Transcription

Practical Design to Eurocode 2The webinar will start at 12.30

Course OutlineLecture DateSpeakerTitle121 Sep Jenny Burridge2328 Sep Charles Goodchild EC2 Background, Materials, Coverand effective spans5 Oct Paul GregoryBending and Shear in Beams412 Oct Charles Goodchild Analysis519 Oct Paul Gregory626 Oct Charles Goodchild Deflection and Crack Control72 NovPaul GregoryColumns89 NovJenny BurridgeFire916 Nov Paul Gregory23 Nov Jenny Burridge10Introduction, Background and CodesSlabs and Flat SlabsDetailingFoundations

FoundationsLecture 1023rd November 2017

Lecture 9 ExerciseLap length for column longitudinal bars

Column lap length exerciseDesign informationH25’s C40/50 concrete 400 mm square columnLap 45mm nominal cover to main bars Longitudinal bars are in compression Maximum ultimate stress in the barsis 390 MPaExercise:Calculate the minimum lap lengthusing EC2 equation 8.10:H32’s

Column lap length exerciseProcedure Determine the ultimate bond stress, fbdEC2 Equ. 8.2 Determine the basic anchorage length, lb,reqEC2 Equ. 8.3 Determine the design anchorage length, lbdEC2 Equ. 8.4 Determine the lap length, l0 anchorage length x α6

Model AnswersLap length for column longitudinal bars

Column lap length exerciseDesign informationH25’s C40/50 concrete 400 mm square columnLap 45mm nominal cover to main bars Longitudinal bars are in compression Maximum ultimate stress in the barsis 390 MPaExercise:Calculate the minimum lap lengthusing EC2 equation 8.10:H32’s

Column lap length exerciseProcedure Determine the ultimate bond stress, fbdEC2 Equ. 8.2 Determine the basic anchorage length, lb,reqEC2 Equ. 8.3 Determine the design anchorage length, lbdEC2 Equ. 8.4 Determine the lap length, l0 anchorage length x α6

Solution - Column lap lengthDetermine the ultimate bond stress, fbdfbd 2.25 η1 η2 fctdEC2 Equ. 8.2η1 1.0 ‘Good’ bond conditionsη2 1.0 bar size 32fctd αct fctk,0,05/γcαct 1.0EC2 cl 3.1.6(2), Equ 3.16γc 1.5fctk,0,05 0.7 x 0.3 fck2/3EC2 Table 3.1 0.21 x 402/3 2.456 MPafctd αct fctk,0,05/γc 2.456/1.5 1.637fbd 2.25 x 1.637 3.684 MPa

Solution - Column lap lengthDetermine the basic anchorage length, lb,reqlb,req (Ø/4) ( σsd/fbd)EC2 Equ 8.3Max ultimate stress in the bar, σsd 390 MPa.lb,req (Ø/4) ( 390/3.684) 26.47 ØFor concrete class C40/50

Solution - Column lap lengthDetermine the design anchorage length, lbdlbd α1 α2 α3 α4 α5 lb,req lb,minlbd α1 α2 α3 α4 α5 (26.47Ø)Equ. 8.4For concrete class C40/50For bars in compression α1 α2 α3 α4 α5 1.0Hence lbd 26.47Ø

Solution - Column lap lengthDetermine the lap length, l0 anchorage length x α6All the bars are being lapped at the same section, α6 1.5A lap length is based on the smallest bar in the lap, 25mmHence,l0 lbd x α6l0 26.47 Ø x 1.5l0 39.71 Ø 39.71 x 25l0 993 mm

Foundations

Outline – Week 10, FoundationsWe will look at the following topics: Eurocode 7: Geotechnical design –Partial factors, spreadfoundations. Pad foundation – Worked example & workshop Retaining walls Piles

Eurocode 76 (p43 et seq)Eurocode 7 has two parts:Part 1: General RulesPlus NAPart 2: Ground Investigation and testingPlus NA

Eurocode 7How to 6. FoundationsThe essential features of EC7, Pt 1 relatingto foundation design are discussed.Note:This publication covers only the design ofsimple foundations, which are a small partof EC7.It should not be relied on for generalguidance on EC7.

Limit StatesThe following ultimate limit states apply to foundationdesign:EQU: Loss of equilibrium of the structureSTR: Internal failure or excessive deformation of thestructure or structural memberGEO: Failure due to excessive deformation of the groundUPL: Loss of equilibrium due to uplift by water pressureHYD: Failure caused by hydraulic gradients

Categories of StructuresCategory DescriptionRisk of geoExamples fromtechnical failure EC71Small and relativelysimple structuresNegligibleNone given2Conventional types ofstructure – no difficultgroundNo exceptionalriskSpreadfoundations3All other structuresAbnormal risksLarge or unusualstructures

EC7 – ULS DesignEC7 provides for three Design ApproachesUK National Annex - Use Design Approach 1 – DA1For DA1 (except piles and anchorage design) there are twosets of combinations to use for the STR and GEO limitstates.Combination 1 – generally governs structural resistanceCombination 2 – generally governs sizing of foundations

STR/GEO ULS –Actions partial factorsPermanent Accompanying variableactionsExp 6.101.35Gk1.0Gk1.5QkExp 6.10a1.35Gk1.0GkExp 0,iQkMainOthersCombination 11.5ψ0,iQk1.5ψ0,1Qk1.5ψ0,iQkCombination 2Exp 6.10Notes:If the variation in permanent action is significant, use Gk,j,sup and Gk,j,infIf the action is favourable, γQ,i 0 and the variable actions should be ignored

Factors for EQU, UPL and HYDLimitstatePermanent ActionsVariable eEQU1.10.91.50UPL1.10.91.50HYD1.350.91.50

Partial factors – material propertiesParameterSymbolCombination Combination12EQUAngle of shearingresistanceγφ1.01.251.1Effective cohesionγc’1.01.251.1Undrained shearstrengthγcu1.01.41.2Unconfined strengthγqu1.01.41.2Bulk densityγγ1.01.01.0

Geotechnical ReportThe 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,adviceFoundation design recommendations should state: bearing resistances, characteristic values of soil parameters and whether values are SLS or ULS , Combination 1 orCombination 2 values

Spread FoundationsEC7 Section 6Three methods for design: Direct method – check all limit states: Load and partial factor combinations (as before) qult c’Ncscdcicgcbc q’Nqsqdqiqgqbq ––c cohesionq overburdenγ body-weightNi bearing capacity factorssi shape factorsdi depth factorsii inclination factorsgi ground inclination factorsbi base inclination factors“We just bung it in aspreadsheet”Settlement often criticalSee Decoding Eurocode 7 by ABond & A Harris, Taylor & Francis

Spread FoundationsEC7 Section 6Three methods for design: Direct method – check all limit states Indirect method – experience and testingused to determine SLS parameters thatalso satisfy ULS Prescriptive methods – use presumedbearing resistance (BS8004 quoted in NA).Used in subsequent slides).

Spread FoundationsDesign procedures in:

Fig 6/1 (p46)Procedure for depth ofspread foundations

Pressure distributionsSLS pressuredistributionsULS pressuredistribution

Load casesEQU : 0.9 Gk 1.5 Qk (assuming variable action isdestabilising e.g. wind, andpermanent action is stabilizing)STR : 1.35 Gk 1.5 Qk (Using (6.10). Worse case of Exp(6.10a) or (6.10b) could be used)

Plain Concrete Strip Footings &Pad Foundations:Cl. 12.9.3, Exp (12.13)0,85 hF (3σgd/fctd,pl)ahFwhere:σgd is the design value of the ground pressure as a simplification hf/a 2 may be usedaabF

Plain Concrete Strip Footings &Pad Foundationsallowablepressure C30/37hF /ahF /ahF 0.780.951.101.23e.g. cavity wall 300 wide carrying 80 kN/m onto 100 kN/m2ground:bf 800 mma 250 mmhf say assuming C20/25 concrete0.85 x 250 213 say 225 mm0.520.740.901.041.17hFaabF

Reinforced Concrete Bases Check critical bending moments at column faces Check beam shear and punching shearFor punching shear theground reaction withinthe perimeter may bededucted from thecolumn load

Pad foundationWorked example

Worked ExampleDesign a square pad footing for a 350 350 mm columncarrying Gk 600 kN and Qk 505 kN. The presumedallowable bearing pressure of the non-aggressive soil is200 kN/m2.Answer:Category 2. So using prescriptive methods:Base area: (600 505)/200 5.525m2 2.4 x 2.4 base x 0.5m (say) deep.

Worked ExampleLoading 1.35 x 600 1.5 x 505 1567.5 kNULS bearing pressure 1567.5/2.42 272 kN/m2Critical section at face of columnMEd 272 x 2.4 x 1.0252 / 2 343 kNmd 500 – 50 – 16 434 mmUse C30/37 concreteK 343 x106/(2400 x 4342 x30) 0.025

Worked Example z 0.95d 0.95 x 434 412mm As MEd/fydz 343 x 106 / (435 x 412) 1914mm2 Provide 10H16 @ 250 c/c b.w (2010 mm2) (804 mm2/m)Beam shear:Check critical section d away from column faceVEd 272 x (1.025 – 0.434) 161kN/mvEd 161 / 434 0.37MPaρ 2010/ (434 x 2400) 0.0019 0.19%vRd,c (from table) 0.42MPa beam shear ok.6/Table 6 (p47)Concise Table 15.6

Worked ExamplePunching shear:Basic control perimeter at 2d from face of columnvEd βVEd / uid vRd,cβ 1, ui (350 x 4 434 x 2 x 2 x π) 6854mmVEd load minus net upward force within the area of thecontrol perimeter) 1567.5 – 272 x (0.352 π x .8682 .868 x .35 x 4) 560kNvEd 0.188 MPa; vRd,c 0.42 (as before) ok

Retaining WallsChapter 9

Ultimate Limit Statesfor the design ofretaining walls

Calculation Model ARankine theoryModel applies if bh ha tan (45 - ϕ’d/2)

Calculation Model BInclined ‘ virtual’ plane theoryModel applies to walls of all shapes and sizes

General expressionsGeneralModel AWs b sHγ k,ch tb H bh tanβWb tbBγ k,c b tanβ Wf bh H h γk,f2 bLf bt bs h2Ω βLvp Bbh B b s b tL s bt Lb B2bs2Model B

9 (Figure 4)Overall designprocedure

Initial sizingbs tb h/10 to h/15B 0.5h to 0.7hbt B/4 to B/3

9 (Figure 4)Overall designprocedure

9 (Figure 6)Figure 6 for overalldesign procedure

Soil DensitiesEx Concrete Basements

Design value of effective angleof shearing resistance, φ’dtan φ’d tan (φ’k/γφ)whereφ’k φ’max for granular soils and φ’ for clay soils,φ’max and φ’ are as defined as followsγφ 1.0 or 1.25 dependent on theCombination being considered.Ex Concrete Basements

Angle of shearing resistanceGranular SoilsEstimated peak effective angle of shearing resistance,φ’max 30 A B CEstimated critical state angle of shearing resistance,φ’crit 30 A B, which is the upper limiting value.Ex Concrete Basements

Clay soilsLong term Granular SoilsEx Concrete Basements

Calcs –Material properties & earth pressures9 (Panel 2)9 Panel 22’(p71)

9 (Figure 4)Overall designprocedure

9 (Figure 7)Design againstsliding(Figure 7)

Sliding Resistance9 (Panel 3)

9 (Figure 4)Overall designprocedure

Design against Toppling9 (Figure 9)

9 (Figure 4)Overall designprocedure

Design against bearing failure9 (Figure 10)

Expressions for bearing resistance9 (Panel 4,Figure 11)

9 (Figure 4)Overall designprocedure

Structural design9 (Figure 13)

Remember: Load and PartialFactor CombinationsParameterSymbolComb. 1Comb. 2γG,unfav1.351.0γG,fav1.001.00γQ1.501.30Angle of shearing resistanceγφ1.01.25Effective cohesionγc’1.01.25Undrained shear strengthγcu1.01.4Unconfined strengthγqu1.01.4Bulk densityγγ1.01.0ActionsPermanent action : unfavourablePermanent action: favourableVariable actionSoil Properties

Piles

Flexural and axial resistanceof piles‘Uncertainties related to the cross-section of cast in place piles andconcreting procedures shall be allowed for in design’‘In the absence of other provisions’, the design diameter of cast in placepiles without permanent casing is less than the nominal diameter Dnom: Dd Dnom – 20 mm for Dnom 400 mm Dd 0.95 Dnom for 400 Dnom 1000 mm Dd Dnom – 50 mm for Dnom 1000 mmICE Specification for piling and embedded retaining walls (ICE SPERW)B1.10.2 states ‘The dimensions of a constructed pile or wall element shallnot be less than the specified dimensions’. A tolerance of 5% on augerdiameter, casing diameter, and grab length and width is permissible.

Flexural and axial resistanceof piles The partial factor for concrete, γc, should be multiplied by afactor, kf, for calculation of design resistance of cast in placepiles without permanent casing. The UK value of kf 1.1, therefore γc,pile 1.65 “If the width of the compression zone decreases in the directionof the extreme compression fibre, the value η fcd should bereduced by 10%”

Bored pilesReinforcement should be detailed for free flow ofconcrete.Minimum diameter of long. reinforcement 16mmMinimum number of longitudinal bars 6[BUT – BS EN 1536 Execution of special geotechnical work Bored Pilessays 12 mm and 4 bars!]Minimum areas:Pile crosssection: AcAc 0.5 m20.5 m2 Ac 1.0 m2Ac 1.0 m2Min area of long.rebar, As,bpminPilediameters 0.5% Ac 800 mm 2500 mm2 0.25% Ac 1130 mm

Minimum area of reinforcement,As,bpmin (mm2)Minimum 00002004006008001000Pile diameter, mm120014001600

WorkshopDesign a pad foundation for a 300mm square columntakingGk 600kN, Qk 350kN.Permissible bearing stress 225kPa.Concrete for base C30/37.Work out size of base, tension reinforcement and any shear reinforcement.

Workshop ProblemCategory 2, using prescriptive methodsBase size: (Gk Qk)/bearing stress m2 x base x mm deep (choose size of pad)Use C 30/37 (concrete)Loading γg x Gk γq x Qk kNULS bearing pressure / 2 kN/m2Critical section at face of columnMEd x x 2 / 2 kNmd – cover – assumed ø mmK M/bd2fck

Workshop Problem z d x mmTable 15.5 As MEd/fydz mm2 Provide H @ c/c ( mm2)Check minimum steel100As,prov/bd For C30/37 concrete As,min OK/not OKBeam shearCheck critical section d away from column faceVEd x kN/mvEd VEd / d M