Eurocode 2: Design of concrete structures EN1992-1-1 - Eurocodes

Requirements to a code1. Scientifically well founded, consistent and coherent2. Transparent3. New developments reckognized as much as possible4. Open minded: models with different degree of complexity allowed5. As simple as possible, but not simplier6. In harmony with other codes22 February 2008 2

EC-2: Concrete StructuresEC2: General rules and rules for buildingsFireBridgesContainersMaterialsExecutionPrecast elementsConcretePrestressing steelCommon rulesReinforcing steelProduct standards22 February 2008 3

EC-2: Concrete StructuresEC2: General rules and rules for buildingsFireBridgesContainment structuresMaterialsExecutionPrecast elementsConcretePrestressing steelCommon rulesReinforcing steelProduct standards22 February 2008 4

EN 1992-1-1 “Concrete structures” (1)Content:1. General2. Basics3. Materials4. Durability and cover5. Structural analysis6. Ultimate limit states7. Serviceability limit states8. Detailing of reinforcement9. Detailing of members and particular rules10. Additional rules for precast concrete elements and structures11. Lightweight aggregate concrete structures12. Plain and lightly reinforced concrete structures22 February 2008 5

EN 1992-1-1 “Concrete structures” (2)Annexes:A. Modifications of safety factor (I)B. Formulas for creep and shrinkage (I)C. Properties of reinforcement (N)D. Prestressing steel relaxation losses (I)E. Indicative strength classes for durability (I)F. In-plane stress conditions (I)G. Soil structure interaction (I)H. Global second order effects in structures (I)I. Analysis of flat slabs and shear walls (I)J. Detailing rules for particular situations (I)I = InformativeN = Normative22 February 2008 6

EN 1992-1-1 “Concrete structures” (3)In EC-2 “Design of concrete structures –Part 1: General rules and rules for buildings109 national choices are possible22 February 2008 7

Chapter: 3 MaterialsJ.C. Walraven22 February 20088Vermelding onderdeel organisatie

Concrete strength classesConcrete strength class C8/10 tot C100/115.(Characteristic cylinder strength / char. cube strength)22 February 2008 9

Concrete strength classes and propertiesStrength classes for concretef ck (MPa) 12 16 20 25 30 35 40 45 50 55 60 70 80 90f ck,cube 15 20 25 30 37 45 50 55 60 67 75 85 95 105(MPa)f cm 20 24 28 33 38 43 48 53 58 63 68 78 88 98(MPa)f ctm 1,6 1,9 2,2 2,6 2,9 3,2 3,5 3,8 4,1 4,2 4,4 4,6 4,8 5,0(MPa)f ctk,0,05 1 1 1,3 1,5 1,8 2,0 2,2 2,5 2,7 2,9 3,0 3,1 3,2 3,4 3,5(MPa)f ctk,0,95 2,0 2,5 2,9 3,3 3,8 4,2 4,6 4,9 5,3 5,5 5,7 6,0 6,3 6,6(MPa)E cm 27 29 30 31 32 34 35 36 37 38 39 41 42 44(Gpa)ε c1 (‰) 1,8 1,9 2,0 2,1 2,2 2,25 2,3 2,4 2,45 2,5 2,6 2,7 2,8 2,8ε cu1 (‰) 3,5 3,2 3,0 2,8 2,8 2,8ε c2 (‰) 2,0 2,2 2,3 2,4 2,5 2,6ε cu2 (‰) 3,5 3,1 2,9 2,7 2,6 2,6n 2,0 1,75 1,6 1,45 1,4 1,4ε c3 (‰) 1,75 1,8 1,9 2,0 2,2 2,3ε cu3 (‰) 3,5 3,1 2,9 2,7 2,6 2,622 February 2008 10

Design Strength Values(3.1.6)• Design compressive strength, f cdf cd = α cc f ck /γ c• Design tensile strength, f ctdf ctd = α ct f ctk,0.05 /γ cα cc (= 1,0) and α ct (= 1,0) are coefficients to take accountof long term effects on the compressive and tensilestrengths and of unfavourable effects resulting from theway the load is applied (national choice)22 February 2008 11

Concrete strength at a time t (3.1.2)Expressions are given for the estimation of strengths at times other than 28days for various types of cementf cm(t) = β cc(t) f cmwhere f cm(t) is the mean compressive strength at an age of t daysβ cc(t) = exp {s[1-(28/t) 1/2 ]}The coeeficient s depends on type of cement: s = 0,20 for rapid hardeningcement (Class R), s = 0,25 for normal hardening (Class N) and s = 0,38for Class S (slow hardening) cement. Classes according to EN 197-122 February 2008 12

Elastic deformation (3.1.3)• Values given in EC2 are indicative and vary accordingto type of aggregate• E cm (t) = (f cm (t)/f cm ) 0,3 E cm• Tangent modulus E c may be taken as 1,05 E cm• Poissons ratio: 0,2 for uncracked concrete0 for cracked concrete• Linear coefficient of expansion 10⋅10 -6 K -122 February 2008 13

Concrete stress - strain relations (3.1.5 and 3.1.7)For structural analysisσc“Schematic”σcFor section analysis“Parabola-rectangle”σc“Bi-linear”fcmfckfckfcdfcd0,4 fcmtan α = Ecmαε c1( 0 / 00) = 0,7 f cm0,31ε cu1( 0 / 00) =ε c1ε cu12,8 + 27[(98-f cm)/100] 4 f cm)/100] 4for f ck≥ 50 MPa otherwise 3.5ε c0ε c2ε cu2n⎡ ⎛ ε ⎞ ⎤cσ = f ⎢1 − ⎜1 − ⎟ ⎥ for 0 ≤ ε < ε⎢εc2⎣ ⎝ ⎠ ⎥⎦σ = f for ε ≤ ε ≤ εc cd c c2c cd c2 c cu2n = 1,4 + 23,4 [(90- f ck)/100] 4for f ck≥ 50 MPa otherwise 2,0ε c2 (0 / 00) = 2,0 + 0,085(f ck-50) 0,53ε c0εc3ε cu3ε c3( 0 / 00) = 1,75 + 0,55 [(f ck-50)/40]for f ck≥ 50 MPa otherwise 1,75ε cu3( 0 / 00) =2,6+35[(90-f ck)/100] 4for f ck≥ 50 MPa otherwise 3,5ε cfor f ck≥ 50 MPa otherwise 2,0ε cu2( 0 / 00) = 2,6 + 35 [(90-f ck)/100] 4for f ck≥ 50 MPa otherwise 3,522 February 2008 14

Concrete stress-strain relations- Higher concrete strength show more brittlebehaviour, reflected by shorter horizontal branche22 February 2008 15

Chapter 3.1: ConcreteSimplified σ - ε relation for cross sections with non rectangularcross-sectionλ= 0,8 for f ck ≤ 50 MPaλ = 0,8 – (f ck -50)/400 for50 ≤ fck ≤ 90 MPaη = 1,0 for f ck ≤ 50 MPaη = 1,0 – (f ck -50)/200 for50 ≤ fck ≤ 90 MPa22 February 2008 16

Shrinkage (3.1.4)• The shrinkage strain ε cs is composed of two components:ε cs = ε cd + ε cawhere- drying shrinkage strainε cd (t) = β ds (t, t s )⋅k h ⋅ε cd,0 where ε cd,0 is the basic drying shrinkagestrain- autogenous shrinkage strainε ca (t) = β as (t)⋅ε ca (∞)22 February 2008 17

Autogenous shrinkageStichtse Bridge, 1997:Autogenous shrinkage20.10 -3 after 2 dayslConcrete strength f c =90 MPa22 February 2008 18

Shrinkage (3.1.4)βds( t,ts)=( t− ts( t − ts)) + 0,04h30where t = age of concrete at time considered, t s= age at beginning ofdrying shrinkage (mostly end of curing)εca( t)β ( t)ε ( ∞)where=as ca−60,5εca( ∞)= 2,5( fck−10)⋅10and β ( t)= 1−exp( −0,2t)as22 February 2008 19

Creep of concrete (3.1.4)123t 0SNInside conditions – RH = 50%Example: 600 mm thick slab, loading at 30 days, C30/37 - ϕ = 1,8R5102030C20/25C25/30C30/37C35/45C40/50C45/55C50/60 C55/67C60/75 C70/85C80/95C90/105501007,06,0ϕ (∞, t 0)5,04,03,0 2,0 1,0 0 100 300 500 700 900 1100 1300 1500h 0 (mm)h 0= 2A c/u where A cis the cross-section area andu is perimeter of the member in contact with the atmosphere22 February 2008 20

Confined Concrete (3.1.9)σ1 = fck,cσcfck,cfckfcd,cσ2 σ3 ( = σ2)A0εcuεc2,cεcu2,cεcf ck,c= f ck(1.000 + 5.0 σ 2/f ck)for σ 2≤ 0.05f ck= f ck(1.125 + 2.50 σ 2/f ck) for σ 2> 0.05f ckε c2,c= ε c2(f ck,c/f ck)2ε cu2,c= ε cu2+ 0,2 σ 2/f ck22 February 2008 21

Stress-strain relations for reinforcing steel22 February 2008 22

Reinforcement (2) – From Annex CProduct form Bars and de-coiled rods Wire FabricsClass A B C A B CCharacteristic yieldstrength f yk or f 0,2k (MPa)cold worked400 to 600hot rolledseismick = (f t /f y ) k ≥1,05 ≥1,08 ≥1,15

Idealized and design stress strain relations for reinforcing steelAlternative design stress/strain relationships are permitted:- inclined top branch with a limit to the ultimate strain horizontal- horizontal top branch with no strain limitσIdealisedkfykfykkfyk/γsfyd = fyk/γsDesignk = (f t /f y ) kfyd/ Esε udε ukεε ud = 0.9 ε uk22 February 2008 24

Durability and coverProf.dr.ir. J.C. Walraven22 February 200825Group Concrete Structures

Penetration of corrosion stimulatingcomponents in concrete22 February 2008 26

Deterioration of concreteCorrosion of reinforcement by chloride penetration22 February 2008 27

Deterioration of concrete structuresCorrosion of reinforcement by chloride attack in amarine environment22 February 2008 28

Avoiding corrosion of steel in concreteDesign criteria- Aggressivity of environment- Specified service lifeDesign measures- Sufficient cover thickness- Sufficiently low permeability of concrete (in combination with coverthickness)- Avoiding harmfull cracks parallel to reinforcing bars- Other measures like: stainless steel, cathodic protection, coatings, etc.22 February 2008 29

Aggressivity of the environmentMain exposure classes:• The exposure classes are defined in EN206-1. The main classes are:• XO – no risk of corrosion or attack• XC – risk of carbonation induced corrosion• XD – risk of chloride-induced corrosion (other than sea water)• XS – risk of chloride induced corrosion (sea water)• XF – risk of freeze thaw attack• XA – Chemical attack22 February 2008 30

Cover to reinforcement, required to fulfillservice life demandsDefinition of concrete coverOn drawings the nominal cover should be specified. It isdefined as a minimum cover c min plus an allowance in designfor deviation Δc dev, soc nom = c min +Δc dev22 February 2008 32

Procedure to determine c min,durEC-2 leaves the choice of c min,dur to the countries, but gives thefollowing recommendation:The value c min,dur depends on the “structural class”, which has to bedetermined first. If the specified service life is 50 years, the structuralclass is defined as 4. The “structural class” can be modified in case ofthe following conditions:-The service life is 100 years in stead of 50 years-The concrete strength is higher than necessary- Slabs (position of reinforcement not affected by construction process- Special quality control measures applyThe finally applying service class can be calculated with Table 4.3N22 February 2008 34

Table for determining final Structural Class22 February 2008 35

Final determination of c min,dur (1)The value c min,dur is finally determined as a function of the structuralclass and the exposure class:22 February 2008 36

Special considerationsIn case of stainless steel the minimum cover may be reduced.The value of the reduction is left to the decision of the countries(0 if no further specification).22 February 2008 37

Structural Analysis22 February 2008 38

Methods to analyse structuresLinear elastic analysis1. Suitable for ULS and SLS2. Assumptions:- uncracked cross-sections- linear σ - ε relations-meanE-modulus3. Effect of imposed deformationsin ULS to be calculated withreduced stiffnesses and creep22 February 2008 39

Geometric Imperfections (5.2)• Deviations in cross-sectiondimensions are normally taken intoaccount in the material factors andshould not be included in structuralanalysisθ iHiNbNal• Imperfections need not beconsidered for SLS• Out-of-plumb is represented by an inclination, θ lθ l = θ 0 α h α m where θ 0 = l/200α h = 2/√l; 2/3 ≤ α h ≤ 1α m = √(0,5(1+1/m)l is the height of member (m)m is the number of vert. members22 February 2008 40

Forces due to geometric imperfections onstructures(5.2)θ iHiNaNblθi/2θ iHiNaθi/2NbBracing System Floor Diaphragm RoofH i = θ i (N b -N a )H i = θ i (N b +N a )/2H i = θ i N a22 February 2008 41

Methods to analyse structures5.5 Linear elastic analysis with limited redistribution1. Valid for 0,5 ≤ l 1 / l 2 ≤ 2,02. Ratio of redistribution δ, withδ ≥ k 1 + k 2 x u /d for f ck ≤ 50 MPaδ ≥ k 3 + k 4 x u /d for f ck > 50 MPaM 2M 1δ ≥ k 5 for reinforcement class B or Cδ ≥ k 6 for reinforcement class Al 1 l 222 February 2008 42

Redistribution limits for Class B & C steel353025% redist201510500.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60x /dfck =70 fck =60 fck =5022 February 2008 43

Methods to analyse structures5.6 Plastic methods of analysis(a) Yield line analysis(b) Strut and tie analysis(lower bound)- Suitable for ULS- Suitable for SLS if compatibilityis ensured (direction of strutsoriented to compression in elasticanalysis22 February 2008 44

Methods to analyse structuresCh. 5.7 Nonlinear analysis“Nonlinear analysis may be usedfor both ULS and SLS, providedthat equilibrium and compatibilityare satisfied and an adequate nonlinearbehaviour for materials isassumed. The analysis may be firstor second order”.22 February 2008 45

Chapter 5 “Structural analysis”5.8 Second order effects with axial loads- Slenderness criteria for isolated membersand buildings (when is 2 nd order analysis required?)- Methods of second order analysis• General method based on nonlinearbehaviour, including geometric nonlinearity• Analysis based on nominal stiffness• Analysis based on moment magnification factor• Analysis based on nominal curvatureExtended calculation tools are given22 February 2008 46

Methods of analysisBiaxial bendingM Rdz/yM Rdz/ydesign moment aroundrespective axismoment resistance inrespective directionFor circular and elliptical cross-sectiona = 2.For rectangular cross section, see tableN E /N Rd 0,1 0,7 1,0a 1,0 1,5 2,022 February 2008 47

Methods of analysisLateral buckling of beamsNo lateral buckling if:22 February 2008 48

Bending with or without axial forceProf.dr.ir. J.C. Walraven22 February 200849Group Concrete Structures

Concrete design stress - strain relations (3.1.5 and 3.1.7)for section analysisσcfck“Parabola-rectangle”σcfck“Bi-linear”fcdfcd0ε c2n⎡ ⎛ ε ⎞ ⎤cσ = f ⎢1 − ⎜1 − ⎟ ⎥ for 0 ≤ ε < ε⎢εc2⎣ ⎝ ⎠ ⎥⎦σ = f for ε ≤ ε ≤ εc cd c c2c cd c2 c cu2n = 1,4 + 23,4 [(90- f ck)/100] 4for f ck≥ 50 MPa otherwise 2,0ε cu2ε c2 (0 / 00) = 2,0 + 0,085(f ck-50) 0,53for f ck≥ 50 MPa otherwise 2,0ε cu2( 0 / 00) = 2,6 + 35 [(90-f ck)/100] 4for f ck≥ 50 MPa otherwise 3,5ε c0εc3ε cu3ε c3( 0 / 00) = 1,75 + 0,55 [(f ck-50)/40]for f ck≥ 50 MPa otherwise 1,75ε cu3( 0 / 00) =2,6+35[(90-f ck)/100] 4for f ck≥ 50 MPa otherwise 3,5ε c22 February 2008 50

Concrete design stress strain relations fordifferent strength classes- Higher concrete strength shows more brittle behaviour,reflected by shorter horizontal branche22 February 2008 51

Simplified concrete design stress blockεcu3η fcdAcxλxFcdAsFsεsλ = 0,8= 0,8 −for f ck≤ 50 MPa(fck −40050)for 50 < f ck≤ 90 MPaη = 1,0for f ck≤ 50 MPa= 1,0 – (f ck– 50)/200 for 50 < f ck≤ 90 MPa22 February 2008 52

Simplified factors for flexure (1)Factors for NA depth (n) and lever arm (=z) for concrete grade ≤ 50 MPa1.201.00lever arm0.80Factor0.600.40NA depth0.200.000.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10 0.11 0.12 0.13 0.14 0.15 0.16 0.17n 0.02 0.04 0.07 0.09 0.12 0.14 0.17 0.19 0.22 0.24 0.27 0.30 0.33 0.36 0.39 0.43 0.46z 0.99 0.98 0.97 0.96 0.95 0.94 0.93 0.92 0.91 0.90 0.89 0.88 0.87 0.86 0.84 0.83 0.82M/bd 2 f ck22 February 2008 53

Simplified factors for flexure (2)Factors for NA depth (=n) and lever arm (=z) for concrete grade 70 MPa1.201.00lever arm0.80Factor0.600.40NA depth0.200.000.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10 0.11 0.12 0.13 0.14 0.15 0.16 0.17n 0.03 0.05 0.08 0.11 0.14 0.17 0.20 0.23 0.26 0.29 0.33z 0.99 0.98 0.97 0.96 0.95 0.94 0.93 0.91 0.90 0.89 0.88M/bd 2 f ck22 February 2008 54

Column design chart for f ck ≤ 50 MPa21.8bd 1Nd/bhfcd1.61.41.210.80.600.10.20.3A sf yk/bhf ck1.00.8 0.90.6 0.70.4 0.5d 1d 1 /h = 0.05f ck

Column design chart for f ck = 70 MPa21.8d 1bNd/bhfcd1.61.41.210.80.60.40.200.10.20.3A sf yk/bhf ck1.00.8 0.90.6 0.70.4 0.5d 1d 1 /h = 0.1f ck = 90h00 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6M d /bh 2 f cd22 February 2008 56

ShearProf.dr.ir. J.C. Walraven22 February 200857Group Concrete Structures

Principles of shear control in EC-2Until a certain shear force V Rd,c no calculated shear reinforcementis necessary (only in beams minimum shear reinforcement isprescribed)If the design shear force is larger than this value V Rd,c shearreinforcement is necessary for the full design shear force. Thisshear reinforcement is calculated with the variable inclination trussanalogy. To this aim the strut inclination may be chosen betweentwo values (recommended range 1≤ cot θ ≤ 2,5)The shear reinforcement may not exceed a defined maximumvalue to ensure yielding of the shear reinforcement22 February 2008 58

Concrete slabs without shearreinforcementShear resistance V Rd,c governed by shear flexurefailure: shear crack develops from flexural crack22 February 2008 59

Concrete slabs without shearreinforcementPrestressed hollow core slabShear resistance V Rd,c governed by shear tension failure:crack occurs in web in region uncracked in flexure22 February 2008 60

Concrete beam reinforced in shearShear failure introduced by yielding of stirrups,followed by strut rotation until web crushing22 February 2008 61

Principle of variabletruss actionApproach “Variable inclination struts”:a realisticStage 1: web uncracked in shearStage 2: inclined cracks occurStage 3: stabilized inclined cracksStage 4: yielding of stirrups,further rotation, finallyweb crushingStrut rotation as measured in tests(TU Delft)22 February 2008 62

Principles of variable angle trussStrut rotation, followed by new cracks under lower angle, even inhigh strength concrete (Tests TU Delft)22 February 2008 63

Web crushing in concrete beamWeb crushing providesmaximum to shear resistanceAt web crushing:V Rd,max = b w z υ f cd /(cotθ + tanθ)22 February 2008 64

Advantage of variable angle truss analogy-Freedom of design:• low angle θ leads to low shear reinforcement• High angle θ leads to thin webs, saving concreteand dead weightOptimum choice depends on type of structure- Transparent equilibrium model, easy in use22 February 2008 65

Shear design value under which no shearreinforcement is necessary in elementsunreinforced in shear (general limit)VRd, c= CRd, ck( 100ρlfck)1/3bwdC Rd,c coefficient derived from tests (recommended 0,12)k size factor = 1 + √(200/d) with d in meterρ l longitudinal reinforcement ratio ( ≤0,02)f ck characteristic concrete compressive strengthb w smallest web widthd effective height of cross section22 February 2008 66

Shear design value under which no shearreinforcement is necessary in elementsunreinforced in shear (general limit)Minimum value for V Rd,c:V Rd,c = v min b w dValues for v min (N/mm 2 )d=200 d=400 d=600 d=800C20 0,44 0,35 0,25 0,29C40 0,63 0,49 0,44 0,41C60 0,77 0,61 0,54 0,50C80 0,89 0,70 0,62 0,5822 February 2008 67

Shear design value under which no shearreinforcement is necessary in elementsunreinforced in shear (special case of sheartension)22 February 2008 68

Special case of shear tension (examplehollow core slabs)VIb wSf ctdα lσ cpI⋅b=w( f )2Rd , cctdS+αmoment of inertiasmallest web widthsection modulusdesign tensile strength of concretereduction factor for prestress in case ofprestressing strands or wires in ends of memberconcrete compressive stress at centroidal axis iforfor fully developed prestress22 February 2008 69lσcpfctd

sσc=zθθ Afswyw= fVzcot θVu,3u,2Design of members if shearreinforcement is needed (V E,d >V Rd,c )V u,3θsAz cot θfsw ywzσ c = f c1V u,2θ= fυ cFor most cases:-Assume cot θ = 2,5 (θ = 21,8 0 )-Calculate necessary shear reinforcement-Check if web crushing capacity is not exceeded (V Ed >V Rd,s )-If web crushing capacity is exceeded, enlarge web width or calculatethe value of cot θ for which V Ed = V Rd,c and repeat the calculation22 February 2008 70

sσc=zθθ Afswyw= fVzcot θVu,3u,2Upper limit of shear capacity reached dueto web crushingV u,3θsAz cot θfsw ywzσ c = f c1V u,2θ= fυ cFor yielding shear reinforcement:V Rd,s = (A sw /s) z f ywd cotθθ from 45 0 to 21,8 02,5 times larger capacityAt web crushing:V Rd,max = b w z υ f cd /(cotθ + tanθ)θ from 21,8 0 to 45 01,45 times larger capacity22 February 2008 71

Special case of loads near to supportsdda va vFor a v ≤ 2d the contribution of the point load to the shear force V Ed may bereduced by a factor a v /2d where 0.5 ≤ a v ≤ 2d provided that the longitudinalreinforcement is fully anchored at the support. However, the conditionV Ed ≤ 0,5b w dυf cdshould always be fulfilled22 February 2008 72

Influence of prestressing on shearresistance (1)1. Prestressing introduces a set of loads on the beam22 February 2008 73

Influence of prestressing on shearresistance (2)Prestressing increases the load V Rc,dbelow which no calculatedshear reinforcement is required1/3VRd , c= [ CRd, ck(100ρ lfck) + k1σcp] bwdk 1 coefficient, with recommended value 0,15σ cp concrete compressive stress at centroidal axis dueto axial loading or prestressing22 February 2008 74

Influence of prestressing on shearresistance (3)1. Prestressing increases the web crushing capacityVRd, max= αcwbwzνfcd/(cotθ+ tanθ)α cw factor depending on prestressing forceα cw= 1 for non prestressed structures(1+σ cp/f cd) for 0,25 < σ cp < 0,25f cd1,25 for 0,25f cd

Increase of web crushing capacity byprestressing (4)22 February 2008 76

Influence of prestressing on shearresistance (4)Reducing effect of prestressing duct (with or without tendon)on web crushing capacityGrouted ductsb w,nom = b w - ΣφUngrouted ducts b w,nom = b w –1,2 Σφ22 February 2008 77

Shear between web and flanges of T-sectionsStrut angle θ:1,0 ≤ cot θ f ≤ 2,0 for compression flanges (45 0 ≥θ f ≥ 26,5 01,0 ≤ cot θ f ≤ 1,25 for tension flanges (45 0 ≥θ f ≥ 38,6 0 )No transverse tension ties required if shear stress in interfacev Ed = ΔF d /(h f·Δx) ≤ kf ctd (recommended k = 0,4)22 February 2008 78

Shear at the interface between concretescast at different timesInterface shear models based on shear friction principle22 February 2008 79

Shear at the interface between concretescast at different times22 February 2008 80

Shear at the interface between concrete’scast at different times (Eurocode 2, Clause6.5.2)v Rdi = c⋅f ctd + μ⋅σ n + ρ⋅f yd (μ⋅sin β + cos β) ≤ 0,5 ν⋅f cdf ctd=concrete design tensile strengthσ n = eventual confining stress, notfrom reinforcementρ= reinforcement ratioβ = inclination between reinforcementand concrete surfacef cd= concrete design compressivestrengthυ = 0,6 for f ck≤ 60 MPa= 0,9 – f ck/200≥0,5 for f ck≥ 60 MPaVery smoothcsmooth 0,35rough 0,45indented 0,50μ0,25 0,5(=tan α)0,60,70,822 February 2008 81

TorsionProf.dr.ir. J.C. Walraven22 February 200882Group Concrete Structures

Modeling solid cross sections byequivalent thin-walled cross sectionsCentre-lineziCoverTEdOuter edge of effectivecrossection, circumference utef/2tefEffective wall-thickness follows from t ef,i =A/u, where;A = total area of cross section within outer circumference, including hollow areasU = outer circumference of the cross section22 February 2008 83

Design procedure for torsion (1)Shear flow in any wall followsfrom:τt, itef, i=whereTEd2Akτ t,It ef,IT EdA ktorsional shear stress in wall Ieffective wall thickness (A/u)applied torsional momentarea enclosed by centre linesof connecting walls, includinghollow areas22 February 2008 84

Design procedure for torsion (2)Shear force V Ed in wall i due totorsion is:VEd , i= τt,itef, iziwhereτ t,It ef,IZ itorsional shear stress in wall ieffective wall thickness (A/u)inside length of wall I definedby distance of intersectionpoints with adjacent walls22 February 2008 85

Design procedure for torsion (3)The shear reinforcement in any wall can now be designed like a beamusing the variable angle truss analogy, with 1≤ cot θ ≤ 2,522 February 2008 86

Design procedure for torsion (4)The longitudinal reinforcement in any wall follows from:ΣAslukfyd=TEdA2 kcotθwhereu kf ykθperimeter of area A kdesign yield stress of steelangle of compression struts22 February 2008 87

Punching shearProf.dr.ir. J.C. Walraven22 February 200888Group Concrete Structures

Design for punching shearMost important aspects:- Control perimeter- Edge and corner columns- Simplified versus advancedcontrol methods22 February 2008 89

Definition of control perimeter22 February 2008 90

Definition of control perimetersThe basic control perimeter u 1 is taken at a distance 2,0d fromthe loaded area and should be constructed as to minimise its length22 February 2008 91

Limit values for design punchingshear stress in designThe following limit values for the punching shear stress are used indesign:IfvEd≤ v Rd , cno punching shear reinforcement requiredwhere:v1/3Rd , c= CRd, ck( 100ρ lfck) + 0,10σcp≥ ( vmin+ 0,10σcp)22 February 2008 92

How to take account of eccentricityMore sophisticated method for internal columns:zc12dyc22de y and e zb y and b zeccentricities M Ed /V Ed along y and z axesdimensions of control perimeter22 February 2008 93

How to take account of eccentricityVOr, how to determine β in equationEdvEd= βu dCiβ = 1,5B Aβ = 1,4β = 1,15For structures where lateralstability does not depend onframe action and whereadjacent spans do not differby more than 25% theapproximate values for βshown below may be used:22 February 2008 94

How to take account of eccentricityAlternative for edge and corner columns: use perimeter u 1* in stead offull perimeter and assume uniform distribution of punching force22 February 2008 95

Design of punching shear reinforcementIf v Ed ≥ v Rd,c shear reinforcementis required.The steel contribution comesfrom the shear reinforcementcrossing a surface at 1,5d fromthe edge of the loaded area, toensure some anchorage at theupper end. The concretecomponent of resistance istaken 75% of the designstrength of a slab without shearreinforcement22 February 2008 96

Punching shear reinforcementCapacity with punching shear reinforcementV u = 0,75V Rd,c + V SShear reinforcement within 1,5d from column is accounted for withf y,red = 250 + 0,25d(mm)≤f ywd22 February 2008 97

Punching shear reinforcementOuter controlperimeterAkdOuter perimeter of shearreinforcement0.5d0.75d1.5d (2d if > 2d fromcolumn)AThe outer control perimeter at whichshear reinforcement is not required,should be calculated from:u out,ef= V Ed/ (v Rd,cd)0.5d0.75dOuter controlperimeterkdThe outermost perimeter of shearreinforcement should be placed at adistance not greater than kd (k =1.5) within the outer controlperimeter.Section A - A22 February 2008 98

Special types of punching shear reinforcementDowel strips22 February 2008 99

Punching shear reinforcementWhere proprietary systems are used the control perimeter at which shearreinforcement is not required, u outor u out,ef(see Figure) should be calculated fromthe following expression:u out,ef= V Ed/ (v Rd,cd)uout,efuout> 2d2d1,5d1,5dd22 February 2008 100d

Punching shear• Column bases; critical parameters possible at a

Design with strut and tie modelsProf.dr.ir. J.C. Walraven22 February 2008102Group Concrete Structures

General idea behind strut and tie modelsStructures can be subdivided into regions with a steady state ofthe stresses (B-regions, where “B” stands for “Bernoulii” and inregions with a nonlinear flow of stresses (D-regions, where “D”stands for “Discontinuity”22 February 2008 103

D-region: stress trajectories and strut andtie modelSteps in design:1. Define geometry of D-region(Length of D-region is equalto maximum width of spread)2. Sketch stress trajectories3. Orient struts to compressiontrajectories4. Find equilibrium model byadding tensile ties5. Calculate tie forces6. Calculate cross section of tie7. Detail reinforcement22 February 2008 104

Examples of D-regions in structures22 February 2008 105

Design of struts, ties and nodesStruts with transverse compression stress or zero stress:σ Rd,max = f cd22 February 2008 106

Design of struts, ties and nodesStruts in cracked compression zones, with transverse tensionσ Rd,max = υf cdRecommended value υ = 0,60 (1 – f ck /250)22 February 2008 107

Design of struts, ties and nodesCompression nodes without tieσ Rd,max = k 1 υ’ f cdwhereυ’ = 0,60 (1 – f ck /250)Recommended valueK 1 = 1,022 February 2008 108

Design of struts, ties and nodesCompression-Compression-Tension (CTT) nodeσ Rd,max = k 2 υ’ f cdwhereυ’ = 0,60 (1 – f ck /250)Recommended valuek 2 = 0,8522 February 2008 109

Design of struts, ties and nodesCompression-Tension-Tension (CTT) nodeσ Rd,max = k 3 υ’ f cdwhereυ’ = 0,60 (1 – f ck /250)Recommended valuek 3 = 0,7522 February 2008 110

Example of detailing based on strut and tiesolutionStress - strain relation for confined concrete (dotted line)22 February 2008 111

Bearing capacity of partially loaded areas22 February 2008 112

Crack width control in concrete structuresProf.dr.ir. J.C. Walraven22 February 2008113Group Concrete Structures

Theory of crack width control (4)NWhen more cracks occur, more disturbed regions are found in theconcrete tensile bar. In the N-ε relation this stage (the “crackformation stage” is characterized by a “zig-zag”-line (N r,1 -N r,2 ). At acertain strain of the bar, the disturbed areas start to overlap.If no intermediate areas are left, the concrete cannot reach thetensile strength anymore, so that no new cracks can occur. The“crack formation stage” is ended and the stabilized cracking stagestarts. No new cracks occur, but existing cracks widen. t2 . t 2. t 2. t tNNN rN r,1N r,2N 0disturbed area22 February 2008 114ε

EC-formulae for crack width control (1)For the calculation of the maximum (or characteristic) crack width,the difference between steel and concrete deformation has to becalculated for the largest crack distance, which is s r,max = 2l t . Sowk=sr,max( )ε − εsmcmEq. (7.8) tσ srσ sesteel stress tf ctmwwheres r,max is the maximum crack distanceand(ε sm - ε cm ) is the difference in deformation betweensteel and concrete over the maximum crack distance.Accurate formulations for s r,max and (ε sm -ε cm ) will be givenconcrete stress22 February 2008 115

EC-2 formulae for crack width control(2)fct,effσs− kt( 1+αeρp,eff)ρp,effσsεsm− εcm=≥ 0, 6EEssEq. 7.0where: σ s is the stress in the steel assuming a cracked sectionα eis the ratio E s /E cmρ p,eff = (A s + ξA p )/A c,eff(effective reinforcement ratioξk tincluding eventual prestressing steel A pis bond factor for prestressing strands or wiresis a factor depending on the duration of loading(0,6 for short and 0,4 for long term loading)22 February 2008 116

EC-3 formulae for crack width control (4)Maximum final crack spacing s r,maxsr, max 3.4c+ 0. 425 k1k2ρ φ= (Eq. 7.11)p,effwhere c is the concrete coverΦ is the bar diameterk 1 bond factor (0,8 for high bond bars, 1,6 for barswith an effectively plain surface (e.g.prestressing tendons)k 2 strain distribution coefficient (1,0 for tensionand 0,5 for bending: intermediate values van beused)22 February 2008 117

EC-2 requirements for crack width control(recommended values)Exposure classRC or unbondedPSC membersQuasi-permanentloadPrestressedmembers withbonded tendonsFrequent loadX0,XC1 0.3XC2,XC3,XC4XD1,XD2,XS1,XS2,XS30.30.2Decompression22 February 2008 118

EC-2 formulae for crack width control (5)In order to be able to applythe crack width formulae,basically valid for a concretetensile bar, to a structureloaded in bending, adefinition of the “effectivetensile bar height” isnecessary. The effectiveheight h c,ef is the minimumof:2,5 (h-d)(h-x)/3h/2ab2.5 (h-d) < h-x e3beamceff. crosssectionslabdgravity lineof steelφhcelement loadedin tensiontcsmallest value ofφ2.5 . (c + /2) of t/2smallest value ofφ2.5 . (c + /2)of(h - x e)/322 February 2008 119

Maximum bar diameters for crackcontrol (simplified approach 7.3.3)50maximum bar diameter (mm)40302010w k = 0.4w k =0.3 mmw k =0.2 mm0100 150 200 250 300 350 400 450 500Reinforcement stress, σ s (N/mm 2 )22 February 2008 120

Maximum bar spacing for crack control(simplified approach 7.3.3)300Maximum bar spacing (mm)25020015010050w k = 0.2w k = 0.4w k = 0.30150 200 250 300 350 400stress in reinforcement (MPa)22 February 2008 121

Example (1)Continuous concrete roadData: Concrete C20/25, f ctm = 2,2 MPa, shrinkage ε sh =0,25·10 -3 ,temperature difference in relation to construction situation ΔT=25 0 .Max. crack width allowed = 0,2mm.CalculationThe maximum imposed deformation (shrinkage + temperature) isε tot = 0,50·10 -3 . Loading is slow, so E c,∞ =E c /(1+ϕ) ≅ 30.000/(1+2) =10.000 MPa. At ε tot = 0,50 ·10 -3 a concrete tensile strength of 5MPa applies, so the road is cracked.Cont. →22 February 2008 122

Example (1, cont.)For imposed deformation the “crackformation stage” applies. So, the load willnot exceed the cracking load, which is N cr= A c(1+nρ)f ctm≅ 1,1A cf ctm= 330 kN for b =1m. From the diagram at the right it isfound that a diameter of 12mm wouldrequire a steel stress not larger than 225MPa. To meet this requirement d =12mmbars at distances 150mm, both at top andbottom, are required.maximum bar diameter (mm)50w k = 0.44030w k =0.3 mm20w k =0.2 mm100100 150 200 250 300 350 400 450 500Reinforcement stress, σ s (N/mm 2 )22 February 2008 123

Example (2)q=4kN/mm2q6000275φ12-17515A slab bearing into one direction is subjected to a maximumvariable load of 4KN/m 2 . It should be demonstrated that themaximum crack width under the quasi permanent loadcombination is not larger than 0,4mm. (The floor is a part of ashopping centre: the environmental class is X0) (cont.→)22 February 2008 124

Example (2)q=4kN/mm2q6000275φ12-17515The governing load for the quasi-permanent load combination is:q = q g + ψ 2·q var. = (0.275·2500) + 0,6·400 = 928 kg/m 2 . Themaximum bending moment is then M = 9,28·6 2 /8=41,8 kNm/m ’ .For this bending moment the stress in the steel is calculated asσ s =289 MPa.Cont.→22 February 2008 125

Example (2)77.3hiddentieφ12The effective height of the tensile tie is the minimum of 2,5(c+ φ/2) of (hx)/3,where x = height of compression zone, calculated as 44mm. So, thegoverning value is (h-x)/3 = 77 mm. The effective reinforcement ratio isthen ρ eff= (113/0,175)/(77·1000)=0,83·10 -2. The crack distance s r,amx(Eq. 7.11) is found to be 245mm. For the term (ε sm-ε cm) a value 1,0·10 -3is found. This leads to a cracks width equal to w k= 0,25 mm, which hissmaller than the required 0,4mm.22 February 2008 126

Example (2)q=4kN/mm2q6000275φ12-17515A slab bearing into one direction is subjected to a maximumvariable load of 4KN/m 2 . It should be demonstrated that themaximum crack width under the quasi permanent loadcombination is not larger than 0,4mm. (The floor is a part of ashopping centre: the environmental class is X0) (cont.→)22 February 2008 127

Deformation of concrete structuresProf.dr.ir. J.C. Walraven22 February 2008128Group Concrete Structures

Deformation of concreteReason to worry orchallenge for the future?Deflection of ECC specimen, V. Li, Universityof MichiganDamage in masonry wall due to excessivedeflection of lintel22 February 2008 129

Reasons for controling deflections (1)AppearanceDeflections of such amagnitude that membersappear visibly to sag willupset the owners oroccupiers of structures. It isgenerally accepted that adeflection larger thanspan/250 should be avoidedfrom the appearance pointof view. A survey ofstructures in Germany thathad given rise to complaintsproduced 50 examples. Themeasured sag was less thanspan/250 in only two of these.22 February 2008 130

Reasons for controling deflections (2)Damage to non-structuralMembersAn important consequence of excessivedeformation is damage to nonstructural members, like partition walls.Since partition walls are unreinforcedand brittle, cracks can be large (severalmillimeters). The most commonlyspecified limit deflection is span/500,for deflection occurring afterconstruction of the partitions. It shouldbe assumed that all quasi permanentloading starts at the same time.22 February 2008 131

Reasons for controling deflections (3)CollapseIn recent years many cases ofcollapse of flat roofs havebeen noted. If the rainwaterpipes have a too low capacity,often caused by pollution andfinally stoppage, the roofdeflects more and more underthe weight of the water andfinally collapses. This occurspredominantly with light roofs.Concrete roofs are lesssusceptible for this type ofdamage22 February 2008 132

EC-2 Control of deflectionsDeflection limits according tochapter 7.4.1• Under the quasi permanent loadthe deflection should not exceedspan/250, in order to avoidimpairment of appearance andgeneral utility• Under the quasi permanentloads the deflection should belimited to span/500 afterconstruction to avoid damage toadjacent parts of the structure22 February 2008 133

EC-2: SLS - Control of deflectionsControl of deflection can bedone in two ways- By calculation- By tabulated values22 February 2008 134

Calculating the deflection of a concrete memberThe deflection follows from:δ = ζ δ II + (1 - ζ)δ Iδδ Iδ IIζdeflectiondeflection fully crackeddeflection uncrackedcoefficient for tension stiffening (transition coefficient)ζ = 1 - β (σ sr /σ s ) 2σ srσ sβsteel stress at first crackingsteel stress at quasi permanent service load1,0 for single short-term loading0,5 for sustained loads or repeated loading22 February 2008 135

Calculating the deflection of a concretememberThe transition from the uncracked state (I) to the cracked state (II) does notoccur abruptly, but gradually. From the appearance of the first crack,realistically, a parabolic curve can be followed which approaches the line forthe cracked state (II).22 February 2008 136

Calculating the deflection of a concrete memberFor pure bending the transitionfactorξ = 1− β ( σ s/ σrcan as well be written as2)ξ = 1− β ( / MM cr2)where M cris the cracking moment and M is the applied moment22 February 2008 137

Calculating the deflection of a concrete member7.4.3 (7)“The most rigorous method of assessingdeflections using the method givenbefore is to compute the curvatures atfrequent locations along the memberand then calculate the deflection bynumerical integration.In most cases it will be acceptableto compute the deflection twice,assuming the whole member to be inthe uncracked and fully cracked conditionin turn, and then interpolate using theexpression:ξ = 1− β ( / MM cr2)22 February 2008 138

Cases where detailed calculation may beomittedIn order to simplify the design,expressions have been derived, givinglimits of l/d for which no detailedcalculation of the deflection has to becarried out.These expressions are the results of anextended parameter analysis with themethod of deflection calculation asgiven before. The slenderness limitshave been determined with the criteriaδ

Calculating the deflection of a concrete memberFor span-depth ratios below the following limits no further checks is neededl3⎡2 ⎤⎢ρ0⎛ ρ0⎞= K 11 + 1,5 f + ⎜ − ⎟ ⎥ck3,2 fck1 if ρ ≤ ρ (7.16.a)d ⎢ρ⎥⎣⎝ ρ0 ⎠⎦d⎡= K⎢11+1,5⎣fckρ01+ρ − ρ'12fckρ'⎤⎥ρ0⎦l if ρ > ρ 0 (7.16.b)l/d is the limit span/depthK is the factor to take into account the different structural systemsρ 0 is the reference reinforcement ratio = √f ck 10 -3ρ is the required tension reinforcement ratio at mid-span to resist the momentdue to the design loads (at support for cantilevers)ρ’ is the required compression reinforcement ratio at mid-span to resist themoment due to design loads (at support for cantilevers)22 February 2008 140

Previous expressions in a graphical form (Eq. 7.16):60f ck =30 40 50 60 70 80 9050limiting span/depth ratio4030201000 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2Reinforcement percentage (A s /bd)22 February 2008 141

Limit values for l/d below which no calculatedverification of the deflection is necessaryThe table below gives the values of K (Eq.7.16), corresponding tothe structural system. The table furthermore gives limit l/d valuesfor a relatively high (ρ=1,5%) and low (ρ=0,5%) longitudinalreinforcement ratio. These values are calculated for concrete C30and σ s= 310 MPa and satisfy the deflection limits given in 7.4.1 (4)and (5).Structural system K ρ = 0,5% ρ = 1,5%Simply supported slab/beam1,0l/d=14l/d=20End span1,3l/d=18l/d=26Interior span1,5l/d=20l/d=30Flat slab1,2l/d=17l/d=24Cantilever0,4l/d= 6l/d=822 February 2008 142

Bond and anchorage22 February 2008 143

Ultimate Bond Stress, f bd (8.4.2)• The design value of the ultimate bond stress, f bd= 2,25 η 1η 2f ctdwheref ctdshould be limited to C60/75η 1=1 for ‘good’ and 0,7 for ‘poor’ bond conditionsη 2= 1 for φ≤32, otherwise (132- φ)/100Direction of concretingDirection of concretingα250a) 45º ≤ α ≤ 90º c) h > 250 mmDirection of concretingDirection of concretingh≥ 300hb) h ≤ 250 mm d) h > 600 mma) & b) ‘good’ bondconditions for all barsc) & d) unhatched zone – ‘good’ bond conditionshatched zone - ‘poor’ bond conditions22 February 2008 144

Basic Required Anchorage Length, l b,rqd(8.4.3)l b,rqd = (φ / 4) (σ sd / f bd )where σ sd is the design stress of the bar at theposition from where the anchorage is measured• For bent bars l b,rqd should be measured along thecentreline of the bar• Where pairs of wires/bars form welded fabrics φshould be replaced by φ n = φ√222 February 2008 145

Design Anchorage Length, l bd (8.4.4)l bd = α 1 α 2 α 3 α 4 α 5l b,rqd ≥ l b,minα 1α 2α 3effect of bendseffect of concrete coverFor straight bars α 1= 1.0, otherwise 0.7α 2= 1- 0.15(cover - φ)/φ ≥ 0.7 and ≤ 1.0effect of confinement by transverse reinforcement (not welded)α 3= 1- Kλ ≥ 0.7 and ≤ 1.0 where λ = (ΣA st- ΣA st,min)/A sA sφ t , A stA s t , A stA s φ t, Astα 4K = 0.1 K = 0.05 K = 0effect of confinement by welded transverse reinforcementα 4= 0.7α 5effect of confinement by transverse pressureα 5= 1 - 0.04p ≥ 0.7 and ≤ 1.0where p is the transverse pressure (MPa) at ULS along l bd(α 2α 3α 5) ≥ 0.7 l b,min> max(0.3l b; 15φ, 100mm)22 February 2008 146

Design Lap Length, l 0 (8.7.3)l 0 = α 1 α 2 α 3 α 5 α 6 l b,rqd ≥ l 0,minα 1 α 2α 3α 5are as defined for anchorage lengthα 6 = (ρ 1 /25) 0,5 but between 1,0 and 1,5where ρ 1 is the % of reinforcement lapped within 0,65l 0the centre of the lapfromPercentage of lapped bars relative tothe total cross-section area< 25% 33% 50% >50%α 61 1,15 1,4 1,5Note: Intermediate values may be determined by interpolation.l 0,min ≥ max{0,3 α 6 l b,rqd ; 15φ; 200}22 February 2008 147

Anchorage of Bottom Reinforcement atIntermediate Supports(9.2.1.5)lbdlbdφmφφl ≥ 10φl ≥ dml ≥ 10φ• Anchorage length, l, ≥ 10φ for straight bars≥φ m for hooks and bends with φ ≥16mm≥ 2φ m for hooks and bends with φ < 16mm• Continuity through the support may be required for robustness(Job specification)22 February 2008 148

Supporting Reinforcement at ‘Indirect’ Supports(9.2.5)Asupporting beam with height h 1BBsupported beam with height h 2(h 1≥ h 2)≤ h 2 /3≤ h 2 /2• The supporting reinforcement is in additionto that required for other reasons≤ h 1 /3≤ h 1 /2A• The supporting links may be placed in a zone beyond theintersection of beams22 February 2008 149

Columns (2)(9.5.3)≤ 150mms cl,tmax• s cl,tmax = 20 × φ min ; b; 400mm≤ 150mm• s cl,tmax should be reduced by a factor 0,6:– in sections within h above or below a beam or slab– near lapped joints where φ > 14. A minimum of 3 bars is rqd. inlap length22 February 2008 150

Additional rules for precast concrete22 February 2008 151

Bearing definitions (10.9.5)a 3+ Δa 3a 1aa 2+ Δa 2b 1a 1a = a 1+ a 2+ a 3+Δa2 22+ Δa3a 1a 2a 3Δa 2Δa 3net bearing length = F Ed/ (b 1f Rd), but ≥ min. valueF Eddesign value of support reactionb 1net bearing widthf Rddesign value of bearing strengthdistance assumed ineffective beyond outer end of supporting membersimilar distance for supported memberallowance for tolerances for the distance between supporting members= l n/2500, l nis length of member22 February 2008 152

Bearing definitions (10.9.5)a 3+ Δa 3a 1aa 2+ Δa 2b 1a 1a = a 1+ a 2+ a 3+Δa2 22+ Δa3Minimum value of a 1 in mm22 February 2008 153

Pocket foundations(10.9.6)hM F vsMF vFhl0,1lF 2F 3F h0,1lF 1μF 3μF 2μF 1l ssl ≤ s + l sl ≤ 1.2 hSpecial attention should be paid to:• shear resistance of column ends• detailing of reinforcement for F 1in top of pocket walls• punching resistance of the footing slab under the column force22 February 2008 154

Connections transmitting compressiveforcesFor soft bearings, in theabsence of a more accurateanalysis, the reinforcementmay be taken as:A s = 0,25 (t/h) F ed /f ydConcentratedbearingSoft bearingWhere:t = padding thicknessh = dimension of padding indirection of reinforcementF ed = design compressiveforce on connection22 February 2008 155

Lightweight aggregate concreteProf.dr.ir. J.C. Walraven22 February 2008156Group Concrete Structures

Lightweight concrete structures in the USANappa bridgeCalifornia 1977Oronado bridge San Diego52 m prestressed concrete beams, LafayetteUSA22 February 2008 157

Rilem Standard testRaftsundet Bridge, NorwayAntioch Bridge california22 February 2008 158

Qualification of lightweight aggregateconcrete (LWAC)Lightweight aggregate concrete is aconcrete having a closed structureand an oven dry density of not morethan 2200 kg/m 3 consisting of orcontaining a proportion of artificialor natural lightweight aggregateshaving a density of less than 2000kg/m 322 February 2008 159

Lightweight concrete density classificationDensity classificationDensity class 1,0 1,2 1,4 1,6 1,8 2,0Oven dry density801-1001-1201-1401-1601-1801-(kg/m 3 )100012001400160018002000Density Plain concrete105012501450165018502050(kg/m 3 ) Reinforced concrete11501350155017501950215022 February 2008 160

Conversion factors for mechanicalpropertiesThe material properties of lightweight concrete are related to thecorresponding properties of normal concrete. The following conversionfactors are used:η Econversion factor for the calculation of the modulus ofelasticityη 1coefficient for the determination of the tensile strengthη 2coefficient for the determination of the creep coefficientη 3coefficient for the determination of the drying shrinkageρ oven-dry density of lightweight aggregate concrete in kg/m 3Antioch Bridge, California, 197722 February 2008 161

Design stress strain relations for LWACThe design stress strain relations for LWAC differ in tworespects from those for NDC.• The advisory value for thestrength is lower than for NDC(sustained loading factor 0,85 instead of 1,0)•The ultimate strain ε l,cuis reducedwith a factor η 1=0,40+0,60ρ/220022 February 2008 162

Shrinkage of LWACThe drying shrinkage values for lightweight concrete (concreteclass ≥ LC20/25) can be obtained by multiplying the values fornormal density concrete for NDC with a factor η 3 =1,2The values for autogenous shrinkage of NDC represent a lowerlimit for those of LWAC, where no supply of water from theaggregate to the drying microstructure ispossible. If water-saturated, or even partiallysaturated lightweight concrete is used, theautogenous shrinkage values will considerablybe reduced (water stored in LWAC particles isextracted from aggregate particles into matrix,reducing the effect of self-dessication22 February 2008 163

Shear capacity of LWAC membersThe shear resistance of members without shearreinforcement is calculated by:1/3VlRd ct= {(0,15/ γc) η k(100ρlflck) + 0,15cp] b, 1σwdwhere the factor η 1 =0,40+0,60ρ/2200 is the onlydifference with the relation for NDC22 February 2008 164

Punching shear resistanceLike in the case for shear of LWAC members, also the punchingshear resistance of LWAC slab is obtained using the reductionfactor η 1= 0,4 + 0,6ρ/2200. the punching shear resistance ofa lightweight concrete slab follows from:V1/3Rd , c= ( ClRd, ck1(100ρlflck) + 0,08σcp≥η1vlmin+η 0, 08σcpwhere C lRd,c= 0,15/γ c(in stead of the 0,18/γ cfor NDC)22 February 2008 165

Plain and lightly reinforced concreteProf.dr.ir. J.C. Walraven22 February 2008166Group Concrete Structures

Field of applicationMembers for which the effect of dynamic action may beignored• Members mainly subjected to compression other than due toprestressing, e.g. walls, columns, arches, vaults and tunnels• Strip and pad footings for foundations• Retaining walls• Piles whose diameter is ≥ 600mm and where N ed /A c ≤ 0,3f ck22 February 2008 167

Additional design assumptions12.3.1 Due to the less ductile properties of plain concrete, thedesign values should be reduced. The advisory reductionfactor is 0,822 February 2008 168

ULS: design resistance to bending andaxial failureThe axial resistance N Rd, of a rectangularcross-section with a uniaxial eccentricitye, in the direction of h w, may be taken as:N Rd=ηf cdbh(1-2e/h w)whereηf cdis the design compressivestrength belonging to the blockshaped stress-strain relation22 February 2008 169

Shear12.6.3 (1): “In plain concrete members account may be takenof the concrete tensile strength in the ultimate limit state forshear, provided that either by calculation or by experiencebrittle failure can be excluded and adequate resistance can beensured”Using Mohr’s circle itshould be demonstratedthat nowhere in thestructure the principalconcrete tensile stress ofthe concrete exceeds thedesign tensile strength f ctk22 February 2008 170

Simplified design method for walls andcolumnsIn the absence of a more rigorous approach, the design resistance interms of axial force slender wall or column in plain concrete may becalculated as follows:N Rd=b·h w·f cd·φwhereN Rdis the axial resistanceb is the overall width of the cross-sectionh wis the overall depth of the cross-sectionφ is a factor taking account eccentricity, including secondorder effectsφ = 1,14·(1-2e tot/h w) – 0,02 l 0/h w≤ (1 – 2e tot/h w)22 February 2008 171

Eurocodes: a big step forward:• The rules within the Eurocode are very wide ranging (muchbetter than all existing national codes)• The fact that the Eurocodes cover a range of structuralmaterials is an advantage to designers• The use of common loading suggests a logical andeconomical approach to design• The Eurocodes are written in a way that allows thedesigner to adopt the most modern design techniques• The Eurocodes are unique among modern codes in thatthey allow for local variations in climate and custom, andcan thus easily be adopted for safe and economic use22 February 2008 172

Eurocode: only for Europe?22 February 2008 173