Session 3. Calculation Models. EQU, UPL and HYD. Design of Pile Foundations

Transcription

1 1 Session 3 Calculation Models EQU, UPL and HYD Design of Pile Foundations (Killiney Bay)

2 2 SLS and Settlement Calculations (Dalkey Island)

3 Summary of ULS Designs Undrained width (m) Drained width (m) ULS Design Width DA1.C1 (1.32) (1.62) DA1.C DA DA 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 3

4 SLS Design 4 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 (OFS u ) 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 OFS u < 3, one should calculate settlement If OFS u < 2, one should calculate settlement accounting for non-linear stiffness For undrained designs of foundations with permanent structural loads only DA1 give OFS u = 1.4 while DA2 and DA3 give OFS u = Hence settlement calculations are needed

5 OFS u Ratios ULS design (drained) width (m) OFS u using drained width = R u,k / V k = B 2 ULS undrained width (m) OFS u using undrained width = R u,k / V k = γ F x γ M/R DA DA DA Calculating the undrained OFS ratio using the design width i.e. the drained design width: OFS u = R u,k / V k = A ( (p + 2) c u,k b c s c ic + qc ) / Vc = B2 x ( 5.14 x 200 x 1.0 x 1.2 x x 0.8) / ( ) = B2 x ( ) / 1500 = B In this example, using design (i.e. drained) widths: For DA1, OFS = u 3.60 ( > 3 ) Settlement need not be calculated For DA2, OFS u = 2.91 ( < 3 ) Settlement should be calculated For DA3, OFS u = 4.37 ( > 3 ) Settlement need not be calculated But using the undrained widths OFS u values are all less than 2.0, - much lower than value of 3 often used in traditional designs 5

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 / E m where: E m = design value of the modulus of elasticity f = settlement coefficient p = bearing pressure Assume E m = E = 1.5N = 1.5 x 40 = 60 MPa f = (1 ν 2 ) I where ν = 0.25 and I = 0.95 for square flexible uniformly loaded foundation Then f = ( ) x 0.95 = p = (G k + Q k )/B 2 = ( ) / B 2 = 1500 / B 2 Hence settlement: where B is in m s = p B f / E m = (1500 / B 2 ) x B x x 1000 / = / B mm 6

7 7 Calculated Settlements ULS design width (m) OFS u Settlement ( mm ) s = / B DA DA DA 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 E d C d is fulfilled as for each DA, s < 25 mm Note words of caution in EN : Settlement calculations should not be regarded as accurate but as merely providing an approximate indication

8 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 R u,k / V k for the ULS drained design widths is greater than 3 for DA1 and DA3 so settlement calculations are not required

9 Any Questions? 9

10 10 Session 3a Calculation Models (Killiney Hill and Dalkey Island)

11 Need for Calculation Models 11 Equations: ULS E d R d SLS E d C d For a ULS, calculation models are generally not needed to determine E d e.g. the design load on a foundation is obtained by multiplying the characteristic applied load by the partial factor: E d = F d = γ F F k 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. F d = γ F f{c k, φ k, F k } Calculation models are always required to determine the design resistance, R d, e.g. the bearing or sliding resistance of a spread foundation For an SLS, calculations are always required to determine E d, e.g. the settlement of a foundation Values of C d, 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 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 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 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 K a and K p 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. K a values in figure below:

15 Analytical Earth Pressures 15 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 K a and K p is provided based on the method of characteristics with slip line fields e = c K c + q K q + γ d K γ Equations for the earth pressure factors K c, K q, and K γ 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 N b instead of 0.5N γ Exact solutions for N c and N q factors were originally derived by Prandtl (1920 and 1921) and published in German 16

17 17 Undrained Bearing Resistance Undrained bearing resistance R/A = (π + 2)c u b c s c i c + q where the dimensionless factors for: Inclination α of the foundation base b c = 1 2α / (π + 2) The shape of the foundation s c = 1+ 0/2 (B /L ) for a rectangular foundation s c = 1.2 for a square or circular shape The inclination of the load, caused by a horizontal load H 1(11)c= 2+ A'cwith H A c u Hiu

18 Drained Bearing Resistance Drained bearing resistance R/A = c' N c b c s c i c + q 'N q b q s q i q + 0.5γ'B 'N γ b γ s γ i γ with the values of the dimensionless factors for: the bearing resistance: N q = e π tanϕ' tan 2 (45 + φ'/2) N c = (N q -1) cot φ N γ = 2 (N q -1) tanφ', where δ φ'/2 (rough base) the inclination of the foundation base: b c = b q -(1-b q ) / (N c tanφ ) b q = b γ = (1 - α tanφ ) the shape of foundation: s q = 1 + (B' / L' ) sinφ', for a rectangular shape s q = 1 + sinφ', for a square or circular shape s γ = 1 0,3 (B'/L ), for a rectangular shape s γ = 0,7, for a square or circular shape s c = (s q N q -1)/(N q - 1) for rectangular, square or circular shape 18

19 19 Characteristic and Factored N c Values Bearing resistance factor, Nc, Nc,DA1.C2 Nck Nc,DA Characteristic angle of shearing resistance, φ'k ( o )

20 20 Characteristic and Factored N q Values 120 Bearing resistance factor, Nq N q,k Nq,DA2 N q,da1.c Characteristic angle of shearing resistance, φ' k ( o )

21 21 Characteristic and Factored N γ Values 120 Bearing resistance factors, Nc, Nq, N γ N γ,k N γ,da2 N γ,da1.c Characteristic angle of shearing resistance, φ' k ( o )

22 Comparison of N γ Values 22 Exact solutions are not available for N γ and a number of different equations have been proposed: Caquot and Kerisel (1953): N γ = 2(N q + 1) tan φ Meyerhof (1963): N γ = (N q - 1) tan (1.4φ ) Brinch Hansen (1970): N γ = 1.5 (N q - 1) tan φ Equation for N γ adopted in Eurocode 7: N γ = 2 (N q - 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 N γ to which Salgado (2008) has fitted the following eqn: N γ = (N q - 1) tan (1.32 φ ) Nγ Eurocode 7 Caquot & Kerisel Meyerhof Salgado Brinch Hansen Effective angle of shearing resistance, φ' Salgado s eqn. is closest to Brinch Hansen s eqn. and appears to indicate that the Eurocode 7 equation for N g may be unconservative (i.e. unsafe)

23 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 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 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 = σ 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

25 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 ) /E m

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, q c on the abscissa and discontinuity spacing, d c on 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 26

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 27

28 28 Time for Discussion Any questions

29 Session 3b 29 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: Fulcrum Beam T Tension pile For what ULS do you design the pile EQU or GEO? For DA1 T = 0! 30

31 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 heave 3. Failure by internal erosion 4. 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 32

33 Hydraulic Failures Figures from EN showing Hydraulic Failures Conditions that may cause Uplift Conditions that may cause heave Conditions that may cause piping 33

34 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 Equilibrium One equation given: V dst;d G stb;d + R d 2.8 where: V dst;d = design vertical disturbing load = G dst;d (design perm. load) + Q dst;d (design var. load) G dst;d R d = b x u dst;d (design uplift water pressure force) = T d (design wall friction force) 35

36 HYD Equilibrium Equations Hydraulic head h Groundwater level at ground surface Design effective soil weight, G' stb,d Standpipe d Design seepage force, S dst,d Relevant soil column Design total vertical stress, σ stb,d Design total pore water pressure, u dst,d Two equations are given for HYD equilibrium First eqn: u dst;d σ std;d 2.9a (total stress eqn. - only equation in Eurocode 7 in terms of stress) Second eqn: S dst;d G stb;d 2.9b (seepage force and submerged weight eqn.) i.e. (γ w i Vol) d (γ Vol) d where i = h / d (γ w h/d) d (γ ) d (γ w h) d (γ d) d u dst;d σ stb;d (effective stress eqn.) 36

37 Recommended UPL and HYD Partial Factors Partial factors UPL HYD Actions, γ F γ G;dst γ G;stb γ Q;dst33 Material properties, γ M plus pile tensile resistance and anchorage resistance γ φ γ c γ c u γ s;t γ a 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 37

38 38 Overall Factor of Safety (OFS) for Uplift Equation 2.8: V dst;d G stb;d + R d For no R d (i.e. soil resistance on side of buried structure ignored) γ G;dst V dst;k = γ G;stb G stb;k Overall factor of safety (OFS) = G stb;k / V dst;k = γ G;dst / γ G;stb Applying recommended partial factors γ G;dst /γ G;stb = 1.0/0.9 = 1.11 Hence OFS = 1.11

39 OFS against Heave using EC7 Equations ydraulic Potential head Standpipe h Groundwater level at ground surface d Design effective soil weight, G' stb;d Design seepage force, S dst;d Relevant soil column Design total vertical stress, σ stb;d Design total pore water pressure, u dst;d Equation 2.9b S dst;d G stb;d γ G;dst S dst;k γ G;stb G stb;k γ G;dst γ w i V γ G;stb γ V OFS (b) = G stb;k / S dst;k = γ G;dst / γ G;stb = γ / (γ w i) = i c /i = critical hydraulic gradient / actual hydraulic gradient OFS (b) = γ dst /γ stb = 1.35/0.9 = 1.5 Equation 2.9a u dst;d σ stb;d γ G;dst γ w d + γ G;dst γ w h γ G;stb γ'd + γ G;stb γ w d OFS (a) = γ G;dst / γ G;stb = (γ d + γ w d) / (γ w h + γ w d) = (i c + 1)/(i + 1) = 1.5 i c /i = /i if i = 0.5 then i c /i = OFS (b) = 2.5 i.e. more cautious than using Eqn. 2.9b because γ w d occurs on both sides of equation and is multiplied by different γ F values 39

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 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. i c value depends on the design conditions EN 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 40

41 Uplift Design Example (Issued for Workshop on the Evaluation of Eurocode 7 in Dublin in 2005) Structural loading g k = 40kPa 5.0m R 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 = 35 o, g = 20 kn/m 3 - Concrete weight density = 24 kn/m3 Base thickness, D requested T 15.0m U Forces U = Uplift water pressure force = γ w 15 (5 + T G = Weight of basement plus structural load R = Resisting force from soil on side walls A wide range of design values obtained for D = m Why? 41

42 Model for UPL Equilibrium Calculation Model Assumptions Include or ignore R? R = Aτ = Aσ h tanφ = AKσ v tanδ where A = sidewall area What value for K? K is a function of φ and δ. Should K = K 0 or K a? 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 R d? No UPL resistance factors are provided in EN 1997 to obtain R d from R k i.e. there is no UPL equivalent to DA2 Could design according to EN and assume σ h = K a σ v 1) With R k = AK a;k σ v tanδ k apply partial factor γ M to φ k to obtain K a;d and δ as for DA1.C2 and hence get R d d (Clause (1)) 2) Treat R κ as a permanent stabilising vertical action and apply γ G;stb to R k to obtain R d (Clause (2)) 42

43 43 Determination of Design Value of R d Assume K = K a and is obtained from EN for δ = 2/3φ 1) Clause (1): Apply partial factor γ M to φ k to obtain K a;d and δ and hence R d d No factors applied: φ k = 35 o and δ = 2/3φ K a;k = 0.23 δ k = 2/3φ k = 23.3 o R k = AK a;k σ v tanδ k = 0.099Aσ v a) γ = 1.25 applied to obtain φ M d and δ d by reducing φ k and hence δ k φ d = 29.3 o and δ = 2/3φ K a;d = 0.29 δ d = 2/3φ d = 19.5 o R d = AK a;d σ v tanδ d = 0.103Aσ v Since R is a resistance, need R d < R k : R d = (0.103/0.099)R k R d = 1.04 R k unsafe b) γ M applied to increase φ k but to reduce δ φ d = 41.2 o and δ d = 19.0 o δ d /φ = d 19.0/41.2 = 0.46 K a;d = 0.18 R d = AK a;d σ v tanδ d = 0.062Aσ v R d = (0.062/0.099)R k R d = 0.69 R k safe 2) Clause (2): Treat R κ as a permanent stabilising vertical action and apply γ G;stb to R k to obtain R d R d = γ G;stb R k R d = 0.9 R k safe

44 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 σ h tanδ = A Kσ v tanδ K = K 0 or K a δ = 0.5φ or (2/3)φ How R d 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 GWL - Sheet pile wall H =? 7.0m 1.0m Water - Pile penetration 3m below excavation level m water in excavation - Weight density of sand = 20 kn/m 3 Sand γ = 20kN/m 3 3.0m Require H - Height of GWL behind wall above excavation level A very wide range of design values obtained: H = m Why? 45

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 46

47 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 48 Discussion Any Questions?

49 49 Session 3c Design of Pile Foundations (Killarney Waterfall)

50 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 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 ) Execution standard Execution of Special Geotechnical Works EN 12699:2000 Displacement piles EN 12063: Sheet pile walls EN 1536: Bored Piles EN 14199: Micro-piles

52 52 Limit State Checklist for Pile Design Limit states to be considered Checked Loss of overall stability Bearing resistance failure of the pile foundation Uplift or insufficient tensile resistance of the pile foundation Failure in the ground due to transverse loading of the pile foundation Structural failure of the pile in compression, tension, bending, buckling or shear Combined failure in the ground and in the pile foundation Combined failure in the ground and in the structure Note change from handout Excessive settlement Excessive heave Excessive lateral movement Unacceptable vibrations

53 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 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 55 Design considerations 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

56 56 Checklist for Selection of Pile Type SELECTION OF PILE TYPE Checked The ground and ground-water conditions, including the presence or possibility of obstructions in the ground. The stresses generated in the pile during installation The possibility of preserving and checking the integrity of the pile being installed The effect of the method and sequence of pile installation on piles, which have already been installed and on adjacent structures or services. The tolerances within which the pile can be installed reliably The deleterious effects of chemicals in the ground The possibility of connecting different ground-water regimes The handling and transportation of piles

57 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 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 59 Equilibrium equation: where: F c;d < R c;d F c;d is the ULS design axial compression load F c;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 F c;d and R c;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. R c;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 60 Recommended Partial Factor Values

61 61 Design of a Compression Pile from Pile Load Tests F c;d is calculated in the normal way with γ F factors applied to applied loads R c;d must be determined from the measured pile resistance R c;m R c;m can be based on the total resistance or can be separated into R b;m (base) and R s;m (shaft) DA3 not used for pile design from load tests

62 62 Determination of Characteristic Resistance Characteristic resistance for DA1 & DA2 R c;k = Min {(R c;m ) mean /ξ 1 ; (R c;m ) min /ξ 2 } Characteristic values determined directly not estimated ξ for n = ξ ξ Table A.9

63 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 64 Example 1 Pile Design from Pile Load Tests F Characteristic Loads G k = 1200 kn Loose fill 2.0m Q k = 200 kn Glacial till CFA pile 600 mm diameter 9.0m Pile Load Test Results No of test = 2 Max Applied Load = 4000kN (same on both piles) Max load on base = 600kN

65 DA1.C1 DA1.C1 (A1 + M1 + R1) F c;d = 1.35* *200 = 1920 kn Note weight of pile not included) R c;m = 4000 kn for both R c;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 R b;k = (600 / 000) R c;k = 462 kn; R s;k = 2615 kn R c;d = R b;k / R s;k / 1.0 = 462 / / 1.0 = 3035 kn F c;d (1920 kn) < R c;d (3035kN) DA1.C1 is satisfied. 65

66 66 DA1.C2 DA1.C2 (A2 + M1 or M2 + R4) F c;d = 1.0* *200 = 1460 kn Note weight of pile not included R b;k = 462 kn; R s;k = 2615 kn as before for DA1.C1 R c;d = R b;k / R s;k / 1.3 = 462 / / 1.3 = 2615 kn F c;d (1460 kn) < R c;d (2615kN) DA1.C2 is satisfied DA1.C1 and DA1.C2 are both satisfied, therefore DA1 is satisfied

67 67 DA2 DA2 (A1 + M1 + R2) F c;d = 1.35* *200 = 1920 kn Note weight of pile not included R c;m = 4000 kn for both R c;k = lesser of 4000 / 1.3 or 4000 / 1.2 = 3077kN Use measured toe force to give ratio R b;k = (600 / 4000) R c;k = 462 kn; R s;k = 2615 kn as before R c;d = R b;k / R s;k / 1.1 = 462 / / 1.1 = 2797 kn F c;d (1920 kn) < R c;d (2797kN) DA2 is satisfied

68 68 Dynamic Pile Load Tests 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

69 69 Design of a Compression Pile from Ground Test results F c;d is determined in the normal way by factoring the loads R b;k and R s;k or R t;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 tests R b,d = R b,k / γ b and R s,d = R s,k / γ s where the γ b, γ b and γ t values are given in Table A.6, A.7 and A.8 (same as for design from pile load tests)

70 Alternative Procedure to Determine R s,k and R b,k from Ground Strength Parameters Calculate the characteristic base resistance (q b;k ) and shaft resistances (q s;k ) using characteristic values of ground parameters [C (8)] and hence: R b;k = A b q b;k and R sk = Σq si;k A si where A b = the nominal plan area of the base of the pile A si = 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 γ b and γ s 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 γ b and γ s or γ t when using this approach 70

71 71 Example 2 Pile Design using Ground Tests F Loose fill 2.0m Glacial till CFA pile 600 mm diameter 9.0m

72 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 c uk 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/m 3, c k = 0, φ s,k = 36 o for shaft resistance and φ b,k = 34 o 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/m 3 and 24kN/m 3, 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 γ b and γ s or γ t (γ R values)

73 73 Ground Parameters Only undrained conditions considered Assume q b = 9c u + σ v0 q s = α c u where α = 0.4 W P (weight of pile) = π * * 11.0 * 24.0 / 4 = 74.6 kn

74 74 DA1 q b;k = (9c u + σ v ) = (9* *18 + 9*22) = 5634 kpa q s;k = αc u = 0.4*{( )/2} = 140 kpa R b,k = A b q b,k = (π * / 4) * 5634 = 1593 kn R s,k = ΣA s q s,k = (π * 0.6 * 9 * 140 = 2375 kn DA1.C1: (A1+M1+R1) F c;d = 1.35( ) + 1.5*200 = kn R c;d = R b,k /(1.0*1.75) + R s,k /(1.0*1.75) = = kn OK DA1.C2 (A2+(M1 or M2) +R4) F c;d = 1.0( ) + 1.3*200 = kn R c;d = R bk /(1.45*1.75) + R sk /(1.3*1.75) = = kn OK

75 75 DA2 F c;d = 1.35( ) + 1.5*200 = kn R b;k = kn ; R s;k = kn as for DA1 R c;d = R b;k /(1.1*1.75) + R s;k /(1.1*1.75) = = kn F c;d < R c;d ( < ) so inequality is satisfied for DA2

76 76 DA3 F c;d as per DA1.C2 but partial factors on structural actions are 1.35 and 1.5 Design value of resistance obtained by using γ m (Table A4) parameters on soil properties and in Ireland a model factor of 1.75 is applied on γ b and γ s

77 77 DA3 F c;d = 1.35( ) + 1.5*200 = kn Note: increased partial factors compared with DA1:C2 q b;d = (9c u / γ m + σ v ) = (9*600/(1.4*1.75) + 2*18 + 9*22) = kpa q s;d = α c u / γ m = 0.4*{(100/(1.4*1.75) + 600/(1.4*1.75))/2} = 57.1 kpa R b;d = A b q b;d = (π*0.6 2 /4)*2438 = kn R s;d = ΣA s q s;d = π*0.6*9*57.1 = kn F c;d = R c;d = = kn kn Inequality F c;d R c;d Not satisfied

78 78 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 Must consider F t;d R t;d 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 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 R t;d = R t;k /γ s;t R t;k is determined from R t;m using ξ values in same manner as for a compression test ξ values from Table A.9 as before γ 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

80 esign of a Tension Pile from Ground Test Results Established from pile load test and from comparable experience R t;d = R t;k /γ s;t where R t;k = R s;k R t;k obtained from: R t;k = Min {(R s;calc ) mean / ξ 3 ; (R s;calc ) min / ξ 3 } or using the alternative procedure from R t;k = Σq si;k A si ξ values from Table A.10 γ s,t from Tables A.6, A.7 & A.8 In Ireland a model factor of 1.75 on γ s,t from Tables A.6, A.7 & A.8 if alternative procedure is adopted 80

81 81 Discussion Any Questions