Practical Design to Eurocode 2

Transcription

1 Practical Design to Eurocode 2 The webinar will start at Course Outline Lecture Date Speaker Title 1 21 Sep Charles Goodchild Introduction, Background and Codes 2 28 Sep Charles Goodchild EC2 Background, Materials, Cover and effective spans 3 5 Oct Paul Gregory Bending and Shear in Beams 4 12 Oct Charles Goodchild Analysis 5 19 Oct Paul Gregory Slabs and Flat Slabs 6 26 Oct Charles Goodchild Deflection and Crack Control 7 2 Nov Paul Gregory Detailing 8 9 Nov Jenny Burridge Columns 9 16 Nov Jenny Burridge Fire Nov Jenny Burridge Foundations Lecture 10: Foundations 1

2 Foundations Lecture th November 2015 Model Answers Lecture 9 Exercise: Fire resistance of Column Lecture 10: Foundations 2

3 Fire resistance of column Using Equation 5.7, work out the fire resistance of a 250 x 750 column with an axial capacity of 3750kN and an axial load in cold conditions of 3500kN. The column is on the ground floor of a three storey building and the length is 4.5m. The cover is 30mm, main bars are 20mm and the links are 10mm diameter. Design Exercise Answers =120( ). 120 μ fi = 0.7 x 3500/3750 = 0.65 R ηfi = 83(1-0.65) = 28.8 a = = 50mm R a = 1.6(50-30) = 32 l 0,fi = 0.5 x 4.5 = 2.25m R l = 9.6(5-2.25) = 26.4 b = 1.2 x 250 = 300mm R b = 0.09 x 300 = 27 n > 4 R n = 12 R = 120((R ηfi + R a +R l + R b + R n )/120) 1.8 = 131 minutes Lecture 10: Foundations 3

4 Foundations Outline Week 10, Foundations We will look at the following topics: Eurocode 7: Geotechnical design Partial factors, spread foundations. Pad foundation Worked example & workshop Retaining walls Piles Lecture 10: Foundations 4

5 Eurocode 7 Eurocode 7 has two parts: Part 1: General Rules Plus NA Part 2: Ground Investigation and testing Plus NA 6 (p43 et seq) Eurocode 7 How to 6. Foundations The essential features of EC7, Pt 1 relating to foundation design are discussed. Note: This publication covers only the design of simple foundations, which are a small part of EC7. It should not be relied on for general guidance on EC7. Lecture 10: Foundations 5

6 Limit States The following ultimate limit states apply to foundation design: EQU: Loss of equilibrium of the structure STR: Internal failure or excessive deformation of the structure or structural member GEO: Failure due to excessive deformation of the ground UPL: Loss of equilibrium due to uplift by water pressure HYD: Failure caused by hydraulic gradients Categories of Structures Category Description 1 Small and relatively simple structures 2 Conventional types of structure no difficult ground Risk of geotechnical failure Negligible No exceptional risk Examples from EC7 None given Spread foundations 3 All other structures Abnormal risks Large or unusual structures Lecture 10: Foundations 6

7 EC7 ULS Design EC7 provides for three Design Approaches UK National Annex - Use Design Approach 1 DA1 For DA1 (except piles and anchorage design) there are two sets of combinations to use for the STR and GEO limit states. Combination 1 generally governs structural resistance Combination 2 generally governs sizing of foundations STR/GEO ULS Actions partial factors Combination 1 Permanent Actions Leading variable action Accompanying variable actions Unfavourable Favourable Main Others Exp G k 1.0G k 1.5Q k 1.5ψ 0,i Q k Exp 6.10a 1.35G k 1.0G k 1.5ψ 0,1 Q k 1.5ψ 0,i Q k Exp 6.10b 1.25G k 1.0G k 1.5Q k 1.5ψ 0,i Q k Combination 2 Exp G k 1.0G k 1.3Q k 1.3ψ 0,i Q k Notes: If the variation in permanent action is significant, use G k,j,sup and G k,j,inf If the action is favourable, γ Q,i = 0 and the variable actions should be ignored Lecture 10: Foundations 7

8 Factors for EQU, UPL and HYD Limit state Permanent Actions Variable Actions Unfavourable Favourable Unfavourable Favourable EQU UPL HYD Partial factors material properties Parameter Angle of shearing resistance Symbol Combination 1 Combination 2 EQU γ φ Effective cohesion γ c Undrained shear strength γ cu Unconfined strength γ qu Bulk density γ γ Lecture 10: Foundations 8

9 Geotechnical Report The Geotechnical Report should: be produced for each project (if even just a single sheet) contain details of: the site, interpretation of ground investigation report, geotechnical recommendations, advice Foundation design recommendations should state: bearing resistances, characteristic values of soil parameters and whether values are SLS or ULS, Combination 1 or Combination 2 values Spread Foundations EC7 Section 6 Three methods for design: Direct method check all limit states: Load and partial factor combinations (as before) q ult =c N c s c d c i c g c b c + q N q s q d q i q g q b q +γ BN γ s γ d γ i γ g γ b γ /2 where c = cohesion q = overburden γ = body-weight N i = bearing capacity factors s i = shape factors d i = depth factors i i = inclination factors g i = ground inclination factors b i = base inclination factors We just bung it in a spreadsheet Settlement often critical See Decoding Eurocode 7 by A Bond & A Harris, Taylor & Francis Lecture 10: Foundations 9

10 Spread Foundations EC7 Section 6 Three methods for design: Direct method check all limit states Indirect method experience and testing used to determine SLS parameters that also satisfy ULS Prescriptive methods use presumed bearing resistance (BS8004 quoted in NA). Used in subsequent slides). Spread Foundations Design procedures in: Lecture 10: Foundations 10

11 Fig 6/1 (p46) Procedure for depth of spread foundations Pressure distributions SLS pressure distributions ULS pressure distribution Lecture 10: Foundations 11

12 Load cases EQU : 0.9 G k Q k (assuming variable action is destabilising e.g. wind, and permanent action is stabilizing) STR : 1.35 G k Q k (Using (6.10). Worse case of Exp (6.10a) or (6.10b) could be used) Plain Concrete Strip Footings & Pad Foundations: Cl , Exp (12.13) 0,85 h a F (3σ gd /f ctd,pl ) where: σ gd is the design value of the ground pressure as a simplification h f /a 2 may be used a a h F b F Lecture 10: Foundations 12

13 Plain Concrete Strip Footings & Pad Foundations C16/20 C20/25 C25/30 C30/37 allowable pressure σ gd h F /a h F /a h F /a h F /a e.g. cavity wall 300 wide carrying 80 kn/m onto 100 kn/m 2 ground: b f = 800 mm a = 250 mm h f = say assuming C20/25 concrete 0.85 x 250 = 213 say 225 mm a b F a h F Reinforced Concrete Bases Check critical bending moments at column faces Check beam shear and punching shear For punching shear the ground reaction within the perimeter may be deducted from the column load Lecture 10: Foundations 13

14 Pad foundation Worked example Worked Example Design a square pad footing for a mm column carrying G k = 600 kn and Q k = 505 kn. The presumed allowable bearing pressure of the non-aggressive soil is 200 kn/m 2. Answer: Category 2. So using prescriptive methods: Base area: ( )/200 = 5.525m 2 => 2.4 x 2.4 base x 0.5m (say) deep. Lecture 10: Foundations 14

15 Worked Example Use C30/37 concrete Loading = 1.35 x x 505 = kn ULS bearing pressure = /2.4 2 = 272 kn/m 2 Critical section at face of column M Ed = 272 x 2.4 x / 2 = 343 knm d = = 434 mm K = 343 x10 6 /(2400 x x30) = Worked Example z = 0.95d = 0.95 x 434 = 412mm A s = M Ed /f yd z = 343 x 10 6 / (435 x 412) = 1914mm 2 Provide 250 c/c b.w (2010 mm 2 ) (804 mm 2 /m) Beam shear: Check critical section d away from column face V Ed = 272 x ( ) = 161kN/m v Ed = 161 / 434 = 0.37MPa ρ = 2010/ (434 x 2400) = = 0.19% v Rd,c (from table) = 0.42MPa => beam shear ok. 6/Table 6 (p47) Concise Table 15.6 Lecture 10: Foundations 15

16 Worked Example Punching shear: Basic control perimeter at 2d from face of column v Ed = βv Ed / u i d < v Rd,c β = 1, u i = (350 x x 2 x 2 x π) = 6854mm V Ed = load minus net upward force within the area of the control perimeter) = x ( π x x.35 x 4) = 560kN v Ed = MPa; v Rd,c = 0.42 (as before) => ok Retaining Walls Chapter 9 Lecture 10: Foundations 16

17 Ultimate Limit States for the design of retaining walls Calculation Model A Rankine theory Model applies if b h h a tan (45 - ϕ d /2) Lecture 10: Foundations 17

18 Lecture 10: Foundations 18 Calculation Model B Inclined virtual plane theory Model applies to walls of all shapes and sizes General Model A Model B 2 B L 2 b b L b b B b B t W b H W b s t s t s h k,c b b k,c s s = + = = γ = γ = B L 2 b b b L 2 tan b H b W tan b H t h vp h s t f k,f h h f h b = Ω =β + + γ β + = β + + = General expressions

19 9 (Figure 4) Overall design procedure Initial sizing b s t b h/10 to h/15 B 0.5h to 0.7h b t B/4 to B/3 Lecture 10: Foundations 19

20 9 (Figure 4) Overall design procedure 9 (Figure 6) Figure 6 for overall design procedure Lecture 10: Foundations 20

21 Soil Densities Ex Concrete Basements Design value of effective angle of shearing resistance, φ d tan φ d = tan (φ k /γ φ ) where φ k = φ max for granular soils and = φ for clay soils, φ max and φ are as defined as follows γ φ = 1.0 or 1.25 dependent on the Combination being considered. Ex Concrete Basements Lecture 10: Foundations 21

22 Angle of shearing resistance Granular Soils Estimated peak effective angle of shearing resistance, φ max = 30 + A + B + C Estimated critical state angle of shearing resistance, φ crit = 30 + A + B, which is the upper limiting value. Ex Concrete Basements Clay soils Long term Granular Soils Ex Concrete Basements Lecture 10: Foundations 22

23 Calcs Material properties & earth pressures 9 (Panel 2) 9 Panel 2 (p71) 2 9 (Figure 4) Overall design procedure Lecture 10: Foundations 23

24 9 (Figure 7) Design against sliding (Figure 7) Sliding Resistance 9 (Panel 3) Lecture 10: Foundations 24

25 9 (Figure 4) Overall design procedure Design against Toppling 9 (Figure 9) Lecture 10: Foundations 25

26 9 (Figure 4) Overall design procedure Design against bearing failure 9 (Figure 10) Lecture 10: Foundations 26

27 Expressions for bearing resistance 9 (Panel 4,Figure 11) 9 (Figure 4) Overall design procedure Lecture 10: Foundations 27

28 Structural design 9 (Figure 13) Remember: Load and Partial Factor Combinations Parameter Symbol Comb. 1 Comb. 2 Actions Permanent action : unfavourable γ G,unfav Permanent action: favourable γ G,fav Variable action γ Q Soil Properties Angle of shearing resistance γ φ Effective cohesion γ c Undrained shear strength γ cu Unconfined strength γ qu Bulk density γ γ Lecture 10: Foundations 28

29 Piles Flexural and axial resistance of piles Uncertainties related to the cross-section of cast in place piles and concreting procedures shall be allowed for in design In the absence of other provisions, the design diameter of cast in place piles without permanent casing is less than the nominal diameter D nom : D d = D nom 20 mm for D nom < 400 mm D d = 0.95 D nom for 400 D nom 1000 mm D d = D nom 50 mm for D nom > 1000 mm ICE Specification for piling and embedded retaining walls (ICE SPERW) B states The dimensions of a constructed pile or wall element shall not be less than the specified dimensions. A tolerance of 5% on auger diameter, casing diameter, and grab length and width is permissible. Lecture 10: Foundations 29

30 Flexural and axial resistance of piles The partial factor for concrete, γ c, should be multiplied by a factor, k f, for calculation of design resistance of cast in place piles without permanent casing. The UK value of k f = 1.1, therefore γ c,pile = 1.65 If the width of the compression zone decreases in the direction of the extreme compression fibre, the value η f cd should be reduced by 10% Bored piles Reinforcement should be detailed for free flow of concrete. Minimum diameter of long. reinforcement = 16mm Minimum number of longitudinal bars = 6 [BUT BS EN 1536 Execution of special geotechnical work Bored Piles says 12 mm and 4 bars!] Minimum areas: Pile cross Min area of long. Pile section: A c rebar, A s,bpmin diameters A c 0.5 m 2 0.5% A c < 800 mm 0.5 m 2 < A c 1.0 m mm 2 A c > 1.0 m % A c >1130 mm Lecture 10: Foundations 30

31 Minimum reinforcement Minimum area of reinforcement, A s,bpmin (mm 2 ) Pile diameter, mm Workshop Design a pad foundation for a 300mm square column taking G k = 600kN, Q k = 350kN. Permissible bearing stress = 225kPa. Concrete for base C30/37. Work out size of base, tension reinforcement and any shear reinforcement. Lecture 10: Foundations 31

32 Workshop Problem Category 2, using prescriptive methods Base size: (G k + Q k )/bearing stress = = _m 2 x base x mm deep (choose size of pad) Use C 30/37 (concrete) Loading = γ g x G k + γ q x Q k = = kn ULS bearing pressure = / 2 = kn/m 2 Critical section at face of column M Ed = x x 2 / 2 = knm d = cover assumed ø = = mm K = M/bd 2 f ck = Workshop Problem z = d = x = mm A s = M Ed /f yd z = mm 2 Provide c/c ( mm 2 ) Check minimum steel 100A s,prov /bd = For C30/37 concrete A s,min = OK/not OK Beam shear Check critical section d away from column face V Ed = x = kn/m v Ed = V Ed / d = MPa ρ = / ( x ) = = % v Rd,c (from table) = MPa beam shear OK/not OK. Table Lecture 10: Foundations 32

33 Workshop Problem Punching shear Basic control perimeter at 2d from face of column v Ed = βv Ed / u i d < v Rd,c β = 1, u i = = mm V Ed = load minus net upward force within the area of the control perimeter) = x ( ) = kn v Ed = MPa; v Rd,c = (as before) => ok/not ok End of Lecture 10 (and the course!) s: pgregory@concretecentre.com cgoodchild@concretecentre.com jburridge@concretecentre.com Lecture 10: Foundations 33