Tension Pile

8/2/2019 Tension Pile

1/81

1

Session 3

Calculation Models

EQU, UPL and HYD

Design of Pile Foundations

(Killiney Bay)


2/81

SLS and Settlement Calculations

(Dalkey Island)


3/81

Summary of ULS Designs

2.292.291.56DA3

1.871.871.57DA22.082.081.39DA1.C2

(1.62)(1.32)DA1.C1

ULSDesign Width

Drainedwidth (m)

Undrainedwidth (m)

For this example

Drained conditions give the larger design widths

Considering undrained and drained conditions (design width):

DA1.C2 is larger than DA1.C1 DA3 gives largest width for ULS (2.29m)

DA2 gives smallest width for ULS (1.87m)

Considering just undrained conditions:

DA2 gives the largest width for ULS (1.57m)

DA1 gives smallest width for ULS


4/81

SLS Design

Is it always necessary to calculate the settlement to check the SLS?In SLS Application Rules, Eurocode 7 states that:

For spread foundations on stiff and firm clays calculations of vertical

displacements (settlements) should usually be undertaken

For conventional structures founded on clays, the ratio of the bearing capacity

of the ground, at its initial undrained shear strength, to the applied

serviceability loading(OFSu) should be calculated If this ratio is less than 3,

calculations of settlements should always be undertaken. If the ratio is less

than 2, the calculations should take account ofnon-linear stiffness effects in

the ground

i.e. if OFSu < 3, one should calculate settlement

If OFSu < 2, one should calculate settlement accounting for non-linear stiffness

For undrained designs of foundations with permanent structural loads only

DA1 give OFSu = 1.4 while DA2 and DA3 give OFSu = 1.89. Hence settlemen

calculations are needed


5/81

OFSu Ratios

Calculating the undrained OFS ratio using the design width i.e. the draineddesign width:

OFSu = Ru,k / Vk = A ( (p + 2) cu,k bc sc ic + qc ) / Vc

= B2 x ( 5.14 x 200 x 1.0 x 1.2 x 1.0 + 20.0 x 0.8) / ( 900 + 600 )

= B2 x ( 1234.0 + 16.0 ) / 1500 = 0.833 B

In this example, using design (i.e. drained) widths:

For DA1, OFSu = 3.60 ( > 3 ) Settlement need not be calculated

For DA2, OFSu = 2.91 ( < 3 ) Settlement should be calculated

For DA3, OFSu = 4.37 ( > 3 ) Settlement need not be calculatedBut using the undrained widths OFSu values are all less than 2.0,

- much lower than value of 3 often used in traditional designs

1.971.564.372.29DA3

1.991.572.911.87DA2

1.571.393.602.08DA1

OFSu using undrainedwidth

= Ru,k / Vk = F x M/R

ULSundrainedwidth (m)

OFSu using drained width= Ru,k / Vk = 0.833 B

2ULS design(drained)width (m)


6/81

Settlement Calculations

Components of settlement to consider on saturated soils: Undrained settlements (due to shear deformation with no volume change) Consolidation settlements

Creep settlements

The form of an equation to evaluate the total settlement of a foundation on

cohesive or non-cohesive soil using elasticity theory, referred to as the adjustedelasticity method, is given in Annex F:

s = p B f / Emwhere:

Em

= design value of the modulus of elasticity

f = settlement coefficient

p = bearing pressure

Assume Em = E = 1.5N = 1.5 x 40 = 60 MPa f = (1 2) I where = 0.25 and I = 0.95 for square flexible uniformly loaded foundation

Then f = (1 0.252) x 0.95 = 0.891

p = (Gk + Qk)/B2 = (900 + 600) / B2 = 1500 / B2

Hence settlement:

s = p B f / Em = (1500 / B2 ) x B x 0.891 x 1000 / 60000 = 22.28 / B mm

where B is in m


7/81

Calculated Settlements

9.74.372.29DA3

11.92.911.87DA2

10.73.602.08DA1

Settlement ( mm )

s = 22.28 / B

OFSuULS design

width (m)

In this example, using adjusted elasticity method and ULS design widths, the

calculated settlements, s for all the Design Approaches are less than 25 mm

The SLS design requirement Ed Cd is fulfilled as for each DA, s < 25 mm

Note words of caution in EN 1997-1:

Settlement calculations should not be regarded as accurate but as merely

providing an approximate indication


8/81

Conclusions

In the example considered:

ULS design: For each Design Approach, the drained condition determines

the foundation width

SLS design: The calculated settlements are less than the allowable

settlement of 25mm, so that the SLS condition is satisfied using the designwidths obtained using all the Design Approaches

The ratio Ru,k / Vk for the ULS drained design widths is greater than 3 for DA1

and DA3 so settlement calculations are not required


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Any Questions?


10/81

1

Session 3a

Calculation Models

(Killiney Hill and Dalkey Island)

Need for Calculation Models


11/81

11

Need for Calculation Models

Equations: ULS Ed Rd SLS Ed Cd

For a ULS, calculation models are generally not needed to determine Ed e.g. the design

load on a foundation is obtained by multiplying the characteristic applied load by the

partial factor:

Ed = Fd = F Fk

An exception is where the action is due to the soil, e.g. the earth pressure on a retaining

structure. In such situations a calculation model involving the soil strength as well as the

applied load is required to obtain the earth pressure

e.g. Fd = F f{ck, k, Fk} Calculation models are always required to determine the design resistance, Rd, e.g. the

bearing or sliding resistance of a spread foundation

For an SLS, calculations are always required to determine Ed, e.g. the settlement of a

foundation

Values of Cd, the limiting design value of the effect of an action, e.g. the maximum

allowable settlement, are provided and so no calculation model is required


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1

Calculation Models in Eurocode 7

Since it was decided that the code text of Eurocode 7 should focus on

the principles and not be prescriptive, the calculation models have been

placed in the following informative Annexes

Annex C: Samples procedures to determine limit values of earth

pressures on walls

Annex D: A sample analytical method for bearing resistance

calculation

Annex E: A sample semi-empirical method for bearing resistance

estimation

Annex F: Sample methods for settlement evaluation Annex G: A sample method for determining presumed bearing

resistances for spread foundations on rock


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1

Status of Calculation Models in Eurocode 7

As the calculation models in Eurocode 7 are in Annexes, they

are optional, not mandatory

Each country has to decide if:

The calculation models in the Annexes are to be used in its

jurisdiction

Or if alternative calculation models, more suited to its soil

conditions, climate and testing methods, are to be used

However, if a design is carried out using an alternative

calculation rule it cannot claim to be fully in accordance with

Eurocode 7


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1

Annex C: Earth PressuresTwo methods are provided to determine the earth pressures on walls

The first method is a graphical method giving graphs of horizontal components of Kand Kp for different values, wall friction and slope angles, of ground behind the

wall

These are taken from BS 8002 (BSI 1994) and are based on work by Kerisel and

Absi (1990), e.g. Ka values in figure below:

Analytical Earth Pressures


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1

Analytical Earth Pressures

Since graphical earth pressures require the visual selection of a value and since

numerical methods, such as finite element analyses, require analytical values ofthe earth pressure, it was decided also to provide an analytical method to

determine the earth pressure

Following general equation for Ka and Kp is provided based on the method of

characteristics with slip line fieldse = c Kc + q Kq + d K

Equations for the earth pressure factors Kc, Kq, and K in Eurocode 7 are the

same as those in the Danish Code DS 415 (1984)

Eurocode 7 equation is based on equations by Ktter (1903) in Berlin for the

stress on curved slip surfaces


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Annex D: Bearing Resistance Calculation

Annex D provides equations for calculating the

bearing resistance of spread foundations for

undrained and drained conditions

Bearing resistance acts over effective foundationarea A, which in the case of eccentric loads is

defined as the reduced area of the foundation base

through the centroid of which the resulting loads act

Factors are given for foundation base inclination (b),shape (s) and load inclination (i), but not for

embedment depth (d)

Factors based on equations in DIN 4017: Parts 1and 2:1979

Only difference is that DIN has Nb instead of 0.5N

Exact solutions for Nc

and Nq

factors were originally

derived by Prandtl (1920 and 1921) and published in

German


17/81

1

Undrained Bearing Resistance

Undrained bearing resistance

R/A = ( + 2)cubcscic + q

where the dimensionless factors for:

Inclination of the foundation base bc = 1 2 / ( + 2)

The shape of the foundation

sc = 1+ 0/2 (B/L) for a rectangular foundationsc = 1.2 for a square or circular shape

The inclination of the load, caused by a horizontal load H

with H Acu

)cA'H1(1

21i

u

c +=

D i d B i R i t


18/81

1

Drained Bearing Resistance

Drained bearing resistance

R/A = c' Ncbcscic + q 'Nqbqsqiq + 0.5'B 'Nbsi

with the values of the dimensionless factors for:

the bearing resistance:

Nq = e tan' tan2(45 + '/2)Nc = (Nq - 1) cot

N = 2 (Nq- 1) tan', where '/2 (rough base)

the inclination of the foundation base:

bc = bq - (1 - bq) / (Nc tan)

bq = b = (1 - tan)

the shape of foundation:

sq = 1 + (B' / L' ) sin', for a rectangular shape

sq = 1 + sin', for a square or circular shape

s = 1 0,3 (B'/L), for a rectangular shape

s = 0,7, for a square or circular shapesc = (sqNq -1)/(Nq - 1) for rectangular, square or circular shape


19/81

1

Characteristic and Factored Nc Values

0

20

40

60

80

100

120

20 25 30 35 40

Characteristic angle of shearing resistance, 'k(o)

Bearingresistancefactor,

Nc,

ck

c,DA2c,DA1.C2


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Characteristic and Factored Nq Values

0

20

40

60

80

100

120

20 25 30 35 40


Bearingresistancefactor,

Nq

q,k

q,DA1.C2

q,DA2


21/81

21

Characteristic and Factored N Values

0

20

40

60

80

100

120

20 25 30 35 40


Bearingresist

ancefactors,

Nc,

Nq,

N

,DA2

,DA1.C2

,k

Comparison of N Values


22/81

2

Comparison of N Values

0.0

20.0

40.0

60.0

80.0

100.0

120.0

15.0 20.0 25.0 30.0 35.0 40.0

Eurocode 7

Caquot & Kerisel

Salgado

Brinch Hansen

Meyerhof

N

Effective angle of shearing resistance,'

Exact solutions are not available for N

and a number of different equations havebeen proposed:

Caquot and Kerisel (1953):

N = 2(Nq + 1) tan

Meyerhof (1963):

N = (Nq - 1) tan (1.4)

Brinch Hansen (1970):

N = 1.5 (Nq - 1) tan

Equation for Nadopted in Eurocode 7:

N = 2 (Nq - 1) tan

EC7 eqn. was obtained by Vesic (1973)and adopted in Eurocode 7 as anupdating of Brinch Hansens eqn. on basisof tests carried out Muhs (1973) in Berlin

Recently Martin in Oxford has obtainedexact solutions for N to which Salgado(2008) has fitted the following eqn:

N = (Nq - 1) tan (1.32 )

Salgados eqn. is closest toBrinch Hansens eqn. and

appears to indicate that the

Eurocode 7 equation for Ng may

be unconservative (i.e. unsafe)


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2

b, s and i factors

The source of the b, s and i factors in Eurocode 7 are:

For undrained conditions:

The shape factor was proposed by Skempton (1951) The load inclination factor is an algebraic fit to an exact solution by Green

(1954)

For drained conditions:

The base inclination factors, b for drained conditions were proposed by

Brinch Hansen (1970)

The shape factors are taken from DIN 4017. They are less conservative byabout 10 20% than the values proposed by de Beer (1970)

The load inclination factors, i for drained conditions were proposed by

Vesic (1975)


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2

Bearing Resistance Estimation

Annex F provides the following semi-empirical equation for estimating

the bearing resistance of a spread foundation from pressuremeter test

results:

R / A = v0 + k p*le

where k is the bearing resistance factor, and

p*le is the net effective limit pressure from a pressuremeter test

This equation is taken from the French rules, MELT 1993

No values for the k factors are given in Eurocode 7


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Settlement Evaluation

Annex F has 5 sections outlining the principles and methods for

evaluating the settlement of a foundation

Section F.2, titled Adjusted Elasticity Method, states that the total

settlement of a foundation on cohesive or non-cohesive soil may

be evaluated using elasticity theory and an equation of the form:

s = ( p b f ) /Em


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2

Bearing Resistance for Foundations on Rock

Annex G provides a sample method for

deriving presumed bearing resistance

for spread foundations on rock It has four figures, taken from BS 8004

(BSI, 1986), with contours of presumed

bearing resistance plotted against the

uniaxial strength of rock, qc on theabscissa and discontinuity spacing, dcon the ordinate

Contours are plotted for four groups ofrock, ranging from Group 1 rocks,

defined as pure limestones and

carbonate sandstones of low porosity,

to Group 4 rocks, defined asuncemented mudstones and shales

Conclusions


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2

Conclusions

Eurocode 7 developed to provide the principles for geotechnical design

Emphasises that knowledge of ground conditions and control of workmanship

have greater significance to fulfilling the fundamental requirements than applying

sophisticated calculation models

Variety of calculation models used for geotechnical design in different countries

due to different soil types, climatic conditions and testing methods

Calculations models are informative not mandatory and placed in Annexes

Calculation models in Annexes from many sources, times and countries

The various models in Eurocode 7 have been reviewed by many geotechnical

engineers during the 29 years since the first EC7 meeting

They represent a synthesis of geotechnical knowledge, design practice and

experience

Should provide a sound basis for harmonised geotechnical design in Europe usin

Eurocode 7


28/81

2

Time for Discussion

Any questions

Session 3b


29/81

2

EQU, UPL and HYD Ultimate Limit States

(Dalkey Island)

EQU UPL and HYD Ultimate Limit Sates


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3

EQU, UPL and HYD Ultimate Limit Sates

All involve stability of structures subjected to forces with no or very

little soil or structural resistance

Relevance of EQU for geotechnical and structural design a topic for

discussion at present UPL and HYD are only relevant for geotechnical designs and can be

very important

Example of a design situation:

Beam

Tension pile

For what ULS do you design the pile EQU or GEO?

For DA1 T = 0 !

T

Fulcrum

S ti 10 H d li F il


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31

Section 10 Hydraulic Failure

Eurocode 7 is mainly concerned with GEO ULS failure modesinvolving the strength of the ground and SLS involving soilstiffness or compressibility e.g.

Bearing resistance failure of spread foundations

Failure by rotation of embedded retaining walls, and

Excessive settlement of spread foundations

Section 10 of Eurocode 7 is concerned with hydraulic failure

where the strength of the ground is not significantin providingresistance and where failure is induced by excessive pore-waterpressures or seepage

The hydraulic modes of failure include:

1. Failure by uplift (buoyancy)

2. Failure by heave

3. Failure by internal erosion

4. Failure by piping

UPL d HYD Ulti t Li it St t


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3

UPL and HYD Ultimate Limit States

In Eurocode 7 the hydraulic failure ULSs are divided into UPL and HYD and

recommended partial factor values are provided for each

A UPL ultimate limit state is loss of equilibrium of a structure or the grounddue to uplift by water pressure (buoyancy) or other vertical actions

A typical UPL situation is uplift of a deep basement due to hydrostatic

static groundwater pressure

An HYD ultimate limit state is hydraulic heave, internal erosion and piping in

the ground caused by hydraulic gradients

A typical HYD situation is heave of the base of a deep excavation due to

seepage around a retaining wall

Since the strength of the ground is not significant in UPL or HYD situations,

only one set of recommended partial factors is provided for each of these

ULSs, not three Design Approaches as for GEO ULSs

Hydraulic Failures


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3

yd au c a u es

Figures from EN 1997-1 showing Hydraulic Failures

Conditions that may cause piping

Conditions that may cause Uplift Conditions that may cause heave

Stabilising Forces in UPL and HYD


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Stabilising Forces in UPL and HYD

For both UPL and HYD ultimate limit states one needs to check there is not

loss of equilibrium with regard to stabilising and destabilising forces

The stabilising force in UPL is mainly due to the self-weight of structure, butsome stabilising force is provided by the ground resistance on the side of

structure due to the strength of the ground

HYD failure occurs when, due to the hydraulic gradient, the pore waterpressure at a point in the soil exceeds the total stress or the upward

seepage force on a column of soil exceeds the effective weight of the soil

Stabilising force in HYD is provided entirely by the weight of the soil

The strength of the ground is not considered to be involved at all in HYD in

resisting the force of the seeping water

UPL Equilibrium Equation


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q q

UPL Equilibrium

One equation given:

Vdst;d Gstb;d + Rd 2.8

where:

Vdst;d = design vertical disturbing load

= Gdst;d (design perm. load) + Qdst;d (design var. load)

Gdst;d = b x udst;d (design uplift water pressure force)

Rd = Td (design wall friction force)

HYD Equilibrium Equations


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Two equations are given for HYD equilibrium

First eqn: udst;d std;d 2.9

(total stress eqn. - only equation in Eurocode 7 in terms of stress)

Second eqn: Sdst;d Gstb;d 2.9

(seepage force and submerged weight eqn.)

i.e. (w i Vol)d ( Vol)d where i = h / d

(w h/d)d ()d

(w h)d (d)dudst;d stb;d (effective stress eqn.)

=

d

h

Relevant soilcolumn

Groundwater level at ground

surface

Standpipe

Hydraulic head

Design effective soil weight, G'stb,d

Design seepage force, Sdst,d

Design total pore water pressure, udst,d

Design total vertical stress, stb,d

surface

Recommended UPL and HYD Partial Factors


37/81

3

a

s;t

cu

c

Material properties, M plus pile tensile

resistance and anchorage resistance

Q;dst33

G;stb

G;dst

Actions, F

Partial factors

1.4

1.4

1.4

1.25

1.25

1.5

0.9

1.0

UPL

-

-

-

-

-

1.5

0.9

1.35

HYD

Note:

In UPL, a factor of 1.0 is recommended for destabilising permanent actions, e.g. uplift

water pressures. The required safety is thus obtained by factoring stabilising permanent

actions by 0.9 and the soil strength or resistance

In HYD, no partial material factors are provided as no soil strength is involved

Overall Factor of Safety (OFS) for Uplift


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Overall Factor of Safety (OFS) for Uplift

Equation 2.8:

Vdst;d Gstb;d + Rd

For no Rd (i.e. soil resistance on side of buried structure ignored)

G;dst Vdst;k = G;stb Gstb;k

Overall factor of safety (OFS) = Gstb;k / Vdst;k = G;dst / G;stb

Applying recommended partial factors

G;dst/G;stb = 1.0/0.9 = 1.11

Hence OFS = 1.11

OFS against Heave using EC7 Equations


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3

Equation 2.9bSdst;d Gstb;d

G;dst Sdst;k G;stb Gstb;k

G;dst w i V G;stb V

OFS(b) = Gstb;k / Sdst;k = G;dst / G;stb

= / (wi) = ic/i = critical hydraulic

gradient / actual hydraulic gradient

OFS(b) = dst/stb = 1.35/0.9 = 1.5

d

h

Relevant soil column

Groundwater level at ground

surface

Standpipe

ydraulicPotential head

Design effective soil weight,

G'stb;dDesign seepage force, Sdst;d

Design total pore water pressure, udst;d

Design total vertical stress, stb;d

Equation 2.9audst;d stb;d

G;dstwd + G;dstwh G;stb'd + G;stbwd

OFS(a) = G;dst / G;stb = (d + wd) / (wh + wd)

= (ic + 1)/(i + 1) = 1.5ic/i = 1.5 + 0.5/i

if i = 0.5 then ic/i = OFS(b) = 2.5

i.e. more cautious than using Eqn. 2.9b because

w

d occurs on both sides of equation and ismultiplied by different F values

Comment on HYD


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4

HYD ultimate limit states include internal erosion and piping as well as heave

The OFS value traditionally used to avoid piping is often very much greater than

the 1.5 provided by the HYD partial factors; e.g. 4.0

Hence, EN 1997-1 gives additional provisions to avoid the occurrence of internalerosion or piping

For internal erosion, it states that:

Filter criteria shall be used to limit the danger of material transport by internal erosion

Measures such as filter protection shall be applied at the free surface of the ground

Alternatively, artificial sheets such as geotextiles may be used

If the filter criteria are not satisfied, it shall be verified that the design value of the

hydraulic gradient is well below the critical hydraulic gradient at which soil particlesbegin to move. ic value depends on the design conditions

EN 1997-1 states thatpiping shall be prevented by providing sufficient resistance

against internal soil erosion through by providing:

- Sufficient safety against heave

- Sufficient stability of the surface layers

Uplift Design Example(Issued for Workshop on the Evaluation of Eurocode 7


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41

(Issued for Workshop on the Evaluation of Eurocode 7

in Dublin in 2005)

T

15.0m

Structural loading gk

= 40kPa

5.0m R

U

G

Design situation given:

- Long basement, 15m wide

- Sidewall thickness = 0.3m

- Characteristic structural loading = 40 kPa- Groundwater can rise to ground surface

- Soil is sand with k = 35o, g = 20 kN/m3

- Concrete weight density = 24 kN/m3

Base thickness, D requested

Forces

U = Uplift water pressure force = w15 (5 +

G = Weight of basement plus structural loa

R = Resisting force from soil on side walls

A wide range of design values obtained for D = 0.42 0.85m

Why?

Model for UPL Equilibrium Calculation


42/81

4

Model Assumptions

Include or ignore R ?

R = A = Ahtan = AKvtan where A = sidewall area

What value for K? K is a function of and . Should K = K0 or Ka ?

What value for wall friction ?

Is a function of? Should = or 2/3 ?

How should partial factors be applied to obtain Rd?

No UPL resistance factors are provided in EN 1997 to obtain Rd from Rk

i.e. there is no UPL equivalent to DA2

Could design according to EN 1997-1 and assume h = Kav1) With Rk =AKa;kvtank apply partial factorM to k to obtain Ka;d

and d as for DA1.C2 and hence get Rd (Clause 2.4.7.4(1))

2) Treat R as a permanent stabilising vertical action and apply G;stb to Rk

to obtain Rd (Clause 2.4.7.4(2))

Determination of Design Value of Rd


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Assume K = Ka and is obtained from EN 1997-1 for = 2/3

1) Clause 2.4.7.4(1): Apply partial factorM to k to obtain Ka;d and d and hence Rd

No factors applied: k = 35o and = 2/3 Ka;k = 0.23

k = 2/3k = 23.3o Rk = AKa;kvtank = 0.099Av

a) M = 1.25 applied to obtain d and d by reducing k and hence k

d = 29.3o and = 2/3 Ka;d = 0.29

d = 2/3d = 19.5o Rd = AKa;dvtand = 0.103Av

Since R is a resistance, need Rd < Rk: Rd = (0.103/0.099)Rk Rd = 1.04 Rk unsafe

b) M applied to increase k but to reduce

d = 41.2o and d = 19.0

o d/d = 19.0/41.2 = 0.46 Ka;d = 0.18

Rd = AKa;dvtand = 0.062Av

Rd = (0.062/0.099)Rk Rd = 0.69 Rk safe

2) Clause 2.4.7.4(2): Treat R as a permanent stabilising vertical action and apply G;stb to

Rk to obtain Rd

Rd = G;stbRk Rd = 0.9 Rk safe

Comments on Uplift Design Example


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4

p g p

Reasons for Range of Solutions Obtained for Uplift Design Example:

Whether R Ignored or included

Model chosen for R = A h tan = A Kv tan

K = K0 or Ka

= 0.5 or (2/3)

How Rd is obtained

Treated as a resistance or a stabilising action

How partial factors are applied

What partial factors are applied

Heave Design Example

(I d f W k h th E l ti f E d 7


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4

(Issued for Workshop on the Evaluation of Eurocode 7

in Dublin in 2005)

Design Situation

- 7m deep excavation

- Sheet pile wall- Pile penetration 3m below excavation

level

- 1.0 m water in excavation

- Weight density of sand = 20 kN/m3

Require H

- Height of GWL behind wall above

excavation level

GWL

1.0m

Sand = 20kN/m3 3.0m

7.0m

H = ?Water

A very wide range of design values obtained: H = 1.7 6.6m

Why?

Reasons for Range of Solutions to Heave Example


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4

Assumption Regarding PWP distribution around the wall (i.e. pwp at toeof wall)

Some used equation for pwp at toe from EAU Recommendations

Some obtained pwp at toe from flownet

Some assumed a linear dissipation of pwp around wall - this gives leastconservative designs

Choice of Equilibrium Equation

Some used Equation 2.9a with partial factors applied to total pwp and totalstress. This involves applying different partial factors to hydrostatic pwp on

either side of equation and gave an overall factor of safety that is 1.5d/h

greater than Equation 2.9b

Most design solutions were based on Equation 2.9b i.e. comparing seepage

force and effective soil weight

Treatment of Seepage Force

Some treated seepage force as a variable action

Most considered it a permanent action

Conclusions on EQU, UPL and HYD Ultimate

Limit States


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Limit States

EQU, UPL and HYD are all ultimate limit states involving the equilibrium of

forces (actions) with little or no resistance forces

EQU is rarely relevant for geotechnical designs

EQU is being debated within the Eurocodes at present

Uplift and heave ultimate limit states involving failure due to water pressures

and seepage are important in geotechnical design and are different from

geotechnical designs involving soil strength Need to clearly identify the stabilising and destabilising actions

This is best achieved by working in terms of actions (forces) rather than

stresses

Need to apply partial factors appropriately to get the design stabilising and

destabilising actions for both uplift and heave design situations

Designs against uplift and heave failure are clarified using the equilibrium

equations and partial factors provided in Eurocode 7

Discussion


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Discussion

Any Questions?


49/81

4

Session 3c

Design of PileFoundations

(Killarney Waterfall)

Scope


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Scope

Applies to end-bearing piles, friction piles, tension piles and

transversely loaded piles installed by driving, jacking, and by

screwing or boring with or without grouting

Not to be applied directly to the design of piles that are intended to

act as settlement reducers

Relevant CEN Standards


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Reference is made in Eurocode 7 to other CEN standards that are

relevant to the design of pile foundations

Eurocode 3, Part 5: Design of steel Structures Piling

(EN 1993-5)

Execution standard Execution of Special Geotechnical Works

EN 12699:2000 Displacement piles

EN 12063:2000 - Sheet pile walls

EN 1536:1999 - Bored Piles

EN 14199:2005 - Micro-piles

Limit State Checklist for Pile Design


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Unacceptable vibrations

Excessive lateral movement

Excessive heave

Excessive settlement

Combined failure in the ground and in the structure Note change from handout

Combined failure in the ground and in the pile foundation

Structural failure of the pile in compression, tension, bending, buckling or shear

Failure in the ground due to transverse loading of the pile foundation

Uplift or insufficient tensile resistance of the pile foundation

Bearing resistance failure of the pile foundation

Loss of overall stability

CheckedLimit states to be considered

Actions due to Ground Movements


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Effects on piles of ground movements due to consolidation, swelling,adjacent loads, creeping soil, landslides, earthquakes are treated as actions

Give rise to downdrag (negative skin friction), heave, stretching, transverse

loading and displacement

Usually design values of strength and stiffness are upper values

Two approaches

Ground displacements considered as an action and an interaction

analysis carried out to determine the forces

An upper bound force, which the ground could transmit to the pile, is

introduced as a design action

Downdrag An upper bound on a group of piles may be calculated from theweight of surcharge causing the movement

Transverse loading is normally evaluated by considering the interaction

between piles, treated as stiff or flexible beams, and the moving soil mass

Pile Design Methods


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g

Static load tests which have been demonstrated to be consistent

with other relevant experience

Empirical or analytical calculation whose validity has been

demonstrated by static tests in comparable situations

Dynamic tests - whose validity has been demonstrated by statictests in comparable situations

Observed performance of a comparable foundation provided thisapproach is supported by the results of site investigations and

ground testing

Design considerations


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Stiffness and strength of the structure connecting the piles

Duration and variation in time of the loading when selecting the

calculation method and parameter values and in using load test

results

Planned future changes in overburden or potential changes in the

ground water level

Checklist for Selection of Pile Type


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The handling and transportation of piles

The possibility of connecting different ground-water regimes

The deleterious effects of chemicals in the ground

The tolerances within which the pile can be installed reliably

The effect of the method and sequence of pile installation on piles, which

have already been installed and on adjacent structures or services.

The possibility of preserving and checking the integrity of the pile beinginstalled

The stresses generated in the pile during installation

The ground and ground-water conditions, including the presence or

possibility of obstructions in the ground.

CheckedSELECTION OF PILE TYPE

Special Features of Pile Design to Eurocode 7


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Gives method of determining characteristic values directly from the

results of pile load tests or from profiles of tests etc. using values

DA1 is a resistance factor approach

DA3 not used for design from pile load tests

Pile Load Tests


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Can be on TRIAL PILES or on WORKING PILES

If one pile test is carried out, normally located in the most adverseground conditions

Adequate time to ensure required strength of pile material and that

pore-water pressures have regained their initial values

Axially Loaded Piles in Compression

Equilibrium equation:


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Equilibrium equation:

Fc;d < Rc;dwhere:

Fc;d is the ULS design axial compression load

Fc;d

is determined using partial factors applied to the characteristic loads relevant

to the DA being used

Self weight of pile should be included, along with downdrag, heave or

transverse loading, however the common practice of assuming that the

weight of the pile is balanced by that of the overburden allowing both to beexcluded from Fc;d and Rc;d is permitted, where appropriate

The pile weight may not cancel the weight of the overburden if a) downdrag is significant

b) the soil is light c) the pile extends above the ground surface.

Rc;d is ULS design bearing resistance and is the sum of all the bearing

resistance components against axial loads, taking into account the effect of

any inclined or eccentric loads


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RecommendedPartial Factor

Values

Design of a Compression Pile

from Pile Load Tests


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from Pile Load Tests

Fc;d is calculated in the normal way with F factors applied to applied

loads

Rc;d must be determined from the measured pile resistance Rc;m

Rc;m can be based on the total resistance or can be separated into

Rb;m (base) and Rs;m (shaft)

DA3 not used for pile design from load tests

Determination of Characteristic Resistance


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Characteristic resistance for DA1 & DA2

Rc;k = Min {(Rc;m) mean /1; (Rc;m) min /2 }

Characteristic values determined directly not estimated

1.01.01.051.21.42

1.01.11.21.31.41

54321 for n =

Table A.9

Note


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For structures which have sufficient stiffness to transfer loads from

weak to strong piles, may be divided by 1.1 provided it is not less

than 1.0

Example 1

Pil D i f Pil L d T t


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Pile Design from Pile Load Tests

CFA pile

600 mm diameter

Glacial till

Loose fill

F

9.0m

2.0m

Characteristic Loads

Gk= 1200 kN

Qk = 200 kN

Pile Load Test Results

No of test = 2

Max Applied Load = 4000kN

(same on both piles)

Max load on base = 600kN

DA1.C1


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DA1.C1 (A1 + M1 + R1) Fc;d = 1.35*1200+1.5*200 = 1920 kN

Note weight of pile not included)

Rc;m = 4000 kN for both

Rc;k= lesser of 4000/1.3 or 4000/1.2 = 3077 kN

Use measured toe force to give shaft force from ratio of total load to base load

Rb;k= (600 / 000) Rc;k = 462 kN; Rs;k= 2615 kN

Rc;d = Rb;k / 1.1 + Rs;k / 1.0 = 462 / 1.10 + 2615 / 1.0 = 3035 kN

Fc;d (1920 kN) < Rc;d (3035kN)

DA1.C1 is satisfied.

DA1.C2


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DA1.C2 (A2 + M1 or M2 + R4)

Fc;d = 1.0*1200+1.3*200 = 1460 kN

Note weight of pile not included

Rb;k = 462 kN; Rs;k = 2615 kN as before for DA1.C1

Rc;d

= Rb;k

/ 1.45 + Rs;k

/ 1.3 = 462 / 1.45 + 2615 / 1.3 = 2615 kN

Fc;d (1460 kN) < Rc;d (2615kN)

DA1.C2 is satisfied

DA1.C1 and DA1.C2 are both satisfied, therefore DA1 is satisfied

DA2 (A1 + M1 + R2)

DA2


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DA2 (A1 + M1 + R2)

Fc;d = 1.35*1200+1.5*200 = 1920 kN

Note weight of pile not included

Rc;m = 4000 kN for both

Rc;k= lesser of 4000 / 1.3 or 4000 / 1.2 = 3077kN

Use measured toe force to give ratio Rb;k= (600 / 4000) Rc;k = 462 kN; Rs;k= 2615 kN as before

Rc;d = Rb;k / 1.1 + Rs;k/ 1.1 = 462 / 1.1 + 2615 / 1.1 = 2797 kN

Fc;d (1920 kN) < Rc;d (2797kN)

DA2 is satisfied

Dynamic Pile Load Tests


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Provided an adequate site investigation has been carried out and

the method has been calibrated against static load tests on same

type of pile, of similar length and cross-section and in comparable

ground conditions

Dynamic load tests may be used as an indicator of the consistency

of piles and to detect weak piles

Design of a Compression Pile

from Ground Test results


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Fc;d is determined in the normal way by factoring the loads

Rb;k and Rs;k or Rt;k must be determined from:

Number of profiles of tests and applying factors(different to values of factors for design from pile load tests)

or

an alternative procedure see next slide

Then as for design from pile load tests

Rb,d = Rb,k / b and Rs,d = Rs,k / s

where the b , b and t values are given in Table A.6, A.7 and A.8

(same as for design from pile load tests)

Alternative Procedure to Determine Rs,k and

Rb,k from Ground Strength Parameters


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Calculate the characteristic base resistance (qb;k) and shaft

resistances (qs;k) using characteristic values of ground parameters

[C7.6.2.3(8)] and hence:

Rb;k = Abqb;k and Rsk = qsi;kAsi

where

Ab = the nominal plan area of the base of the pile

Asi = the nominal surface area of the pile in soil layer i

A Note in Eurocode 7 states: If this alternative procedure is applied,

the values of the partial factors b ands recommended in Annex A

may need to be corrected by a model factor larger than 1,0. The

value of the model factor may be set by the National annex

In Ireland a model factor of 1.75 is applied to b and s ort when

using this approach

Example 2

Pile Design using Ground Tests


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CFA pile600 mm diameter

Glacial till

Loose fill

F

9.0m

2.0m

Pile Design using Ground Tests

Pile Design from Ground Test Results

Th pil d d diti f thi pl th th f


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The piles and ground conditions for this example are the same as those for

Example 1, where the pile design is based on pile load tests. It is required in this

example to verify that, on the basis of the ground properties, 600mm diameter

CFA piles will support the characteristic permanent and variable vertical loads of

1200kN and 200kN, respectively The cuk value for the very stiff glacial till increases linearly from 100kPa at 2m

depth (top of the till) to 600kPa at 11m (bottom of the pile). The properties of the

glacial till are = 22kN/m3, ck

= 0, s,k

= 36o for shaft resistance and b,k

= 34o for

the base resistance (the reduction in ' is to allow for the higher stress levels at

the base). The unit weights of the fill and concrete are 18kN/m3 and 24kN/m3,

respectively. The water table is at a depth of 2.0m, which coincides with the top

of the till

As the characteristic values of the soil properties are given, the characteristic

resistances are to be determined using these values. In Ireland a model factor of

1.75 is applied to b and s ort (R values)

Ground Parameters


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Only undrained conditions considered

Assume

qb = 9cu + v0

qs = cu where = 0.4

WP (weight of pile) = * 0.62 * 11.0 * 24.0 / 4 = 74.6 kN

DA1

qb k = (9c + ) = (9*600 + 2*18 + 9*22) = 5634 kPa


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qb;k (9cu + v) (9 600 + 2 18 + 9 22) 5634 kPa

qs;k = cu = 0.4*{(100 + 600)/2} = 140 kPa

Rb,k = Abqb,k = ( * 0.62 / 4) * 5634 = 1593 kN

Rs,k = Asqs,k = ( * 0.6 * 9 * 140 = 2375 kN

DA1.C1: (A1+M1+R1)

Fc;d = 1.35(1200 + 74.6) + 1.5*200 = 2020.7 kN

Rc;d = Rb,k /(1.0*1.75) + Rs,k /(1.0*1.75) = 910.3 + 1357.1 = 2267.4 kN OK

DA1.C2 (A2+(M1 or M2) +R4)

Fc;d = 1.0(1200 + 74.6) + 1.3*200 = 1534.6 kN

Rc;d = Rbk /(1.45*1.75) + Rsk /(1.3*1.75) = 700.2 + 1044.0 = 1744.2 kN OK

DA2


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Fc;d = 1.35(1200 + 74.6) + 1.5*200 = 2020.7 kN

Rb;k = 1593.0 kN ; Rs;k = 2375.0 kN as for DA1

Rc;d = Rb;k /(1.1*1.75) + Rs;k /(1.1*1.75)

= 827.5 + 1233.8 = 2061.3 kN

Fc;d < Rc;d (2020.7 < 2061.3) so inequality is satisfied for DA2

DA3


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Fc;d as per DA1.C2 but partial factors on structural actions are 1.35

and 1.5

Design value of resistance obtained by using m (Table A4)

parameters on soil properties and in Ireland a model factor of 1.75 is

applied on b and s

DA3


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qb;d = (9cu / m + v) = (9*600/(1.4*1.75) + 2*18 + 9*22) = 2438.1 kPa

qs;d = cu / m = 0.4*{(100/(1.4*1.75) + 600/(1.4*1.75))/2} = 57.1 kPa

Fc;d

= 2020.7 kN

Rc;d = 689.4 + 968.7 = 1658.1 kN

Inequality Fc;d Rc;d Not satisfied

Rb;d = Abqb;d = (*0.62/4)*2438 = 689.4 kN

Rs;d = Asqs;d = *0.6*9*57.1 = 968.7 kN

Fc;d = 1.35(1200 + 74.6) + 1.5*200 = 2020.7 kNNote: increased partial factors compared with DA1:C2

Piles in Tension


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Design of piles in tension is same as the design of piles in compression

except a greater margin of safety required and there is no base resistance

Ft;d R

t;d

Must consider

Pull-out of piles from the ground mass

Uplift of block of ground (or cone)

Group effect shall consider reduction in vertical effective stress

The severe adverse effect of cyclic loading and reversal of loading shall be

considered

Design of a Pile from Tension Load Tests


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Extrapolation of the load-displacement curve from pile load tests should

not be used for tension tests

Rt;d

= Rt;k

/s;t

Rt;k is determined from Rt;m using values in same manner as for a

compression test

values from Table A.9 as before

s,t from Tables A.6, A.7 & A.8

Normally it should be specified that more than one pile should be tested,

or 2% if a large number of piles

Design of a Tension Pile from Ground Test Results


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Established from pile load test and from comparable experience

Rt;d = Rt;k/s;t where Rt;k = Rs;k

Rt;k obtained from:

Rt;k = Min {(Rs;calc)mean / 3; (Rs;calc)min / 3}

or using the alternative procedure from Rt;k = qsi;k Asi

values from TableA.10

s,t from Tables A.6, A.7 & A.8

In Ireland a model factor of 1.75 on s,t from Tables A.6, A.7 & A.8 if

alternative procedure is adopted


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