laterally loaded piles.pdf

Draft on Laterally Loaded Piles M. Budhu April, 2002

Laterally loaded piles Structures founded on piles are often subjected to lateral loads and moments in addition to vertical loads. Lateral loads may come from wind, traffic, seismic events, waves and earth pressures. Moments may come from the eccentricity of the vertical force, fixity of the superstructure to the pile or piles and the location of the lateral forces on the pile with reference to the ground surface. When a pile is subjected to lateral forces and moments, the pile tends to bend or deflect as illustrated in Figure 1. The deflection of the pile causes strains in the soil mass. To satisfy equilibrium, the soil must provide reactions along the length of the pile to balance the applied loads and moments.

Figure 1 Pile – soil response to lateral loads and moments Because soil is a non-linear material, the soil reaction is not linearly related to the pile deflection. Consequently at every point along the length of the pile, a non-linear relationship between soil resistance (p) and pile deflection (y) exists as illustrated in Figure 1. In designing laterally loaded piles, we need to know the pile deflection, particularly the pile head deflection, to satisfy serviceability requirement and the bending moments for sizing the pile. The pile head deflection depends on soil type, pile installation, pile flexibility (or pile stiffness), loading condition and on how the pile is attached to the superstructure and pile cap. A


pile that is attached to the pile cap such that no rotation occurs is called a fixed head pile (Figure 2a). A pile that is attached to the pile cap such that rotation is unrestricted is called a free head pile (Figure 2b). The mechanism of failure depends on the length to diameter or width ratio, soil type and the fixity of the pile head. Free head piles tend to fail by rotation. Lateral loads and moments applied to a free head pile is initial resisted by the soil near ground level. For very small pile deflections, the soil behaves elastically and as the deflection increases the soil yields and then permanent soil displacement occurs. The soil resistance is shifted to the lower part of the pile as yielding progressively occurs from the top to the bottom of the pile. Fixed head piles tend to fail by translation. Piles in general are neither fixed head nor free head. They have undermined fixity somewhere between free head and fixed head conditions. You can view fixed head and free head as two limiting conditions in which piles in practice will respond somewhere within these limits.

(b) (a)

Passive soil pressure


Yield and failure

Yield and failure

(d) (c) (a) Short, free head (b) Short, fixed head (c) Long, free head (d) Long, fixed head Figure 2 Possible failure modes in short and long piles. Two types of piles are normally defined to distinguish failure mechanisms. One is called a short pile that is characterized by length to diameter (or width) ratio of less than 10 (sometimes a ratio of up to 15 is quoted in the literature). Short, free head piles tend to fail by rotation about a

Passive soilpressure


Translation of a short, fixed head pile


point near the pile tip (or base). Rotation occurs when the sum of the soil resistance (called passive pressures) at the top part of the pile is lower than the sum of the soil resistance at the lower part of the pile. Short, fixed head piles tend to fail by translation. The failure of short piles is due to soil failure. Long piles tend to fail by structural failure near ground level rather than by soil failure.

Figure 3 Group piles subjected to lateral loads Laterally loaded pile, particularly group piles are particularly difficult to analyze mainly because of the complexity of the soil–structure (pile) interaction. The displacement and rotation are in the directions of the resultant lateral load and resultant moment (Figure 3). Outer piles in a group are subjected uplift (pull) and compressive (push) forces while the piles in the center translate at the level of the superstructure connection. The response of a pile group to lateral loads and moments is influenced by

1. Geometry of the group 2. Pile-soil interaction 3. Stiffness or flexibility of the piles 4. Load conditions 5. Individual pile response 6. Pile group response resulting from individual pile responses. Analyses of lateral loaded piles Basic structural mechanics Consider an element of thickness, dz, of the pile shown in Figure 1 at a depth z. The free body diagram for this element is shown in Figure 4. We will assume that the pile is symmetrical about the z-axis, the loads lie in the yz plane, deflection of the pile occurs only along the y-axis, i.e., no out of plane deflection, and shear deflection is negligible.


Figure 4 Free body diagram of a pile section under lateral loads and moments. Taking moments about O, we get

( ) zdzM dM M P dy Vdz p dz 02

+ − + − + = Eq.1

Neglecting higher terms (e.g. dz2) and rearranging Eq.1 gives

zdM dyP Vdz dz

+ − = 0 Eq. 2

Let us differentiate equation with respect to z, then 2 2

z2 2

d M d y dVPdz dz dz

+ − 0= Eq. 3

From basic mechanics, we have the following identities 2 4

p p2

d M d yE Idz dz

= 4 Eq. 4

dV pdz

= Eq. 5

p k= − y Eq. 6 where Pz is the vertical load, p is the soil resistance, y is the soil compression (or pile deflection), Ep is the elastic modulus of the pile, k is a soil (stiffness) parameter, and Ip is the second moment of area of the pile. Substituting the above Eqs. 4-6 into Eq. 3 gives the governing equation for a laterally loaded elastic pile as

4 2

p p z4 2

d y d yE I P ky 0dz dz

+ + = Eq. 7

If the pile above the ground level is subjected to a distributed load, for example from water, then the governing equation becomes

4 2

p p z4 2

d y d yE I P ky W 0dz dz

+ + − = Eq. 8


where W is the resultant of the distributed load. Eq. 8 can be solved using numerical methods such as the finite difference method or finite element method. The finite difference equations are shown in Appendix D. The term, ky, in the governing equation changes depending on the assumption made regarding the soil response. One can assume an elastic soil or an elasto-plastic soil or some other type of load-deformation response. The solution of Eq. 8 therefore varies on the assumption of how the soil will respond. Three methods of obtaining solutions to the governing equation based on three different types of soil responses are described below. . 1. Elastic beam on elastic foundation method In this method, the pile is assumed to be an elastic beam that is attached to discrete springs representing the soil (Figure 5). The soil is assumed to be elastic but discontinuous. The soil stiffness parameter, k, is taken as the lateral subgrade modulus of the soil. It is assumed to be a constant value for fine-grained soils and linearly increasing with depth for coarse-grained soils. The lateral subgrade modulus, Kh, is the ratio of the horizontal soil reaction per unit area to the lateral soil displacement. The lateral subgrade modulus can be obtained from field tests using instrumented test piles. The lateral soil reactions and the bending moments are inferred from grain gauges attached to the test pile. The determination of the lateral subgrade modulus is often difficult and expensive. Empirical relationships are often used to relate results from simpler and routine tests such as SPT, cone tests and or laboratory shear tests to Kh. A relationship between Kh and elastic properties was proposed by Valsangkar et. al. (1973) as shown in Table 1. The main problem with the elastic beam on elastic foundation approach is that it neglects soil continuity and soil shearing resistance. This method is satisfactory for very small strain levels (<0.001%)

Discrete springs representing adjacent soil mass

Figure 4 Simulation of adjacent soil mass as a set of discrete springs 2. p-y analysis


The p-y (soil resistance – pile deflection) method (Matlock and Ripperger, 1958; Reese, Cox and Koop, 1975 and others) is conceptually similar to the beam on elastic foundation method except that the soil stiffness parameter k is not constant i.e. there is a non-linear relationship between the soil resistance and its displacement (equal to the pile deflection. The non-linear relationships are obtained from results of well-instrumented test piles. General procedures have been developed to construct p-y curves for many soil types lessening the need to conduct lateral load tests, at least, in the preliminary stages of design. However, it is recommended that pile load tests be conducted whenever such tests are feasible and economical. Computer programs (e.g. Com 624) are available for routine use. Like the elastic beam on elastic foundation approach, the p-y method neglects soil continuity and soil shearing resistance. 3. Continuum Analysis In this method of analysis, soil is treated as a continuous media with assumptions made on its stress-strain behavior or constitutive relationships. The simplest stress-strain behavior is elastic described by Hooke’s law. The solution gives the load-deformation response of the pile for small strain levels (Poulos,1971; Randolph, 1981). Soil yielding or pile yielding cannot be obtained from this analysis. To account for soil yielding, the soil can be assumed to be an elasto-plastic material (Davies and Budhu, 1986; Budhu and Davies,1986, 1987).

Soil active stress at back face

Shear stress on side faces

Soil resistance on bearing face

H

e Figure 6 Soil stresses on a pile segment resulting from lateral loads that are taken into account in Davies and Budhu method We will adopt the elasto-(rigid) plastic solution using boundary element method proposed by Davies and Budhu (1986) and Budhu and Davies (1986, 1987) for the general analysis of lateral loaded single piles. The soil is assumed to behave linearly elastic at small strain levels. Yielding occurs when the normal and shear stresses on the pile reach values determined from a limiting equilibrium analysis. Bearing failure on the front face, shear stresses along the sides and tension


failure along the back face (Figure 6) are incorporated into the analysis. Failure on the bearing face occurs when

bf c uN sσ = where σbf is the limiting compressive stress, Nc is the bearing capacity factor that is assumed to vary from 2 at the surface to a constant value of 9 at a depth of three pile diameters and below. The limiting shear stress at the sides of the pile is

ss usuτ α= where the adhesion factor, αu, is taken as 1 for soft clays. On the back face or tension face, the normal stress cannot exceed the in situ lateral effective stress, i.e.,

bf is zoKσ σ ′≤ where Kis is the lateral earth pressure coefficient that accounts for pile installation. For soft clays, Kis, is usually 1.0 or greater. The important findings from the continuum analysis are as follows. 1. The analysis is valid only for a pile with a length greater than its effective length. The

effective length is the length of pile that is effective in resisting the loads and moments. Piles longer than their effective lengths will behave identically.

2. The maximum bending moment in free head piles would normally occur at depth less than 5 diameters below ground level.

3. Eccentric lateral loads increase the maximum bending moment (regardless of the type of analysis).

4. Fixed head piles reduce the pile head deflection by about 50%. 5. Laterally loaded piles, in general, fail by yielding of the pile section well before the soil fails. 6. The lower the flexibility of the pile the greater the pile deflection and the greater the

maximum bending moment. Flexibility, K, is the ratio of the elastic modulus of the pile to the elastic modulus of the soil.

The predictions from boundary element analyses are curve-fitted to give equations that can be easily used in practice. A summary of the relevant equations for single, solid piles embedded in homogenous stiff clays (undrained shear strength constant with depth), homogenous soft clays (undrained shear strength varies linearly with depth) and sand are summarized in Table 1. Table 1 Design equations for laterally loaded piles

Soft Clay

Free head pile Fixed head pile

Effective pile length (Le) Le = 1.3 K 9

2

; K = mDE p Le = 1.3 K 9

2

su Es

mc

z

z


Collapse load ( )c ru2

H W L1.2 1 0.25 0.008g 0.88 DcDL

⎡ ⎤= + α +⎢ ⎥+ ⎣ ⎦

2g3

∞ > ≥ − ; u0 1≤ α ≤

( )c ru2

H W L1.2 4.5 0.025g 0.88 DcDL

⎡ ⎤= + α +⎢ ⎥+ ⎣ ⎦

2g3

∞ > ≥ − ; u0 1≤ α ≤

Yield load/Working load

2y 3

e3

3 14 4

e

H0.5n

cD

exp( 2f n )−

= β

β = −

Replace Hy by Hw to get working load

2y 33

H0.8n

cD=

Pile deflection (elastic) at ground surface

uE = IuH 2mD

H + Ium 3

MmD

IuH = 3.2K 31

−

IuM = 5.0 K59−

uE = IuH 2mD

H

IuH = 1.4K 31

−

Pile deflection including yield at ground surface

u = Iuy uE

Iuy = 1 + 0.32

n n0.53n

h 1.4k40k−

3=nHh

cD; n

Kk1000

=

u = Iuy uE

Iuy = 1 + 0.43

n n0.54n

h 32k105k−

3=nHh

cD; n

Kk1000

=

Pile rotation (elastic) at ground surface

θE = IθH 3mD

H + Iθm 4

MmD

IθH = 5.0 K 95

−

Iθm = 13.6 K 97

−

Iθy = 1 + 0.32

n n0.53n

h 14 k54 k−

(None)

Pile rotation including yield at ground surface

θ = Iθy θE

Iθy = 1 + 0.32

n n0.53n

h 14 k54 k−

(None)

Maximum bending moment (elastic)

ME = IMH HD

IMH = 0.3 K 92

for e = 0 For eccentric loading ImH = aKb

a = 0.6f b = 0.17f –0.3

8 > f> 0.5

ME = -IMH HD

IMH = 0.4 K 92

Maximum bending moment including yield

Mm = IMy ME

IMy = 1 + 0.32

0.48

896−n n

n

h kk

Mm = IMy ME

IMy = 1 + 0.53

0.56

30312−n n

n

h kk


Maximum bending moment occurs at depth, Lm, from ground surface

Lm = 0.53 K 92

D; L > Le

Stiff Clay

Free head pile Fixed head pile

Effective pile length (Le)

4110.5

Le KD=

; p

sE=

EK

Collapse load ( )c ru

u

H W L4.1 1 0.25 0.008s DL 1.2g 1 D

⎡ ⎤= + α +⎢ ⎥+ ⎣ ⎦

2g3

∞ > ≥ − ; 0 1u

z

z

su Es

≤ α ≤

( )c ru2

H W L1.2 4.5 0.025g 0.88 DcDL

⎡ ⎤= + α +⎢ ⎥+ ⎣ ⎦

2g3

∞ > ≥ − ; u0 1≤ α ≤

Yield load/Working load

2y 3

e3

3 14 4

e

H0.5n

cD

exp( 2f n )−

= β

β = −

Replace Hy by Hw to get working load

2y 33

H0.8n

cD=

Pile deflection (elastic) at ground surface

uE = IuH 2mD

H + Ium 3

MmD

IuH = 3.2K 31

−

IuM = 5.0 K59−

uE = IuH 2mD

H

IuH = 1.4K 31

−

Pile deflection including yield at ground surface

u = Iuy uE

Iuy = 1 + 0.32

n n0.53n

h 1.4k40k− ;

u = Iuy uE

Iuy = 1 + 0.43

n n0.54n

h 32k105k− ;

Pile rotation (elastic) at ground surface

θE = IθH 3mD

H + Iθm 4

MmD

(None)


IθH = 5.0 K 95

−

Iθm = 13.6 K 97

−

Iθy = 1 + 0.32

n n0.53n

h 14 k54 k−

Pile rotation including yield at ground surface

θ = Iθy θE

Iθy = 1 + 0.32

n n0.53n

h 14 k54 k−

(None)

Maximum bending moment (elastic)

ME = IMH HD

IMH = 0.3 K 92

for e = 0 For eccentric loading ImH = aKb

a = 0.6f b = 0.17f –0.3

8 > f> 0.5

ME = -IMH HD

IMH = 0.4 K 92

Maximum bending moment including yield

Mm = IMy ME

IMy = 1 + 0.32

0.48

896−n n

n

h kk

Mm = IMy ME

IMy = 1 + 0.53

0.56

30312−n n

n

h kk

Maximum bending moment occurs at depth Lm, from ground surface

Lm = 0.53 K 92

D; L > Le

K = mDE p , Ep = elastic modulus of pile, m = slope of line depicting the linear variation of the

elastic modulus of the soil with depth, c = slope of line depicting the linear variation of the undrained shear strength ( us ) with depth and D is the diameter of the pile. For non-cylindrical

and annual piles, the equivalent pile stiffness is, Ep = 4

64

R Rp pE IDπ⎛ ⎞

⎜ ⎟⎝ ⎠

where Ip is the second moment of

area and the superscript R refers to the real pile.

nKk

1000= , 3=n

HhcD

, where c slope of the undrained shear strength with depth, or

KW

cγ

= , γ = soil

unit weight (use effective weight for submerged section), Ko is the lateral earth pressure coefficient, which should account for the method of pile installation ( a value of 1 is reasonable

for soft clays), egL

= and efD

= . The above equations are for solid, cylindrical piles of diameter,


D, with ncDσ

= where σ is either the yield stress or working stress. For other types of geometry,

an equivalent n value can be calculated from 3

Iy

nDcD32

⎛ ⎞⎜ ⎟σ ⎝ ⎠=π

where I is second moment of area, D is

the projected width and y is the distance from the neutral axis to the extreme fibers. Procedures

1. Obtain soils information for the site, particularly the variation of undrained shear strength with depth.

2. Select pile (geometry and material, e.g. concrete). 3. Obtain loads and moments 4. Determine fixity condition – free head or fixed head. 5. Determine effective length and check that embedded length is greater than the effective

length. 6. Determine K, f, g, Wr, n, kn, hn 7. Calculate collapse load, yield load and working load. 8. Calculate elastic influence factors, IuH etc. 9. Calculate elastic ground surface deflection and rotation. The latter for only free head

pile. 10. Calculate elastic maximum bending moment. 11. Calculate yield influence factors, Iuy etc 12. Calculate total ground surface deflection and rotation. 13. Calculate total maximum bending moment. 14. Scale ground surface deflection as follows. Neglecting the bending of the free-standing portion of the pile, the pile head deflection = u + eθ

ILLUSTRATIVE EXAMPLE A 460-mm diameter timber pile of total length 9.75 m is embedded to a depth, L, of 9.14 m in a soft, normally consolidated, clay deposit of medium plasticity. The working stress, σw, for timber of this quality is 11 MN/m2, and its Young’s modulus of elasticity is 10.3 GN/m2. A lateral load is to be applied at a height, e, of 690 mm above ground level. Required are: (1) The working load; (2) the lateral deflection; and (3) the maximum bending moment. Assume: αu = 1; Ko = 1; γ = 9.4 kN/m3; c = 2.7 kN/m3; and m = 2.7 MN/m3, i.e., m/c =1,000. SOLUTION Step 1: Calculate normalized values

or

'K 9.4 1Wc 2.7

γ ×= = = 3.5 , L/D = 20, Wr L/D = 70, pE

K 8mD

= = 300 , kn = 8300/1000 = 8.3, g = 0.075, f

= 1.5, n = σy/cD = 8900, 3 14 4

e exp( 2f n ) 0.76−

β = − = Step 2: Calculate collapse load and working load


( )c ru2

H W L1.2 1 0.25 0.008g 0.88 DcDL

⎡ ⎤= + α +⎢ ⎥+ ⎣ ⎦; Hc = 236 kN

2w 3

e w3

H0.5n ;H 42kN

cD= β =

Step 3: Calculate working moment M = Hw e = 29 kNm. Step 4: Calculate elastic influence factors IUH = 0.16, IθH = 0.033, and IθM = 0.012 Step 5: Calculate elastic deflection and rotation Ground level:

uE = IuH w2

HmD

+ Ium 3

MmD

; uE = 16 mm

θE = IθH w3

HmD

+ Iθm 4

MmD

; θE = 0.0083 rad.

Step 6: Calculate yield influence factors Iuy = 2.1, Iθy = 1.8 Step 7: Calculate deflection and rotation at pile head Ground level: u = Iuy uE ; u = 33 mm, θ = Iθy θE; θ = .015 rad. Pile head: Neglecting the bending of the free-standing portion of the pile, upile head = 33 + 690 x 0.015 = 43 mm Step 8: Calculate bending moment on pile ImH = aKb; a = 0.6f, b = 0.17f –0.3 ; ImH = 3.5 ME = IMH HwD, ME = 68 kN.m. IMy = 1 +

0.32

0.48

896−n n

n

h kk

; IMY = 1.54

Thus, MMY = 104 kN m.


Appendix D

0=−+ VvdxdyPx

dxdM

2.2

02

2

2

2

=−+dx

dVvdx

ydPxdx

Md 2.3

4

4

2

2

dxydEI

dxMd

= 2.4

pdx

dVv= 2.5

yEp s−= 2.6

02

2

4

4

=++ yEdx

ydPxdx

ydEI s 2.7

SPxSVvVn sincos −= 2.8

dxdyPxVvVn −= 2.9

02

2

2

2

=−++ Wkydx

ydPxdx

Md 2.13

( ) ( )

( ) 41211

1111122

2 122

422hRYRRY

RRRyRRyRydx

Md

mmmmm

mmmmmmmmm

m⎥⎦

⎤⎢⎣

⎡+−−+

+++−−+=⎟⎟

⎠

⎞⎜⎜⎝

⎛

++++

+−−−−− 2.14

( )

211

2

2 2h

yyyPxdx

ydPx mmm

m

+− +−=⎟⎟

⎠

⎞⎜⎜⎝

⎛ 2.15

mmm IER = 2.16

( ) ( )( ) 022

24224

122

11

4211

21112

=−++−−+

+−++++−−+

++++

+−−−−−

hWRyPxhRRy

hKPxhRRRYPxhRRYRY

mmmmmm

mmmmmmmmmm 2.17

00

2

2

=⎟⎟⎠

⎞⎜⎜⎝

⎛dx

yd 2.18

000

2

3

0 =⎟⎠⎞

⎜⎝⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛dxdyP

dxydR x 2.19


101 RRR ==− 2.20

02 101 =+−− yyy 2.21

20

2

10

2

12 22 yR

PxhyR

Pxhyy +⎟⎟⎠

⎞⎜⎜⎝

⎛−−⎟⎟

⎠

⎞⎜⎜⎝

⎛−= −− 2.22

020100 dybyay +−= 2.23

24010

21

0 2222

PxhhkRRhPRR

a xo

−++−+

= 2.24

24010

10 2PxhhkRR

RRb o

−+++

= 2.25

24010

40

0 2PxhhkRRhW

d−++

= 2.26

mmmmmm dybyay +−= ++ 21 2.27

( )m

mmmmmmmmmmm c

bPxhRRbRRbaRba 1

21111211 12222 −+−−−−−− −++−++−

= 2.28

m

mm c

Rb 1+= 2.29

( )

m

mmmmmmmmm c

RdPxhRRRadhWd 12

21121

4 22 −−−−−− −+−−−= 2.30

( )124

111212111

2

242

−

−−−−−−−−−

−−+

−++−−=

mm

mmmmmmmmmmmm

aPxhhk

RaRRaaRbRaRc 2.31

thSJ 21 = 2.32

t

t

RhM

J2

2 = 2.33

t

t

RhP

J3

32

= 2.34


t

t

t SM

RhJ

24 = 2.35

tRPxhE

2−= 2.36

( ) ( ) tttttttt Pyy

hPxyyyy

hR

=−+−+− +−++−− 1121123 222

2 2.37

( ttttt Myyy

hR

=+− +− 112 2 ) 2.38

1

2

QQyt = 2.39

2

1121 G

dyGJy tt

t−

+

−+= 2.40

t

ttttt b

dyyay

+−= +

+1

2 2.41

tt bG

GaGHGHQ 11

2

1

2

2111 ⎟⎟

⎠

⎞⎜⎜⎝

⎛−++= 2.42

( ) ( ) ( ) 221

2

212

2

1232 2 −−−

−− −++++−

+−

+= tttt

tt

t

tt daEdbd

GJdH

GbdJa

JQ 2.43

11 2 −−= taG 2.44

12 1 −−= tbG 2.45

212111 2 −−−−− +−−−= ttttt aabEaaH 2.46

( 11122 122 −−−− ++++−= tttt bEbbaH ) 2.47

111 Jyy tt =− +− 2.48

3

4

QQyt = 2.49


4

1111 G

dJyay ttt

t−−

+

+−= 2.50

t

ttttt b

dyyay

+−= +

+1

2 2.51

tt

ttt

bGbaa

GaH

HQ 1

4

1

4

1213 +−+= −− 2.52

( ) ( ) 221

4

2141

4

2134 2

1−−−

− −−+++−−−

−+= tttt

ttttt daEdGb

HdbdGdJaGHJJQ 2.53

14 1 −+= tbG 2.54

411

11 2J

yyyyy

tt

ttt =−

+−

+−

+− 2.55

( )

( ) ( ) ( )

tt

t

tttt

t

tt

t

bH

ba

HHH

GJGJHd

daEdbtdt

GJGbJda

Jy

1

12

1

3321

442

421221

442

413

+−+

+−

+−−++++−

−=

−−−−

−

2.56

( ) ( ) ( )

442

413

442

411411

11GJGJd

yHGJG

JdaJGyy t

tttt

t +−

−=+

−−+= −−−

+ 2.57

( ttttt

t dyyab

y +−= ++ 121 ) 2.58

442

1413 GJG

aJGH t

++

= − 2.59

tt yy −= 2.60

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