research article analytical solution for stress and ...z-plane (a) 0 0 0 0-plane =1 = p (b) f : (a)...

Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2013 Article ID 937421 11 pageshttpdxdoiorg1011552013937421

Research ArticleAnalytical Solution for Stress and Displacement afterX-Section Cast-in-Place Pile Installation

Hang Zhou Han Long Liu Gangqiang Kong and Zhaohu Cao

College of Civil and Transportation Engineering Hohai University Nanjing Jiangsu 210098 China

Correspondence should be addressed to Gangqiang Kong gqkong1163com

Received 13 October 2013 Accepted 1 November 2013

Academic Editor Yuji Liu

Copyright copy 2013 Hang Zhou et al This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

X-section cast-in-place (referred to as XCC) pile which is one of new pile types developed by Hohai University is widely used forpile foundation and pile-supported embankment over soft ground in China However little research has been carried out on thisnew type pile especially the surrounding soil disturbance under XCC pile installation This paper presents an analytical solutionfor estimating the horizontal stress and displacement of surrounding soil of XCC pile after XCC pile installation The reliabilityand accuracy of the present solution are verified by comparing them with the field test results Then parametric studies suchas outsourcing diameter (119886) open arc distance (119887) open arc angle (120579) the undrained strength (119878

119906) the limit pile cavity pressure

(119875lim) and the radius of the plastic zone (119877119901) are discussed for the practice engineering designThe results show that the stress anddisplacement distributions of surrounding soil calculated by this paper are in agreement with field test results

1 Introduction

Driven cast-in-place pile which belongs to displacement piletype is widely used in China [1] Surrounding environmentwill be influenced by displacement pile installation obviouslyIf handed improperly it will cause the uplift or subsidenceof the ground and even engineering accidents Therefore itis essential to predict the horizontal stress and displacementinduced by the pile installation Various approaches havebeen used to study the horizontal stress and displacementincluding cavity expansion method (CEM) [2ndash7] strain pathmethod (SPM) [8] and modified SPM (SSPM) [9 10] CEMwas proposed by Bishop et al firstly and was used to solvethe metal indentation problems Subsequently CEM wasapplied to solve the geotechnical problems such as the pilepenetration and the bearing capacity of deep foundation(Vesic (1972)) [2] In this method a cylindrical (or spherical)cavity of zero radius was assumed in soil located near thetip of pile The pressure around the tip of a pile to causepenetration is the limit pressure required to expand the cavityfrom an initial radius to the radius of the pile The limitpressure for the expansion of the cavity is a function of theshear strength and compressibility of the soil Based on thefluidmechanics the strain pathmethod (SPM) was proposed

by Baligh (1985) [8] andmodified by Sagaseta et al (1997) [9]The process of the pile penetration is assumed as a steady flowof soil around the pile rather than an expansion of a cavity insoil Although the CEM and SPM are simple and easy to usethey can only solve the axisymmetric or spherical symmetricproblem However for the special shaped pile installationsuch as XCC pile rectangular cross-section pile (barrette)they are unavailable

As a new immersed tube pile XCC pile is developedby Hohai University China [11] The pile is formed byinstallation of the X cross-section steel mold which is pro-tected by valve pile shoe or precast pile tip The installationprocedure includes immersing the tube pouring concretevibratory extubation and curing the concrete XCC pile isone of new displacement piles and widely used in prac-tice engineering However the theoretical research are farbehind the application especially the horizontal stress anddisplacement induced by pile installation In this paper ananalytical solution is provided to study the horizontal stressand displacement distribution of surrounding soil Thenbased on the Fourth Yangtze River Bridgersquos north-line softsoil treatment engineering in Nanjing the analytical solutionwas comparedwith the field test results Finally the geometricparameters of XCC pile (outsourcing diameter a open arc

2 Mathematical Problems in Engineering

distance b and open arc angle 120579) the undrained strength 119878119906

and the pile hole pressure 119875 were discussed

2 Mathematical Model

21 Definition of the Problem and Basic Assumption Figure 1shows that the elastoplastic soil which is described by theTresca model is under initial stress 120590

0before the XCC

pile installation Then the XCC pile installation progress issimplified as the expansion of a cavity from zero to the X-shaped cavity of the XCC pile cross-section The soil aroundthe pile enter into the passive limit balance state after theXCC pile installation Thus the X-shaped cavity internalpressure after XCC pile installation can be assumed as 119875lim =1205900tan2(45∘ + 1205932) + 2119888 tan(45∘ + 1205932) where 119888 and 120593 are

the soil cohesion and the internal friction angle respectivelyAdditionally the plain strain condition is assumed in themodel The Cartesian coordinates system is selected for theanalysis The origin of the coordinates is located at the centerof the cavity For the Cartesian coordinates system the 119909-axis and 119910-axis are in the horizontal and vertical directionrespectively The stress and strain are taken as positive in thepositive direction of the coordinates system

22 Basic Governing Equations According to the elasticity[12] the stress around the X-shaped cavity should obey thefourth-order partial differential equation as follows

nabla4

119880 = 0 (1)

where nabla is Laplace operator and U is stress function in theplastic zone

The three stress components in Cartesian coordinatessystem 120590

119909 120590119910and 120591

119909119910 around the X-shaped cavity can be

determined by (1)

120590119909=1205972119880

1205971199102 120590

119910=1205972119880

1205971199092 120591

119909119910= minus

1205972119880

120597119909120597119910 (2)

According to the complex variable elasticity [13ndash17] thestress around the X-shaped cavity can be expressed with twostress functions 120593

1(119911) and 120595

1(119911) as follows

120590119909+ 120590119910= 4Re [1205931015840

1(119911)] (3a)

120590119910minus 120590119909+ 2119894120591119909119910= 2 [119911120593

10158401015840

1(119911) + 120595

1015840

1(119911)] (3b)

119864

1 + 120583(119906 + 119894V) = (3 minus 4120583) 120593

1(119911) minus 119911120593

1015840

1(119911) minus 120595

1(119911) (3c)

where 12059310158401(119911) and 12059310158401015840

1(119911) are the first and second derivative of

the function 1205931(119911) respectively 1205951015840

1(119911) is the first derivative

of the function 1205951(119911) 1205931015840

1(119911) and 120595

1(119911) are the conjugate

complex functions of the 12059310158401(119911) and 120595

1(119911) 119906 and V are the

displacement component acting in 119909-axis and 119910-axis direc-tions respectively 119864 is the Youngrsquos modulus of the soil 120583 isthe Poisson ratio of the soil

For obtaining the solution to calculate the stress anddisplacement distributions a conformal mapping function is

provided to map the outside of the X-shaped cavity in the 119911-plane onto the outside of the unit circle in the phase planenamely 120585-plane (120585 = 120577+119894120578 = 120588119890119894120579) in Figure 2The conformalmapping function can be expressed in a series as follows

119911 = 119908 (120585) = 1198880120585+

119899

sum119896=1

1198882119896minus1

1205851minus2119896

10038161003816100381610038161205851003816100381610038161003816 le 1 (4)

where n is integral number (in this paper 119899 = 7 is selectedfor analysis and it can give enough accuracy) The constantcoefficients 119888

0and 1198882119896minus1

(119896 = 1 2 119899) are real numbers andcan be obtained by the iterative technique [18] 119911 = 119909 + 119894119910 (119909and 119910 are the variables in the Cartesian coordinates system119894 = radicminus1) is complex variable 120585 is complex variable in thephase plane 120585 = 120588119890119894120579

Substituting the conformal mapping function (4) into(3a) (3b) and (3c) leads to the following equations

120590119909+ 120590119910= 4Re[

1205931015840 (120585)

1199081015840 (120585)] (5a)

120590119910minus 120590119909+ 2119894120591119909119910=

2

1199081015840 (120585)[119908 (120585)(

1205931015840 (120585)

1199081015840 (120585))

1015840

+ 1205951015840

(120585)] (5b)

119864

1 + 120583(119906 + 119894V) = 120594120593 (120585) minus

119908 (120585)

1199081015840 (120585)1205931015840

(120585) minus 120595 (120585) (5c)

where 120593(120585) = 1205931(119908(120585)) 120595(120585) = 120595

1(119908(120585))

The two stress functions 1205931(119911) and 120595

1(119911) are transformed

as 120593(120585) and 120595(120585) To solve the stress functions 120593(120585) and120595(120585)the stress boundary conditions should be considered FromFigure 2 it can be seen that the stress boundary condition canbe expressed as

[

[

120593 (120585) +119908 (120585)

1199081015840 (120585)1205931015840

(120585) + 1205951015840

(120585)]

]119904

= minus119901119908 (120585) (6)

where s is the X-shaped cavity boundary curve and P is thepressure at the X-shaped cavity

By the complex elasticity [10 11] the stress functions 120593(120585)and 120595(120585) can be written as follows

120593 (120585) =1

8120587 (1 minus 120583)(119865119909+ 119894 119865119910) ln 120585 + 119861119908 (120585) + 120593

0(120585) (7a)

120595 (120585) = minus3 minus 4120583

8120587 (1 minus 120583)(119865119909minus 119894 119865119910) ln 120585

+ (1198611015840

+ 1198941198621015840

)119908 (120585) + 1205950(120585)

(7b)

where 119865119909and 119865

119910are the composite surface force in 119909 and 119910

direction on the X-shaped cavity boundary respectively Onehas 119861 = (120590

1+ 1205902)4 (120590

1and 120590

2are the principal stress at

infinity) and 1198611015840 + 1198621015840 = minus(12)(1205901minus 1205902)119890minus2119894120572 (120572 is the angle

between the principal stress 1205901and ox-axis)

Mathematical Problems in Engineering 3

σ0

σ0

σ0

σ0 Plimit o x

y

(a)

Open arc angle

Outsourcing diameter

(b)

Figure 1 Mechanics model (a) passive limit balance state after the pile mold installation (b) XCC pile cross-section

σ0

σ0

σ0

σ0 Px

y

Soil

z-plane

(a)

σ0

σ0σ0

σ0

120585-plane

120588 = 1

120578

120588 = infin

120577P 120579

(b)

Figure 2 (a) z-plane containing X-shaped cavity subjected uniform pressure 119875 at the cavity and isotropic initial stress at infinity (b)transformed 120585-plane containing a unit circle subjected uniform pressure at the cavity and isotropic initial stress at infinity


From the mechanics model in Figure 2 the followingequations are established

119865119909= 119865119910= 0 (8a)

119861 = 1205900 (8b)

1198611015840

= 1198621015840

= 0 (8c)

Additionally the two stress functions 1205930(120585) and 120595

0(120585) can

be expressed as the series

1205930(120585) =

119899

sum119896=0

1205722119896minus1

1205852119896minus1

(9a)

1205950(120585) =

119899

sum119896=1

1205732119896minus1

1205852119896minus1

(9b)

where 1205722119896minus1

and 1205732119896minus1

are the coefficients of the complexfunctions 120593

0(120585) and 120595

0(120585) They can be determined by the

boundary conditionsThus (7a) and (7b) can be simplified by substituting (8a)

(8b) (8c) (9a) and (9b) into (7a) and (7b) as

120593 (120585) = 1205900119908 (120585) +

119899

sum119896=0

1205722119896minus1

1205852119896minus1

(10a)

120595 (120585) =

119899

sum119896=1

1205732119896minus1

1205852119896minus1

(10b)

Then (10a) and (10b) are substituted into the stressboundary condition (6) and (6) can be transformed asfollows

1205930(120590) +

119908 (120590)

1199081015840 (120590)1205931015840

0(120590) + 120595

0(120590) = minus (119875 + 120590

0) 119911 (11)

where 120590 = 119890119894120579Equations (10a) and (10b) are conjugated at both sides

1205930(120590) +

119908 (120590)

1199081015840 (120590)1205931015840

0(120590) + 120595

0(120590) = minus (119875 + 120590

0) 119911 (12)

Equations (10a) (10b) and (11) are multiplied by (12120587119894)(119889120590(120590 minus 120585)) and integrated along the cavity boundary 119904 atboth sides

1

2120587119894int1205930(120590) 119889120590 +

1

2120587119894int119908 (120590)

1199081015840 (120590)1205931015840

0(120590) 119889120590

+1

2120587119894int1205950(120590) 119889120590 =

1

2120587119894intminus (119875 + 120590

0) 119908 (120590) 119889120590

(13a)

1

2120587119894int1205930(120590) 119889120590 +

1

2120587119894int119908 (120590)

1199081015840 (120590)1205931015840

0(120590) 119889120590

+1

2120587119894int1205950(120590) 119889120590 =

1

2120587119894intminus (119875 + 120590

0) 119908 (120590) 119889120590

(13b)

According to the principle of the series expansion theterms of 119908(120590)1199081015840(120590) can be expressed as follows

119908 (120590)

1199081015840 (120590)= 1198872119896minus3

1205902119896minus3

+ 1198872119896minus5

1205902119896minus5

+ sdot sdot sdot + 11988711205901

+ 119887minus1120590minus1

+ 119874(1

1205903)

(14)

Equation (4) is substituted into (14) and can be written inmatrix form as follows

119860119861 = 119862 (15)

where

119860 =

[[[[[[[[[[[

[

minus 119888011988813119888351198885sdot sdot sdot sdot sdot sdot (2119899 minus 3) 119888

2119899minus3

minus1198880119888131198883sdot sdot sdot sdot sdot sdot (2119899 minus 5) 119888

2119899minus5


2119899minus7

d sdot sdot sdot sdot sdot sdot

d sdot sdot sdot

minus1198880

1198881

minus1198880

]]]]]]]]]]]

]

119861 = [119887minus1119887111988731198875sdot sdot sdot 1198872119899minus3

]119879

119862 = [1198881119888311988851198887sdot sdot sdot 1198882119899minus1

]119879

(16)

After the coefficients 119887minus1

and 1198872119896minus3

(119896 = 1 2 3 119899)

are determined by solving (15) substituting the expressionof 119908(120590)1199081015840(120590) into (13a) the equation for calculating thecoefficients of the stress functions can be obtained as follows

[119864119899119872

119872 119864119899

] 120572 = 119873 (17)

where

119872 =

[[[[[[

[

1198871

31198873sdot sdot sdot (2119899 minus 3) 119887

2119899minus30

1198873

31198875sdot sdot sdot 0 0

sdot sdot sdot sdot sdot sdot 0 0

1198872119899minus3

0 sdot sdot sdot 0 0

0 0 sdot sdot sdot 0 0

]]]]]]

]

(18a)


12057211205723sdot sdot sdot 1205722119899minus1

]119879

(18b)

119873 = minus (119875 + 1205900) [11988811198883sdot sdot sdot 1198882119899minus1


]119879

(18c)

119872 is the conjugate matrix ofM and 119864119899is n-dimension unit

matrixSimilarly the coefficient 120573

2119896minus1of the stress function120595

0(120585)

can be calculated like 1205722119896minus1

Thus the stress functions 1205930(120585)

and 1205950(120585) are completely determined and the horizontal

stress change and displacement are obtained by solving (5a)(5b) and (5c)

23 Elastoplastic Boundary (EP Boundary) Normally thesurrounding soil will enter into plastic stage after XCC pileinstallation and lead to a formation of a plastic zone around


the X-shaped cavity wall Therefore an elastoplastic analysisis necessary The nonaxisymmetric problem in the originalplane can be transformed into axisymmetric problem in thephase plane by the conformalmapping techniqueThus it canbe easily processed in the phase plane for the axisymmetriccharacteristics Considering an element at a radial distance 120588from the center of the cavity the equation of equilibrium inthe phase plane can be expressed as follows

120597Δ120590120588

120597120588+Δ120590120588minus Δ120590120579

120588= 0 (19)

where Δ120590120588

and Δ120590120579are radial and circumference stress

increment respectively and 120588 is the radial position of the soilparticle

Note that the Tresca yield criteria has the following form

Δ120590120588minus Δ120590120579= 2119878119906 (20)

where 119878119906is the undrained strength of the soil

The stress boundary conditions in the phase plane are

Δ120590120588= 119875 at 120588 = 1 (21a)

Δ120590120588= 0 at 120588 997888rarr infin (21b)

Combining (19) (20) and the stress boundary conditionsthe stress in the plastic zone can be obtained

Δ120590120588

119901

= minus2119878119906ln 120588 minus 119875 (22a)

Δ120590120579

119901

= minus2119878119906(ln 120588 + 1) minus 119875 (22b)

In the elastic zone the stress can be written as

Δ120590120588

119890

= 120582119875

1205882 (23a)

Δ120590120579

119890

= 120582119875

1205882 (23b)

where 120582 is the stress redistribution coefficient in the elasticzone

At the EP boundary the stress in the elastic zone shouldalso obey the Tresca yield criteriaThus the stress redistribu-tion coefficient in the elastic zone can be expressed as followsby substituting (23a) and (23b) into (22a) and (22b)

120590120588= 119878119906

120588119887

2

1205882 (24a)

120590120579= minus119878119906

120588119887

2

1205882 (24b)

where 120588119887is the radius of the plastic zone in the phase plane

At the EP boundary the stress in the plastic zone shouldbe equal to that in the elastic zone Therefore combining(22a) (22b) and (24a) (24b) the relationship of the pressure-plastic zone radius can be expressed as follows

120588119887= 1198901+119875119878

1199062

(25)

Substituting the limit cavity pressure 119875lim into (25) theradius of the plastic zone in 120585-plane after the XCC pileinstallation can be obtained as follows

120588119887= 1198901+119875lim1198781199062 (26)

The radius of the plastic zone in the physical plane can beobtained by combining (4) and (25)

119877119887(120579) =

10038161003816100381610038161003816119908 (120588119887119890119894120579

)10038161003816100381610038161003816 (27)

where the plastic zone 119877119887(120579) is the function of the polar

angleAccording to the above analysis EP boundary is circle

curve with radius equal to 120588119887in the phase plane The real EP

boundary in the physical plane is not circular curve and itcan be calculated by (27) However the EP boundary in thephysical plane is closed to circular curve far away from theX-shaped cavity from (27)Therefore the radius of the plasticzone can be assumed as follows

119877119887= max (10038161003816100381610038161003816119908 (120588119887119890

119894120579

)10038161003816100381610038161003816) (28)

where 119877119887is the maximum radius of the plastic zone in the

physical plane

24 The Horizontal Stress and Displacement Solutions in theElastic Zone After the soil around the cavity wall enters yieldstate the stress in the elastic zone has a redistribution effectand the stress in the elastic zone cannot be calculated bythe elastic analysis directly However the stress redistributioneffect can be considered by introducing a coefficient 120582 intothe elastic analysis In other words the new stress functions120582120593(120585) and 120582120595(120585) instead of 120593(120585) and120595(120585) are introduced intothe governing equations Thus the governing equations (5a)(5b) and (5c) of the elastic zone can be expressed as

120590119909+ 120590119910= 4Re[120582

1205931015840 (120585)

1199081015840 (120585)] (29a)

120590119910minus 120590119909+ 2119894120591119909119910=

2

1199081015840 (120585)[120582119908 (120585)(

1205931015840 (120585)

1199081015840 (120585))

1015840

+ 1205821205951015840

(120585)]

(29b)

2119866 (119906119909+ 119894119906119910) = (3 minus 4120583) 120582120593 (120585) minus 120582

119908 (120585)

1199081015840 (120585)1205931015840

(120585) minus 120582120595 (120585)

(29c)

Under the undrained condition the volume change ofthe X-shaped cavity induced by the XCC pile installation isequivalent to the change in position of the EP boundary Themathematical relation can be expressed as follows

119860119909= 1205871198772

119887minus 120587(119877

119887minus 119906119887)2

(30)

where 119860119909is the area of X-shaped cavity 119877

119887is the radius of

the plastic zone and 119906119901is the radial displacement at the EP

boundary


Table 1 Physical-mechanical properties of soils on site

Soil name ℎ (m) 119908 () 120574 (kNm3) 120592 119864119904(MPa) 119890 119888 (kPa) 120593 (∘)

Silt clay 200 305 1850 03 549 0913 264 151Silt clay 460 414 1790 03 297 1159 108 34Silt soil 340 303 1860 03 1168 0897 127 264Silt clay 370 414 1790 03 297 1159 108 34Silt soil 200 303 1860 03 1168 0897 127 264Silt clay 160 329 1880 03 441 0915 264 161Silt soil 030 303 1860 03 1168 0897 127 264Notes h the thickness of the soil layer w the moisture content 120574 the bulk density 120592 the Poisson ratio 119864

119904 the compression modulus e the void ratio c the

cohesion 120593 the internal friction angle

The 1199062119887is higher order driblet and can be ignored and thus

(28) can be expressed as

119906119887=119860119909

2120587119877119887

(31)

The stress redistribution factor 120582 can be obtained bysolving the coupled equations (28) (29c) and (31) thenthe new stress functions 120582120593(120585) and 120582120595(120585) can be obtainedSubstituting the new stress functions into the governingequation of the elastic zone (29a) (29b) and (29c) thehorizontal stress and displacement in the elastic zone can bedetermined

3 Verification

31 Engineering Description The Fourth Yangtze RiverBridgersquos north-line soft soil treatment field is located inNanjing China The total length of the soft ground improve-ment engineering is 290 km Physical-mechanical propertiesof soils on site are shown in Table 1 The form of plum-shaped layout is carried out in the engineering The pilespacing and length are 22m and 12m respectivelyThe threeparameters of the XCC pile cross-section the outsourcingdiameter (parameter a) the open arc distance (parameter b)and the open arc angle (parameter 120579) are 611mm 120mmand 130∘ respectively (see Figure 3)

The arrangement of the test equipment and measuringpoints are shown in Figure 4 The location of the test instru-ment is concluded as follows (1) Inclinometer tubes wereburied at the distance from the XCC pile center 1m 2mand 35m respectively (2) Pore water pressure gauges wereburied at the depth of 6m and 9m and the distances from thepile center equal 1m 2m and 35m respectively (3) Earthpressure cells were buried at the depth of 3m and 6m andthe distances from the pile center equal 1m 2m and 35mrespectively

32 Comparison on the Theoretical Calculated Results withField Test Results The radial stress and displacement atthe depth of 3 meters are selected for comparison whichare shown in Figure 5 The radial stress and displacementare plotted against the normalized radius rR where thevariable 119903 is the radial position and 119877 is the radius of theoutsourcing round of XCC pile cross-section The stress and

θ = 130∘

xy

Outsourcing round of XCC pile cross-section(cross-section of circular pile B)

The same area with XCC pile cross-section (cross-section

of circular pile A)

Cross-section of XCC pile

a = 611

b = 120

Figure 3 Geometry of XCC pile cross-section

displacement of soil around the XCC pile calculated by thisstudy are similar to those of the measured results on siteTherefore this study can simulate the stress and displacementinduced by XCC pile installation well Additionally the stressand radial displacement decrease rapidly with the distancefrom the pile centerThe radius of the influence zone inducedby the XCC pile installation is about 12R

Figure 5 also gives the comparison between the XCCpile the circular pile A (the same area with XCC pile cross-section) and the circular pile B (outsourcing round of XCCpile cross-section) The results show that the displacementsfor circular pile B which is calculated by cylindrical cavityexpansion method (CEM) are larger than those of thecircular pile A and XCC pile Additionally the displacementof circular pile A and that of XCC pile are almost the sameIn other words the area of the pile cross-section governs thedisplacement induced by the pile installation Therefore itis reasonable to calculate displacement caused by XCC pileinstallationwith circular pile A instead of XCCpile As shown


Earth pressure cellInclinometer tube

Pore water pressure gauge

1000

1000

1500

1000

1000

1500

1000

1000

1500

Burial depth 6m 9m

Burial depth 3m 6m Length of the inclinometertube 15m

Figure 4 Equipment arrangement and measuring points on site

in Figure 5(b) the stress of circular pile B is also larger thanthose of the circular pile A and XCC pile However the stressof circular pile A and XCC pile is different which meansthe stress is related to the pile cross-section Thus it is notaccurate to calculate the stress or excess pore pressure withcircular pile A instead of XCCpileWhen it refers to the stressor excess pore pressure this study should be used

4 Parametric Studies

In order to provide engineers and researchers with cal-culation charts and tables for estimating horizontal stressdisplacements and the radius of the plastic zone inducedby XCC pile installation a parametric study is carried outThe stress displacement and the radius of the plastic zonehave many influence factors This paper focus on the factorsof outsourcing diameter open arc distance open arc anglethe pile hole pressure 119875 and the undrained strength 119878

119906 The

stress displacement along 119909-axis in the elastic zone and theradius of the plastic zone are analyzed The Youngrsquos modulusof the soil is selected for 5MPa and the Poissonrsquos ratio is 03The influence characteristics of the parameters on the stressdisplacement of the soil around the pile and the radius of theplastic zone are obtained by the parametric study

41 The Radius of the Plastic Zone Analysis From (26) and(28) the radius of the plastic zone is the function of the ratioof the limit pressure 119875lim the undrained strength 119875lim119878119906 andthe parameters of theXCCpile cross-section (the outsourcingdiameter the open arc angle and the open distance) Thusthe four parameters are selected for the parametric studies

The plastic zone radius 119877119901is plotted against the variable

of 119875lim119878119906 with different parameters of the XCC pile cross-section in Figure 6 As shown in Figures 6(a) 6(b) and6(c) the higher 119875lim119878119906 develops the larger plastic zoneradius 119877

119901 From Figure 6(a) it can be seen that the out-

sourcing diameter 119886 increases with the increasing plasticzone radius 119877

119901 provided that all other factors are held

2 4 6 8 10 120

5

10

15

20

25

30

35

40

45

50

u r(m

m)

This study XCC pileMeasured XCC pile

CEM circular pile A CEM circular pile B

rR

(a)

2 4 6 8 10 120

10

20

30

40



σ r(k

Pa)

rR

(b)

Figure 5 Comparison between this study and measured data (a)radial displacement (b) radial stress

constants With the variable of 119886 range from 500mm to1000mm it is found that the increasing amplitude ofthe 119877119901increases with the increasing 119875lim119878119906 provided that

the variables of 119887 and 120579 are constant Figure 6(b) shows thatthe open arc distance 119887 has similar characteristics as theoutsourcing diameter 119886 Figure 6(c) shows that the openarc angle 120579 reduces with the increasing plastic zone radius


0 1 2 3 4 5 60

1

2

3

4

5

6

7

8

9

10

a = 500mma = 600mma = 700mm

a = 800mma = 900mma = 1000mm

RP

(m)

PSu

(a)

0 1 2 3 4 5 60

1

2

3

4

5

6

b = 100mmb = 120mmb = 140mm

b = 160mmb = 180mmb = 200mm

RP

(m)

PSu

(b)

0 1 2 3 4 5 60

1

2

3

4

5

6

120579 = 80∘

120579 = 90∘

120579 = 100∘

120579 = 110∘

120579 = 120∘

120579 = 130∘

RP

(m)

PSu

(c)

Figure 6 Variation of the plastic radius119877119901with different geometric parameters (119886 119887 120579) and 119875lim119878119906 (a) 119886 = 500mm to 1000mm 119887 = 120mm

120579 = 90∘ (b) 119886 = 600mm 119887 = 120mm to 200mm 120579 = 90∘ (c) 119886 = 600mm 119887 = 120mm 120579 = 80∘ to 130∘

0 5 10 15 200

10

20

30

40

50

60

70

80

90

120590 r(k

Pa)

rRP

a = 330mma = 530mm

a = 730mm

(a)

0 5 10 15 200

5

10

15

20

25

a = 330mma = 530mm

a = 730mm

ur

(mm

)

rRP

(b)

Figure 7 Radial stress and displacement distribution of different outsourcing diameter 119886 along the radial direction (b= 110mm 120579= 90∘ 119878119906=

10 kPa) (a) radial stress (b) radial displacement


0 5 10 15 200

10

20

30

40

50

60

70

80

90

b = 90mmb = 110mm

b = 130mm

rRP

120590r

(kPa

)

(a)

0 5 10 15 200

5

10

15

20

25

b = 90mmb = 110mm

b = 130mm

ur

(mm

)

rRP

(b)

Figure 8 Radial stress and displacement distribution of different open arc angle 119887 along the radial direction (a = 530mm 120579 = 90∘ 119878119906=


0 5 10 15 200

10

20

30

40

50

60

70

80

90

120579 = 110∘

120579 = 90∘120579 = 70∘

rRP

120590r

(kPa

)

(a)

0 5 10 15 200

5

10

15

20

25

u r(m

m)

120579 = 110∘

120579 = 90∘120579 = 70∘

rRP

(b)

Figure 9 Radial stress and displacement distribution of different open arc angle 120579 along the radial direction (a = 530mm b = 110mm 119878119906=


119877119901 However the plastic zone radius 119877

119901is not sensitive

to the open arc distance 119887 and open arc angle 120579 In allthe outsourcing diameter 119886 is the most obvious influenceparameter of the radius of the plastic zone among the threegeometric parameters of XCC pile cross-section

42 Stress and Displacement Distribution Analysis Based onthe three geometric parameters of the XCC pile cross-sectionand the undrained strength 119878

119906 the influence characteristics

of the stress changes and displacement in the elastic zone areobtained The limit pressure 119875lim is assumed to be 10 kPa


0 5 10 15 200

20

40

60

80

100

120

140

160

180

200

Su = 10kPaSu = 20kPa

Su = 30kPa

120590r

(kPa

)

rRP

(a)

0 5 10 15 200

5

10

15

20

25

30

35

40


Su = 30kPa

u r(m

m)

rRP

(b)

Figure 10 Radial stress and displacement distribution of different 119878119906along the radial direction (a = 530mm b = 110mm 120579 = 90∘) (a) radial

stress (b) radial displacement

As shown in Figure 7 with the outsourcing diameterrange from 330mm to 730mm the radial displacementincreases with the increasing outsourcing diameter 119886 in theelastic zone provided that all other factors are held constantHowever the stresses do not almost change It can beconcluded that the outsourcing diameter has little influenceon the horizontal stress in the elastic zone It can be seenthat the radius of stress influence zone is about 17119877

119901

which is less than that of the stress influence zone (morethan 17119877

119901) by comparing Figure 7(a) with Figure 7(b) From

Figures 8 and 9 it can be observed that both of the openarc distance and open arc angle have little influence on thehorizontal stress and displacement

Figure 10(a) shows that the stress increased with theincreasing of undrained strength 119878

119906in the elastic zone From

Figure 10(b) it can be seen that the larger the undrainedstrength 119878

119906is the larger the displacement will be It is because

the radius of the plastic zone 119877119901reduces with the increasing

119878119906 However the volume of the plastic zone is constant under

Tresca condition and the volume change induced by the XCCpile installation can only be manifested in the elastic zoneThus the volume change in the elastic zone increases withthe reducing 119877

119901and the displacement in the elastic zone will

increase

5 Conclusions

An analytical solution considering the pile cross-sectionshape for the horizontal stress and displacement of the soilaround the XCC pile after installation is presented in thisstudy An elastoplastic model for calculating the horizontal

stress and displacement is established by complex variablesSome main results can be concluded as follows

(1) Compared with the data of the field test it can be seenthat the elastoplastic model calculation results on thehorizontal stress and displacement of the soil aroundthe XCC pile after installation are in agreement withthose of field results A theoretical method for study-ing the special shaped piles installation is provided inthis paper

(2) The radius of the plastic zone caused by the XCCpile installation can be calculated conveniently bythis study The 119877

119901increased with the increasing

of 119875lim119878119906 outsourcing diameter 119886 and open arcdistance 119887 while it decreases with the increasing ofopen arc distance 120579 The outsourcing diameter 119886 isthe most obvious influence parameter of the radius ofthe plastic zone among the three geometric parame-ters (a b and 120579) of XCC pile cross-section

(3) The radial displacement increases with the increasingof outsourcing diameter 119886 in the elastic zone andthe outsourcing diameter 119886 has little influence onthe horizontal stress in the elastic zone The stressand displacement increased with the increasing ofundrained strength 119878

119906obviously in the elastic zone

Both of the open arc distance 119887 and the open arcangle 120579 have little influence on the horizontal stressand displacement The extent of the displacementinfluence zone is larger than that of the stress influ-ence zone


Acknowledgments

The authors wish to thank the National Science Founda-tion of China (nos 51278170 and U1134207) Program forChangjiang Scholars and Innovative Research Team inHohaiUniversity (no IRT1125) and 111 Project (no B13024) forfinancial support

References

[1] X Xu H Liu and B M Lehane ldquoPipe pile installation effectsin soft clayrdquo Proceedings of the Institution of Civil EngineersGeotechnical Engineering vol 159 no 4 pp 285ndash296 2006

[2] A S Vesic ldquoExpansion of cavities in infinite soil massrdquo Journalof the Soil Mechanics amp Foundations Division vol 98 no 3 pp265ndash290 1972

[3] H S Yu Cavity Expansion Methods in Geomechanics KluwerAcademic Publishers London UK 2000

[4] L F Cao C I Teh and M-F Chang ldquoAnalysis of undrainedcavity expansion in elasto-plastic soils with non-linear elastic-ityrdquo International Journal for Numerical and Analytical Methodsin Geomechanics vol 26 no 1 pp 25ndash52 2002

[5] R Salgado and M F Randolph ldquoAnalysis of cavity expansionin sandrdquo International Journal For Numerical and AnalyticalMethods in Geomechanics vol 26 no 1 pp 175ndash192 2001

[6] Y N Abousleiman and S L Chen ldquoExact undrained elasto-plastic solution for cylindrical cavity expansion in modifiedCam Clay soilrdquo Geotechnique vol 62 no 5 pp 447ndash456 2012

[7] I F Collins and H S Yu ldquoUndrained cavity expansions incritical state soilsrdquo International Journal for Numerical andAnalytical Methods in Geomechanics vol 20 no 7 pp 489ndash5161996

[8] M M Baligh ldquoStrain path methodrdquo Journal of GeotechnicalEngineering vol 111 no 9 pp 1108ndash1136 1985

[9] C Sagaseta A J Whittle and M Santagata ldquoDeformationanalysis of shallow penetration in clayrdquo International Journal forNumerical and Analytical Methods in Geomechanics vol 21 no10 pp 687ndash719 1997

[10] D R Gill and B M Lehane ldquoExtending the strain path methodanalogy for modelling penetrometer installationrdquo InternationalJournal For Numerical and AnalyticalMethods in Geomechanicsvol 24 no 5 pp 175ndash192 2000

[11] Y R Lv H L Liu X M Ding and G Q Kong ldquoField tests onbearing characteristics of X-section pile composite foundationrdquoJournal of Performance of Constructed Facilities vol 26 no 2pp 180ndash189 2012

[12] S P Timoshenko and J N Goodier Theory of ElasticityMcGraw-Hill New York NY USA 1970

[13] N I Muskhelishvili Some Basic Problems of the MathematicalTheory of Elasticity P Noordhoff GroningenThe Netherlands1963

[14] JW BrownComplex Variables and Applications McGraw-HillNew York NY USA 2008

[15] AVerruijt ldquoDeformations of an elastic half planewith a circularcavityrdquo International Journal of Solids and Structures vol 35 no21 pp 2795ndash2804 1998

[16] G H Lei C W W Ng and D B Rigby ldquoStress and displace-ment around an elastic artificial rectangular holerdquo Journal ofEngineering Mechanics vol 127 no 9 pp 880ndash890 2001

[17] O E Strack and A Verruijt ldquoA complex variable solutionfor a deforming buoyant tunnel in a heavy elastic half-planerdquo

International Journal for Numerical and Analytical Methods inGeomechanics vol 26 no 12 pp 1235ndash1252 2002

[18] G E Exadaktylos P A Liolios and M C Stavropoulou ldquoAsemi-analytical elastic stress-displacement solution for notchedcircular openings in rocksrdquo International Journal of Solids andStructures vol 40 no 5 pp 1165ndash1187 2003

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of


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Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of


Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of


Mathematical PhysicsAdvances in

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of


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Journal of


Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences


The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014


Algebra

Discrete Dynamics in Nature and Society



Decision SciencesAdvances in

Discrete MathematicsJournal of


Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of


distance b and open arc angle 120579) the undrained strength 119878119906

and the pile hole pressure 119875 were discussed

2 Mathematical Model

21 Definition of the Problem and Basic Assumption Figure 1shows that the elastoplastic soil which is described by theTresca model is under initial stress 120590

0before the XCC

pile installation Then the XCC pile installation progress issimplified as the expansion of a cavity from zero to the X-shaped cavity of the XCC pile cross-section The soil aroundthe pile enter into the passive limit balance state after theXCC pile installation Thus the X-shaped cavity internalpressure after XCC pile installation can be assumed as 119875lim =1205900tan2(45∘ + 1205932) + 2119888 tan(45∘ + 1205932) where 119888 and 120593 are

the soil cohesion and the internal friction angle respectivelyAdditionally the plain strain condition is assumed in themodel The Cartesian coordinates system is selected for theanalysis The origin of the coordinates is located at the centerof the cavity For the Cartesian coordinates system the 119909-axis and 119910-axis are in the horizontal and vertical directionrespectively The stress and strain are taken as positive in thepositive direction of the coordinates system

22 Basic Governing Equations According to the elasticity[12] the stress around the X-shaped cavity should obey thefourth-order partial differential equation as follows

nabla4

119880 = 0 (1)

where nabla is Laplace operator and U is stress function in theplastic zone

The three stress components in Cartesian coordinatessystem 120590

119909 120590119910and 120591

119909119910 around the X-shaped cavity can be

determined by (1)

120590119909=1205972119880

1205971199102 120590

119910=1205972119880

1205971199092 120591

119909119910= minus

1205972119880

120597119909120597119910 (2)

According to the complex variable elasticity [13ndash17] thestress around the X-shaped cavity can be expressed with twostress functions 120593

1(119911) and 120595

1(119911) as follows

120590119909+ 120590119910= 4Re [1205931015840

1(119911)] (3a)

120590119910minus 120590119909+ 2119894120591119909119910= 2 [119911120593

10158401015840

1(119911) + 120595

1015840

1(119911)] (3b)

119864

1 + 120583(119906 + 119894V) = (3 minus 4120583) 120593

1(119911) minus 119911120593

1015840

1(119911) minus 120595

1(119911) (3c)

where 12059310158401(119911) and 12059310158401015840

1(119911) are the first and second derivative of

the function 1205931(119911) respectively 1205951015840

1(119911) is the first derivative

of the function 1205951(119911) 1205931015840

1(119911) and 120595

1(119911) are the conjugate

complex functions of the 12059310158401(119911) and 120595

1(119911) 119906 and V are the

displacement component acting in 119909-axis and 119910-axis direc-tions respectively 119864 is the Youngrsquos modulus of the soil 120583 isthe Poisson ratio of the soil

For obtaining the solution to calculate the stress anddisplacement distributions a conformal mapping function is

provided to map the outside of the X-shaped cavity in the 119911-plane onto the outside of the unit circle in the phase planenamely 120585-plane (120585 = 120577+119894120578 = 120588119890119894120579) in Figure 2The conformalmapping function can be expressed in a series as follows

119911 = 119908 (120585) = 1198880120585+

119899

sum119896=1

1198882119896minus1

1205851minus2119896

10038161003816100381610038161205851003816100381610038161003816 le 1 (4)

where n is integral number (in this paper 119899 = 7 is selectedfor analysis and it can give enough accuracy) The constantcoefficients 119888

0and 1198882119896minus1

(119896 = 1 2 119899) are real numbers andcan be obtained by the iterative technique [18] 119911 = 119909 + 119894119910 (119909and 119910 are the variables in the Cartesian coordinates system119894 = radicminus1) is complex variable 120585 is complex variable in thephase plane 120585 = 120588119890119894120579

Substituting the conformal mapping function (4) into(3a) (3b) and (3c) leads to the following equations

120590119909+ 120590119910= 4Re[

1205931015840 (120585)

1199081015840 (120585)] (5a)

120590119910minus 120590119909+ 2119894120591119909119910=

2

1199081015840 (120585)[119908 (120585)(

1205931015840 (120585)

1199081015840 (120585))

1015840

+ 1205951015840

(120585)] (5b)

119864

1 + 120583(119906 + 119894V) = 120594120593 (120585) minus

119908 (120585)

1199081015840 (120585)1205931015840

(120585) minus 120595 (120585) (5c)

where 120593(120585) = 1205931(119908(120585)) 120595(120585) = 120595

1(119908(120585))

The two stress functions 1205931(119911) and 120595

1(119911) are transformed

as 120593(120585) and 120595(120585) To solve the stress functions 120593(120585) and120595(120585)the stress boundary conditions should be considered FromFigure 2 it can be seen that the stress boundary condition canbe expressed as

[

[

120593 (120585) +119908 (120585)

1199081015840 (120585)1205931015840

(120585) + 1205951015840

(120585)]

]119904

= minus119901119908 (120585) (6)

where s is the X-shaped cavity boundary curve and P is thepressure at the X-shaped cavity

By the complex elasticity [10 11] the stress functions 120593(120585)and 120595(120585) can be written as follows

120593 (120585) =1

8120587 (1 minus 120583)(119865119909+ 119894 119865119910) ln 120585 + 119861119908 (120585) + 120593

0(120585) (7a)

120595 (120585) = minus3 minus 4120583

8120587 (1 minus 120583)(119865119909minus 119894 119865119910) ln 120585

+ (1198611015840

+ 1198941198621015840

)119908 (120585) + 1205950(120585)

(7b)

where 119865119909and 119865

119910are the composite surface force in 119909 and 119910

direction on the X-shaped cavity boundary respectively Onehas 119861 = (120590

1+ 1205902)4 (120590

1and 120590

2are the principal stress at

infinity) and 1198611015840 + 1198621015840 = minus(12)(1205901minus 1205902)119890minus2119894120572 (120572 is the angle

between the principal stress 1205901and ox-axis)


σ0

σ0

σ0

σ0 Plimit o x

y

(a)

Open arc angle


(b)


σ0

σ0

σ0

σ0 Px

y

Soil

z-plane

(a)

σ0

σ0σ0

σ0

120585-plane

120588 = 1

120578

120588 = infin

120577P 120579

(b)




119865119909= 119865119910= 0 (8a)

119861 = 1205900 (8b)

1198611015840

= 1198621015840

= 0 (8c)


0(120585) can


1205930(120585) =

119899

sum119896=0

1205722119896minus1

1205852119896minus1

(9a)

1205950(120585) =

119899

sum119896=1

1205732119896minus1

1205852119896minus1

(9b)


and 1205732119896minus1


0(120585) and 120595




120593 (120585) = 1205900119908 (120585) +

119899

sum119896=0

1205722119896minus1

1205852119896minus1

(10a)

120595 (120585) =

119899

sum119896=1

1205732119896minus1

1205852119896minus1

(10b)


1205930(120590) +

119908 (120590)

1199081015840 (120590)1205931015840

0(120590) + 120595

0(120590) = minus (119875 + 120590

0) 119911 (11)


1205930(120590) +

119908 (120590)

1199081015840 (120590)1205931015840

0(120590) + 120595

0(120590) = minus (119875 + 120590

0) 119911 (12)


1

2120587119894int1205930(120590) 119889120590 +

1

2120587119894int119908 (120590)

1199081015840 (120590)1205931015840

0(120590) 119889120590

+1

2120587119894int1205950(120590) 119889120590 =

1

2120587119894intminus (119875 + 120590

0) 119908 (120590) 119889120590

(13a)

1

2120587119894int1205930(120590) 119889120590 +

1

2120587119894int119908 (120590)

1199081015840 (120590)1205931015840

0(120590) 119889120590

+1

2120587119894int1205950(120590) 119889120590 =

1

2120587119894intminus (119875 + 120590

0) 119908 (120590) 119889120590

(13b)


119908 (120590)

1199081015840 (120590)= 1198872119896minus3

1205902119896minus3

+ 1198872119896minus5

1205902119896minus5

+ sdot sdot sdot + 11988711205901

+ 119887minus1120590minus1

+ 119874(1

1205903)

(14)


119860119861 = 119862 (15)

where

119860 =

[[[[[[[[[[[

[


2119899minus3


2119899minus5


2119899minus7


d sdot sdot sdot

minus1198880

1198881

minus1198880

]]]]]]]]]]]

]


]119879


]119879

(16)


and 1198872119896minus3

(119896 = 1 2 3 119899)


[119864119899119872

119872 119864119899

] 120572 = 119873 (17)

where

119872 =

[[[[[[

[

1198871


2119899minus30

1198873



1198872119899minus3



]]]]]]

]

(18a)



]119879

(18b)



]119879

(18c)




0(120585)








120597Δ120590120588

120597120588+Δ120590120588minus Δ120590120579

120588= 0 (19)

where Δ120590120588




Δ120590120588minus Δ120590120579= 2119878119906 (20)



Δ120590120588= 119875 at 120588 = 1 (21a)

Δ120590120588= 0 at 120588 997888rarr infin (21b)


Δ120590120588

119901

= minus2119878119906ln 120588 minus 119875 (22a)

Δ120590120579

119901

= minus2119878119906(ln 120588 + 1) minus 119875 (22b)


Δ120590120588

119890

= 120582119875

1205882 (23a)

Δ120590120579

119890

= 120582119875

1205882 (23b)



120590120588= 119878119906

120588119887

2

1205882 (24a)

120590120579= minus119878119906

120588119887

2

1205882 (24b)



120588119887= 1198901+119875119878

1199062

(25)


120588119887= 1198901+119875lim1198781199062 (26)


119877119887(120579) =

10038161003816100381610038161003816119908 (120588119887119890119894120579

)10038161003816100381610038161003816 (27)





119877119887= max (10038161003816100381610038161003816119908 (120588119887119890

119894120579

)10038161003816100381610038161003816) (28)


physical plane


120590119909+ 120590119910= 4Re[120582

1205931015840 (120585)

1199081015840 (120585)] (29a)

120590119910minus 120590119909+ 2119894120591119909119910=

2

1199081015840 (120585)[120582119908 (120585)(

1205931015840 (120585)

1199081015840 (120585))

1015840

+ 1205821205951015840

(120585)]

(29b)

2119866 (119906119909+ 119894119906119910) = (3 minus 4120583) 120582120593 (120585) minus 120582

119908 (120585)

1199081015840 (120585)1205931015840

(120585) minus 120582120595 (120585)

(29c)


119860119909= 1205871198772

119887minus 120587(119877

119887minus 119906119887)2

(30)




boundary









119906119887=119860119909

2120587119877119887

(31)


3 Verification




θ = 130∘

xy



of circular pile A)


a = 611

b = 120







1000

1000

1500

1000

1000

1500

1000

1000

1500

Burial depth 6m 9m






119906 The








2 4 6 8 10 120

5

10

15

20

25

30

35

40

45

50

u r(m

m)



rR

(a)

2 4 6 8 10 120

10

20

30

40



σ r(k

Pa)

rR

(b)





0 1 2 3 4 5 60

1

2

3

4

5

6

7

8

9

10

a = 500mma = 600mma = 700mm

a = 800mma = 900mma = 1000mm

RP

(m)

PSu

(a)

0 1 2 3 4 5 60

1

2

3

4

5

6

b = 100mmb = 120mmb = 140mm

b = 160mmb = 180mmb = 200mm

RP

(m)

PSu

(b)

0 1 2 3 4 5 60

1

2

3

4

5

6

120579 = 80∘

120579 = 90∘

120579 = 100∘

120579 = 110∘

120579 = 120∘

120579 = 130∘

RP

(m)

PSu

(c)



0 5 10 15 200

10

20

30

40

50

60

70

80

90

120590 r(k

Pa)

rRP

a = 330mma = 530mm

a = 730mm

(a)

0 5 10 15 200

5

10

15

20

25

a = 330mma = 530mm

a = 730mm

ur

(mm

)

rRP

(b)




0 5 10 15 200

10

20

30

40

50

60

70

80

90

b = 90mmb = 110mm

b = 130mm

rRP

120590r

(kPa

)

(a)

0 5 10 15 200

5

10

15

20

25

b = 90mmb = 110mm

b = 130mm

ur

(mm

)

rRP

(b)



0 5 10 15 200

10

20

30

40

50

60

70

80

90

120579 = 110∘

120579 = 90∘120579 = 70∘

rRP

120590r

(kPa

)

(a)

0 5 10 15 200

5

10

15

20

25

u r(m

m)

120579 = 110∘

120579 = 90∘120579 = 70∘

rRP

(b)










0 5 10 15 200

20

40

60

80

100

120

140

160

180

200


Su = 30kPa

120590r

(kPa

)

rRP

(a)

0 5 10 15 200

5

10

15

20

25

30

35

40


Su = 30kPa

u r(m

m)

rRP

(b)




119901












increase

5 Conclusions











Acknowledgments


References



























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










σ0

σ0

σ0

σ0 Plimit o x

y

(a)

Open arc angle


(b)


σ0

σ0

σ0

σ0 Px

y

Soil

z-plane

(a)

σ0

σ0σ0

σ0

120585-plane

120588 = 1

120578

120588 = infin

120577P 120579

(b)




119865119909= 119865119910= 0 (8a)

119861 = 1205900 (8b)

1198611015840

= 1198621015840

= 0 (8c)


0(120585) can


1205930(120585) =

119899

sum119896=0

1205722119896minus1

1205852119896minus1

(9a)

1205950(120585) =

119899

sum119896=1

1205732119896minus1

1205852119896minus1

(9b)


and 1205732119896minus1


0(120585) and 120595




120593 (120585) = 1205900119908 (120585) +

119899

sum119896=0

1205722119896minus1

1205852119896minus1

(10a)

120595 (120585) =

119899

sum119896=1

1205732119896minus1

1205852119896minus1

(10b)


1205930(120590) +

119908 (120590)

1199081015840 (120590)1205931015840

0(120590) + 120595

0(120590) = minus (119875 + 120590

0) 119911 (11)


1205930(120590) +

119908 (120590)

1199081015840 (120590)1205931015840

0(120590) + 120595

0(120590) = minus (119875 + 120590

0) 119911 (12)


1

2120587119894int1205930(120590) 119889120590 +

1

2120587119894int119908 (120590)

1199081015840 (120590)1205931015840

0(120590) 119889120590

+1

2120587119894int1205950(120590) 119889120590 =

1

2120587119894intminus (119875 + 120590

0) 119908 (120590) 119889120590

(13a)

1

2120587119894int1205930(120590) 119889120590 +

1

2120587119894int119908 (120590)

1199081015840 (120590)1205931015840

0(120590) 119889120590

+1

2120587119894int1205950(120590) 119889120590 =

1

2120587119894intminus (119875 + 120590

0) 119908 (120590) 119889120590

(13b)


119908 (120590)

1199081015840 (120590)= 1198872119896minus3

1205902119896minus3

+ 1198872119896minus5

1205902119896minus5

+ sdot sdot sdot + 11988711205901

+ 119887minus1120590minus1

+ 119874(1

1205903)

(14)


119860119861 = 119862 (15)

where

119860 =

[[[[[[[[[[[

[


2119899minus3


2119899minus5


2119899minus7


d sdot sdot sdot

minus1198880

1198881

minus1198880

]]]]]]]]]]]

]


]119879


]119879

(16)


and 1198872119896minus3

(119896 = 1 2 3 119899)


[119864119899119872

119872 119864119899

] 120572 = 119873 (17)

where

119872 =

[[[[[[

[

1198871


2119899minus30

1198873



1198872119899minus3



]]]]]]

]

(18a)



]119879

(18b)



]119879

(18c)




0(120585)








120597Δ120590120588

120597120588+Δ120590120588minus Δ120590120579

120588= 0 (19)

where Δ120590120588




Δ120590120588minus Δ120590120579= 2119878119906 (20)



Δ120590120588= 119875 at 120588 = 1 (21a)

Δ120590120588= 0 at 120588 997888rarr infin (21b)


Δ120590120588

119901

= minus2119878119906ln 120588 minus 119875 (22a)

Δ120590120579

119901

= minus2119878119906(ln 120588 + 1) minus 119875 (22b)


Δ120590120588

119890

= 120582119875

1205882 (23a)

Δ120590120579

119890

= 120582119875

1205882 (23b)



120590120588= 119878119906

120588119887

2

1205882 (24a)

120590120579= minus119878119906

120588119887

2

1205882 (24b)



120588119887= 1198901+119875119878

1199062

(25)


120588119887= 1198901+119875lim1198781199062 (26)


119877119887(120579) =

10038161003816100381610038161003816119908 (120588119887119890119894120579

)10038161003816100381610038161003816 (27)





119877119887= max (10038161003816100381610038161003816119908 (120588119887119890

119894120579

)10038161003816100381610038161003816) (28)


physical plane


120590119909+ 120590119910= 4Re[120582

1205931015840 (120585)

1199081015840 (120585)] (29a)

120590119910minus 120590119909+ 2119894120591119909119910=

2

1199081015840 (120585)[120582119908 (120585)(

1205931015840 (120585)

1199081015840 (120585))

1015840

+ 1205821205951015840

(120585)]

(29b)

2119866 (119906119909+ 119894119906119910) = (3 minus 4120583) 120582120593 (120585) minus 120582

119908 (120585)

1199081015840 (120585)1205931015840

(120585) minus 120582120595 (120585)

(29c)


119860119909= 1205871198772

119887minus 120587(119877

119887minus 119906119887)2

(30)




boundary









119906119887=119860119909

2120587119877119887

(31)


3 Verification




θ = 130∘

xy



of circular pile A)


a = 611

b = 120







1000

1000

1500

1000

1000

1500

1000

1000

1500

Burial depth 6m 9m






119906 The








2 4 6 8 10 120

5

10

15

20

25

30

35

40

45

50

u r(m

m)



rR

(a)

2 4 6 8 10 120

10

20

30

40



σ r(k

Pa)

rR

(b)





0 1 2 3 4 5 60

1

2

3

4

5

6

7

8

9

10

a = 500mma = 600mma = 700mm

a = 800mma = 900mma = 1000mm

RP

(m)

PSu

(a)

0 1 2 3 4 5 60

1

2

3

4

5

6

b = 100mmb = 120mmb = 140mm

b = 160mmb = 180mmb = 200mm

RP

(m)

PSu

(b)

0 1 2 3 4 5 60

1

2

3

4

5

6

120579 = 80∘

120579 = 90∘

120579 = 100∘

120579 = 110∘

120579 = 120∘

120579 = 130∘

RP

(m)

PSu

(c)



0 5 10 15 200

10

20

30

40

50

60

70

80

90

120590 r(k

Pa)

rRP

a = 330mma = 530mm

a = 730mm

(a)

0 5 10 15 200

5

10

15

20

25

a = 330mma = 530mm

a = 730mm

ur

(mm

)

rRP

(b)




0 5 10 15 200

10

20

30

40

50

60

70

80

90

b = 90mmb = 110mm

b = 130mm

rRP

120590r

(kPa

)

(a)

0 5 10 15 200

5

10

15

20

25

b = 90mmb = 110mm

b = 130mm

ur

(mm

)

rRP

(b)



0 5 10 15 200

10

20

30

40

50

60

70

80

90

120579 = 110∘

120579 = 90∘120579 = 70∘

rRP

120590r

(kPa

)

(a)

0 5 10 15 200

5

10

15

20

25

u r(m

m)

120579 = 110∘

120579 = 90∘120579 = 70∘

rRP

(b)










0 5 10 15 200

20

40

60

80

100

120

140

160

180

200


Su = 30kPa

120590r

(kPa

)

rRP

(a)

0 5 10 15 200

5

10

15

20

25

30

35

40


Su = 30kPa

u r(m

m)

rRP

(b)




119901












increase

5 Conclusions











Acknowledgments


References



























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











119865119909= 119865119910= 0 (8a)

119861 = 1205900 (8b)

1198611015840

= 1198621015840

= 0 (8c)


0(120585) can


1205930(120585) =

119899

sum119896=0

1205722119896minus1

1205852119896minus1

(9a)

1205950(120585) =

119899

sum119896=1

1205732119896minus1

1205852119896minus1

(9b)


and 1205732119896minus1


0(120585) and 120595




120593 (120585) = 1205900119908 (120585) +

119899

sum119896=0

1205722119896minus1

1205852119896minus1

(10a)

120595 (120585) =

119899

sum119896=1

1205732119896minus1

1205852119896minus1

(10b)


1205930(120590) +

119908 (120590)

1199081015840 (120590)1205931015840

0(120590) + 120595

0(120590) = minus (119875 + 120590

0) 119911 (11)


1205930(120590) +

119908 (120590)

1199081015840 (120590)1205931015840

0(120590) + 120595

0(120590) = minus (119875 + 120590

0) 119911 (12)


1

2120587119894int1205930(120590) 119889120590 +

1

2120587119894int119908 (120590)

1199081015840 (120590)1205931015840

0(120590) 119889120590

+1

2120587119894int1205950(120590) 119889120590 =

1

2120587119894intminus (119875 + 120590

0) 119908 (120590) 119889120590

(13a)

1

2120587119894int1205930(120590) 119889120590 +

1

2120587119894int119908 (120590)

1199081015840 (120590)1205931015840

0(120590) 119889120590

+1

2120587119894int1205950(120590) 119889120590 =

1

2120587119894intminus (119875 + 120590

0) 119908 (120590) 119889120590

(13b)


119908 (120590)

1199081015840 (120590)= 1198872119896minus3

1205902119896minus3

+ 1198872119896minus5

1205902119896minus5

+ sdot sdot sdot + 11988711205901

+ 119887minus1120590minus1

+ 119874(1

1205903)

(14)


119860119861 = 119862 (15)

where

119860 =

[[[[[[[[[[[

[


2119899minus3


2119899minus5


2119899minus7


d sdot sdot sdot

minus1198880

1198881

minus1198880

]]]]]]]]]]]

]


]119879


]119879

(16)


and 1198872119896minus3

(119896 = 1 2 3 119899)


[119864119899119872

119872 119864119899

] 120572 = 119873 (17)

where

119872 =

[[[[[[

[

1198871


2119899minus30

1198873



1198872119899minus3



]]]]]]

]

(18a)



]119879

(18b)



]119879

(18c)




0(120585)








120597Δ120590120588

120597120588+Δ120590120588minus Δ120590120579

120588= 0 (19)

where Δ120590120588




Δ120590120588minus Δ120590120579= 2119878119906 (20)



Δ120590120588= 119875 at 120588 = 1 (21a)

Δ120590120588= 0 at 120588 997888rarr infin (21b)


Δ120590120588

119901

= minus2119878119906ln 120588 minus 119875 (22a)

Δ120590120579

119901

= minus2119878119906(ln 120588 + 1) minus 119875 (22b)


Δ120590120588

119890

= 120582119875

1205882 (23a)

Δ120590120579

119890

= 120582119875

1205882 (23b)



120590120588= 119878119906

120588119887

2

1205882 (24a)

120590120579= minus119878119906

120588119887

2

1205882 (24b)



120588119887= 1198901+119875119878

1199062

(25)


120588119887= 1198901+119875lim1198781199062 (26)


119877119887(120579) =

10038161003816100381610038161003816119908 (120588119887119890119894120579

)10038161003816100381610038161003816 (27)





119877119887= max (10038161003816100381610038161003816119908 (120588119887119890

119894120579

)10038161003816100381610038161003816) (28)


physical plane


120590119909+ 120590119910= 4Re[120582

1205931015840 (120585)

1199081015840 (120585)] (29a)

120590119910minus 120590119909+ 2119894120591119909119910=

2

1199081015840 (120585)[120582119908 (120585)(

1205931015840 (120585)

1199081015840 (120585))

1015840

+ 1205821205951015840

(120585)]

(29b)

2119866 (119906119909+ 119894119906119910) = (3 minus 4120583) 120582120593 (120585) minus 120582

119908 (120585)

1199081015840 (120585)1205931015840

(120585) minus 120582120595 (120585)

(29c)


119860119909= 1205871198772

119887minus 120587(119877

119887minus 119906119887)2

(30)




boundary









119906119887=119860119909

2120587119877119887

(31)


3 Verification




θ = 130∘

xy



of circular pile A)


a = 611

b = 120







1000

1000

1500

1000

1000

1500

1000

1000

1500

Burial depth 6m 9m






119906 The








2 4 6 8 10 120

5

10

15

20

25

30

35

40

45

50

u r(m

m)



rR

(a)

2 4 6 8 10 120

10

20

30

40



σ r(k

Pa)

rR

(b)





0 1 2 3 4 5 60

1

2

3

4

5

6

7

8

9

10

a = 500mma = 600mma = 700mm

a = 800mma = 900mma = 1000mm

RP

(m)

PSu

(a)

0 1 2 3 4 5 60

1

2

3

4

5

6

b = 100mmb = 120mmb = 140mm

b = 160mmb = 180mmb = 200mm

RP

(m)

PSu

(b)

0 1 2 3 4 5 60

1

2

3

4

5

6

120579 = 80∘

120579 = 90∘

120579 = 100∘

120579 = 110∘

120579 = 120∘

120579 = 130∘

RP

(m)

PSu

(c)



0 5 10 15 200

10

20

30

40

50

60

70

80

90

120590 r(k

Pa)

rRP

a = 330mma = 530mm

a = 730mm

(a)

0 5 10 15 200

5

10

15

20

25

a = 330mma = 530mm

a = 730mm

ur

(mm

)

rRP

(b)




0 5 10 15 200

10

20

30

40

50

60

70

80

90

b = 90mmb = 110mm

b = 130mm

rRP

120590r

(kPa

)

(a)

0 5 10 15 200

5

10

15

20

25

b = 90mmb = 110mm

b = 130mm

ur

(mm

)

rRP

(b)



0 5 10 15 200

10

20

30

40

50

60

70

80

90

120579 = 110∘

120579 = 90∘120579 = 70∘

rRP

120590r

(kPa

)

(a)

0 5 10 15 200

5

10

15

20

25

u r(m

m)

120579 = 110∘

120579 = 90∘120579 = 70∘

rRP

(b)










0 5 10 15 200

20

40

60

80

100

120

140

160

180

200


Su = 30kPa

120590r

(kPa

)

rRP

(a)

0 5 10 15 200

5

10

15

20

25

30

35

40


Su = 30kPa

u r(m

m)

rRP

(b)




119901












increase

5 Conclusions











Acknowledgments


References



























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











120597Δ120590120588

120597120588+Δ120590120588minus Δ120590120579

120588= 0 (19)

where Δ120590120588




Δ120590120588minus Δ120590120579= 2119878119906 (20)



Δ120590120588= 119875 at 120588 = 1 (21a)

Δ120590120588= 0 at 120588 997888rarr infin (21b)


Δ120590120588

119901

= minus2119878119906ln 120588 minus 119875 (22a)

Δ120590120579

119901

= minus2119878119906(ln 120588 + 1) minus 119875 (22b)


Δ120590120588

119890

= 120582119875

1205882 (23a)

Δ120590120579

119890

= 120582119875

1205882 (23b)



120590120588= 119878119906

120588119887

2

1205882 (24a)

120590120579= minus119878119906

120588119887

2

1205882 (24b)



120588119887= 1198901+119875119878

1199062

(25)


120588119887= 1198901+119875lim1198781199062 (26)


119877119887(120579) =

10038161003816100381610038161003816119908 (120588119887119890119894120579

)10038161003816100381610038161003816 (27)





119877119887= max (10038161003816100381610038161003816119908 (120588119887119890

119894120579

)10038161003816100381610038161003816) (28)


physical plane


120590119909+ 120590119910= 4Re[120582

1205931015840 (120585)

1199081015840 (120585)] (29a)

120590119910minus 120590119909+ 2119894120591119909119910=

2

1199081015840 (120585)[120582119908 (120585)(

1205931015840 (120585)

1199081015840 (120585))

1015840

+ 1205821205951015840

(120585)]

(29b)

2119866 (119906119909+ 119894119906119910) = (3 minus 4120583) 120582120593 (120585) minus 120582

119908 (120585)

1199081015840 (120585)1205931015840

(120585) minus 120582120595 (120585)

(29c)


119860119909= 1205871198772

119887minus 120587(119877

119887minus 119906119887)2

(30)




boundary









119906119887=119860119909

2120587119877119887

(31)


3 Verification




θ = 130∘

xy



of circular pile A)


a = 611

b = 120







1000

1000

1500

1000

1000

1500

1000

1000

1500

Burial depth 6m 9m






119906 The








2 4 6 8 10 120

5

10

15

20

25

30

35

40

45

50

u r(m

m)



rR

(a)

2 4 6 8 10 120

10

20

30

40



σ r(k

Pa)

rR

(b)





0 1 2 3 4 5 60

1

2

3

4

5

6

7

8

9

10

a = 500mma = 600mma = 700mm

a = 800mma = 900mma = 1000mm

RP

(m)

PSu

(a)

0 1 2 3 4 5 60

1

2

3

4

5

6

b = 100mmb = 120mmb = 140mm

b = 160mmb = 180mmb = 200mm

RP

(m)

PSu

(b)

0 1 2 3 4 5 60

1

2

3

4

5

6

120579 = 80∘

120579 = 90∘

120579 = 100∘

120579 = 110∘

120579 = 120∘

120579 = 130∘

RP

(m)

PSu

(c)



0 5 10 15 200

10

20

30

40

50

60

70

80

90

120590 r(k

Pa)

rRP

a = 330mma = 530mm

a = 730mm

(a)

0 5 10 15 200

5

10

15

20

25

a = 330mma = 530mm

a = 730mm

ur

(mm

)

rRP

(b)




0 5 10 15 200

10

20

30

40

50

60

70

80

90

b = 90mmb = 110mm

b = 130mm

rRP

120590r

(kPa

)

(a)

0 5 10 15 200

5

10

15

20

25

b = 90mmb = 110mm

b = 130mm

ur

(mm

)

rRP

(b)



0 5 10 15 200

10

20

30

40

50

60

70

80

90

120579 = 110∘

120579 = 90∘120579 = 70∘

rRP

120590r

(kPa

)

(a)

0 5 10 15 200

5

10

15

20

25

u r(m

m)

120579 = 110∘

120579 = 90∘120579 = 70∘

rRP

(b)










0 5 10 15 200

20

40

60

80

100

120

140

160

180

200


Su = 30kPa

120590r

(kPa

)

rRP

(a)

0 5 10 15 200

5

10

15

20

25

30

35

40


Su = 30kPa

u r(m

m)

rRP

(b)




119901












increase

5 Conclusions











Acknowledgments


References



























Volume 2014




Journal of











Journal of


Function Spaces






Algebra

















119906119887=119860119909

2120587119877119887

(31)


3 Verification




θ = 130∘

xy



of circular pile A)


a = 611

b = 120







1000

1000

1500

1000

1000

1500

1000

1000

1500

Burial depth 6m 9m






119906 The








2 4 6 8 10 120

5

10

15

20

25

30

35

40

45

50

u r(m

m)



rR

(a)

2 4 6 8 10 120

10

20

30

40



σ r(k

Pa)

rR

(b)





0 1 2 3 4 5 60

1

2

3

4

5

6

7

8

9

10

a = 500mma = 600mma = 700mm

a = 800mma = 900mma = 1000mm

RP

(m)

PSu

(a)

0 1 2 3 4 5 60

1

2

3

4

5

6

b = 100mmb = 120mmb = 140mm

b = 160mmb = 180mmb = 200mm

RP

(m)

PSu

(b)

0 1 2 3 4 5 60

1

2

3

4

5

6

120579 = 80∘

120579 = 90∘

120579 = 100∘

120579 = 110∘

120579 = 120∘

120579 = 130∘

RP

(m)

PSu

(c)



0 5 10 15 200

10

20

30

40

50

60

70

80

90

120590 r(k

Pa)

rRP

a = 330mma = 530mm

a = 730mm

(a)

0 5 10 15 200

5

10

15

20

25

a = 330mma = 530mm

a = 730mm

ur

(mm

)

rRP

(b)




0 5 10 15 200

10

20

30

40

50

60

70

80

90

b = 90mmb = 110mm

b = 130mm

rRP

120590r

(kPa

)

(a)

0 5 10 15 200

5

10

15

20

25

b = 90mmb = 110mm

b = 130mm

ur

(mm

)

rRP

(b)



0 5 10 15 200

10

20

30

40

50

60

70

80

90

120579 = 110∘

120579 = 90∘120579 = 70∘

rRP

120590r

(kPa

)

(a)

0 5 10 15 200

5

10

15

20

25

u r(m

m)

120579 = 110∘

120579 = 90∘120579 = 70∘

rRP

(b)










0 5 10 15 200

20

40

60

80

100

120

140

160

180

200


Su = 30kPa

120590r

(kPa

)

rRP

(a)

0 5 10 15 200

5

10

15

20

25

30

35

40


Su = 30kPa

u r(m

m)

rRP

(b)




119901












increase

5 Conclusions











Acknowledgments


References



























Volume 2014




Journal of











Journal of


Function Spaces






Algebra












1000

1000

1500

1000

1000

1500

1000

1000

1500

Burial depth 6m 9m






119906 The








2 4 6 8 10 120

5

10

15

20

25

30

35

40

45

50

u r(m

m)



rR

(a)

2 4 6 8 10 120

10

20

30

40



σ r(k

Pa)

rR

(b)





0 1 2 3 4 5 60

1

2

3

4

5

6

7

8

9

10

a = 500mma = 600mma = 700mm

a = 800mma = 900mma = 1000mm

RP

(m)

PSu

(a)

0 1 2 3 4 5 60

1

2

3

4

5

6

b = 100mmb = 120mmb = 140mm

b = 160mmb = 180mmb = 200mm

RP

(m)

PSu

(b)

0 1 2 3 4 5 60

1

2

3

4

5

6

120579 = 80∘

120579 = 90∘

120579 = 100∘

120579 = 110∘

120579 = 120∘

120579 = 130∘

RP

(m)

PSu

(c)



0 5 10 15 200

10

20

30

40

50

60

70

80

90

120590 r(k

Pa)

rRP

a = 330mma = 530mm

a = 730mm

(a)

0 5 10 15 200

5

10

15

20

25

a = 330mma = 530mm

a = 730mm

ur

(mm

)

rRP

(b)




0 5 10 15 200

10

20

30

40

50

60

70

80

90

b = 90mmb = 110mm

b = 130mm

rRP

120590r

(kPa

)

(a)

0 5 10 15 200

5

10

15

20

25

b = 90mmb = 110mm

b = 130mm

ur

(mm

)

rRP

(b)



0 5 10 15 200

10

20

30

40

50

60

70

80

90

120579 = 110∘

120579 = 90∘120579 = 70∘

rRP

120590r

(kPa

)

(a)

0 5 10 15 200

5

10

15

20

25

u r(m

m)

120579 = 110∘

120579 = 90∘120579 = 70∘

rRP

(b)










0 5 10 15 200

20

40

60

80

100

120

140

160

180

200


Su = 30kPa

120590r

(kPa

)

rRP

(a)

0 5 10 15 200

5

10

15

20

25

30

35

40


Su = 30kPa

u r(m

m)

rRP

(b)




119901












increase

5 Conclusions











Acknowledgments


References



























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










0 1 2 3 4 5 60

1

2

3

4

5

6

7

8

9

10

a = 500mma = 600mma = 700mm

a = 800mma = 900mma = 1000mm

RP

(m)

PSu

(a)

0 1 2 3 4 5 60

1

2

3

4

5

6

b = 100mmb = 120mmb = 140mm

b = 160mmb = 180mmb = 200mm

RP

(m)

PSu

(b)

0 1 2 3 4 5 60

1

2

3

4

5

6

120579 = 80∘

120579 = 90∘

120579 = 100∘

120579 = 110∘

120579 = 120∘

120579 = 130∘

RP

(m)

PSu

(c)



0 5 10 15 200

10

20

30

40

50

60

70

80

90

120590 r(k

Pa)

rRP

a = 330mma = 530mm

a = 730mm

(a)

0 5 10 15 200

5

10

15

20

25

a = 330mma = 530mm

a = 730mm

ur

(mm

)

rRP

(b)




0 5 10 15 200

10

20

30

40

50

60

70

80

90

b = 90mmb = 110mm

b = 130mm

rRP

120590r

(kPa

)

(a)

0 5 10 15 200

5

10

15

20

25

b = 90mmb = 110mm

b = 130mm

ur

(mm

)

rRP

(b)



0 5 10 15 200

10

20

30

40

50

60

70

80

90

120579 = 110∘

120579 = 90∘120579 = 70∘

rRP

120590r

(kPa

)

(a)

0 5 10 15 200

5

10

15

20

25

u r(m

m)

120579 = 110∘

120579 = 90∘120579 = 70∘

rRP

(b)










0 5 10 15 200

20

40

60

80

100

120

140

160

180

200


Su = 30kPa

120590r

(kPa

)

rRP

(a)

0 5 10 15 200

5

10

15

20

25

30

35

40


Su = 30kPa

u r(m

m)

rRP

(b)




119901












increase

5 Conclusions











Acknowledgments


References



























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










0 5 10 15 200

10

20

30

40

50

60

70

80

90

b = 90mmb = 110mm

b = 130mm

rRP

120590r

(kPa

)

(a)

0 5 10 15 200

5

10

15

20

25

b = 90mmb = 110mm

b = 130mm

ur

(mm

)

rRP

(b)



0 5 10 15 200

10

20

30

40

50

60

70

80

90

120579 = 110∘

120579 = 90∘120579 = 70∘

rRP

120590r

(kPa

)

(a)

0 5 10 15 200

5

10

15

20

25

u r(m

m)

120579 = 110∘

120579 = 90∘120579 = 70∘

rRP

(b)










0 5 10 15 200

20

40

60

80

100

120

140

160

180

200


Su = 30kPa

120590r

(kPa

)

rRP

(a)

0 5 10 15 200

5

10

15

20

25

30

35

40


Su = 30kPa

u r(m

m)

rRP

(b)




119901












increase

5 Conclusions











Acknowledgments


References



























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










0 5 10 15 200

20

40

60

80

100

120

140

160

180

200


Su = 30kPa

120590r

(kPa

)

rRP

(a)

0 5 10 15 200

5

10

15

20

25

30

35

40


Su = 30kPa

u r(m

m)

rRP

(b)




119901












increase

5 Conclusions











Acknowledgments


References



























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










Acknowledgments


References



























Volume 2014




Journal of











Journal of


Function Spaces






Algebra
















Volume 2014




Journal of











Journal of


Function Spaces






Algebra









research article analytical solution for stress and ...z-plane (a) 0 0 0 0-plane =1 = p (b) f : (a)...

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