special cases piles

SPECIAL CASES PILES

ASSOC. PROF. Ir. DR. RAMLI NAZIR

TEL : 013 7927925

OFF: 07 5531722

INSPIRING CREATIVE AND INNOVATIVE MINDS

Ultimate Lateral Resistance Of Pile

• Normally pile resist vertical load. However in some cases horizontal load is more prominent such as gantries, quay, trnsmission tower, unipole structure etc.

• In designing such foundation, two criteria must be satisfied:-

1. Adequate factor of safety against ultimate failure.

2. Acceptable deflection at working loads.

• In many practical cases, the design of piles for lateral loading will be dependant on satisfying a limiting lateral-deflection requirement that may result in the specification of allowable lateral loads much less than the ultimate lateral capacity of the piles.

ULTIMATE SOIL RESISTANCE

• For purely cohesive soil, the ultimate lateral resistance, Pu increases from the surface down to the depth of about 3 pile diameters and remain constant for a greater depth.

• When Pu becomes constant, lateral

failure involves plastic flow of the soil around the pile in horizontal plane only and the value Pu can be determine by plastic theory.

• For more general cases of a c – soil, an alternative derivation of the ultimate lateral soil resistance, based essentially on earth pressure theory has been given by Brinch Hansen (1961)

8 – 12 CuD

2Cu

3D

• He considers the variation of resistance with depth along the pile. The ultimate resistance at any depth, z below the surface is expressed as :-

Pu = qKq + c Kc

Where q = vertical overburden pressure

c = soil cohesion value

Kc and Kq = Factors that are a function of and z/d

BROMS’ THEORY

• Essentially similar as Brinch Hansen method except that simplifications are made to the ultimate soil resistance distribution along the pile.

• Full consideration is given to restrained or fixed head piles as well as unrestrained or free head pile

Unrestrained or Free Headed Pile

dc9Hf

u

u

• For Short Pile (Rigid Pile) • Lateral resistant is dependant wholly on the soil resistance.

• For Long Pile (Flexible Pile) • Lateral resistance is primarily dependant on the yield moment of the pile

itself.

• f defines the location of the maximum moment, and since the shear there is zero,

Eqn 1

• Also taking moments about the maximum moment location,

Mmax = Hu(e + 1.5d + 0.5f) ………………(Eqn 2)

• Also,

Mmax = 2.25dg2cu …………………………( Eqn 3)

• Since L = 1.5d + f + g , eqn (1) and (2) can be solved for the ultimate lateral load, Hu.

• The solution is plotted in terms of L/d and H/cud2 and applies for short piles in which the yield moment My > Mmax

• For long piles, Eqn (3) no longer holds.

• Hu is obtained from Eqn(1) and (2) by setting Mmax equal to the known value of yield moment (My)

• The solution is plotted in terms of :

Hu/cud2 and My/cud2

RESTRAINED / FIXED HEADED PILE

• The change over points from one failure mode to another depend again on the yield moment of the pile.

• It is assumed that moment restrained equal to the moment in the pile just below the cap is available.

• For short piles:

Hu = 9cud(L – 1.5d) ………….Eqn(4)

Mmax = Hu(0.5L + 0.75d) ……………..Eqn(5)

• For intermediate piles Eqn(3) and Eqn(1) holds and taking moments about the surface.

My = 2.25cudg2 – 9cudf(1.5d + 0.5f)………………Eqn(6)

• This equation together with the relationship

L = 1.5d + f + g

May be solved for Hu.

• It is necessary to check the maximum positive moment, at depth

f + 1.5d is less than My.

• Otherwise the failure mechanism for long piles holds.

• For the later mechanism, the following relationship applies :

…………..Eqn(7)

f5.0d5.1

M2H yu

PILES IN COHESIONLESS SOIL

• Following assumptions are made by Broms

1. Active pressure acting on the back of the pile is neglected.

2. Distribution of passive pressure along the front of the pile is equal to three times the Rankine passive pressure.

3. The shape of the pile section has no influence on the distribution of ultimate soil pressure or the ultimate lateral resistance.

4. The full lateral resistance is mobilised at the movement considered.

• The distribution of soil resistance is :

Pu = 3’vKp

Where ’v: effective vertical overburden pressure

Kp: Rankine passive earth pressure coefficient

UNRESTRAINED/FREE HEAD

Le

KdL5.0Hu

p3

• Pile will act as a short pile if the maximum moment is less than the yield moment of the section.

• The rotation is assumed to be about a point close to the tip and a high pressure acting near this point are replaced by a single concentrated force at the tip.

• Taking moment about the toe,

……………….Eqn.(9)

• The relationship can be plotted using L/d and Hu/Kpd3.

• Maximum moment occurs at a distance of below the surface where:-

Hu = (3/2) dKpf2 …………………………..Eqn (10)

p

u

dKH82.0f

Where ……………………….Eqn(11)

The maximum moment is :

Mmax = Hu(e + 2/3(f)) ……………….Eqn(12)

• If after using Eqn(9), the calculated Hu results in Mmax>Myield, then the pile is treated as long pile.

• Hu may then be calculated from Eqn(10) and (12) putting Mmax=Myield.

• Solution of the long pile can be plotted as :

Hu/Kpd3 and My/d4Kp

OTHER METHODS

Deln08.024.0K

where

1DL

KLDM

m

m

3

• Dickin and Wei’s Method.

• Limitation for pile with L/D less than 3.

• L : pile embedment length

• D : pile diameter

• M : Moment at ground level

1

DL7

2S

where

SDe

DLe4M

fm

fm

22

• Dickin’s and Ramli’s method

• Using the integration of plain strain condition to three dimensional projection.

• Apply shape factor as integration coefficient.

• Limitation is for L/D < 7

• If L/D more than 7, Brom’s equation is applicable.

Sfm : Shape factor

M : moment at ground Level

Thus M = Hu x e

RESTRAINED OR FIXED HEAD PILE

• For short piles horizontal equilibrium it gives:-

Hu = 1.5L2dKp ……………………………………………..Eqn(13)

• Maximum moment is :-

Mmax = 2/3 (HuL) ……………………………..Eqn(14).

• If Mmax>Myield, the intermediate pile is relevant by giving ,f (horizontal equilibrium) as:-

f = (3/2)dL2Kp – Hu ………………………..Eqn(15)

• Taking moments about the top of the pile and substituting F from Eqn(15)

Myield = 0.5dL3Kp – HuL …………………….Eqn(16)

Hence Hu can be obtained.

)32(2 feHM uyield

• The equations only holds if the maximum moment at depth f is less than Myield. f is calculated from the earlier equations.

• For long pile where maximum moment reaches Myield at two locations, it is readily found that :-

DETERMINATION OF DISPLACEMENT AT TOP OF PILE

41

pp

h

IEBk

• Displacement at top of the pile can be obtained from Figure A for cohesive soil and Figure B for cohesionless soil.

• Figure A

• Deflection has been plotted as a function of the dimensionless length L where:-

………………………Eqn (1)

Ep : Modulus of elasticity of pile materials

Ip : Moment of inertia of pile in bending plane

B : Pile diameter or width

kh : coefficient of horizontal subgrade reaction

• Much of the accuracy depend on kh value.

• In absence of such information kh can be estimated by :

kh = nhz/B

Where z is the point considered

nh is constant of horizontal subgrade reaction for piles in soil

• Figure B

• For cohesionless soil and the relative stiffness of the pile and the soil are contained in the parameter where :-

….. Eqn(2)

51

pp

h

IEn

• Typical value of nh and kh

nh (MN/m3) for granular soil

kh (MN/m2) for cohesive soil

Soft 8.0

Medium 16.0

Stiff 32.0

Loose Medium Dense

Dry or moist 2.2 6.6 17.5

Submerged 1.25 4.4 10.5

PULL OUT RESISTANCE OF PILE

• The gross uplift resistance of the pile subjected to uplift forces:-

Tug = Tun + W

Where

Tug : Gross uplift capacity

Tun : Net uplift capacity

W : Effective weight of pile

Tug

L Tun

W

D

Pile Embedded in Saturated Clay

• Das and Seeley (1982)

Tun = Lp’cu

L : Length of pile

P : Pile perimeter

’ : adhesion coefficient at soil-pile interface

cu : Undrained cohesion of clay

• For cast insitu piles (concrete)

’ = 0.9 – 0.00625cu (for cu <= 80 kPa)

and

’ = 0.4 (for cu > 80 kPa)

• For pipe piles

’ = 0.715 – 0.0191cu (for cu <= 27 kPa)

and

’ = 0.2 (for cu > 27 kPa)

Pile Embedded in Granular Soil

dz)pf(T u

0

L

un

• fu = Unit Skin Resistance during uplift p = Perimeter of pile cross section • The unit skin friction varies with depth. It increase linearly up to a

depth of z = Lcr. Beyond that it remain constant.

• For z <= Lcr fu = Kuv’tan

Where Ku = uplift coefficient (BREAKOUT FACTOR) v’ = effective vertical stress at depth z. = soil –pile friction angle Values of Lcr and are dependant on relative density.

Procedures for calculating net uplift capacity of piles

1. Determine the relative density of the soil and obtain the value of Lcr.

2. If the pile length L <= Lcr

In dry soil, v’ = z

tanKLp21

dz).tanzK(p

dz).tanK'(pT

dz).tanK'(pdz.fupT

u2

L

0u

L

0uvun

L

0

L

0uvun

3. For case of L > Lcr

In dry soils For estimation of the net allowable uplift capacity, the FOS of 2 to 3 is recommended

Therefore Tu(all) = Tug/FOS

)LL(tanKLptanKLp21T

dz)tanK'(dz)tanK'(p

dzfdzfpdzfpT

crucru2

crun

L

0

L

Lcru)Lcrz(vuv

L

0

L

cuu

L

0uun

cr

cr

crL

RAKING OR BATTERED PILE

• When value of horizontal force acting on a group of vertical piles become too large to accommodate a common solution; raking pile is provided along the vertical one.

• For ease of analysis, the vertical pile resist the vertical load and inclined pile resist horizontal load.

• This analysis is obviously extremely conservative.

• Piles are not generally inserted into ground as battered in excess of 1H : 4V. (Usual design using 1:7 – 1:9)

• Terzaghi(1934) suggests that a graphical method describe by Lohmeyer(1934) known as Culmann’s method to be used in analysis of a mixed pile group.

ANALYSIS

• Often pile group design to resist horizontal forces that can act in either direction

• The example is made up of one set of vertical piles and 2 set of raked pile which are battered in opposite direction.

R

SIMPLIFICATION

• Assuming this is an imaginary piles acting upon by axial force only. They are pin jointed at both ends so it becomes possible to determine the line of action

A B

C

R

FORCE DIAGRAM

• Let QA, QB and QC be the forces per unit length of foundation in the imiginary piles A, B and C.

• Let Resultant QB and QC be R’.

• The R’ must pass through ‘a’, the point of intersection of QA and R (applied load).

a

QA

R’

R

QB

QC

• QA, QB,QC are obtained from the force polygon.

• Therefore the axial force in each pile in set A :

QpA = QA/nA

Where nA = number of pile in set A Similarly : QpB = QB/nB

QpC = QC/nC The value obtain above will be used to

design in similar manner as a vertical pile.

QC

QB

R

R’

QA

END