RC Detailing to Eurocode 2

RC Detailing to Eurocode 2 Jenny Burridge MA CEng MICE MIStructE Head of Structural Engineering Structural Eurocodes BS EN 1990 (EC0): BS EN 1991 (EC1): Basis of structural design Actions on Structures BS EN 1992 (EC2): BS EN 1993 (EC3): BS EN 1994 (EC4): BS EN 1995 (EC5): BS EN 1996 (EC6): BS EN 1999 (EC9): Design of concrete structures Design of steel structures Design of composite steel and concrete structures Design of timber structures Design of masonry structures Design of aluminium structures BS EN 1997 (EC7): BS EN 1998 (EC8): Geotechnical design Design of structures for earthquake resistance

Eurocode 2 - contents General Basis of design Materials Durability and cover to reinforcement Structural analysis Ultimate limit state Serviceability limit state Detailing of reinforcement and prestressing tendons General Detailing of member and particular rules Additional rules for precast concrete elements and structures Lightweight aggregated concrete structures Plain and lightly reinforced concrete structures Eurocode 2 - Annexes A. (Informative) Modification of partial factors for materials B. (Informative) Creep and shrinkage strain C. (Normative) Reinforcement properties D. (Informative) Detailed calculation method for prestressing steel relaxation losses E. (Informative) Indicative Strength Classes for durability F. (Informative) Reinforcement expressions for in-plane stress conditions G. (Informative) Soil structure interaction H. (Informative) Global second order effects in structures I. (Informative) Analysis of flat slabs and shear walls J. (Informative) Examples of regions with discontinuity in geometry or action (Detailing rules for particular situations) EC2 Annex J - replaced by Annex B in PD 6687

Standards BS EN 206-1 Specifying Concrete BS 8500 Specifying Concrete PD 6687-1 (Parts 1 & 3) PD 6687-2 ( Part 2) BS EN 1992 Design of concrete structures Part 1-1: General & buildings Part 1-2: Fire design Part 2: Part 3: Bridges Liquid retaining National Annex BS EN 13670 Execution of Structures NSCS N.A. BS EN 10080 Reinforcing Steels BS 4449 Reinforcing Steels BS 8666 Reinforcement Scheduling Specification NSCS, Finishes NSCS Guidance: 1 Basic 2 Ordinary 3 Plain 4 Special Visual Concrete

Labour and Material (Peri) 24% 58% 18% Rationalisation of Reinforcement Optimum cost depends on: Material cost Labour Plant Preliminaries Finance Team decision required

Detailing Reinforcement

Reinforcement EC2 does not cover the use of plain or mild steel reinforcement Principles and Rules are given for deformed bars, decoiled rods, welded fabric and lattice girders. EN 10080 provides the performance characteristics and testing methods but does not specify the material properties. These are given in Annex C of EC2 Properties of reinforcement (Annex C) Product form Bars and de-coiled rods Wire Fabrics Class A B C A B C Characteristic yield strength f yk or f 0,2k (MPa) cold worked 400 to 600 hot rolled seismic k = (f t /f y ) k 1,05 1,08 1,15 <1,35 1,05 1,08 1,15 <1,35 Characteristic strain at maximum force, ε uk (%) 2,5 5,0 7,5 2,5 5,0 7,5 Fatigue stress range (N = 2 x 10 6 ) (MPa) with an upper limit of 0.6f yk 150 100 The UK has chosen a maximum value of characteristic yield strength, f yk, = 600 MPa, but 500 MPa is the value assumed in BS 4449 and 4483 for normal supply.

Extract BS 8666 www.ukcares.co.uk www.uk-bar.org UK CARES (Certification - Product & Companies) 1. Reinforcing bar and coil 2. Reinforcing fabric 3. Steel wire for direct use of for further processing 4. Cut and bent reinforcement 5. Welding and prefabrication of reinforcing steel

www.ukcares.co.uk www.uk-bar.org A B C Reinforcement supply Coil up to 16mm (2.5T) Bar 12,14,15 and 18m Cut and bent approx 550 to 650/T

Table power bender

Health & Safety Potential Risk factor Low Medium High High Risk 33,51,56,63,64 & 99? SC Detail Comment Designer Fabricator 33 64 Sausage Link. Health and safety risk is high with larger diameter bar. Also the risk increases with small dimensions. When bent on an automatic link bender with small diameter bars the risk is relatively low. When bending on a manual bender the risk is high, especially with larger diameters and non standard formers. Boot Link. Greater risk than shape code 51 as the bars have to cross over twice to achieve the shape. Health and safety risk becomes higher with larger diameter bar. Also the risk increases with small dimensions. See Note SN2. When bent on an automatic link bender with small diameter bars the risk is relatively low. When bending on a manual bender the risk is higher, especially with larger diameters. This shape is designed for producing small to medium sized links in small diameter bar. Do not detail this shape in large diameter bar, try to use an alternative (eg. 2 no. shape code 13 s facing each other to create a shape code 33). See Note SN2. Smaller diameter bars cause less of a problem as they can often be produced on an automatic link bending machine. Larger diameter bars have to be produced on a manual power bender with the potential to trap the operator s fingers. Try to avoid/minimise the use of shapes which cause a scissor action, especially with larger diameter bars. See Note SN2. Great care should be taken when bending this shape. If the operator has concerns when producing this shape he should consult his supervisor. Bending Minimum Bending & projections Minimum Bends 6mm - 16mm = 2x Dia Internal 20mm - 50mm = 3.5x Dia Internal Minimum of 4 x dia between bends End Projection = 5 x Dia from end of bend BS8666, Table 2

Tolerances (not in EC2 BS8666) Minimum Cover for Bond For bars: Bar diameter For post-tensioned tendons: Circular ducts: Duct diameter Rectangular ducts: The greater of: the smaller dimension or half the greater dimension For pre-tensioned tendons: 1.5 x diameter of strand or wire 2.5 x diameter of indented wire

Structural Fire Design BS EN 1992-1-2 Scope: Part 1-2 Structural fire design gives several methods for fire engineering Tabulated data for various elements is given in section 5 Reinforcement cover Axis distance, a, to centre of bar a = c + φ m /2 + φ l a Axis Distance Allowance in Design for Deviation c dev : Allowance for deviation = 10mm A reduction in c dev may be permitted: for a quality assurance system, which includes measuring concrete cover, 10 mm c dev 5 mm where very accurate measurements are taken and non conforming members are rejected (eg precast elements) 10 mm c dev 0 mm

Nominal Cover Nominal cover, c nom Minimum cover, c min c min = max {c min,b ; c min,dur ; 10 mm} Allowance for deviation, c dev Axis distance, a Fire protection Procurement Lead-in times should be 4 weeks for rebar Express reinforcement (and therefore expensive) 1 7 days The more complicated the scheduling the longer for bending

Standard Detailing Practicalities 12m maximum length H20 to H40 (12m H40 = 18 stone/ 118Kg) Health & safety 9m maximum length H16 & H12 6m maximum length H10 & H8 Transport Fixing Control of Cracking EC2: Cl. 7.3 In Eurocode 2 cracking is controlled in the following ways: Minimum areas of reinforcement cl 7.3.2 & Equ 7.1 A s,min σ s = k c kf ct,eff A ct this is the same as Crack width limits (Cl. 7.3.1 and National Annex). These limits can be met by either: direct calculation (Cl. 7.3.4) crack width is W k Used for liquid retaining structures deemed to satisfy rules (Cl. 7.3.3) Note: slabs 200mm depth are OK if A s,min is provided.

Minimum Reinforcement Area EC2: Cl. 9.2.1.1, Eq 9.1N The minimum area of reinforcement for slabs (and beams) is given by: A s,min 0.26f f ctm yk bt d 0.0013b d t Crack Control Without Direct Calculation EC2: Cl. 7.3.3 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 Note: For cracking due to restraint use only max bar size

Spacing of bars EC2: Cl. 8.2 Clear horizontal and vertical distance φ, (d g +5mm) or 20mm For separate horizontal layers the bars in each layer should be located vertically above each other. There should be room to allow access for vibrators and good compaction of concrete. Ultimate bond stress EC2: Cl. 8.4.2 The design value of the ultimate bond stress, f bd = 2.25 η 1 η 2 f ctd where f ctd should be limited to C60/75 η 1 =1 for good and 0.7 for poor bond conditions η 2 = 1 for φ 32, otherwise (132- φ)/100 Direction of concreting Direction of concreting α 250 a) 45º α 90º c) h > 250 mm Direction of concreting Direction of concreting h 300 h b) h 250 mm d) h > 600 mm unhatched zone good bond conditions hatched zone - poor bond conditions

Basic required anchorage length EC2: Cl. 8.4.3 l b,rqd = (φ / 4) (σ sd / f bd ) where σ sd is the design stress of 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 Design Anchorage Length, l bd EC2: Cl. 8.4.4 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 8.2 - C d & K factors EC2: Figure 8.3 EC2: Figure 8.4

Anchorage of links EC2: Cl. 8.5 Design Lap Length, l 0 (8.7.3) EC2: Cl. 8.7.3 l 0 = α 1 α 2 α 3 α 5 α 6 l b,rqd l 0,min 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 6 1 1.15 1.4 1.5 Note: Intermediate values may be determined by interpolation. l 0,min max{0.3α 6 l b,rqd ; 15φ; 200}

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

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 = 1.8 MPa f ctd = α ct f ctk,0,05 /γ c = 1.8/1.5 = 1.2 f bd = 2.25 x 1.2 = 2.7 MPa Basic anchorage length, l b,req l b.req = (Ø/4) ( σ sd /f bd ) EC2 Equ 8.3 Max stress in the bar, σ sd = f yk /γ s = 500/1.15 = 435MPa. l b.req = (Ø/4) ( 435/2.7) = 40.3 Ø For concrete class C25/30

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 (40.3Ø) For concrete class C25/30 Alpha values EC2: Table 8.2 Concise: 11.4.2

Table 8.2 - C d & K factors EC2: Figure 8.3 Concise: Figure 11.3 EC2: Figure 8.4 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 (40.3Ø) 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 α 5 = 1.0 N/A conservative value α 2 = 1.0 0.15 (c d Ø)/Ø α 2 = 1.0 0.15 (25 16)/16 = 0.916 l bd = 0.916 x 40.3Ø = 36.9Ø = 590mm

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 (40.3Ø) 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 = 1.0 0.15 (c d 3Ø)/Ø 1.0 α 2 = 1.0 0.15 (25 48)/16 = 1.25 1.0 l bd = 1.0 x 40.3Ø = 40.3Ø = 645mm Worked example - summary H16 Bars Concrete class C25/30 25 Nominal cover Tension anchorage straight bar l bd = 36.9Ø = 590mm Tension anchorage Other shape bars l bd = 40.3Ø = 645mm 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 = 40.3Ø = 645mm Anchorage for Poor bond conditions = Good /0.7 Lap length = anchorage length x α 6

Anchorage & lap lengths How to design concrete structures using Eurocode 2 Arrangement of Laps EC2: Cl. 8.7.2, Fig 8.7 If more than one layer a maximum of 50% can be lapped

Arrangement of Laps EC2: Cl. 8.7.3, Fig 8.8 Transverse Reinforcement Anchorage of bars F There is transverse tension reinforcement required

Transverse Reinforcement Lapping of bars F tanθ F θ F tanθ F/2 F/2 F There is transverse tension reinforcement required Transverse Reinforcement at Laps Bars in tension EC2: Cl. 8.7.4, Fig 8.9 only if bar Ø 20mm or laps > 25% Where the diameter, φ, of the lapped bars 20 mm, 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 and between that and the surface of the concrete. If more than 50% of the reinforcement is lapped at one point and the distance between adjacent laps at a section is 10 φ transverse bars should be formed by links or U bars anchored into the body of the section. The transverse reinforcement provided as above should be positioned at the outer sections of the lap as shown below. F s ΣA /2 st l 0 /3 ΣA /2 st l 0 /3 150 mm F s l 0

Beams EC2: Cl. 9.2 A s,min = 0,26 (f ctm /f yk )b t d but 0,0013b t d A s,max = 0,04 A c Section at supports 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 Beams EC2: Cl. 9.2 Tension reinforcement in a flanged beam at supports should be spread over the effective width (see 5.3.2.1)

Shear Design: Links EC2: Cl. 6.2.3 Variable strut method allows a shallower strut angle hence activating more links. As strut angle reduces concrete stress increases V s V z d θ z d x Angle = 45 V carried on 3 links x Angle = 21.8 V carried on 6 links Short Shear Spans with Direct Strut Action EC2: Cl. 6.2.3 (8) d d a v a v Where a v 2d the applied shear force, V Ed, for a point load (eg, corbel, pile cap etc) may be reduced by a factor a v /2d where 0.5 a v 2d provided: The longitudinal reinforcement is fully anchored at the support. Only that shear reinforcement provided within the central 0.75a v is included in the resistance. Note: see PD6687-1:2010 Cl 2.14 for more information.

Shear reinforcement EC2: Cl. 9.2.2 Minimum shear reinforcement, ρ w,min = (0,08 f ck )/f yk Maximum longitudinal spacing, s l,max = 0,75d (1 + cotα) For vertical links s l,max = 0,75d Maximum transverse spacing, s t,max = 0,75d 600 mm Shear Design EC2: Cl. 6.2.3 V z d x s V θ z d x

Curtailment of reinforcement EC2: Cl. 9.2.1.3, Fig 9.2 Envelope of (MEd /z +NEd) Acting tensile force lbd lbd Resisting tensile force lbd a l Ftd lbd a l Ftd lbd lbd lbd lbd Shift rule 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 Anchorage of Bottom Reinforcement at End Supports EC2: Cl. 9.2.1.4 Tensile Force Envelope a l Shear shift rule Simple support (indirect) Simple support (direct) 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.

Simplified Detailing Rules for Beams Supporting Reinforcement at Indirect Supports EC2: Cl. 9.2.5 A supporting beam with height h 1 B supported beam with height h 2 (h 1 h 2 ) Plan view B h 2 /3 h 2 /2 The supporting reinforcement is in addition to that required for other reasons h 1 /3 h 1 /2 A The supporting links may be placed in a zone beyond the intersection of beams

Solid slabs EC2: Cl. 9.3 Curtailment as beams except for the Shift rule a l = d may be used Flexural Reinforcement min and max areas as beam Secondary transverse steel not less than 20% main reinforcement Reinforcement at Free Edges Solid slabs EC2: Cl. 9.3 Where partial fixity exists, not taken into account in design: Internal supports: A s,top 0,25A s for M max in adjacent span End supports: A s,top 0,15A s for M max in adjacent span This top reinforcement should extend 0,2 adjacent span

Particular rules for flat slabs Distribution of moments EC2: Table I.1 Particular rules for flat slabs EC2: Cl. 9.4 Arrangement of reinforcement should reflect behaviour under working conditions. At internal columns 0.5A t should be placed in a width = 0.25 panel width. At least two bottom bars should pass through internal columns in each orthogonal directions.

Columns EC2: Cl. 9.5.2 h 4b φ min 12 A s,min = 0,10N Ed /f yd but 0,002 A c A s,max = 0.04 A c (0,08A c at laps) Minimum number of bars in a circular column is 4. Where direction of longitudinal bars changes more than 1:12 the spacing of transverse reinforcement should be calculated. Columns EC2: Cl. 9.5.3 150mm s cl,tmax s cl,tmax = min {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 min of 3 bars is required in lap length s cl,tmax = min {12 φmin; 0.6b ; 240mm}

Walls Vertical Reinforcement A s,vmin = 0,002 A c (half located at each face) A s,vmax = 0.04 A c (0,08A c at laps) s vmax = 3 wall thickness or 400mm Horizontal Reinforcement A s,hmin = 0,25 Vert. Rein. or 0,001A c s hmax = 400mm Transverse Reinforcement Where total vert. rein. exceeds 0,02 A c links required as for columns Where main rein. placed closest to face of wall links are required (at least 4No. m 2 ). [Not required for welded mesh or bars Ø 16mm with cover at least 2Ø.] Detailing Comparisons Beams Main Bars in Tension A s,min A s,max EC2 Clause / Values 9.2.1.1 (1): 0.26 f ctm /f yk bd 0.0013 bd 9.2.1.1 (3): 0.04 bd BS 8110 Values 0.0013 bh 0.04 bh Main Bars in Compression A s,min A s,max Spacing of Main Bars s min S max Links -- 9.2.1.1 (3): 0.04 bd 8.2 (2): d g + 5 mm or φ or 20mm Table 7.3N 0.002 bh 0.04 bh d g + 5 mm or φ Table 3.28 A sw,min 9.2.2 (5): (0.08 b s f ck )/f yk 0.4 b s/0.87 f yv s l,max 9.2.2 (6): 0.75 d 0.75d s t,max 9.2.2 (8): 0.75 d 600 mm d or 150 mm from main bar 9.2.1.2 (3) or 15φ from main bar

Detailing Comparisons Slabs Main Bars in Tension A s,min A s,max Secondary Transverse Bars A s,min A s,max Spacing of Bars s min S max EC2 Clause / Values 9.2.1.1 (1): 0.26 f ctm /f yk bd 0.0013 bd 0.04 bd 9.3.1.1 (2): 0.2A s for single way slabs 9.2.1.1 (3): 0.04 bd 8.2 (2): d g + 5 mm or φ or 20mm 9.3.1.1 (3): main 3h 400 mm secondary: 3.5h 450 mm places of maximum moment: main: 2h 250 mm secondary: 3h 400 mm BS 8110 Values 0.0013 bh 0.04 bh 0.002 bh 0.04 bh d g + 5 mm or φ 3d or 750 mm Detailing Comparisons Punching Shear Links A sw,min Spacing of Links EC2 Clause / Values 9.4.3 (2): Link leg = 0.053 s r s t (f ck )/f yk BS 8110 Values Total = 0.4ud/0.87fyv 0.75d S r 9.4.3 (1): 0.75d S t 9.4.3 (1): 1.5d within 1st control perim.: 1.5d outside 1st control perim.: 2d Columns Main Bars in Compression A s,min A s,max Links Min size S cl,tmax 9.5.2 (2): 0.10N Ed /f yk 0.002bh 9.5.2 (3): 0.04 bh 9.5.3 (1) 0.25φ or 6 mm 9.5.3 (3): min (12φmin; 0.6 b;240 mm) 9.5.3 (6): 150 mm from main bar 0.004 bh 0.06 bh 0.25φ or 6 mm 12φ 150 mm from main bar

How to Compendium Detailing

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