Design of laterally loaded piles

Design of Laterally Loaded Piles

Manjriker Gunaratne

CONTENTS

8.1Introduction (327)

8.2Lateral Load Capacity Based on Strength (328)

8.2.1Ultimate Lateral Resistance of Piles (328)

8.2.1.1Piles in Homogeneous Cohesive Soils (328)

8.2.1.2Piles in Cohesionless Soils (335)

8.3Lateral Load Capacity Based on Deflections (339)

8.3.1Linear Elastic Method (339)

8.3.1.1Free-Headed Piles (339)

8.3.1.2Fixed-Headed Piles (341)

8.3.2Nonlinear Methods (343)

8.3.2.1Stiffness Matrix Analysis Method (343)

8.3.2.2Lateral Pressure–Deflection(p–y)Method of Analysis (348)

8.3.2.3Synthesis of p–y Curves Based on

Pile Instrumentation (350)

8.4Lateral Load Capacity of Pile Groups (355)

8.5Load and Resistance Factor Design for Laterally Loaded Piles (356)

8.6Effect of Pile Jetting on the Lateral Load Capacity (356)

8.7Effect of Preaugering on the Lateral Load Capacity (360)

References (361)

8.1Introduction

Single piles such as sign-posts and lamp-posts and pile groups that support bridge piers and offshore construction operations are constantly subjected to significant natural lateral loads(such as wind loads and wave actions)(Figure8.1).Lateral loads can be also introduced on piles due to artificial causes like ship impacts.Therefore,the lateral load capacity is certainly a significant attribute in the design of piles under certain construction situations.

Unlike in the case of axial load capacity,the lateral load capacity must be determined by considering two different failure mechanisms:(1)structural failure of the pile due to yielding of pile material or shear failure of the confining soil due to yielding of soil,and (2)pile becoming dysfunctional due to excessive lateral deflections.Although passive failure of the confining soil is a potential failure mode,such failure occurs only at relatively large deflections which generally exceed the tolerable movements.

327

One realizes that ‘‘short’’piles embedded in relatively stiffer ground would possibly fail due to yielding of the soil while ‘‘long’’piles embedded in relatively softer ground would produce excessive deflections.In view of the above conditions,this chapter is organized to analyze separately,the two distinct issues presented above.Hence the discussion will deal with two main issues:(1)lateral pile capacity from strength consid-erations,and (2)lateral pile capacity based on deflection limitations.

On the other hand,piles subjected to both axial and lateral loading must be designed for structural resistance of the piles as beam-columns.

8.2

Lateral Load Capacity Based on Strength

8.2.1

Ultimate Lateral Resistance of Piles

Broms (1964a,b)produced simplified solutions for the ultimate lateral load capacity of piles by considering both the ultimate strength of the bearing ground and the yield stress of the pile material.For simplicity,the Broms (1964a,b)solutions are presented separately for different soil types,namely,cohesive soils and cohesionless soils.8.2.1.1

Piles in Homogeneous Cohesive Soils

When a pile is founded in a predominantly fine-grained soil,the most critical design case is the case where soil is in an undrained situation.The maximum load that can be applied on the pile depends on the the following factors:

1.Fixity conditions at the top (i.e.,free piles or fixed piles).Most single piles can be considered as free piles under lateral loading whereas piles clustered in a group by a pile cap must be analyzed as fixed piles.

2.Relative stiffness of the pile compared to the surrounding soil.If the deform-ation conditions are such that the soil yields before the pile material then the pile is classified as a ‘‘short’’pile.Similarly,if the pile material yields first,then the pile is considered a ‘‘long’’pile.8.2.1.1.1Unrestrained or Free-Head Piles

assumed for ‘‘short’’and ‘‘long’’piles,respectively.

The ultimate lateral resistance P u based on the geometrical properties and the undrained soil strength.For short piles,M max ,g ,P u ,and f can be determined from Equations (8.1)to (8.4).

FIGURE 8.1

Laterally loaded pile.

328The Foundation Engineering Handbook

Figure 8.2and Figure 8.3illustrate the respective failure mechanisms that Broms (1964a,b)can be directly determined from Figure 8.4(a)and (b)

Since the shear force is zero at the location of maximum moment,from the area of the soil reaction plot (Figure 8.2)one obtains

f ?

P u 9c u D

(8:1)

Similarly,by taking the first moments of Figure 8.2about the yield point

M max ?2:25Dg 2c u (8:2)M max ?H u (e t1:5D t0:5f )

(8:3)

For the total length of the pile,

L ?g t1:5D tf

(8:4)8.2.1.1.2Restrained or Fixed-Head Piles

According to the Broms (1964a)formulations,restrained piles can reach their ultimate capacity through three separate mechanisms giving rise to (1)short piles,(2)long piles,and (3)intermediate piles.These

failure mechanisms assumed by Broms (1964a)for analytical solutions is that the moment generated on the pile top can be provided by the pile cap to restrain the pile with the boundary condition at the top (i.e.,no rotation).

FIGURE 8.2

Deflection,soil reaction,and bending moment distributions for laterally loaded short piles in cohesive soil.(From Broms,B.,1964a,J.Soil Mech.Found.Div.,ASCE ,90(SM3):27–56.With permission.)

Design of Laterally Loaded Piles

329

restrained piles are illustrated in Figure 8.5(a)–(c).The assumption that leads to the

u The reader would notice that this condition is presented through a

single curve in Figure 8.4(a)due to the insignificance of the e parameter.M max and KP u can also be determined using the following equations:

P u ?9c u D (L à1:5D )(8:5)M max ?P u (0:5L t0:75D )

(8:6)

For long piles,the ultimate lateral load,P u ,can be found from Figure 8.4(b).Then,the following equations can be used to determine f and hence the location of pile yielding:

P u ?

2M y

(1:5d t0:5f )

(8:7)

the basic shear moment and total length consideration in Equations (8.1),(8.4),and (8.8)can be used to obtain P u :

hinge

(a) Deflection

(b) Soil reaction

max

FIGURE 8.3

Failure mechanism for laterally loaded long piles in cohesive soil.(From Broms,B.,1964a,J.Soil Mech.Found.Div.,ASCE ,90(SM3):27–56.With permission.)

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The Foundation Engineering Handbook

The ultimate lateral load,P ,of short piles can be directly obtained from Figure 8.4(a).On the other hand,for ‘‘intermediate’’piles where yielding occurs at the top (Figure 8.5b ),

Embedment length, L /D

010

U l t i m a t e l a t e r a l r e s i s t a n c e , P u l t /c U D 2

312

610

204060100

U l t i

m a t e l a t e r a l r e s i s t a n c e , P u l t /C u D 2

461020Yield moment, M yield /c u D 3

4060100200400600

(a)

(b)

FIGURE 8.4

Ultimate lateral resistance of piles in cohesive soils:(a)short piles and (b)long piles.(From Broms,B.,1964a,J.Soil Mech.Found.Div.,ASCE ,90(SM3):27–56.With permission.)

Design of Laterally Loaded Piles

331

9 c u D

(a) Deflection(b) Soil reaction(C) Bending moment

(a)

u max

(a) Deflection(b) Soil reaction(c) Bending moment

(b)

M yield

(a) Deflection(b) Soil reaction(c) Bending moment

(c)

FIGURE8.5

Failure mechanisms for laterally loaded restrained piles in cohesive soils:(a)short piles,(b)intermediate piles and(c)long piles.(From Broms,B.,1964a,J.Soil Mech.Found.Div.,ASCE,90(SM3):27–56.With permission.) 332The Foundation Engineering Handbook

P u

9c u D

(8:1)

M y?2:25c u Dg2à9c u Df(1:5Dt0:5f)(8:8)

L?gt1:5Dtf(8:4) Example8.1

Estimate the ultimate lateral load that can be applied on the steel H pile(HP250?62) shown in Figure8.6assuming that the pile cap can provide the moment required at the pile top to keep it from rotating.The yield strength of steel is300MPa.The CPT test results(q c)for the site are also plotted in Figure8.6(a).The Atterberg limits for the clay

are:LL?60and PL?25and the saturated unit weight of clay is17.5kN/m3

c (a)

(b)X

FIGURE8.6

(a)Illustration for Example8.1.

(b)HP section.

Design of Laterally Loaded Piles333

S xx ?0.711?10à3m 3,d ?256mm

M y ?S xx 'y ?(0.711)(10à3)(300)MN m ?213.3kN m.From the q c profile in Figure 8.6(a),q c can be expressed as

q c ?4:7t0:04z MPa

From Robertson and Campanella (1983)

S u ?

q T àp 0N KT

q T ?q c tu c (1àa )

From Bowles (1996)

N KT ?13t

5:5PI

where PI is the plasticity index of the soil.

One obtains the following s u profile for PI ?35:

S u ?(1=13:16)[(4:7t0:04z )t0:001f (9:8z )(1à0:5)à(17:5à9:8)z g ]?0:357t0:0028z MPa

s u ranges along the length of the pile from 357to 385kPa showing the linear trend with depth that is typical for clays.Due to its relatively narrow range,it can be reasonably averaged along the pile depth to be about 371kPa

c u ?371kPa

Assume that the ground conditions and the pile stiffness are such that it behaves as a short pile.

39,P u =c u D 2can be extrapolated as P u =c u D 2?337But c u D 2?24:314kN,and hence P u ?8.22MN.

Thus,if the pile does not yield,it can take 8.22MN before the soil fails.

In order to check the maximum moment in the pile,Equation (8.6)can be applied.

M max ?P u (0:5L t0:75D )?8:22(0:5?10t0:75?0:256)MN m ?42:68MN m But M y ?213.3kN m.Hence the pile would yield long before the clay,and the pile has to be reanalyzed as a long pile.

M y

c u D

?(213:3)=(371)=(0:256)3?34:27From Figure 8.4(b),P u =c u D 2%25

But c u D 2?24:314kN P u ?608kN.

Hence,the ultimate lateral load that can be applied on the given pile is about 600kN.

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The Foundation Engineering Handbook

From steel section tables and Figure 8.6(b)

Then from Figure 8.4(a)or Equation (8.5),for an embedment length of 10m/0.256m ?

8.2.1.2Piles in Cohesionless Soils

Based on a number of assumptions,Broms (1964b)formulated analytical methodologies to determine the ultimate lateral load capacity of a pile in cohesionless soils as well.The most significant assumptions were:(1)negligible active earth pressure on the back of the pile due to forward movement of the pile bottom,and (2)tripling of passive earth pressure along the top front of the pile.Hence

p p ?3s 0v K P

(8:9)

where s 0v is the effective vertical overburden pressure and K P ?(1tsin f 0)=(1àsin f 0).f 0is the angle of internal friction (effective stress).

8.2.1.2.1Free-Head Piles

By following terminology similar to that in the case of

cohesive soils,the failure mechan-following equation.

P u ?

0:5g DL 3K P

e tL

(8:10)

Then,the location of the maximum moment (f in Figure 8.7)can be determined by the following equation.

f ?0:82

?????????????P u

DK P g

s (8:11)

Finally,the maximum moment can be estimated by Equation (8.12)

M max

?P u e t2

(8:12)(a) Deflections

(b) Soil reactions

max

FIGURE 8.7

Failure mechanism for laterally loaded short pile in cohesionless soil.(From Broms,B.,1964b,J.Soil Mech.Found.Div.,ASCE ,90(SM3):123–156.With per-mission.)

Design of Laterally Loaded Piles 335

The ultimate lateral load for short piles can be estimated from Figure 8.9(a)or the isms of short and long piles are illustrated in Figure 8.7and Figure 8.8,respectively.

If the M max value computed from Equation (8.12)is larger than M yield for the pile material,then obviously the pile behaves as a long pile and the actual ultimate lateral load P u can be computed from

Equations (8.11)and (8.12)by setting M max ?M yield .

long piles directly.8.2.1.2.2

Restrained or Fixed-Head Piles

P u ?1:5g L 2DK P

(8:13)

Hence P u can be found either from Equation (8.13)or Figure 8.9(a).Also,from Figure 8.10(a)it follows that

M max ?2

P u L

(8:14)

If M max computed from Equation (8.14)is larger that M yield for the pile material,then the failure mechanism in Figure 8.10(b)applies.For this case,the following expression can be written for the moment about the pile bottom from which the ultimate lateral load can be computed:

M y ?(0:5g DL 3K P )àP u L

(8:15)

The above solution only applies if the moment M max at a depth of f computed by

f ?0:82

?????????????P u

DK P g

s (8:11)is less than M yield for the pile material.

FIGURE 8.8

Failure mechanism for laterally loaded long piles in cohesionless soil (From Broms, B.,1964b,J.Soil Mech.Found.Div.,ASCE ,90(SM3):123–156.With per-mission.)

(a) Deflection (b) Soil reaction (c) Bending moment

336The Foundation Engineering Handbook

On the other hand,Figure 8.9(b)enables one to determine the ultimate lateral load for For restrained short piles,consideration of horizontal equilibrium in Figure 8.10(a)yields

Finally,if the above M max is larger than M yield ,then the failure mechanism in equation or its nondimensional form in Figure 8.9(b).

P u

e t23

?2M y

(8:16)

200160

120

A p p l i e d l a t e r a l l o a d , P u l t /K P D 3g

61216

Length, L /D

U l t i m a t e l a t e r a l r e s i s d e n c e , P u l t /K p D 3g

1000

100

110

1.0

10.0100.0

1000.0

10.000.0

Yield moment M yield /D 4g

K P

FIGURE 8.9

Ultimate lateral resistance of piles in cohesionless soils:(a)short piles,(b)long piles.(From Broms,B.,1964b,J.Soil Mech.Found.Div.,ASCE ,90(SM3):123–156.With permission.)

Design of Laterally Loaded Piles

337

Figure 8.10(c)applies.Thus,the ultimate lateral load can be computed from the following

(a) Deflection (b) Soil Reaction

(b)

(a) Deflection

(b) Soil Reaction

(a)

y p

max

(a) Deflection

(b) Soil Reaction (c) Bending Moment

(c)

FIGURE 8.10

Failure mechanisms for restrained piles in cohesionless soils:(a)short piles,(b)intermediate piles,and (c)long piles.(From Broms,B.,1964b,J.Soil Mech.Found.Div.,ASCE ,90(SM3):123–156.With permission.)

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The Foundation Engineering Handbook

8.3Lateral Load Capacity Based on Deflections

The maximum permissible ground line deflection must be compared with the lateral deflection of a laterally loaded pile to fulfill one important criterion of the design proced-ure.A number of commonly adopted methods to determine the lateral deflection are discussed in the ensuing sections.

8.3.1Linear Elastic Method

A laterally loaded pile can be idealized as an infinitely long cylinder laterally deforming in an infinite elastic medium(Pyke and Beikae,1984)with the horizontal deformation governed by the following equation:

p?k h y(8:17) But,from distributed load vs.moment relations,

pB?d2M

d z

?àE P I

d4y

d z

(8:18)

where B is the width of pile and E P I is the pile stiffness.

Then the equation governing the lateral deformation can be expressed by combining (8.17)and(8.18)as

E P I d4y

d z4

tBk h y?0(8:19)

The characteristic coefficient of the solution to y is defined by

k h D

4E p I

1=4

(8:20)

1/b is also known as the nondimensional length,where k h is the coefficient of horizontal subgrade reaction.

Broms(1964a,b)showed that a laterally loaded pile behaves as an infinitely stiff member when the coefficient b is less than2.Further,when b L!4,it was shown to behave as an infinitely long member in which failure occurs when the maximum bending moment exceeds the yield resistance of the pile section.

For the simple situation where k h can be assumed constant along the pile depth, Hetenyi(1946)derived the following closed-form solutions:

8.3.1.1Free-Headed Piles

8.3.1.1.1Case(1):Lateral Deformation due to Load H

Horizontal displacement

D?2H b

k h d

K D H(8:21a)

Design of Laterally Loaded Piles339 The following expressions can be used in conjunction with Figure8.11,for a pile of width d.

Slope

u ?

2H b 2

k h d

K u H (8:21b)

Moment

M ?à

K MH (8:21c)Shear force

V ?àHK VH

(8:21d)

The influence factors K D H ,K u H ,K MH ,and K VH are given in Table 8.1.8.3.1.1.2Case (2):Lateral Deformation due to Moment M Horizontal displacement

D ?

2M 0b 2

k h d

K D M (8:22a)Slope

u ?

2M 0b 3

k h d

K u M (8:22b)Moment

M ?M 0K MM

(8:22c)Shear force

V ?à2M o b K VM

(8:22d)The influence factors K D M ,K u M ,K MM ,and K VM are also given in Table 8.1.

FIGURE

8.11

340The Foundation Engineering Handbook

Aid for using Table 8.1for lateral load.

The following expressions can be used with Figure 8.12.

Design of Laterally Loaded Piles341

TABLE8.1

Influence Factors for the Linear Solution

b L Z/L K(áH)K(u H)K(MH)K(VH)K(D M)K(u M)K(MM)K(VM)

2.00 1.1376 1.134101à1.0762 1.076210

2.00.1250.8586 1.08280.18480.5015à0.65790.83140.93970.2214 2.00.250.60150.96730.2620.1377à0.29820.61330.79590.3387 2.00.3750.37640.83330.2637à0.1054à0.03760.43660.61380.3788 2.00.50.18380.71150.218à0.24420.14630.30680.42620.3639 2.00.6250.01820.61920.1491à0.29370.27670.2220.25640.3101 2.00.75à0.12880.56280.0776à0.26540.37470.17570.12080.2282 2.00.875à0.26590.53890.0222à0.16650.45720.15780.03180.1241

2.01à0.39990.5351000.53510.155100

3.00.1250.64590.89190.25080.3829à0.38540.64330.89130.2514 3.00.250.35150.66980.31840.0141à0.01840.34930.66840.3202 3.00.3750.14440.43940.285à0.16640.16070.14290.4360.2887 3.00.50.01640.25280.2091à0.22230.21620.01680.24580.215 3.00.625à0.05290.12710.1272à0.20570.2011à0.04890.11480.1353 3.00.75à0.08610.05840.0594à0.15190.1524à0.07630.03960.0684 3.00.875à0.10210.03210.0154à0.08070.0916à0.08390.00690.0225

3.01à0.1130.0282000.0282à0.084700

4.00 1.0008 1.00150à0.00000.0282à0.08470.00000

4.00.12500.53230.82470.29070.2411à0.24090.53440.82290.2910 4.00.25000.19790.51010.3093à0.11080.11360.20100.50820.3090 4.00.37500.01400.24030.2226à0.20550.21180.01780.23970.2200 4.00.5000à0.05900.06820.1243à0.17580.1858à0.05580.07200.1176 4.00.6250à0.0687à0.01760.0529à0.10840.1200à0.0696à0.00430.0406 4.00.7500à0.0505à0.04880.0147à0.04750.0538à0.0616à0.0206à0.0025 4.00.8750à0.0239à0.05520.0014à0.0101à0.0033à0.0535à0.0096à0.0148

4.0 1.00000.0038à0.0555à00.0000à0.0555à0.0517à0.0000à0

5.00 1.0003 1.00030 1.0000à1.0003 1.0002 1.00000

5.00.12500.43420.74760.31310.1206à0.12100.43430.74720.3133 5.00.25000.09010.36280.2716à0.18170.18180.09070.36200.2720 5.00.3750à0.04660.10130.1461à0.19190.1930à0.04550.10020.1461 5.00.5000à0.0671à0.01570.0494à0.11330.1163à0.0654à0.01610.0482 5.00.6250à0.0456à0.04350.0026à0.04120.0461à0.0444à0.0409à0.0012 5.00.7500à0.0197à0.0369à0.0088à0.00080.0055à0.0221à0.0276à0.0159 5.00.87500.0002à0.0279à0.00440.0108à0.0139à0.0110à0.0086à0.0125 5.0 1.00000.0167à0.0259à00.0000à0.0259à0.0091à0.0000à0 8.3.1.2Fixed-Headed Piles

Due to the elastic nature of the solution,lateral deformation of the fixed-headed piles can be handled by superimposing the deformations caused by:(1)the known deforming lateral force and the unknown restraining pile head moment,or(2)the known deform-ing moment and the unknown restraining pile head moment.Then,by setting the pile head rotation to zero(for fixed end conditions),the unknown restraining moment and hence the resultant solution can be determined.

Example8.2

The300mm wide steel pile shown in Figure8.13is one member of a group held together by a pile cap that exerts a lateral load of8kN on the given pile and a certain magnitude of a moment required to restrain the rotation at the top.It is given that the coefficient of

FIGURE

8.12

FIGURE8.13 Illustration for Example

8.2.

342The Foundation Engineering Handbook Aid for using Table8.1for moment.

horizontal subgrade modulus is 1000kN/m 3and invariant with the depth.Determine the lateral deflection and the restraining moment at the top.Assume that the second moment of area (I )of the steel section is 2.2?10à6m 4and the elastic modulus of steel to be 2.0?106kPa.

Solution

First determine b ?k h B 4E p I 1=4?(1000)(0:3)

4(2;000;000)(0:0000022) 1=4?2:03m à1(Equation 8.20)

But L ?3.75m,therefore,b L ?7.61.

Then,determine the lateral displacement and the slope due to a force 8kN (Equation 8.21)

D H ?

2H b k h d K D H ?

2(8)(2:03)

(1;000)(0:3)

(1:0)?0:108m ?108mm u H ?2H b 2k h d K u H ?2(8)(2:03)2(1;000)(0:3)

(1:0)?0:219rad

If the restraining moment needed at the top is M ,then the lateral displacement and the

slope due to M are evaluated as follows (Equation 8.22):

D M ?2M 0b 2k h d K D M ?2M (2:03)2

(1000)(0:3)(à1:0)?à0:0275M m

u M ?

2M 0b 3k h d K u M ?2M (2:03)3

(1000)(0:3)

(1:0)?0:056M rad For restrained rotation at the top,

0.056M t0.219?0;M ?à3.93kN m Then D M ?0:108m

Hence,the total lateral displacement is D M tD H ?0:216m.8.3.2

Nonlinear Methods

Several nonlinear numerical methods have become popular nowadays due to the avail-ability of superior computational capabilities.Of them the most widely used ones are the stiffness matrix method of analysis and the lateral force–deflection (p –y )approach.8.3.2.1Stiffness Matrix Analysis Method

This method is also known as the finite element method due to the similarity in the basic formulation of the conventional finite element method and the stiffness matrix analysis method.First,the pile is discretized into a number of one-dimensional (beam)elements.analysis.The following notation applies to Figure 8.14:

1,2,...,N (in bold)—node number

P i (i even)—internal lateral forces on pile elements concentrated (lumped)at the nodes

Design of Laterally Loaded Piles 343

Figure 8.14shows a typical discretization of a pile in preparation for load–deflection

P i (i odd)—internal moments on pile elements concentrated (lumped)at the nodes X i (i even)—nodal deflection of each pile element X i (i odd)—nodal rotation of each pile element

K j —lateral soil resistance represented by an equivalent spring stiffness (kN/m)Based on slope–deflection relations in structural analysis,the following stiffness relation can be written for a free pile element (i.e.,1,2):

P 1P 2P 3P 42

6643775?4EI =L 6EI =L 2

2EI =L à6EI =L 2

6EI =L 212EI =L 36EI =L 2à12EI =L 32EI =L 6EI =L 24EI =L à6EI =L 2à6EI =L 2

à12EI =L 3

à6EI =L 2

12EI =L 3

266

43775X 1X 2X 3X 42664377

(8:23)

where EI is the stiffness of the pile and L is the length of each pile element.

If the pile is assumed to be a beam on an elastic foundation,then the modulus of lateral subgrade reaction k h at any depth can be related to the lateral pile deflection at that depth by the following expression:

p ?k h y

(8:24)

Hence the spring stiffness K j can be expressed conveniently in terms of the modulus of lateral subgrade reaction k h as follows:For buried nodes:

K j ?LBk h

(8:25)

FIGURE 8.14

Stiffness matrix method of analyzing laterally loaded

piles.

344The Foundation Engineering Handbook

For surface node:

K j ?0:5LBk h

(8:26)

where B is the pile width (or the diameter).

8.3.2.1.1Estimation of the Modulus of Horizontal (Lateral)Subgrade Reaction k h

Bowles (1996)suggests the use of the following relation to evaluate k h (at different nodes)corresponding to different depths.

k h ?A h tB h Z n

(8:27)

where A h and B h are evaluated using the bearing capacity expressions as follows:

A h ?F w 1C m C (cN c t0:5g BN g )

(8:28a)B h ?F w 2C m C g N q

(8:28b)where Z is the depth of the evaluated location.

The following values are suggested by Bowles (1996)for the above constants:C ?40when using units of kN/m 3,C m ?1.5–2.0,n ?0.4–0.6,and

F w1,F w2?1.0for square and HP piles and in cohesive soils.F w1?1.3–1.7;F w2?2.0–4.4for round piles.

Thus,Equations (8.25)–(8.28)can be used to evaluate k h and hence K j at each relevant node.This is illustrated in the following example.

Example 8.3

Evaluate the equivalent spring stiffness at each node of the 300mm ?300mm square pile the unit weight of the sand layer is 15and 16.5kN/m 3,respectively.

Solution

c ?42,N q ?29,an

d N g ?29.

From Equations (8.28)

A h ?F w 1C m C (cN c t0:5g BN g )?(1:0)(1:5)(40)(0:5)(16:5)(0:3)(29)?4306:5

B h ?F w 2

C m C g N q ?1:0(1:5)(40)(16:5)(29)?28,710

Applying Equation (8.27)k h ?A s tB h Z n ?4307t28710Z 0:5kN =m 3.

Figure 8.15also shows the k h distribution with the depth and the equivalent spring stiffness corresponding to each node.

It must be noted that Equations (8.25)and (8.26)have been applied to determine the K values.

The element stiffness matrices given by expressions such as Equation (8.23)can be assembled to produce the global stiffness matrix [K ]using basic principles of structural analysis.During the assembling process,the spring stiffness K j of each underground node can be added to the corresponding diagonal element of [K ]

[P ]?[K ][X ]

(8:29)

Then,knowing the global force vector one can solve Equation (8.29)to obtain the global deflection vector.

Design of Laterally Loaded Piles 345

For N ?15,from Equation (3.23),F %348.Also from Table 3.1,N shown in Figure 8.15.Assume that the overburden corrected average SPT (N )value and

Example 8.4

by a pile cap that exerts a lateral

load of 8kN on the given pile and a moment of certain magnitude required to restrain the rotation at the top.It is given that the coefficient of horizontal subgrade modulus is 1000kN/m 3and invariant with the depth.Determine the relevant force and deflection vectors assuming that the total number of nodes is 6.Also illustrate the solution procedure to obtain the lateral deflection of the pile and the moment required at the cap.Assume that the second moment of area (I )of the steel section is 2.2?10à6m 4and the elastic modulus of steel is 2.0?106kPa.

Solution

The equivalent spring stiffness has been computed as in Example 8.3and indicated in Figure 8.16.As shown in Figure 8.16,the only external forces applied on the pile are the ones applied by the pile cap and the soil reactions at the bottom that assure fixity.It is also noted that the spring associated with the bottom-most node has been added to the unknown force P 12.

Hence,the external force vector is given by the following equation:

P ? ?M 1

M 11

P 12? T

(8:30)

On the other hand,the deflection vector is given by the following equation

X ? ?0

u 2

D 2

u 3

D 3

u 4

D 4

u 5

D 5

(8:31)

in which it is assumed that the rotation at the top is restrained due to the pile cap (i.e.,X 1?u 1?0)and the translation as well as the rotation at the bottom are retrained by the ground fixity (i.e.,X 12?D 6?0and X 11?u 6?0).The required lateral deflection is D .

4307 kN/m 3

33,017 kN/m 344,909 kN/m 3

54,034 kN/m 3

k S values

FIGURE 8.15

Illustration for Example 8.3.

346The Foundation Engineering Handbook

The 300mm wide steel pile shown in Figure 8.16is one member of a group held together