RC Detailing to EC2 Webinar August RC Detailing to Eurocode 2. RC Detailing to Eurocode 2

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1 Webinar from The Concrete Centre The presentation is based on Chapter 10 Detailing in The Concrete Centre s publication How to Design Concrete Structures using Eurocode 2. Paul Gregory Structural Engineer MSc MBA CEng MICE MIStructE pgregory@concretecentre.com The publication is available from and 2 nd Edition Feb 2018 Detailing introduction EC2 Outline Detailing introduction EC2 (BS EN 1992 Part 1-1) Crack control without direct calculation EC2 Ch 7.3 Anchorage and Laps EC2 Chapter 8.4 & 8.7 Detailing of Members EC2 Chapter 9 Resources The first part of the presentation will look at topics in EC2 that are relevant to detailing: Reinforcement EC2 cl 3.2 and Annex C Concrete cover EC2 cl Tension due to loading & tension due to restraint Confined concrete EC2 cl Strut and tie models EC2 cl Minimum spacing of bars EC2 cl 8.2 Reinforcement EC2: cl 3.2 Reinforcing steel EC2 applies to ribbed and weldable reinforcement. Reinforcement EC2: ANNEX C (Normative) Extract from Table C.1: Properties of reinforcement Design and detailing are valid for f yk = 400 to 600 MPa. f yk is the characteristic yield strength of reinforcement BS 4449 bars and BS 4483 fabric have adopted 500 MPa cold worked hot rolled seismic i.e In the UK f yk = 500 MPa Three classes of reinforcement A, B and C Difference is ductility C is more ductile than B At least a class B is required for max 30% redistribution of elastic BM s. 1

2 Reinforcement Question: a) How in the UK does the designer communicate to site what class of steel reinforcement is required? Reinforcement BS 8666 Scheduling, bending reinforcement Answer: a) Use the notation in BS 8666 Scheduling, bending reinforcement. e.g. H20 Question: b) What is written on the RC detail drawings if the designer wants class C reinforcement? Answer: b) C and the bar diameter. e.g. C20 For buildings the usual notation to use is H. e.g. H16 & the contractor has to supply either B or C class 500 MPa steel Concrete cover EC2: cl Nominal cover is specified on the drawings. Nominal cover, c nom = c min + c dev Concrete Cover Nominal cover, c nom = c min + c dev Nominal cover, c nom c min depends on durability (BS 8500) and bond (the bar diameter) Δc dev is an allowance for deviation & the recommended value is 10mm. Δc dev may be reduced in certain situations e.g. for a site quality assurance system Δc dev may be 5mm. Fire resistance tables can also affect the specified cover. The fire tables require that a nominal distance from the face of the section to the axis of the reinforcement is achieved. Procedure: Determine nominal cover for durability and bond and check that this gives the required axis distance. Minimum cover, c min c min = max {c min,dur ; c min,b ; 10 mm} Durability as per BS 8500 Bond φ Allowance for deviation, c dev 10 mm Recommended Tables in Section 5 of EC2 part 1-2 Axis distance, a Fire protection Tension due loading & Tension due to restraint Confined concrete EC2: Cl , Fig 3.6 Tension can occur due to loading e.g Simply supported beam & Cracks in the bottom f ck is the characteristic cylinder strength of concrete at 28 days Compressing the cylinder causes the diameter to increase in size. by restraining the free movement of a member e.g. wall panel with edge restraint. Wall panel edge restraint If the lateral expansion is prevented by confining the concrete the vertical compression strength will be greater than f ck and is f ck,c Standard cylinder test No confinement 2

3 Confined concrete EC2: Cl , Fig 3.6 Confinement of concrete results in higher strength. Confinement can be generated by adequately closed links. Anchorage bond stress depends on confinement of surrounding concrete EC2: Cl 6.6 (1). Affects anchorage coefficients α 3 & α 4. Transverse pressure can reduce anchorage lengths, EC2: Cl (3). e.g. bearing pressure. Affects anchorage coefficient α 5. Figure 3.6 Strut-and-tie models EC2: Cl & 6.5 Concise: Strut-and-tie models (STM) are trusses consisting of: struts ties nodes STM model for a deep beam STM is a lower bound plastic theory and can be used to analyse many concrete elements. STM is usually adopted to design non-standard elements such as: Pile caps Corbels Deep beams Any situation where normal beam theory does not apply Strut-and-tie models STM models help us understand: Variable strut inclination method for shear link design Why tension steel needs anchoring at a simple end support The shift rule for curtailment of longitudinal tension reinforcement EC2 cl (2) Shear force 21.8 < θ < 45 Strut-and-tie models STM models help us understand: The anchorage of bars F Shear Strut inclination method There is transverse tension Minimum spacing of bars EC2: Cl. 8.2 Concise: 11.2 Clear horizontal & vertical distance φ, (d g +5mm) or 20mm φ is the bar diameter d g is the maximum size of aggregate Outline Detailing introduction EC2 Crack control without direct calculation EC2 Ch 7.3 Anchorage and Laps EC2 Chapter 8.4 & 8.7 Detailing of Members EC2 Chapter 9 Detail Min 75 mm gaps in the top reinforcement to allow the concrete to be placed in the mould. Resources 3

4 Crack Control Without Direct Calculation EC2: Cl For concrete in buildings the serviceability limit of cracking can be satisfied without the direct calculation of crack widths. The limiting value is w max. A procedure for crack control in buildings: Establish the limiting crack width, w max. Typical value is 0.3mm (see NA Table NA.4) for RC concrete in all exposure classes under the quasi-permanent load combination. Determine the minimum amount of reinforcement, A s,min, to control cracking in beams and slabs. EC2 cl (1) equation 9.1.N. For slabs depths 200mm or less containing A s,min no further measures to control cracking are required. EC2 cl (1). For other slabs and beams, determine the service stress in the reinforcement, σ s Control cracking by limiting the bar dia or bar spacing, EC2 Tables 7.2N & 7.3N. (σ s is an input to the tables) Crack control Design yield strength of reinforcement, f yd = f yk /1.15 Ultimate load combination, γ G G k + γ Q Q k = 500/ 1.15 = 435 MPa Design service load is the Quasi-permanent combination = G k + ψ 2 Q k Service stress is σ s =? Minimum Reinforcement Area EC2: Cl , Eq 9.1N The minimum area of reinforcement for slabs (and beams) is given by: 0.26fctm bt d As,min bt d fyk Crack Control Without Direct Calculation EC2: Cl Provide minimum reinforcement. Crack control may be achieved in two ways: limiting the maximum bar diameter using Table 7.2N limiting the maximum bar spacing using Table 7.3N Service stress, σ s Note: For cracking due to restraint use only max bar size Crack Control Without Direct Calculation EC2: Cl σ s Determination of Steel Stress Ratio G k /Q k = 8.5/4.0 = 2.13 x 1/δ δ is the redistribution ratio Example Check cracking in bottom of a slab. G k = 8.5 kn/m 2 Q k = 4.0 kn/m 2 Ψ 2 = 0.3 (office loading) γ G = 1.25 A s,req = 1324 mm 2 /m Unmodified steel stress, σ su 252 σ su = 252 MPa Try 100 A s,prov = 2010 mm 2 /m Ratio G k /Q k Chapter 2 4

5 Crack Control - Example Check cracking in bottom of a slab From graph σ su = 252 MPa σ s = (σ su A s,req ) / (δ A s,prov ) σ s = (252 x 1324)/(1.0 x 2010) = 166 MPa For 100 c/c (A s,prov = 2010 mm 2 /m) Design meets both criteria Maximum bar size or spacing to limit crack width Steel stress (σ s ) MPa Maximum bar size (mm) w max = 0.3 mm OR Maximum bar spacing (mm) Outline Detailing introduction EC2 Crack control without direct calculation EC2 Ch 7.3 Anchorage and Laps EC2 Chapter 8.4 & 8.7 Detailing of Members EC2 Chapter 9 Resources δ is the ratio of the redistributed moment to the elastic bending moment Anchorage and Laps EC2: Chapter 8 Anchorage of reinforcement Anchorage of reinforcement EC2: Cl. 8.4 Lapping of bars Worked example Ultimate bond stress EC2: Cl Concise: 11.5 The design value of the ultimate bond stress, Ultimate bond stress EC2: Cl , Fig 8.2 Concise: 11.5 Good and bad bond conditions f bd = 2.25 η 1 η 2 f ctd where f ctd should be limited to C60/75 Top is poor Bond condition η 1 = 1 for good and 0.7 for poor bond conditions η 2 = 1 for φ 32, otherwise (132- φ)/100 unhatched zone good bond conditions hatched zone - poor bond conditions 5

6 Ultimate bond stress EC2: Cl , Fig 8.2 Concise: 11.5 Good and bad bond conditions Question: The slip form formwork system is often used to construct tall concrete cores. Do you use good or poor bond conditions? Basic required anchorage length EC2: Cl Concise: l b,rqd = (φ / 4) (σ sd / f bd ) where σ sd is the design stress in the bar at the position from where the anchorage is measured. For bent bars l b,rqd should be measured along the centreline of the bar Answer: Poor and η 1 = 0.7 Slip form Formwork system EC2 Figure 8.1 Design Anchorage Length, l bd EC2: Cl Concise: Table C d & K factors EC2: Figure 8.3 Concise: Figure 11.3 l bd = α 1 α 2 α 3 α 4 α 5 l b,rqd l b,min However: (α 2 α 3 α 5 ) 0.7 l b,min > max(0.3l b,rqd ; 10φ, 100mm) Alpha values EC2: Table 8.2 Table requires values for: C d Value depends on cover and bar spacing, see Figure 8.3 K λ Factor depends on position of confinement reinforcement, see Figure 8.4 = ( A st Ast,min )/ A s Where A st is area of transverse reinf. EC2: Figure 8.4 Beam corner bar? Table Other shapes EC2: Figure 8.1 Concise: Figure 11.1 Alpha values EC2: Table 8.2 Concise: α 1 α 2 α 3 α 4 α 5 6

7 Design Lap Length, l 0 (8.7.3) EC2: Cl Concise: l 0 = α 1 α 2 α 3 α 5 α 6 l b,rqd l 0,min Laps EC2: Cl. 8.7 α 1 α 2 α 3 α 5 are as defined for anchorage length α 6 = (ρ 1 /25) 0,5 but between 1.0 and 1.5 where ρ 1 is the % of reinforcement lapped within 0.65l 0 from the centre of the lap Percentage of lapped bars relative to the total crosssection area < 25% 33% 50% >50% α Note: Intermediate values may be determined by interpolation. l 0,min max{0.3α 6 l b,rqd ; 15φ; 200} Arrangement of Laps EC2: Cl , Fig 8.8 Worked example Anchorage and lap lengths Anchorage Worked Example Calculate the tension anchorage for an H16 bar in the bottom of a slab: a) Straight bars b) Other shape bars (Fig 8.1 b, c and d) Concrete strength class is C25/30 Nominal cover is 25mm Assume maximum design stress in the bar Bond stress, f bd f bd = 2.25 η 1 η 2 f ctd EC2 Equ. 8.2 η 1 = 1.0 Good bond conditions η 2 = 1.0 bar size 32 f ctd = α ct f ctk,0,05 /γ c EC2 cl 3.1.6(2), Equ 3.16 α ct = 1.0 γ c = 1.5 f ctk,0,05 = 0.7 x 0.3 f 2/3 ck EC2 Table 3.1 = 0.21 x 25 2/3 = MPa f ctd = α ct f ctk,0,05 /γ c = 1.795/1.5 = What is the value of f ck? Answer: f ck = 25 MPa f bd = 2.25 x = MPa 7

8 Basic anchorage length, l b,req Design anchorage length, l bd l b.req = (Ø/4) ( σ sd /f bd ) EC2 Equ 8.3 l bd = α 1 α 2 α 3 α 4 α 5 l b.req l b,min Max stress in the bar, σ sd = f yk /γ s = 500/1.15 l bd = α 1 α 2 α 3 α 4 α 5 (40.36Ø) For concrete class C25/30 = 435MPa. l b.req = (Ø/4) ( 435/2.693) = Ø For concrete class C25/30 Alpha values EC2: Table 8.2 Concise: Table C d & K factors EC2: Figure 8.3 Concise: Figure 11.3 EC2: Figure 8.4 Design anchorage length, l bd Design anchorage length, l bd l bd = α 1 α 2 α 3 α 4 α 5 l b.req l b,min l bd = α 1 α 2 α 3 α 4 α 5 l b.req l b,min l bd = α 1 α 2 α 3 α 4 α 5 (40.36Ø) For concrete class C25/30 a) Tension anchorage straight bar α 1 = 1.0 α 3 = 1.0 conservative value with K= 0 α 4 = 1.0 N/A α 5 = 1.0 conservative value α 2 = (C d Ø)/Ø α 2 = (25 16)/16 = l bd = x 40.36Ø = 36.97Ø = 592mm l bd = α 1 α 2 α 3 α 4 α 5 (40.36Ø) For concrete class C25/30 b) Tension anchorage Other shape bars α 1 = 1.0 C d = 25 is 3 Ø = 3 x 16 = 48 α 3 = 1.0 conservative value with K= 0 α 4 = 1.0 N/A α 5 = 1.0 conservative value α 2 = (C d 3Ø)/Ø 1.0 α 2 = (25 48)/16 = l bd = 1.0 x 40.36Ø = 40.36Ø = 646mm 8

9 Worked example - summary H16 Bars Concrete class C25/30 25 Nominal cover Anchorage & lap lengths How to design concrete structures using Eurocode 2 Tension anchorage straight bar l bd = 36.97Ø = 592mm Tension anchorage Other shape bars l bd = 40.36Ø = 646mm l bd is measured along the centreline of the bar Compression anchorage (α 1 = α 2 = α 3 = α 4 = α 5 = 1.0) l bd = 40.36Ø = 646mm Anchorage for Poor bond conditions = Good /0.7 Lap length = anchorage length x α 6 Column lap length for 100% laps & grade C40/50 = 0.73 x 61Ф = 44.5 Ф Terminology Verbal forms BS EN ISO 9001:2015, section 0.1 ISO International Organization for Standardization ISO 9001:2015 Quality management systems, Requirements Arrangement of Laps EC2: Cl Concise: Cl 11.6 Laps between bars should normally be staggered and not located in regions of high stress. Arrangement of laps should comply with Figure 8.7: In this ISO standard, the following verbal forms are used: shall indicates a requirement; Distance a is used in cl (3), Transverse reinf. should indicates a recommendation; may indicates a permission; can indicates a possibility or a capability. All bars in compression and secondary (distribution) reinforcement may be lapped in one section. Transverse Reinforcement at Laps Bars in tension Concise: Cl EC2: Cl , Fig 8.9 Transverse reinforcement is required in the lap zone to resist transverse tension forces. Any Transverse reinforcement provided for other reasons will be sufficient if the lapped bar Ø < 20mm or laps< 25% Transverse Reinforcement at Laps Bars in tension Concise: Cl EC2: Cl , Fig 8.9 Also, if the lapped bar Ø 20mm and more than 50% of the reinforcement is lapped at one point and the distance between adjacent laps at a section, a, 10φ, then transverse bars should be formed by links or U bars anchored into the body of the section. If the lapped bar Ø 20mm the transverse reinforcement should have a total area, ΣA st 1,0A s of one spliced bar. It should be placed perpendicular to the direction of the lapped reinforcement. Also it should be positioned at the outer sections of the lap as shown. F s ΣA st /2 l 0 /3 ΣA /2 st l 0 /3 150 mm F s l0 Figure 8.9 (a) - bars in tension 9

10 Outline Detailing introduction EC2 Crack control without direct calculation EC2 Ch 7.3 Anchorage and Laps EC2 Chapter 8.4 & 8.7 Detailing of Members EC2 Chapter 9 Resources Detailing of Members EC2: Chapter 9 Beams Solid slabs - one-way and two-way spanning Columns Flat slabs Walls Deep beams Foundations Tying systems Beams EC2: Cl. 9.2 A s,min = 0,26 (f ctm /f yk )b t d but 0,0013b t d Beams EC2: Cl. 9.2 Tension reinforcement in a flanged beam at supports should be spread over the effective width (see ) A s,max = 0,04 A c Section at supports in monolithic construction should be designed for a hogging moment 0,25 max. span moment Any design compression reinforcement (φ) should be held by transverse reinforcement with spacing 15 φ Shear reinforcement EC2: Cl Curtailment of reinforcement EC2: Cl , Fig 9.2 Concise: Minimum shear reinforcement, ρ w,min = (0,08 (f ck )/f yk Envelope of (M Ed /z +NEd) Acting tensile force Resisting tensile force Maximum longitudinal spacing, s l,max = 0,75d (1 + cotα) a l a l Ftd For vertical links s l,max = 0,75d a l Ftd Shift rule Maximum transverse spacing, s t,max = 0,75d 600 mm For members without shear reinforcement this is satisfied with a l = d For members with shear reinforcement: a l = 0.5 z Cot θ But it is always conservative to use a l = 1.125d (for θ = 45 o, a l = 0.45d) 10

11 Anchorage of Bottom Reinforcement at End Supports EC2: Cl Simplified Detailing Rules for Beams Simple support (indirect) Simple support (direct) Ch 10, Detailing A s bottom steel at support 0.25 A s provided in the span l bd is required from the line of contact of the support. Transverse pressure may only be taken into account with a direct support. α 5 anchorage coefficient Solid slabs EC2: Cl. 9.3 Solid Slabs One-way and two-way spanning Flexural Reinforcement min and max areas as beam Secondary transverse steel not less than 20% main reinforcement Reinforcement at Free Edges EC2: Cl 9.3 Columns EC2: Cl Columns h 4b φ min 12 A s,min = 0,10N Ed /f yd but 0,002 A c EC2: Cl 9.5 A s,max = 0.04A c (0,08A c at laps) 11

12 Columns Links Diameter ¼ x largest main bar diameter & 8mm Spacing - can conservatively be taken as not greater than any of the following; - 12 x smallest main bar dia x smallest column dimension - 240mm Spacing can be increased by a factor of 1.67 when at a distance more than the larger dimension of the column above or below a beam or slab. i.e. near the mid height of the column Detailing Issues EC2 Clause Lap lengths Table (3) & Fig 8.7 Issue Possible resolve in 2023? α 6 varies depending on amount staggered 0.3 l o gap between ends of lapped bars is onerous. α 6 should always = 1.5. Staggering doesn t help at ULS For ULS, there is no advantage in staggering bars( fib bulletin Mar 2014). For SLS staggering at say 0.5 l o might be helpful. Detailing Issues EC2 Clause Table 8.2 Issue Possible resolve in 2023? α 2 for compression bars Should be the same as for tension. Initial test suggests α 2 = 0.7 Table 8.2 α 2 for bent bars Currently, anchorage worse than for straight bars (4) & Fig 8.9 Fig 9.3 Requirements for transverse bars are impractical l bd anchorage into support 6.4 Numbers of perimeters of punching shear links Requirement only makes 10-15% difference in strength of lap (Corrigendum 1 no longer requires transverse bars to be between lapped bar and surface.) May be OTT as compression forces increase bond strength. Issue about anchorage beyond CL of support Work of CEN TC 250.SC2/WG1/TG4 Outline Detailing introduction EC2 Crack control without direct calculation EC2 Ch 7.3 Anchorage and Laps EC2 Chapter 8.4 & 8.7 Detailing of Members EC2 Chapter 9 Resources Standard Method of Detailing Structural Concrete How to Design Concrete Structures using Eurocode 2 Third edition Published June 2006 Second edition February 2018 Information and advice is based on Eurocodes. Chapter 10 - Detailing Referenced as an NCCI in EC2 12

13 Designed and Detailed Strut-and-tie Models Good Concrete Guide 9 Published February 2015 Concrete Society Publication, January 2017 S&T article in the April 2015 edition of The Structural Engineer End Thank you for your attention Questions pgregory@concretecentre.com Angel Building, London 13