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RC Detailing to Eurocode 2 Jenny Burridge MA CEng MICE MIStructE Head of Structural Engineering

Structural Eurocodes BS EN 1990 (EC0):

Basis of structural design

BS EN 1991 (EC1):

Actions on Structures

BS EN 1992 (EC2):

Design of concrete structures

BS EN 1993 (EC3):

Design of steel structures

BS EN 1994 (EC4):

Design of composite steel and concrete structures

BS EN 1995 (EC5):

Design of timber structures

BS EN 1996 (EC6):

Design of masonry structures

BS EN 1999 (EC9):

Design of aluminium structures

BS EN 1997 (EC7):

Geotechnical design

BS EN 1998 (EC8):

Design of structures for earthquake resistance

Eurocode 2 - contents •


Basis of design


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 13670 Execution of Structures N.A. BS EN 206-1 Specifying Concrete

Part 1-1: BS 8500 Specifying Concrete


BS EN 1992 Design of concrete structures

BS EN 10080 Reinforcing Steels

General & buildings

Part 1-2:

Fire design

Part 2:


Part 3:

Liquid retaining

BS 4449 Reinforcing Steels

PD 6687-1 (Parts 1 & 3) PD 6687-2 ( Part 2)

National Annex 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%


Rationalisation of Reinforcement

Optimum cost depends on: • Material cost • Labour • Plant • Preliminaries • Finance

Team decision required




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

Class Characteristic yield strength fyk or f0,2k (MPa)

Bars and de-coiled rods



cold worked


Wire Fabrics


400 to 600 hot rolled




k = (ft/fy)k



≥1,15 <1,35



≥1,15 <1,35

Characteristic strain at maximum force, εuk (%)







Fatigue stress range (N = 2 x 106) (MPa) with an upper limit of 0.6fyk



The UK has chosen a maximum value of characteristic yield strength, fyk, = 600 MPa, but 500 MPa is the value assumed in BS 4449 and 4483 for normal supply.

Extract BS 8666 UK CARES (Certification - Product & Companies) 1. 2. 3. 4. 5.

Reinforcing bar and coil Reinforcing fabric Steel wire for direct use of for further processing Cut and bent reinforcement Welding and prefabrication of reinforcing

steel A



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 High Risk

Potential Risk factor Low

33,51,56,63,64 & 99?

Medium High







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.

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 13â&#x20AC;&#x2122;s facing each other to create a shape code 33). See Note SN2.

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.


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.

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 operatorâ&#x20AC;&#x2122;s fingers. Try to avoid/minimise the use of shapes which cause a scissor action, especially with larger diameter bars.

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 EC2â&#x20AC;&#x201D;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 + φm/2 + φl


Axis Distance

Allowance in Design for Deviation ∆cdev: Allowance for deviation = 10mm A reduction in ∆cdev may be permitted: • for a quality assurance system, which includes measuring concrete cover, 10 mm ≥ ∆cdev ≥ 5 mm • where very accurate measurements are taken and non conforming members are rejected (eg precast elements) 10 mm ≥ ∆cdev ≥ 0 mm

Nominal Cover

Nominal cover, cnom Minimum cover, cmin cmin = max {cmin,b; cmin,dur ; 10 mm}

Allowance for deviation, â&#x2C6;&#x2020;cdev

Axis distance, a Fire protection


Lead-in times should be 4 weeks for rebar

Express reinforcement (and therefore expensive) 1 â&#x20AC;&#x201C; 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 As,minσs = kckfct,effAct 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 Wk – Used for liquid retaining structures – ‘deemed to satisfy’ rules (Cl. 7.3.3) Note: slabs ≤ 200mm depth are OK if As,min is provided.

Minimum Reinforcement Area EC2: Cl., Eq 9.1N The minimum area of reinforcement for slabs (and beams) is given by:

A s,min ≥

0.26 fctm bt d ≥ 0.0013 b t d fyk

Crack Control Without Direct EC2: Cl. 7.3.3 Calculation 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, (dg +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, fbd = 2.25 η1η2fctd where fctd should be limited to C60/75 η1 =1 for ‘good’ and 0.7 for ‘poor’ bond conditions η2 = 1 for φ ≤ 32, otherwise (132- φ)/100 Direction of concreting


a) 45º ≤ α ≤ 90º Direction of concreting

Direction of concreting


c) h > 250 mm Direction of concreting ≥ 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

lb,rqd = (φ / 4) (σsd / fbd) where σsd is the design stress of the bar at the position from where the anchorage is measured. • For bent bars lb,rqd should be measured along the centreline of the bar

Design Anchorage Length, lbd EC2: Cl. 8.4.4 lbd = α1 α2 α3 α4 α5 lb,rqd ≥ lb,min However: (α2 α3 α5) ≥ 0.7 lb,min > max(0.3lb,rqd ; 10φ, 100mm)

Alpha values EC2: Table 8.2

Table 8.2 - Cd & K factors EC2: Figure 8.3

EC2: Figure 8.4

Anchorage of links EC2: Cl. 8.5

Design Lap Length, l0 (8.7.3) EC2: Cl. 8.7.3 l0 = α1 α2 α3 α5 α6 lb,rqd ≥ l0,min α1 α2 α3 α5 are as defined for anchorage length α6 = (ρ1/25)0,5 but between 1.0 and 1.5 where ρ1 is the % of reinforcement lapped within 0.65l0 from the centre of the lap Percentage of lapped bars relative to the total crosssection area α6

< 25%








Note: Intermediate values may be determined by interpolation. l0,min ≥ max{0.3 α6 lb,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, fbd fbd = 2.25 η1 η2 fctd

EC2 Equ. 8.2

η1 = 1.0 ‘Good’ bond conditions η2 = 1.0 bar size ≤ 32 fctd = αct fctk,0,05/γc αct = 1.0

EC2 cl 3.1.6(2), Equ 3.16 γc = 1.5

fctk,0,05 = 0.7 x 0.3 fck2/3

EC2 Table 3.1

= 0.21 x 252/3 = 1.8 MPa fctd = αct fctk,0,05/γc = 1.8/1.5 = 1.2  fbd = 2.25 x 1.2 = 2.7 MPa

Basic anchorage length, lb,req lb.req

= (Ø/4) ( σsd/fbd)

EC2 Equ 8.3

Max stress in the bar, σsd = fyk/γs = 500/1.15 = 435MPa. lb.req

= (Ø/4) ( 435/2.7) = 40.3 Ø For concrete class C25/30

Design anchorage length, lbd lbd = α1 α2 α3 α4 α5 lb.req ≥ lb,min lbd = α1 α2 α3 α4 α5 (40.3Ø)

For concrete class C25/30

Alpha values EC2: Table 8.2

Concise: 11.4.2

Table 8.2 - Cd & K factors EC2: Figure 8.3

Concise: Figure 11.3

EC2: Figure 8.4

Design anchorage length, lbd lbd = α1 α2 α3 α4 α5 lb.req ≥ lb,min lbd = α1 α2 α3 α4 α5 (40.3Ø)

For concrete class C25/30

a) Tension anchorage – straight bar α1 = 1.0 α3 = 1.0

conservative value with K= 0

α4 = 1.0


α5 = 1.0

conservative value

α2 = 1.0 – 0.15 (cd – Ø)/Ø α2 = 1.0 – 0.15 (25 – 16)/16 = 0.916 lbd = 0.916 x 40.3Ø = 36.9Ø = 590mm

Design anchorage length, lbd lbd = α1 α2 α3 α4 α5 lb.req ≥ lb,min lbd = α1 α2 α3 α4 α5 (40.3Ø)

For concrete class C25/30

b) Tension anchorage – Other shape bars α1 = 1.0

cd = 25 is ≤ 3 Ø = 3 x 16 = 48

α3 = 1.0

conservative value with K= 0

α4 = 1.0


α5 = 1.0

conservative value

α2 = 1.0 – 0.15 (cd – 3Ø)/Ø ≤ 1.0 α2 = 1.0 – 0.15 (25 – 48)/16 = 1.25 ≤ 1.0 lbd = 1.0 x 40.3Ø = 40.3Ø = 645mm

Worked example - summary H16 Bars – Concrete class C25/30 – 25 Nominal cover Tension anchorage – straight bar

lbd = 36.9Ø = 590mm

Tension anchorage – Other shape bars

lbd = 40.3Ø = 645mm

lbd is measured along the centreline of the bar Compression anchorage (α1 = α2 = α3 = α4 = α5 = 1.0) lbd = 40.3Ø = 645mm Anchorage for ‘Poor’ bond conditions = ‘Good’/0.7 Lap length = anchorage length x α6

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


There is transverse tension â&#x20AC;&#x201C; reinforcement required

Transverse Reinforcement Lapping of bars

F tanθ




F tanθ F/2


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, ΣAst ≥ 1,0As 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.

ΣAst /2

ΣAst /2

l 0 /3

l 0 /3

≤150 mm F s



Beams EC2: Cl. 9.2

• As,min = 0,26 (fctm/fyk)btd but ≥ 0,0013btd

• As,max = 0,04 Ac • 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

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 z


s d






Angle = 45° V carried on 3 links

Angle = 21.8° V carried on 6 links

Short Shear Spans with Direct Strut Action EC2: Cl. 6.2.3 (8) d av

d av

• Where av ≤ 2d the applied shear force, VEd, for a point load (eg, corbel, pile cap etc) may be reduced by a factor av/2d where 0.5 ≤ av ≤ 2d provided: − The longitudinal reinforcement is fully anchored at the support. − Only that shear reinforcement provided within the central 0.75av 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, ρw,min = (0,08√fck)/fyk

• Maximum longitudinal spacing, sl,max = 0,75d (1 + cotα) For vertical links sl,max = 0,75d • Maximum transverse spacing, st,max = 0,75d ≤ 600 mm

Shear Design EC2: Cl. 6.2.3

V z


x s

V θ x



Curtailment of reinforcement EC2: Cl., Fig 9.2 Envelope of (M Ed /z +N Ed )

lbd lbd

Acting tensile force Resisting tensile force





al ∆Ftd lbd


“Shift rule”



• For members without shear reinforcement this is satisfied with al = d • For members with shear reinforcement: al = 0.5 z Cot θ But it is always conservative to use al = 1.125d

Anchorage of Bottom Reinforcement at End Supports EC2: Cl. Tensile Force Envelope al

Shear shift rule

Simple support (indirect)

Simple support (direct)

• As bottom steel at support ≥ 0.25 As provided in the span • lbd 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 B


supporting beam with height h1 supported beam with height h2 (h1 ≥ h2)

≤ h 2 /3

≤ h 2 /2

Plan view • The supporting reinforcement is in addition to that required for other reasons

≤ h 1 /3


≤ h 1 /2

• 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 al = 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: As,top ≥ 0,25As for Mmax in adjacent span End supports: As,top ≥ 0,15As for Mmax 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 0.5At 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 • φmin ≥ 12 • As,min = 0,10NEd/fyd but ≥ 0,002 Ac • As,max = 0.04 Ac

(0,08Ac 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


≤ 150mm

• scl,tmax = min {20 φmin; b ; 400mm} • scl,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 scl,tmax = min {12 φmin; 0.6b ; 240mm}

Walls Vertical Reinforcement • As,vmin = 0,002 Ac (half located at each face) • As,vmax = 0.04 Ac

(0,08Ac at laps)

• svmax = 3 × wall thickness or 400mm Horizontal Reinforcement • As,hmin = 0,25 Vert. Rein. or 0,001Ac • shmax = 400mm Transverse Reinforcement • Where total vert. rein. exceeds 0,02 Ac links required as for columns • Where main rein. placed closest to face of wall links are required (at least 4No. m2). [Not required for welded mesh or bars Ø ≤ 16mm with cover at least 2Ø.]

Detailing Comparisons Beams


BS 8110

Main Bars in Tension

Clause / Values

As,min (1): 0.0013 bd

0.26 fctm/fykbd ≥

0.0013 bh

As,max (3):

0.04 bd

0.04 bh


Main Bars in Compression As,min


0.002 bh

As,max (3):

0.04 bd

0.04 bh

Spacing of Main Bars dg + 5 mm or φ or 20mm


8.2 (2):


Table 7.3N

dg + 5 mm or φ Table 3.28

Links Asw,min

9.2.2 (5):

(0.08 b s √fck)/fyk

0.4 b s/0.87 fyv


9.2.2 (6):

0.75 d



9.2.2 (8):

0.75 d ≤ 600 mm

d or 150 mm from main bar (3) or 15φ from main bar

Detailing Comparisons Slabs


BS 8110

Main Bars in Tension

Clause / Values

As,min (1):


0.04 bd

0.26 fctm/fykbd ≥ 0.0013 bd

Values 0.0013 bh 0.04 bh

Secondary Transverse Bars As,min (2):

0.2As for single way slabs

0.002 bh

As,max (3):

0.04 bd

0.04 bh

Spacing of Bars smin

8.2 (2):

dg + 5 mm or φ or 20mm

dg + 5 mm or φ (3): main 3h ≤ 400 mm Smax

secondary: 3.5h ≤ 450 mm

3d or 750 mm

places of maximum moment: main:

2h ≤ 250 mm

secondary: 3h ≤ 400 mm

Detailing Comparisons Punching Shear


BS 8110


Clause / Values



9.4.3 (2): √(fck)/fyk

Link leg = 0.053 sr st

Total = 0.4ud/0.87fyv


9.4.3 (1):




9.4.3 (1):

Spacing of Links


within 1st control perim.: 1.5d outside 1st control perim.: 2d Columns Main Bars in Compression As,min

9.5.2 (2): 0.10NEd/fyk ≤ 0.002bh

0.004 bh


9.5.2 (3):

0.06 bh

0.04 bh

Links Min size

9.5.3 (1) 0.25φ or 6 mm

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

150 mm from main bar

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