Practical Design to Eurocode 2

1 Practical Design to Eurocode 2 The webinar will start at SpeakerTitle121 Sep Jenny Burridge Introduction, Background and Codes228 Sep Charles GoodchildEC2 Background, Materials, Cover and effective spans35 Oct Paul GregoryBending and Shear in Beams412 Oct Charles Goodchild Analysis519 Oct Paul Gregory Slabs and Flat Slabs 626 Oct Charles Goodchild Deflection and Crack Control72 Nov Paul GregoryColumns89 Nov Jenny Burridge Fire916 NovPaul GregoryDetailing1023 Nov Jenny Burridge FoundationsCourse OutlineLecture 1023rdNovember 2017 FoundationsLecture 9 ExerciseLap length for column longitudinal barsColumn lap length exerciseH25 sH32 sLapDesign information C40/50 concrete 400 mm square column 45mm nominal cover to main bars Longitudinal bars are in compression Maximum ultimate stress in the bars is 390 MPaExercise:Calculate the minimum lap length using EC2 equation.

2 Column lap length exerciseProcedure Determine the ultimate bond stress, fbdEC2 Equ. Determine the basic anchorage length, lb,reqEC2 Equ. Determine the Design anchorage length, lbdEC2 Equ. Determine the lap length, l0= anchorage length x 6 Model Answers Lap length for column longitudinal barsColumn lap length exerciseH25 sH32 sLapDesign information C40/50 concrete 400 mm square column 45mm nominal cover to main bars Longitudinal bars are in compression Maximum ultimate stress in the bars is 390 MPaExercise:Calculate the minimum lap length using EC2 equation :Column lap length exerciseProcedure Determine the ultimate bond stress, fbdEC2 Equ. Determine the basic anchorage length, lb,reqEC2 Equ. Determine the Design anchorage length, lbdEC2 Equ. Determine the lap length, l0= anchorage length x 6 Solution - Column lap lengthDetermine the ultimate bond stress, fbdfbd= 1 2fctdEC2 Equ.

3 1= Good bond conditions 2= bar size 32 fctd= ctfctk,0,05/ c EC2 cl (2), Equ ct= c = ,0,05 = x fck2/3EC2 Table x 402/3= MPafctd= ctfctk,0,05/ c = = x = MPaSolution - Column lap lengthDetermine the basic anchorage length, lb,reqlb,req= ( /4) ( sd/fbd)EC2 Equ ultimate stress in the bar, sd= 390 ,req= ( /4) ( 390 ) = For concrete class C40/50 Solution - Column lap lengthDetermine the Design anchorage length, lbdlbd= 1 2 3 4 5lb,req lb,minEqu. 1 2 3 4 5( ) For concrete class C40/50 For bars in compression 1= 2= 3 = 4= 5= lbd= Solution - Column lap lengthDetermine the lap length, l0= anchorage length x 6 All the bars are being lapped at the same section, 6= lap length is based on the smallest bar in the lap, 25mmHence,l0= lbdx 6l0= x = x 25l0= 993 mmFoundationsOutline Week 10, foundations We will look at the following topics: Eurocode 7: Geotechnical Design Partial factors, spread foundations .

4 Pad foundation Worked example & workshop Retaining walls PilesEurocode 7 Eurocode 7 has two parts:Part 1: General RulesPart 2: Ground Investigation and testing6 (p43 et seq)Plus NAPlus NAEurocode 7 How 6. FoundationsThe essential features of EC7, Pt 1 relating to foundation Design are :This publication covers only the Design of simple foundations , which are a small part of should not be relied on for general guidance on EC7. Limit StatesThe following ultimate limit states apply to foundation Design :EQU: Loss of equilibrium of the structureSTR: Internal failure or excessive deformation of the structure or structural memberGEO: Failure due to excessive deformation of the groundUPL: Loss of equilibrium due to uplift by water pressureHYD: Failure caused by hydraulic gradientsCategories of StructuresCategory DescriptionRisk of geo-technical failureExamples from EC71 Small and relatively simple structuresNegligibleNone given2 Conventional types of structure no difficult groundNo exceptional riskSpread foundations3 All other structures Abnormal risks Large or unusual structuresEC7 ULS DesignEC7 provides for three Design ApproachesUK National Annex - Use Design Approach 1 DA1 For DA1 (except piles and anchorage Design )

5 There are two sets of combinations to use for the STR and GEO limit 1 generally governs structural resistance Combination 2 generally governs sizing of foundationsSTR/GEO ULS Actions partial factorsPermanent ActionsLeading variable actionAccompanying variable actionsUnfavourable FavourableMainOthersCombination 1 Exp 0,iQkExp 0, 0,iQkExp 0,iQkCombination 2 Exp 0,iQkNotes:If the variation in permanent action is significant, use Gk,j,supand Gk,j,infIf the action is favourable, Q,i= 0 and the variable actions should be ignoredFactors for EQU, UPL and HYDL imit statePermanent ActionsVariable ActionsUnfavourable Favourable Unfavourable factors material propertiesParameterSymbolCombination 1 Combination 2 EQUA ngle of shearing resistance cohesion c shear strength strength density 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.

6 Bearing resistances, characteristic values of soil parameters and whether values are SLS or ULS , Combination 1 or Combination 2 values Spread FoundationsEC7 Section 6 Three methods for Design : Direct method check all limit states: Load and partial factor combinations (as before) qult=c Ncscdcicgcbc+ q Nqsqdqiqgqbq+ BN s d i g b /2where c = cohesion q = overburden = body-weight Ni= bearing capacity factors si= shape factors di= depth factors ii= inclination factors gi= ground inclination factors bi= base inclination factors We just bung it in a spreadsheet Settlement often criticalSee Decoding Eurocode 7 by A Bond & A Harris, Taylor & FrancisSpread FoundationsEC7 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).

7 Used in sub-sequent slides).Spread FoundationsDesign procedures in:Fig 6/1 (p46)Procedure for depth of spread foundationsPressure distributionsSLS pressure distributionsULS pressure distributionLoad casesEQU : Gk+ Qk(assuming variable action is destabilising wind, and permanent action is stabilizing)STR : Gk+ Qk(Using ( ). Worse case of Exp ( ) or ( ) could be used) aabFhF (3 gd/fctd,pl)haF0,85 where: gdis the Design value of the ground pressure as a simplification hf/a 2 may be usedPlain Concrete Strip Footings & Pad foundations :Cl. , Exp ( ) aabFhFC16/20 C20/25 C25/30 C30/37allowable pressure gdhF/ahF/ahF/ Concrete Strip Footings & Pad cavity wall 300 wide carrying 80 kN/m onto 100 kN/m2ground:bf= 800 mma = 250 mmhf= say assuming C20/25 x 250 = 213 say 225 mmReinforced Concrete Bases Check critical bending moments at column faces Check beam shear and punching shearFor punching shear the ground reaction within the perimeter may be deducted from the column loadPad foundationWorked exampleWorked ExampleDesign a square pad footing for a 350 350 mm column carrying Gk= 600 kN and Qk= 505 kN.

8 The presumed allowable bearing pressure of the non-aggressive soil is 200 :Category 2. So using prescriptive methods:Base area: (600 + 505)/200 = > x base x (say) deep. Worked ExampleLoading = x 600 + x 505= kNULS bearing pressure = 272 kN/m2 Critical section at face of columnMEd= 272 x x 2= 343 kNmd = 500 50 16 = 434 mmK = 343 x106/(2400 x 4342x30)= C30/37 concreteWorked Example z = x 434 = 412mm As= MEd/fydz = 343 x 106 / (435 x 412) = 1914mm2 Provide 10H16 @ 250 c/c (2010 mm2) (804 mm2/m)Beam shear:Check critical section d away from column faceVEd= 272 x ( ) = 161kN/mvEd= 161 / 434 = = 2010/ (434 x 2400) = = ,c(from table) = => beam shear 6 (p47)Concise Table ExamplePunching shear:Basic control perimeter at 2d from face of columnvEd= VEd/ uid < vRd,c = 1, ui= (350 x 4 + 434 x 2 x 2 x ) = 6854mmVEd= load minus net upward force within the area of the control perimeter)= 272 x ( + x.

9 8682+ .868 x .35 x 4)= 560kNvEd= MPa; vRd,c= (as before) => okRetaining WallsChapter 9 Ultimate Limit States for the Design of retaining wallsCalculation Model AModel applies if bh hatan (45 - d/2)Rankine theoryCalculation Model BInclined virtual plane theoryModel applies to walls of all shapes and sizesGeneralModel AModel B2BL2bbLbbBbBtWHbWbststshc,kbbc,kss=+= = = =BL2bbbL2tanbHbWtanbHthvphstff,khhfhb= = ++ += ++=General expressionsOverall Design procedure9 (Figure 4)Initial sizingbs tb h/10 to h/15B to B/4 to B/3 Overall Design procedure9 (Figure 4)Figure 6 for overall Design procedure9 (Figure 6)Soil DensitiesEx Concrete BasementsDesign value of effective angle of shearing resistance, dtan d= tan ( k/ )where k= maxfor granular soils and= for clay soils, maxand are as defined as follows = or dependent on the Combination being considered.

10 Ex Concrete BasementsAngle of shearing resistanceEstimated peak effective angle of shearing resistance, max= 30 + A + B + CEstimated critical state angle of shearing resistance, crit= 30 + A + B, which is the upper limiting Concrete BasementsGranular SoilsLong term Granular SoilsClay soilsEx Concrete BasementsCalcs Material properties & earth pressures9 Panel 2 (p71)9 (Panel 2)2 Overall Design procedure9 (Figure 4) Design against sliding (Figure 7)9 (Figure 7)9 (Panel 3)Sliding ResistanceOverall Design procedure9 (Figure 4) Design against Toppling 9 (Figure 9)Overall Design procedure9 (Figure 4) Design against bearing failure9 (Figure 10)Expressions for bearing resistance9 (Panel 4,Figure 11)Overall Design procedure9 (Figure 4)Structural design9 (Figure 13)Remember: Load and Partial Factor CombinationsParameterSymbol Comb.