

5

0

No seepage

Considering seepage




Dep

th (m

)



40

35

30

25

20

15

10

5

0

Dep

th (m

)


No seepage

Considering seepage






140 120 100 80 60 40 20 0

40

35

30

25

20

15

10

5

0

No seepage

Considering seepage




Dep

th (m

)



40

35

30

25

20

15

10

5

0

Dep

th (m

)


No seepage

Considering seepage






0 30 60 90 120 150 180 210

40

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)





0 30 60 90 120 150 180 210

40

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)







350 300 250 200 150 100 50 0

40

35

30

25

20

15

10

5

0

Soil

dept

hz i

(m)




350 300 250 200 150 100 50 0

40

35

30

25

20

15

10

5

0

Soil

dept

hz i

(m)








Groundwater

36800sand8 -1


silty clay6 -2




1



9000sandy silt2 -3

2 -1 sandy silt5000



Excavation surface

Ground surface

16800



6 Conclusions


119904



level





35

30

25

20

15

10

5

0 0 500 1000 1500 2000 2500 3000 3500


m (kNmiddotmminus4)

Soil

dept

hz

(m)



0 10 20 30 40 50 60



35

30

25

20

15

10

5

0


Dep

th (m

)



0 30 60 90 120 150

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)








119894of


(1205741015840

119904minus 120574119908








increase


Acknowledgment



References





























Volume 2014




Journal of











Journal of


Function Spaces






Algebra












15 m

5 m

l = 10m and 40m L = 100mL = 100m


empty1m

(a) Plane graph



Excavationsurface

l = 10m and 40m L = 100mL = 100m



h=20

m20

m

4 m

(b) Sectional graph



119870119900= 120572119900sdot 119864119904= 120572119900sdot 120582 sdot 120590

1015840

119911= 120572119900sdot 120582 sdot (120574

1015840

119904+ 120574119908sdot 119894119900) sdot 119911119900

(19)




119875119900119904

= (1205741015840

119904+ 120574119908119894119900) sdot 1198700sdot 119911119900

earth pressure

119875119900119908

= 120574119908(1 minus 119894119900) sdot 119911119900

water pressure

119875119900= (1205741015840





Soil spring inside

pit

Active pressure



119875119894119904

= (1205741015840

119904minus 120574119908119894119894) sdot 1198700sdot 119911119894

earth pressure

119875119894119908

= 120574119908(1 + 119894119894) sdot 119911119894

water pressure

119875119894= (1205741015840





Soil spring outside





Groundwater outside


inside

5lowast4=20

m20

m

5sim20m 100 m

10 m


4 Analysis Example





119904= 190 kNm3



1015840

= 5 kPa and 1205931015840

= 30∘ lateral














Parameters





4000



seepage


seepage 119870119900= 05


119894


path


model




119888kPa 120593∘ 119870119881

119870119867






pressure 119875119900119908



5 Case Study






=




119894119900for the





140 120 100 80 60 40 20 0

40

35

30

25

20

15

10

5

0

No seepage

Considering seepage




Dep

th (m

)



40

35

30

25

20

15

10

5

0

Dep

th (m

)


No seepage

Considering seepage






140 120 100 80 60 40 20 0

40

35

30

25

20

15

10

5

0

No seepage

Considering seepage




Dep

th (m

)



40

35

30

25

20

15

10

5

0

Dep

th (m

)


No seepage

Considering seepage






0 30 60 90 120 150 180 210

40

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)





0 30 60 90 120 150 180 210

40

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)







350 300 250 200 150 100 50 0

40

35

30

25

20

15

10

5

0

Soil

dept

hz i

(m)




350 300 250 200 150 100 50 0

40

35

30

25

20

15

10

5

0

Soil

dept

hz i

(m)








Groundwater

36800sand8 -1


silty clay6 -2




1



9000sandy silt2 -3

2 -1 sandy silt5000



Excavation surface

Ground surface

16800



6 Conclusions


119904



level





35

30

25

20

15

10

5

0 0 500 1000 1500 2000 2500 3000 3500


m (kNmiddotmminus4)

Soil

dept

hz

(m)



0 10 20 30 40 50 60



35

30

25

20

15

10

5

0


Dep

th (m

)



0 30 60 90 120 150

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)








119894of


(1205741015840

119904minus 120574119908








increase


Acknowledgment



References





























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











Soil spring outside





Groundwater outside


inside

5lowast4=20

m20

m

5sim20m 100 m

10 m


4 Analysis Example





119904= 190 kNm3



1015840

= 5 kPa and 1205931015840

= 30∘ lateral














Parameters





4000



seepage


seepage 119870119900= 05


119894


path


model




119888kPa 120593∘ 119870119881

119870119867






pressure 119875119900119908



5 Case Study






=




119894119900for the





140 120 100 80 60 40 20 0

40

35

30

25

20

15

10

5

0

No seepage

Considering seepage




Dep

th (m

)



40

35

30

25

20

15

10

5

0

Dep

th (m

)


No seepage

Considering seepage






140 120 100 80 60 40 20 0

40

35

30

25

20

15

10

5

0

No seepage

Considering seepage




Dep

th (m

)



40

35

30

25

20

15

10

5

0

Dep

th (m

)


No seepage

Considering seepage






0 30 60 90 120 150 180 210

40

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)





0 30 60 90 120 150 180 210

40

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)







350 300 250 200 150 100 50 0

40

35

30

25

20

15

10

5

0

Soil

dept

hz i

(m)




350 300 250 200 150 100 50 0

40

35

30

25

20

15

10

5

0

Soil

dept

hz i

(m)








Groundwater

36800sand8 -1


silty clay6 -2




1



9000sandy silt2 -3

2 -1 sandy silt5000



Excavation surface

Ground surface

16800



6 Conclusions


119904



level





35

30

25

20

15

10

5

0 0 500 1000 1500 2000 2500 3000 3500


m (kNmiddotmminus4)

Soil

dept

hz

(m)



0 10 20 30 40 50 60



35

30

25

20

15

10

5

0


Dep

th (m

)



0 30 60 90 120 150

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)








119894of


(1205741015840

119904minus 120574119908








increase


Acknowledgment



References





























Volume 2014




Journal of











Journal of


Function Spaces






Algebra












Parameters





4000



seepage


seepage 119870119900= 05


119894


path


model




119888kPa 120593∘ 119870119881

119870119867






pressure 119875119900119908



5 Case Study






=




119894119900for the





140 120 100 80 60 40 20 0

40

35

30

25

20

15

10

5

0

No seepage

Considering seepage




Dep

th (m

)



40

35

30

25

20

15

10

5

0

Dep

th (m

)


No seepage

Considering seepage






140 120 100 80 60 40 20 0

40

35

30

25

20

15

10

5

0

No seepage

Considering seepage




Dep

th (m

)



40

35

30

25

20

15

10

5

0

Dep

th (m

)


No seepage

Considering seepage






0 30 60 90 120 150 180 210

40

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)





0 30 60 90 120 150 180 210

40

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)







350 300 250 200 150 100 50 0

40

35

30

25

20

15

10

5

0

Soil

dept

hz i

(m)




350 300 250 200 150 100 50 0

40

35

30

25

20

15

10

5

0

Soil

dept

hz i

(m)








Groundwater

36800sand8 -1


silty clay6 -2




1



9000sandy silt2 -3

2 -1 sandy silt5000



Excavation surface

Ground surface

16800



6 Conclusions


119904



level





35

30

25

20

15

10

5

0 0 500 1000 1500 2000 2500 3000 3500


m (kNmiddotmminus4)

Soil

dept

hz

(m)



0 10 20 30 40 50 60



35

30

25

20

15

10

5

0


Dep

th (m

)



0 30 60 90 120 150

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)








119894of


(1205741015840

119904minus 120574119908








increase


Acknowledgment



References





























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










140 120 100 80 60 40 20 0

40

35

30

25

20

15

10

5

0

No seepage

Considering seepage




Dep

th (m

)



40

35

30

25

20

15

10

5

0

Dep

th (m

)


No seepage

Considering seepage






140 120 100 80 60 40 20 0

40

35

30

25

20

15

10

5

0

No seepage

Considering seepage




Dep

th (m

)



40

35

30

25

20

15

10

5

0

Dep

th (m

)


No seepage

Considering seepage






0 30 60 90 120 150 180 210

40

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)





0 30 60 90 120 150 180 210

40

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)







350 300 250 200 150 100 50 0

40

35

30

25

20

15

10

5

0

Soil

dept

hz i

(m)




350 300 250 200 150 100 50 0

40

35

30

25

20

15

10

5

0

Soil

dept

hz i

(m)








Groundwater

36800sand8 -1


silty clay6 -2




1



9000sandy silt2 -3

2 -1 sandy silt5000



Excavation surface

Ground surface

16800



6 Conclusions


119904



level





35

30

25

20

15

10

5

0 0 500 1000 1500 2000 2500 3000 3500


m (kNmiddotmminus4)

Soil

dept

hz

(m)



0 10 20 30 40 50 60



35

30

25

20

15

10

5

0


Dep

th (m

)



0 30 60 90 120 150

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)








119894of


(1205741015840

119904minus 120574119908








increase


Acknowledgment



References





























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










0 30 60 90 120 150 180 210

40

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)





0 30 60 90 120 150 180 210

40

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)







350 300 250 200 150 100 50 0

40

35

30

25

20

15

10

5

0

Soil

dept

hz i

(m)




350 300 250 200 150 100 50 0

40

35

30

25

20

15

10

5

0

Soil

dept

hz i

(m)








Groundwater

36800sand8 -1


silty clay6 -2




1



9000sandy silt2 -3

2 -1 sandy silt5000



Excavation surface

Ground surface

16800



6 Conclusions


119904



level





35

30

25

20

15

10

5

0 0 500 1000 1500 2000 2500 3000 3500


m (kNmiddotmminus4)

Soil

dept

hz

(m)



0 10 20 30 40 50 60



35

30

25

20

15

10

5

0


Dep

th (m

)



0 30 60 90 120 150

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)








119894of


(1205741015840

119904minus 120574119908








increase


Acknowledgment



References





























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











Groundwater

36800sand8 -1


silty clay6 -2




1



9000sandy silt2 -3

2 -1 sandy silt5000



Excavation surface

Ground surface

16800



6 Conclusions


119904



level





35

30

25

20

15

10

5

0 0 500 1000 1500 2000 2500 3000 3500


m (kNmiddotmminus4)

Soil

dept

hz

(m)



0 10 20 30 40 50 60



35

30

25

20

15

10

5

0


Dep

th (m

)



0 30 60 90 120 150

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)








119894of


(1205741015840

119904minus 120574119908








increase


Acknowledgment



References





























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










35

30

25

20

15

10

5

0 0 500 1000 1500 2000 2500 3000 3500


m (kNmiddotmminus4)

Soil

dept

hz

(m)



0 10 20 30 40 50 60



35

30

25

20

15

10

5

0


Dep

th (m

)



0 30 60 90 120 150

35

30

25

20

15

10

5

0

Soil

dept

hz o

(m)








119894of


(1205741015840

119904minus 120574119908








increase


Acknowledgment



References





























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










References





























Volume 2014




Journal of











Journal of


Function Spaces






Algebra
















Volume 2014




Journal of











Journal of


Function Spaces






Algebra









research article retaining structure force-deformation

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