Reinforced Concrete detailing to EC2.pdf

RC Detailing to Eurocode 2

Jenny Burridge

MA CEng MICE MIStructE

Head of Structural Engineering

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

Structural Eurocodes

• 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 - contents

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)

Eurocode 2 - Annexes

EC2 Annex J - replaced by Annex B in PD 6687

BS EN 1992Design of concrete structures

Part 1-1: General & buildings

Part 1-2: Fire design

Part 2: Bridges

Part 3: Liquid retaining

Standards

BS EN 13670Execution of Structures

BS 4449Reinforcing

Steels

BS EN 10080Reinforcing

SteelsBS 8500Specifying Concrete

BS EN 206-1Specifying Concrete

NSCS

BS 8666Reinforcement Scheduling

National AnnexPD 6687-1 (Parts 1 & 3)

PD 6687-2 ( Part 2)

N.A.

Specification – NSCS, Finishes

NSCS Guidance:

1 Basic

2 Ordinary

3 Plain

4 Special –Visual Concrete

Labour and Material (Peri)

18%

24%

58%

Rationalisation of Reinforcement

Optimum cost depends

on:

• Material cost

• Labour

• Plant

• Preliminaries

• Finance

Team decision required

Detailing

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

Reinforcement

Product form Bars and de-coiled rods Wire Fabrics

Class

A

B

C

A

B

C

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

400 to 600

k = (ft/fy)k

≥1,05

≥1,08

≥1,15 <1,35

≥1,05

≥1,08

≥1,15 <1,35

Characteristic strain at

maximum force, εεεεuk (%)

≥2,5

≥5,0

≥7,5

≥2,5

≥5,0

≥7,5

Fatigue stress range

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

150

100

cold worked seismichot rolled

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.

Properties of reinforcement(Annex C)

Extract BS 8666

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

www.ukcares.co.uk www.uk-bar.org

A

B

C

Coil up to 16mm (2.5T)

Bar – 12,14,15 and 18m

Cut and bent – approx £550 to £650/T

Reinforcement supply

Table power bender

High

Medium

Low

Potential Risk factor

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.

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.

64

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.

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.

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.

33

FabricatorDesignerCommentDetailSC

High Risk

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

Health & Safety

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

Bending

BS8666, Table 2

Tolerances (not in EC2—BS8666)

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 wire2.5 x diameter of indented wire

Minimum Cover for Bond

a Axis

Distance

Reinforcement cover

Axis distance, a, to centre of bar

a = c + φφφφm/2 + φφφφl

Scope:

Part 1-2 Structural fire design gives several methods for fire engineering

Tabulated data for various elements is given in section 5

Structural Fire DesignBS EN 1992-1-2

∆∆∆∆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

Allowance in Design for Deviation

Nominal cover, cnom

Minimum cover, cmin

cmin = max {cmin,b; cmin,dur ; 10 mm}

Axis distance, aFire protection

Allowance for deviation, ∆cdev

Nominal Cover

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

Procurement

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

Standard Detailing

Control of Cracking

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.

EC2: Cl. 7.3

Minimum Reinforcement Area

The minimum area of reinforcement for slabs (and beams) is given by:

db0013.0f

dbf26.0A t

yk

tctm

min,s≥≥

EC2: Cl. 9.2.1.1, Eq 9.1N

Crack Control Without Direct 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

EC2: Cl. 7.3.3

Note: For cracking due to restraint use only max bar size

• 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.

Spacing of barsEC2: Cl. 8.2

The design value of the ultimate bond stress, fbd = 2.25 η1η2fctdwhere 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

a) 45º ≤≤≤≤ αααα ≤≤≤≤ 90º c) h > 250 mm

h

Direction of concreting

≥ 300

h


b) h ≤≤≤≤ 250 mm d) h > 600 mm

unhatched zone – ‘good’ bond conditions

hatched zone - ‘poor’ bond conditions

α


250


Ultimate bond stress

EC2: Cl. 8.4.2

lb,rqd = (φ φ φ φ / 4) (σσσσsd / fbd)

where σsd is the design stress of the bar at the position from where the anchorage is measured.

Basic required anchorage length

EC2: Cl. 8.4.3

• For bent bars lb,rqd should be measured along the centreline of the bar

lbd = α1 α2 α3 α4 α5 lb,rqd ≥≥≥≥ lb,min

However:

(α2 α3 α5) ≥≥≥≥ 0.7

lb,min > max(0.3lb,rqd ; 10φφφφ, 100mm)

Design Anchorage Length, lbd

EC2: Cl. 8.4.4

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

l0 = α1 α2 α3 α5 α6 lb,rqd ≥≥≥≥ l0,min

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 cross-

section area

< 25% 33% 50% >50%

α6 1 1.15 1.4 1.5

Note: Intermediate values may be determined by interpolation.

α1 α2 α3 α5 are as defined for anchorage length

l0,min ≥ max{0.3 α6 lb,rqd; 15φ; 200}

Design Lap Length, l0 (8.7.3)

EC2: Cl. 8.7.3

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, fbdfbd = 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 EC2 cl 3.1.6(2), Equ 3.16

αct = 1.0 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 valuesEC2: Table 8.2 Concise: 11.4.2

Table 8.2 - Cd & K factorsConcise: Figure 11.3EC2: Figure 8.3

EC2: Figure 8.4

Design anchorage length, lbdlbd = α1 α2 α3 α4 α5 lb.req ≥ lb,min


a) Tension anchorage – straight bar

α1 = 1.0

α3 = 1.0 conservative value with K= 0

α4 = 1.0 N/A

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


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 N/A

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

How to design concrete structures using Eurocode 2

Anchorage & lap lengths

Arrangement of LapsEC2: Cl. 8.7.2, Fig 8.7

If more than one layer a maximum of 50% can be lapped

Arrangement of LapsEC2: Cl. 8.7.3, Fig 8.8

Anchorage of bars

F

Transverse Reinforcement

There is transverse tension – reinforcement required

F/2 F/2

θ

F tanθ

F tanθ

F F

Lapping of bars


There is transverse tension – reinforcement required

• Where the diameter, φφφφ, 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 φφφφ 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.

l /30

ΣA /2st

ΣA /2st

l /30

FsFs

≤150 mm

l0

Transverse Reinforcement at LapsBars in tension

EC2: Cl. 8.7.4, Fig 8.9 only if bar Ø ≥ 20mm or laps > 25%

• 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

BeamsEC2: Cl. 9.2

• Tension reinforcement in a flanged beam at

supports should be spread over the effective width

(see 5.3.2.1)

BeamsEC2: Cl. 9.2

Shear Design: Links

Variable strut method allows a shallower strut angle –

hence activating more links.

As strut angle reduces concrete stress increases

Angle = 45°V carried on 3 links Angle = 21.8° V carried on 6 links

d

V

z

x

d

x

V

θz

s

EC2: Cl. 6.2.3

• 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:

dd

av av

− The longitudinal reinforcement is fully anchored at the support.

− Only that shear reinforcement provided within the central 0.75av is

included in the resistance.

Short Shear Spans with Direct Strut Action

EC2: Cl. 6.2.3 (8)

Note: see PD6687-1:2010 Cl 2.14 for more information.

Shear reinforcement

• Minimum shear reinforcement, ρw,min = (0,08√fck)/fyk

• Maximum longitudinal spacing, sl,max = 0,75d (1 + cotα)

• Maximum transverse spacing, st,max = 0,75d ≤ 600 mm

EC2: Cl. 9.2.2

For vertical links sl,max = 0,75d

Shear Design

d

V

z

x

d

x

V

θz

s

EC2: Cl. 6.2.3

• For members without shear reinforcement this is satisfied with al = d

a l

∆Ftd

a l

Envelope of (MEd /z +NEd)

Acting tensile force

Resisting tensile force

lbd

lbd

lbd

lbd

lbd lbd

lbd

lbd

∆Ftd

“Shift rule”

Curtailment of reinforcement

EC2: Cl. 9.2.1.3, Fig 9.2

• For members with shear reinforcement: al = 0.5 z Cot θBut it is always conservative to use al = 1.125d

• lbd is required from the line of contact of the support.

Simple support (indirect) Simple support (direct)

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

• Transverse pressure may only be taken into account with

a ‘direct’ support.

Shear shift rule

al

Tensile Force Envelope

Anchorage of Bottom Reinforcement at End Supports

EC2: Cl. 9.2.1.4

Simplified Detailing Rules for Beams

≤ h /31

≤ h /21

B

A

≤ h /32

≤ h /22

supporting beam with height h1

supported beam with height h2 (h1 ≥ h2)

• The supporting reinforcement is in

addition to that required for other

reasons

A

B

• The supporting links may be placed in a zone beyond

the intersection of beams

Supporting Reinforcement at ‘Indirect’ Supports

Plan view

EC2: Cl. 9.2.5

• 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 slabsEC2: 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

Solid slabsEC2: Cl. 9.3

Distribution of moments

EC2: Table I.1

Particular rules for flat slabs

• 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.

Particular rules for flat slabs

EC2: Cl. 9.4

• 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. 9.5.2

• scl,tmax = min {20 φmin; b ; 400mm}

≤ 150mm

≤ 150mm

scl,tmax

• 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}

Columns

EC2: Cl. 9.5.3

Walls

• 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

Vertical Reinforcement

Horizontal Reinforcement

• As,hmin = 0,25 Vert. Rein. or 0,001Ac

• shmax = 400mm


• 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

d or 150 mm from main bar9.2.2 (8): 0.75 d ≤ 600 mm

9.2.1.2 (3) or 15φ from main bar

st,max

0.75d9.2.2 (6): 0.75 dsl,max

0.4 b s/0.87 fyv9.2.2 (5): (0.08 b s √fck)/fykAsw,min

Links

Table 3.28Table 7.3NSmax

dg + 5 mm or φ8.2 (2): dg + 5 mm or φ or 20mmsmin

Spacing of Main Bars

0.04 bh9.2.1.1 (3): 0.04 bdAs,max

0.002 bh--As,min

Main Bars in Compression

0.04 bh9.2.1.1 (3): 0.04 bdAs,max

0.0013 bh9.2.1.1 (1): 0.26 fctm/fykbd ≥0.0013 bd

As,min

ValuesClause / ValuesMain Bars in Tension

BS 8110EC2Beams


places of maximum moment:

main: 2h ≤ 250 mm

secondary: 3h ≤ 400 mm

3d or 750 mmsecondary: 3.5h ≤ 450 mmSmax

dg + 5 mm or φ8.2 (2): dg + 5 mm or φ or 20mm

9.3.1.1 (3): main 3h ≤ 400 mm

smin

Spacing of Bars

0.04 bh9.2.1.1 (3): 0.04 bdAs,max

0.002 bh9.3.1.1 (2): 0.2As for single way

slabs

As,min

Secondary Transverse Bars

0.04 bh0.04 bdAs,max

0.0013 bh9.2.1.1 (1): 0.26 fctm/fykbd ≥0.0013 bd

As,min

ValuesClause / ValuesMain Bars in Tension

BS 8110EC2Slabs


Columns

150 mm from main bar9.5.3 (6): 150 mm from main bar

12φ9.5.3 (3): min (12φmin; 0.6 b;240 mm)Scl,tmax

0.25φ or 6 mm9.5.3 (1) 0.25φ or 6 mmMin size

Links

0.06 bh9.5.2 (3): 0.04 bhAs,max

0.004 bh9.5.2 (2): 0.10NEd/fyk ≤ 0.002bhAs,min

Main Bars in Compression

1.5d9.4.3 (1):

within 1st control perim.: 1.5d

outside 1st control perim.: 2d

St

0.75d9.4.3 (1): 0.75dSr

Spacing of Links

Total = 0.4ud/0.87fyv9.4.3 (2): Link leg = 0.053 sr st√(fck)/fyk

Asw,min

ValuesClause / ValuesLinks

BS 8110EC2Punching Shear

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Reinforced Concrete detailing to EC2.pdf