RC Detailing to Eurocode 2

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

• 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

BS EN 1992 Design of concrete structures Part 1-1: General & buildings Part 1-2: Fire design Part 2: Bridges Part 3: Liquid retaining

Standards

Specification – NSCS, Finishes

Labour and Material (Peri)

18% 24%

58%

Rationalisation of Reinforcement

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

Detailing

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 hot rolled seismic

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. Reinforcingbar and coil 2. Reinforcing fabric

3. Steel wire for direct use of for further processing

4. Cut and bent reinforcement

5. Welding and prefabricationof reinforcing steel

www.ukcares.co.uk

www.uk-bar.org

www.ukcares.co.uk

www.uk-bar.org

A

B

C

Cut and bent – approx £550 to £650/T

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 Fabricator Designer Comment Detail SC 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

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 wire 2.5 x diameter of indented wire

a Axis Distance Reinforcement cover

Axis distance, a, to centre of bar

a = c + φφφφ_m/2 + φφφφ_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 Design

BS EN 1992-1-2

∆∆∆∆cdev: Allowance for deviation = 10mm A reduction in ∆∆∆∆c_devmay 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

Allowance in Design for

Deviation

Nominal cover, c_nom

Minimum cover, c_min

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

Axis distance, a

Fire protection

Allowance for deviation, ∆c_dev

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

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

A_s,minσ_s= k_ckf_ct,effA_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,minis provided.

Minimum Reinforcement Area

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

d b 0013 . 0 f d b f 26 . 0 A _t yk t ctm 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

• 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.

Spacing of bars

EC2: Cl. 8.2

The design value of the ultimate bond stress, fbd= 2.25 η1η2fctd

where f_ctdshould 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 Direction of concreting b) h ≤≤≤≤250 mm d) h > 600 mm unhatched zone – ‘good’ bond conditions hatched zone - ‘poor’ bond conditions

α Direction of concreting

250

Direction of concreting

Ultimate bond stress

l_b,rqd= (φ φ φ φ / 4) (σσσσ_sd / f_bd)

where σ_sdis 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,rqdshould be measured along the

centreline of the bar

lbd= α1α2α3α4α5 lb,rqd≥≥≥≥lb,min

However: (α₂α₃α₅) ≥≥≥≥0.7

lb,min> max(0.3lb,rqd; 10φφφφ, 100mm)

Design Anchorage Length, l

Alpha values

EC2: Table 8.2

Table 8.2 - C

& K factors

EC2: Figure 8.3

Anchorage of links

EC2: Cl. 8.5

l0= α1α2α3α5α6lb,rqd≥≥≥≥l0,min

α_{6 = (}ρ_1/25)0,5_{but between 1.0 and 1.5}

where ρ1is the % of reinforcement lapped within 0.65l0from the

centre of the lap

Percentage of lapped bars relative to the total cross-section area

< 25% 33% 50% >50%

α

6 1 1.15 1.4 1.5

Note: Intermediate values may be determined by interpolation. α₁α₂α₃α₅are as defined for anchorage length

l_0,min≥max{0.3α₆ l_b,rqd; 15φ; 200}

Design Lap Length, l

(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, f

f_bd= 2.25 η₁η₂f_ctd EC2 Equ. 8.2

η1= 1.0 ‘Good’ bond conditions η₂= 1.0 bar size ≤ 32 f_ctd= α_ctf_ctk,0,05/γ_c EC2 cl 3.1.6(2), Equ 3.16 αct= 1.0 γc = 1.5 f_ctk,0,05 = 0.7 x 0.3 f_ck2/3 _{EC2 Table 3.1} = 0.21 x 252/3 = 1.8 MPa f_ctd= α_ctf_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

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

Design anchorage length, l

l_bd= α₁ α₂ α₃ α₄α₅l_b.req ≥l_b,min

l_bd= α₁ α₂ α₃ α₄α₅(40.3Ø) For concrete class C25/30

Alpha values

Table 8.2 - C

& K factors

Concise: Figure 11.3

EC2: Figure 8.3

EC2: Figure 8.4

Design anchorage length, l

l_bd= α₁ α₂ α₃ α₄ α₅l_b.req ≥l_b,min

l_bd= 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 N/A

α₅= 1.0 conservative value

α₂= 1.0 – 0.15 (c_d– Ø)/Ø

α₂= 1.0 – 0.15 (25 – 16)/16 = 0.916

Design anchorage length, l

l_bd= α₁ α₂ α₃ α₄ α₅l_b.req ≥l_b,min

l_bd= α₁ α₂ α₃ α₄ α₅(40.3Ø) For concrete class C25/30

b) Tension anchorage – Other shape bars

α₁= 1.0 c_d= 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 (c_d– 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 l_bd= 36.9Ø = 590mm

Tension anchorage – Other shape bars l_bd= 40.3Ø = 645mm

l_bdis measured along the centreline of the bar

Compression anchorage (α₁ = α₂ = α₃ = α₄ = α₅ = 1.0)

l_bd= 40.3Ø = 645mm

Anchorage for ‘Poor’ bond conditions = ‘Good’/0.7

How to design concrete structures using Eurocode 2

Anchorage & lap lengths

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

Anchorage of bars

F

Transverse Reinforcement

F/2 F/2 θ F tanθ F tanθ F F Lapping of bars

Transverse Reinforcement

There is transverse tension – reinforcement required

• Where the diameter, φφφφ, 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 φφφφtransverse 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 /2_st ΣA /2st l /30 F_s Fs ≤150 mm l0

Transverse Reinforcement at Laps

Bars in tension

• A_s,min = 0,26 (f_ctm/f_yk)b_td but ≥0,0013b_td

• A_s,max= 0,04 A_c

• Section at supports should be designed for a hogging moment ≥0,25max. span moment

• Any design compression reinforcement (φ) should be held by transverse reinforcement with spacing ≤15φ

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)

Beams

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 θ z 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 a_v/2d

where 0.5 ≤a_v≤ 2d provided:

d d

a_v a_v

− The longitudinal reinforcement is fully anchored at the support.

− Only that shear reinforcement provided within the central 0.75avis included in the resistance.

Short Shear Spans with Direct

Strut Action

EC2: Cl. 6.2.3 (8)

Shear reinforcement

• Minimum shear reinforcement, ρ_w,min= (0,08√f_ck)/f_yk

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

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

EC2: Cl. 9.2.2

For vertical links s_l,max= 0,75d

Shear Design

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

∆Ftd

al

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

•For members with shear reinforcement: a_l = 0.5 z Cotθ But it is always conservative to use a_l= 1.125d

• lbdis required from the line of contact of the support.

Simple support (indirect) Simple support (direct)

• A_sbottom steel at support ≥0.25 A_sprovided in the span

• Transverse pressure may only be taken into account with a ‘direct’ support.

Shear shift rule a_l Tensile Force Envelope

Anchorage of Bottom

Reinforcement at End Supports

Simplified Detailing Rules for

Beams

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 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_sfor M_maxin adjacent span

End supports: A_s,top≥0,15A_sfor M_maxin adjacent span

• This top reinforcement should extend ≥0,2 adjacent span

Solid slabs

Distribution of moments

EC2: Table I.1

Particular rules for flat slabs

• Arrangement of reinforcement should reflect behaviour under working conditions.

• At internal columns 0.5A_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.

Particular rules for flat slabs

• 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_cat 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

s_cl,tmax

• s_cl,tmaxshould 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

Walls

• A_s,vmin= 0,002 A_c (half located at each face)

• A_s,vmax= 0.04 A_c (0,08A_cat laps) • s_vmax= 3 ×wall thickness or 400mm

Vertical Reinforcement

Horizontal Reinforcement

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

• 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. m2_{). [Not required for welded mesh or bars}

Ø≤16mm with cover at least 2Ø.]

Detailing Comparisons

d or 150 mm from main bar

9.2.2 (8): 0.75 d ≤600 mm

9.2.1.2 (3) or 15φfrom main bar

st,max 0.75d 9.2.2 (6): 0.75 d sl,max 0.4 b s/0.87 fyv 9.2.2 (5): (0.08 b s √fck)/fyk Asw,min Links Table 3.28 Table 7.3N Smax dg+ 5 mm or φ 8.2 (2): dg+ 5 mm or φor 20mm smin

Spacing of Main Bars

0.04 bh 9.2.1.1 (3): 0.04 bd As,max 0.002 bh --As,min

Main Bars in Compression

0.04 bh 9.2.1.1 (3): 0.04 bd As,max 0.0013 bh 9.2.1.1 (1): 0.26 fctm/fykbd ≥ 0.0013 bd As,min Values Clause / Values

Main Bars in Tension

BS 8110 EC2

Detailing Comparisons

places of maximum moment:

main: 2h≤250 mm secondary: 3h≤400 mm 3d or 750 mm secondary: 3.5h≤450 mm Smax dg+ 5 mm or φ 8.2 (2): dg+ 5 mm or φor 20mm 9.3.1.1 (3): main 3h≤400 mm smin Spacing of Bars 0.04 bh 9.2.1.1 (3): 0.04 bd As,max 0.002 bh

9.3.1.1 (2): 0.2Asfor single way

slabs As,min

Secondary Transverse Bars

0.04 bh 0.04 bd As,max 0.0013 bh 9.2.1.1 (1): 0.26 fctm/fykbd ≥ 0.0013 bd As,min Values Clause / Values

Main Bars in Tension

BS 8110 EC2

Slabs

Detailing Comparisons

Columns

150 mm from main bar 9.5.3 (6): 150 mm from main bar

12φ 9.5.3 (3): min (12φmin; 0.6 b;240 mm) Scl,tmax 0.25φor 6 mm 9.5.3 (1) 0.25φor 6 mm Min size Links 0.06 bh 9.5.2 (3): 0.04 bh As,max 0.004 bh 9.5.2 (2): 0.10NEd/fyk≤0.002bh As,min

Main Bars in Compression

1.5d 9.4.3 (1):

within 1st control perim.: 1.5d outside 1st control perim.: 2d

St 0.75d 9.4.3 (1): 0.75d Sr Spacing of Links Total = 0.4ud/0.87fyv 9.4.3 (2): Link leg = 0.053 srst √(fck)/fyk Asw,min Values Clause / Values Links BS 8110 EC2 Punching Shear

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