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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
Direction of concreting
b) h 250 mm d) h > 600 mm
unhatched zone good bond conditions
hatched zone - poor bond conditions
Direction of concreting
250
Direction of concreting
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
Transverse Reinforcement
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
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 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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Detailing Comparisons
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
Detailing Comparisons
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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How toCompendium
Detailing
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