This document provides information on reinforcement detailing according to Eurocode 2 (EC2). It begins with an overview of the structural Eurocodes and the contents of EC2. Key topics covered in more detail include reinforcement properties, minimum cover requirements, crack control, bar spacing, bond stress calculations, and the design of anchorage and lap lengths. Worked examples are provided to demonstrate how to calculate the design anchorage length for tension reinforcement according to the equations and factors specified in EC2. In summary, the document outlines the main requirements for reinforcement detailing in concrete structures as defined by EC2.

1. 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

2. • 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

3. 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
BS EN 13670
Execution of
Structures
BS 4449
Reinforcing
Steels
BS EN 10080
Reinforcing
SteelsBS 8500
Specifying
Concrete
BS EN 206-1
Specifying
Concrete
NSCS
BS 8666
Reinforcement
Scheduling
National Annex
PD 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

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

5. Detailing
Reinforcement

6. 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)

7. 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

8. 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

9. Table power bender

10. 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

11. 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
Minimum Cover for Bond

12. 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 Design
BS 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

13. Nominal cover, cnom
Minimum cover, cmin
cmin = max {cmin,b; cmin,dur ; 10 mm}
Axis distance, a
Fire 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

14. 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

15. Minimum Reinforcement Area
The minimum area of reinforcement for slabs (and beams) is given by:
db0013.0
f
dbf26.0
A 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

16. • 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 bars
EC2: Cl. 8.2
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
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
EC2: Cl. 8.4.2

17. lb,rqd = (φφφφ / 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

18. Alpha values
EC2: Table 8.2
Table 8.2 - Cd & K factors
EC2: Figure 8.3
EC2: Figure 8.4

19. 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

20. 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

21. 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 EC2 cl 3.1.6(2), Equ 3.16
αct = 1.0 γc = 1.5
fctk,0,05 = 0.7 x 0.3 fck
2/3 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

22. 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

23. Table 8.2 - Cd & K factors
Concise: Figure 11.3EC2: Figure 8.3
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 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

24. 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 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

25. 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

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

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

28. • 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 5.3.2.1)
Beams
EC2: Cl. 9.2

29. 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 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.

30. 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
θ
z
s
EC2: Cl. 6.2.3

31. • For members without shear reinforcement this is satisfied with al = 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
EC2: Cl. 9.2.1.3, Fig 9.2
• For members with shear reinforcement: al = 0.5 z Cot θ
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 ≥ 0.25 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

32. 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

33. • 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
Solid slabs
EC2: Cl. 9.3

34. 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

35. • 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

36. 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
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
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

37. Detailing Comparisons
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
Detailing Comparisons
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

38. How to…Compendium
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