rc detailing to eurocode 2

7/26/2019 RC Detailing to Eurocode 2

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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


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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 steelrelaxation losses

E. (Informative) Indicative Strength Classes for durability

F. (Informative) Reinforcement expressions for in-plane stressconditions

G. (Informative) Soil structure interaction

H. (Informative) Global second order effects in structuresI. (Informative) Analysis of flat slabs and shear walls

J. (Informative) Examples of regions with discontinuity in geometry oraction (Detailing rules for particular situations)

Eurocode 2 - Annexes

EC2 Annex J - replaced by Annex B in PD 6687


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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 13670

Execution ofStructures

BS 4449Reinforcing

Steels

BS EN 10080Reinforcing

SteelsBS 8500SpecifyingConcrete

BS EN 206-1SpecifyingConcrete

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


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Labour and Material (Peri)

18%

24%

58%

Rationalisation of Reinforcement

Optimum cost depends

on:

Material cost

Labour

Plant

Preliminaries

Finance

Team decision required


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Detailing

Reinforcement


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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 yieldstrength fyk or f0,2k(MPa)

400 to 600

k= (ft/fy)k 1,05 1,08 1,15


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Extract BS 8666

UK CARES (Certification - Product & Companies)

1. Reinforcing bar and coil2. Reinforcing fabric3. Steel wire for direct use of for further

processing4. Cut and bent reinforcement

5. Welding and prefabrication of reinforcingsteel

www.ukcares.co.uk

www.uk-bar.org


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www.ukcares.co.ukwww.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


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Table power bender


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High

Medium

Low

Potential Risk factor

Smaller diameter bars causeless of a problem as theycan often be produced on

an automatic link bending

machine. Larger diameterbars have to be produced ona manual power bender withthe potential to trap the

operators fingers. Try toavoid/minimise the use ofshapes which cause a scissoraction, especially with

larger diameter bars.

Boot Link.Greater risk than shape code 51 as thebars have to cross over twice to

achieve the shape.

Health and safety risk becomes higherwith larger diameter bar.Also the risk increases with smalldimensions.

See Note SN2.When bent on an automatic link benderwith small diameter bars the risk isrelatively 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. Ifthe operator has concernswhen producing this shapehe should consult hissupervisor.

This shape is designed forproducing small to medium

sized links in small diameterbar.Do not detail this shape inlarge diameter bar, try touse an alternative (eg. 2 no.

shape code 13s facing eachother to create a s hapecode 33).See Note SN2.

Sausage Link.Health and safety risk is high with

larger diameter bar.Also the risk increases with smalldimensions.When bent on an automatic link benderwith small diameter bars the risk is

relatively low. When bending on amanual bender the risk is high,especially with larger diameters andnon 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


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Tolerances (not in EC2BS8666)

For bars: Bar diameter

For post-tensioned tendons:

Circular ducts: Duct diameter

Rectangular ducts: The greater of:the smaller dimension orhalf 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


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a AxisDistance

Reinforcement cover

Axis distance, a, to centre of bar

a = c + m/2 + l

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 concretecover,

10 mm cdev 5 mm

where very accurate measurements are taken and non conformingmembers are rejected (eg precast elements)

10 mm cdev

0 mm

Allowance in Design for

Deviation


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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


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Practicalities

12m maximum length H20 to H40

(12m H40 = 18 stone/ 118Kg)

Health & safety

9m maximum length H16 & H12

6m maximum length H10 & H8

TransportFixing

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,mins = kckfct,effAct this is the same as

Crack width limits (Cl. 7.3.1 and National Annex). Theselimits 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


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Minimum Reinforcement Area

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

db0013.0f

dbf26.0A t

yk

tctmmin,s

EC2: Cl. 9.2.1.1, Eq 9.1N

Crack Control Without Direct

CalculationProvide 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


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Clear horizontal and vertical distance , (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 12fctdwherefctd should be limited to C60/75

1 =1 for good and 0.7 for poor bond conditions2 = 1 for 32, otherwise (132-)/100

a) 45 90 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


250


Ultimate bond stress

EC2: Cl. 8.4.2


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lb,rqd = ( / 4) (sd /fbd)

where sd is the design stress of the bar at the positionfrom where the anchorage is measured.

Basic required anchorage length

EC2: Cl. 8.4.3

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

lbd = 1 2 3 4 5 lb,rqd lb,min

However:

(2 3 5) 0.7

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

Design Anchorage Length, lbd

EC2: Cl. 8.4.4


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Alpha values

EC2: Table 8.2

Table 8.2 - Cd & K factors

EC2: Figure 8.3

EC2: Figure 8.4


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Anchorage of linksEC2: Cl. 8.5

l0 = 1 2 3 5 6 lb,rqd l0,min

6 = (1/25)0,5 but between 1.0 and 1.5

where1 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%

6

1 1.15 1.4 1.5

Note: Intermediate values may be determined by interpolation.

1 2 3 5 are as defined for anchorage length

l0,min max{0.3 6 lb,rqd; 15; 200}

Design Lap Length, l0 (8.7.3)

EC2: Cl. 8.7.3


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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


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Bond stress, fbdfbd

= 2.25 1

2

fctd

EC2 Equ. 8.2

1 = 1.0 Good bond conditions

2 = 1.0 bar size 32

fctd = ct fctk,0,05/c EC2 cl 3.1.6(2), Equ 3.16

ct = 1.0 c = 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 = ct fctk,0,05/c = 1.8/1.5 = 1.2

fbd = 2.25 x 1.2 = 2.7 MPa

Basic anchorage length, lb,req

lb.req = (/4) ( sd/fbd) EC2 Equ 8.3

Max stress in the bar, sd = fyk/s = 500/1.15

= 435MPa.

lb.req = (/4) ( 435/2.7)

= 40.3

For concrete class C25/30


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Design anchorage length, lbd

lbd = 123 4 5 lb.req lb,min

lbd = 123 4 5 (40.3) For concrete class C25/30

Alpha valuesEC2: Table 8.2 Concise: 11.4.2


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

EC2: Figure 8.4

Design anchorage length, lbdlbd = 1 2 3 45 lb.req lb,min

lbd = 1 2 3 45 (40.3) For concrete class C25/30

a) Tension anchorage straight bar

1 = 1.0

3 = 1.0 conservative value with K= 0

4 = 1.0 N/A

5

= 1.0 conservative value

2 = 1.0 0.15 (cd )/

2 = 1.0 0.15 (25 16)/16 = 0.916

lbd = 0.916 x 40.3 = 36.9 = 590mm


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Design anchorage length, lbd

lbd = 1 2 3 45 lb.req lb,minlbd = 1 2 3 45 (40.3) For concrete class C25/30

b) Tension anchorage Other shape bars

1 = 1.0 cd = 25 is 3 = 3 x 16 = 48

3 = 1.0 conservative value with K= 0

4 = 1.0 N/A

5 = 1.0 conservative value

2 = 1.0 0.15 (cd 3)/ 1.0

2 = 1.0 0.15 (25 48)/16 = 1.25 1.0

lbd = 1.0 x 40.3 = 40.3 = 645mm

Worked example - summary

H16 Bars Concrete class C25/30 25 Nominal cover

Tension anchorage straight bar lbd = 36.9 = 590mm

Tension anchorage Other shape bars lbd = 40.3 = 645mm

lbd is measured along the centreline of the bar

Compression anchorage (1 = 2 = 3 = 4 = 5 = 1.0)

lbd = 40.3 = 645mm

Anchorage for Poor bond conditions = Good/0.7

Lap length = anchorage length x 6


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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 maximumof 50% can be lapped


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Arrangement of LapsEC2: Cl. 8.7.3, Fig 8.8

Anchorage of bars

F

Transverse Reinforcement

There is transverse tension reinforcement required


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F/2 F/2

F tan

F tan

F F

Lapping of bars


There is transverse tension reinforcement required

Where the diameter, , of the lapped bars 20 mm, the transversereinforcement should have a total area, Ast 1,0As of one spliced bar. Itshould 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 transverse bars shouldbe 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 /30A /2

st

A /2st

l /30FsFs

150 mm

l0

Transverse Reinforcement at Laps

Bars in tensionEC2: Cl. 8.7.4, Fig 8.9 only if bar 20mm or laps > 25%


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

As,max = 0,04Ac

Section at supports should be designed for a

hogging moment 0,25 max. span moment

Any design compression reinforcement () should be

held by transverse reinforcement with spacing 15

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


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Shear Design: Links

Variable strut methodallows a shallower strut angle

hence activating more links.

As strut angle reduces concrete stress increases

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

d

V

z

x

d

x

V

z

s

EC2: Cl. 6.2.3

Where av 2dthe 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 isincluded in the resistance.

Short Shear Spans with Direct

Strut ActionEC2: Cl. 6.2.3 (8)

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


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Shear reinforcement

Minimum shear reinforcement, w,min = (0,08fck)/fyk

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

Maximum transverse spacing,s

t,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

z

s

EC2: Cl. 6.2.3


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

alFtd

al

Envelope of (MEd/z+NEd)

Acting tensile force

Resisting tensile force

lbd

lbd

lbd

lbd

lbd lbd

lbd

lbdFtd

Shift rule

Curtailment of reinforcement

EC2: Cl. 9.2.1.3, Fig 9.2

For members with shear reinforcement: al = 0.5 zCot But 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 0.25As 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 SupportsEC2: Cl. 9.2.1.4


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Simplified Detailing Rules forBeams

h /31

h /21

B

A

h /32h /22

supporting beam with height h1

supported beam with height h2 (h1 h2)

The supporting reinforcement is in

addition to that required for otherreasons

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


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Curtailment as beams except for the Shift rule al = dmay 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 spanEnd supports: As,top 0,15As for Mmax in adjacent span

This top reinforcement should extend 0,2 adjacent span

Solid slabsEC2: Cl. 9.3


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Distribution of momentsEC2: Table I.1

Particular rules for flat slabs

Arrangement of reinforcement should reflect behaviourunder 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 internalcolumns in each orthogonal directions.

Particular rules for flat slabs

EC2: Cl. 9.4


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h 4b

min 12

As,min = 0,10NEd/fyd but 0,002 Ac

As,max= 0.04Ac (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 min; 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 > 14.

A min of 3 bars is required in lap length

scl,tmax = min {12 min; 0.6b ; 240mm}

ColumnsEC2: Cl. 9.5.3


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Walls

A

s,vmin = 0,002A

c (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 16mm with cover at least 2.]

Detailing Comparisons

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

9.2.1.2 (3) or 15 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

LinksTable 3.28Table 7.3NSmax

dg + 5 mm or 8.2 (2): dg + 5 mm or 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): 0.26fctm/fykbd 0.0013 bd

As,min

ValuesClause / ValuesMain Bars in Tension

BS 8110EC2Beams


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places of maximum moment:

main: 2h 250 mm

secondary: 3h 400 mm

3dor 750 mmsecondary: 3.5h 450 mmSmax

dg + 5 mm or 8.2 (2): dg + 5 mm or 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): 0.26fctm/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

129.5.3 (3): min (12min; 0.6 b;240 mm)Scl,tmax

0.25 or 6 mm9.5.3 (1) 0.25 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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