1

Session 3

Calculation Models

EQU, UPL and HYD

Design of Pile Foundations

(Killiney Bay)

2

SLS and Settlement Calculations

(Dalkey Island)

3

Summary of ULS Designs

2.292.291.56DA3

1.871.871.57DA2

2.082.081.39DA1.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

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 of non-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 settlement calculations are needed

5

OFSu Ratios

Calculating the undrained OFS ratio using the design width – i.e. the drained design 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 undrained width

= Ru,k / Vk = γF x γM/R

ULS undrained width (m)

OFSu using drained width= Ru,k / Vk = 0.833 B2

ULS design (drained) width (m)

6

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 adjusted elasticity 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 – ν2) I where ν = 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

Calculated Settlements

9.74.372.29DA3

11.92.911.87DA2

10.73.602.08DA1

Settlement ( mm )s = 22.28 / B

OFSuULS designwidth (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

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 design widths 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

9

Any Questions?

10

Session 3a

Calculation Models

(Killiney Hill and Dalkey Island)

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 = γF 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 = γF f{c’k, φ’ 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

12

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

13

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

14

Annex C: Earth Pressures• Two methods are provided to determine the earth pressures on walls

• The first method is a graphical method giving graphs of horizontal components of Kaand Kp for different φ’ values, wall friction and slope angles, β 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:

15

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 of the 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 fields

e = c’ Kc + q Kq + γ’ d Kγ

• Equations for the earth pressure factors Kc, Kq, and Kγ in Eurocode 7 are the same as those in the Danish Code DS 415 (1984)

• Eurocode 7 equation is based on equations by Kötter (1903) in Berlin for the stress on curved slip surfaces

16

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 foundation area 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 1 and 2:1979

• Only difference is that DIN has Nb instead of 0.5Nγ

• Exact solutions for Nc and Nq factors were originally derived by Prandtl (1920 and 1921) and published in German

17

Undrained Bearing Resistance

Undrained bearing resistanceR/A’ = (π + 2)cubcscic + q

where the dimensionless factors for:– Inclination α of the foundation base bc = 1 – 2α / (π + 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 ≤ A’cu

)cA'

H1(1

2

1i

uc −+=

18

Drained Bearing ResistanceDrained bearing resistance

R/A’ = c' Ncbcscic + q 'Nqbqsqiq + 0.5γ'B 'Nγbγsγiγwith the values of the dimensionless factors for:

– the bearing resistance:Nq = eπ tanϕ' tan2(45 + φ'/2)Nc = (Nq - 1) cot φ‘Nγ = 2 (Nq- 1) tanφ', where δ ≥ φ'/2 (rough base)

– the inclination of the foundation base:bc = bq - (1 - bq) / (Nc tanφ’ )bq = bγ = (1 - α ⋅tanφ’)

– the shape of foundation:sq = 1 + (B' / L' ) sinφ', for a rectangular shapesq = 1 + sinφ', for a square or circular shape

sγ = 1 – 0,3 (B'/L‘ ), for a rectangular shapesγ = 0,7, for a square or circular shapesc = (sqNq -1)/(Nq - 1) for rectangular, square or circular shape

19

Characteristic and Factored Nc Values

0

20

40

60

80

100

120

20 25 30 35 40

Characteristic angle of shearing resistance, φ 'k (o)

Bea

ring

resis

tanc

e fa

ctor

, Nc,

Nck

Nc,DA2Nc,DA1.C2

20

Characteristic and Factored Nq Values

0

20

40

60

80

100

120

20 25 30 35 40Characteristic angle of shearing resistance, φ'k (

o)

Bea

ring

resis

tanc

e fa

ctor

, Nq

Nq,k

Nq,DA1.C2

Nq,DA2

21

Characteristic and Factored Nγ Values

0

20

40

60

80

100

120

20 25 30 35 40Characteristic angle of shearing resistance, φ 'k (

o)

Bea

ring

resi

stan

ce fa

ctor

s, N c

, Nq,

Nγ

Nγ,DA2

Nγ,DA1.C2

Nγ,k

22

Comparison of Nγ 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

Nγ

Effective angle of shearing resistance, φ'

• Exact solutions are not available for Nγand a number of different equations have been proposed:

– Caquot and Kerisel (1953):Nγ = 2(Nq + 1) tan φ’

– Meyerhof (1963): Nγ = (Nq - 1) tan (1.4φ’)

– Brinch Hansen (1970): Nγ = 1.5 (Nq - 1) tan φ’

• Equation for Nγ adopted in Eurocode 7: Nγ = 2 (Nq - 1) tan φ’

• EC7 eqn. was obtained by Vesic (1973) and adopted in Eurocode 7 as an updating of Brinch Hansen’s eqn. on basis of tests carried out Muhs (1973) in Berlin

• Recently Martin in Oxford has obtained exact solutions for Nγ to which Salgado (2008) has fitted the following eqn:

Nγ = (Nq - 1) tan (1.32 φ’)

Salgado’s eqn. is closest to Brinch Hansen’s eqn. and appears to indicate that the Eurocode 7 equation for Ng may be unconservative (i.e. unsafe)

23

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 by

about 10 – 20% than the values proposed by de Beer (1970)– The load inclination factors, i for drained conditions were proposed by

Vesic (1975)

24

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 = σv0 + 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

25

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

26

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 the abscissa and discontinuity spacing, dcon the ordinate

• Contours are plotted for four groups of rock, ranging from Group 1 rocks, defined as pure limestones and carbonate sandstones of low porosity, to Group 4 rocks, defined as uncemented mudstones and shales

27

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 using Eurocode 7

28

Time for Discussion

Any questions

29

Session 3b

EQU, UPL and HYD Ultimate Limit States

(Dalkey Island)

30

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

31

Section 10 – Hydraulic Failure

• Eurocode 7 is mainly concerned with GEO ULS failure modes involving the strength of the ground and SLS involving soil stiffness 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 significant in providing resistance and where failure is induced by excessive pore-water pressures or seepage

• The hydraulic modes of failure include:1. Failure by uplift (buoyancy)2. Failure by heave3. Failure by internal erosion4. Failure by piping

32

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 ground due 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

33

Hydraulic Failures

• Figures from EN 1997-1 showing Hydraulic Failures

Conditions that may cause piping

Conditions that may cause Uplift Conditions that may cause heave

34

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, but some 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 water pressure 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

35

UPL Equilibrium Equation

UPL EquilibriumOne equation given:

Vdst;d ≤ Gstb;d + Rd 2.8where:

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)

36

HYD Equilibrium Equations

• Two equations are given for HYD equilibrium• First eqn: udst;d ≤ σstd;d 2.9a(total stress eqn. - only equation in Eurocode 7 in terms of stress)

• Second eqn: Sdst;d ≤ G’stb;d 2.9b

(seepage force and submerged weight eqn.)• i.e. (γw i Vol)d ≤ (γ’ Vol)d where i = ∆h / d • (γw ∆h/d)d ≤ (γ’)d• (γw ∆h)d ≤ (γ’d)d ∆udst;d ≤ σ’stb;d (effective stress eqn.)

∆udst;d = design seepage pore water pressure

d

∆h

Relevant soil column

Groundwater level at groundsurface

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, σstb,d

surface

37

Recommended UPL and HYD Partial Factors

γa

γs;t’

γcu

γc’

γφ’

Material properties, γM plus pile tensile resistance and anchorage resistance

γQ;dst33

γG;stb

γG;dst

Actions, γF

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

38

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)• γG;dst Vdst;k = γG;stb Gstb;k

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

• Applying recommended partial factors

• γG;dst/γG;stb = 1.0/0.9 = 1.11

• Hence OFS = 1.11

39

OFS against Heave using EC7 Equations

Equation 2.9bSdst;d ≤ Gstb;d

γG;dst Sdst;k ≤ γG;stb G’stb;k

γG;dst γw i V ≤ γG;stb γ’ V

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

= γ’/ (γwi) = ic/i = critical hydraulic gradient / actual hydraulic gradient

OFS(b) = γdst/γstb = 1.35/0.9 = 1.5

d

∆h

Relevant soil column

Groundwater level at ground surface

Standpipe

ydraulic Potential head

Design effective soil weight, G'stb;d

Design seepage force, Sdst;d

Design total pore water pressure, udst;d

Design total vertical stress, σstb;d

Equation 2.9audst;d ≤ σstb;d

γG;dstγwd + γG;dstγw∆h ≤ γG;stbγ'd + γG;stbγwdOFS(a) = γG;dst / γG;stb = (γ’d + γwd) / (γw∆h + γwd)

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

if i = 0.5 then ic/i = OFS(b) = 2.5i.e. more cautious than using Eqn. 2.9b because

γwd occurs on both sides of equation and is multiplied by different γF values

40

Comment on HYD

• 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 internal erosion 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 particles begin to move. ic value depends on the design conditions

• EN 1997-1 states that piping shall be prevented by providing sufficient resistance against internal soil erosion through by providing:- Sufficient safety against heave- Sufficient stability of the surface layers

41

Uplift Design Example(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 φ’k = 35o, g = 20 kN/m3

- Concrete weight density = 24 kN/m3

Base thickness, D requested

Forces

U = Uplift water pressure force = γw15 (5 + T)G = Weight of basement plus structural loadR = Resisting force from soil on side walls

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

Why?

42

Model for UPL Equilibrium Calculation

• Model Assumptions– Include or ignore R ?– R = Aτ = Aσh’tanφ’ = AKσv’tanδ’ where A = sidewall area

• What value for K? – K is a function of φ’ and δ. Should K = K0 or Ka ?

• What value for wall friction δ?– Is δ a function of φ’? Should δ = φ’ or 2/3φ’ ?

• 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 σh’ = Kaσv’1) With Rk = AKa;kσv’tanδ’k apply partial factor γM to φ’k to obtain Ka;d

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

2) Treat Rκ as a permanent stabilising vertical action and apply γG;stb to Rk

to obtain Rd (Clause 2.4.7.4(2))

43

Determination of Design Value of Rd

• Assume K = Ka and is obtained from EN 1997-1 for δ = 2/3φ’• 1) Clause 2.4.7.4(1): Apply partial factor γM to φ’k to obtain Ka;d and δd and hence Rd

• No factors applied: φ’k = 35o and δ = 2/3φ’ → Ka;k = 0.23• δk = 2/3φ’k = 23.3o Rk = AKa;kσv’tanδk = 0.099Aσv’• a) γM = 1.25 applied to obtain φ’d and δd by reducing φ’k and hence δk

• φ’d = 29.3o and δ = 2/3φ’ → Ka;d = 0.29• δd = 2/3φ’d = 19.5o → Rd = AKa;dσv’tanδd = 0.103Aσv’

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

b) γM applied to increase φ’k but to reduce δ φ’d = 41.2o and δd = 19.0o δd/φ’d = 19.0/41.2 = 0.46 → Ka;d = 0.18

Rd = AKa;dσv’tanδd = 0.062Aσv’ Rd = (0.062/0.099)Rk Rd = 0.69 Rk – safe

• 2) Clause 2.4.7.4(2): Treat Rκ as a permanent stabilising vertical action and apply γG;stb to Rk to obtain Rd

Rd = γG;stbRk Rd = 0.9 Rk – safe

44

Comments on Uplift Design Example

Reasons for Range of Solutions Obtained for Uplift Design Example:

• Whether R Ignored or included

• Model chosen for R = A σh’ tanδ = A Kσv’ tanδ

– K = K0 or Ka

– δ = 0.5φ’ or (2/3)φ’

• How Rd is obtained

– Treated as a resistance or a stabilising action

– How partial factors are applied

– What partial factors are applied

45

Heave Design Example(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 γ = 20kN/m3 3.0m

7.0m

H = ? Water

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

Why?

46

Reasons for Range of Solutions to Heave Example

• Assumption Regarding PWP distribution around the wall (i.e. pwp at toe of 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 least

conservative designs

• Choice of Equilibrium Equation– Some used Equation 2.9a with partial factors applied to total pwp and total

stress. 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

47

Conclusions on EQU, UPL and HYD Ultimate 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

48

Discussion

Any Questions?

49

Session 3c

Design of Pile Foundations

(Killarney Waterfall)

50

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

51

Relevant CEN Standards

• 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

52

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

Limit State Checklist for Pile Design

53

Actions due to Ground Movements

• 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 the weight 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

54

Pile Design Methods

• 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 static tests in comparable situations

• Observed performance of a comparable foundation – provided this approach is supported by the results of site investigations and ground testing

55

Design considerations

• Stiffness and strength of the structure connecting the piles

• Duration and variation in time of the loading when selecting thecalculation method and parameter values and in using load test results

• Planned future changes in overburden or potential changes in theground water level

56

Table 9 2

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 being installed

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

Table 9 2

Checklist for Selection of Pile Type

57

Special Features of Pile Design to Eurocode 7

• Gives method of determining characteristic values directly from the results of pile load tests or from ‘profiles of tests’ etc. using ξ values

• DA1 is a resistance factor approach

• DA3 not used for design from pile load tests

58

Pile Load Tests

• Can be on TRIAL PILES or on WORKING PILES

• If one pile test is carried out, normally located in the most adverse ground conditions

• Adequate time to ensure required strength of pile material and that pore-water pressures have regained their initial values

59

Axially Loaded Piles in Compression• 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 be excluded 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

60

Recommended

Partial Factor

Values

61

Design of a Compression Pile from Pile Load Tests

• Fc;d is calculated in the normal way with γF 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

62

• Characteristic resistance for DA1 & DA2

• Rc;k = Min {(Rc;m) mean /ξ1; (Rc;m) min /ξ2 }

• Characteristic values determined directly – not estimated

1.01.01.051.21.4ξ2

1.01.11.21.31.4ξ1

≥ 54321ξ for n =

Table A.9

Determination of Characteristic Resistance

63

Note

• For structures which have sufficient stiffness to transfer loads from ‘weak’ to ‘strong’ piles, ξ may be divided by 1.1 provided it is not less than 1.0

64

Example 1

Pile Design from Pile Load Tests

CFA pile600 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

65

DA1.C1 (A1 + M1 + R1)

• Fc;d = 1.35*1200+1.5*200 = 1920 kNNote 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.C1

66

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

• Fc;d = 1.0*1200+1.3*200 = 1460 kNNote 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

DA1.C2

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DA2 (A1 + M1 + R2)

• Fc;d = 1.35*1200+1.5*200 = 1920 kNNote weight of pile not included

• Rc;m = 4000 kN for both

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

DA2

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Dynamic Pile Load Tests

• Provided an adequate site investigation has been carried out andthe method has been calibrated against static load tests on sametype 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

69

Design of a Compression Pile from Ground Test results

• 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 ξ factors

(different to values of ξ factors for design from pile load tests)or• an alternative procedure – see next slide

• Then as for design from pile load testsRb,d = Rb,k / γb and Rs,d = Rs,k / γs

where the γb , γb and γt values are given in Table A.6, A.7 and A.8 (same as for design from pile load tests)

70

• 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 = Σqsi;kAsi

whereAb = the nominal plan area of the base of the pileAsi = 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 γb and γs 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 γb and γs or γt when using this approach

Alternative Procedure to Determine Rs,k and Rb,k from Ground Strength Parameters

71

CFA pile600 mm diameter

Glacial till

Loose fill

F

9.0m

2.0m

Example 2

Pile Design using Ground Tests

72

Pile Design from Ground Test Results

• 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 γ = 22kN/m3, c‘k = 0, φs,k = 36o for shaft resistance and φb,k = 34o for the base resistance (the reduction in φ' 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 γb and γs or γt (γR values)

73

• Only undrained conditions considered

• Assume

– qb = 9cu + σv0

– qs = α cu where α = 0.4

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

Ground Parameters

74

DA1

qb;k = (9cu + σv) = (9*600 + 2*18 + 9*22) = 5634 kPa

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

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

Rs,k = ΣAsqs,k = (π * 0.6 * 9 * 140 = 2375 kN

DA1.C1: (A1+M1+R1)Fc;d = 1.35(1200 + 74.6) + 1.5*200 = 2020.7 kNRc;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 kNRc;d = Rbk /(1.45*1.75) + Rsk /(1.3*1.75) = 700.2 + 1044.0 = 1744.2 kN OK

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DA2

• 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

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DA3

• 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 γm (Table A4) parameters on soil properties and in Ireland a model factor of 1.75 is applied on γb and γs

77

DA3

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

qs;d = α cu / γm = 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 = (π*0.62/4)*2438 = 689.4 kN

Rs;d = ΣAsqs;d = π*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

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Piles in Tension

• 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 ≤ Rt;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

79

Design of a Pile from Tension Load Tests

• Extrapolation of the load-displacement curve from pile load tests should not be used for tension tests

• Rt;d = Rt;k/γs;t

• Rt;k is determined from Rt;m using ξ values in same manner as for a compression test

• ξ values from Table A.9 as before

• γs,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

80

Design of a Tension Pile from Ground Test Results

Established from pile load test and from comparable experience

• Rt;d = Rt;k/γs;t where Rt;k = Rs;k

• Rt;k obtained from:

– Rt;k = Min {(Rs;calc)mean / ξ3; (Rs;calc)min / ξ3}

– or using the alternative procedure from Rt;k = Σqsi;k Asi

• ξ values from Table A.10

• γs,t from Tables A.6, A.7 & A.8

• In Ireland a model factor of 1.75 on γs,t from Tables A.6, A.7 & A.8 if alternative procedure is adopted

81

Discussion

Any Questions