Introduction to Eurocodes - The Concrete Centre
-
Upload
vladstoica -
Category
Documents
-
view
342 -
download
6
Transcript of Introduction to Eurocodes - The Concrete Centre
Introduction to Eurocodes
Owen BrookerSenior Structural Engineer
• Introduction to the Eurocodes
• Eurocode
• Eurocode 1
• Eurocode 2· Materials· Cover· Flexure· Shear· Deflection· Axial
• Further Information
• Set of ten European standards containing rules for design and construction of structures in all materials
• Codes are integrated i.e. no repetition of rules
• Codes are managed by CEN (European Committee for standardization)
• Will be used in 28 European countries
• Will supersede British Standards, which will be withdrawn in 2010 at the latest
• Some supporting Codes are already in use• BS EN 206-1/BS 8500• BS 4449 / BS 8666
What are the Eurocodes?
• 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 1997 (EC7) : Geotechnical design
• BS EN 1998 (EC8) : Design of structures for earthquake resistance
• BS EN 1999 (EC9) : Design of aluminium structures
The Eurocodes
• 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 1997 (EC7): Geotechnical design• BS EN 1998 (EC8): Design of structures for earthquake resistance
• BS EN 1999 (EC9): Design of aluminium structures
The Eurocodes
Each Eurocode Contains:
a. National front cover
b. National forward
c. CEN front cover
d. Main text and annexes (which must be as produced by CEN)
e. Annexes - can by normative and/or informative
f. National Annex (NA).
Format of the Eurocodes
National Annex
The National Annex provides:
• Values of Nationally Determined Parameters (NDPs) (NDPs have been allowed for reasons of safety, economy and durability)
• The decision where main text allows alternatives
• The choice to adopt informative annexes
• Non-contradictory complementary information (NCCI)
• The Eurocodes contain Principles (P) which comprise:
◦ General statements and definitions for which there is no alternative, as well as:
◦ Requirements and analytical models for which no alternative is permitted
• They also contain Application Rules, which are generally rules which comply with the Principles
• The Eurocodes also use a comma (,) as the decimal marker
Features of the Eurocodes
EN 1990Basis of Design
EN 1991Actions on Structures
EN 1992 ConcreteEN 1993 SteelEN 1994 CompositeEN 1995 TimberEN 1996 MasonryEN 1999 Aluminium
EN 1997Geotechnical
Design
EN 1998Seismic Design
Structural safety, serviceability and durability
Design and detailing
Geotechnical & seismic design
Actions on structures
Eurocode Hierarchy
• Introduction to the Eurocodes
• Eurocode
• Eurocode 1
• Eurocode 2· Materials· Cover· Flexure· Shear· Deflection· Axial
• Further Information
Published 27 July 2002
Structures are to be designed, executed and maintained so that, with appropriate forms of reliability, they will:
• Perform adequately under all expected actions
• Withstand all actions and other influences likely to occur during construction and use
• Have adequate durability in relation to the cost
• Not be damaged disproportionately by exceptional hazards
Eurocode
The code sets out the following:
• Basis for calculating design resistance of materials
• Combinations for ultimate limit state
Either:
Σ γG, j⋅Gk,j + γQ,1 Qk,1 + ΣγQ,I⋅ψ0,I⋅Qk,I Exp. (6.10)
Or the more adverse of:
Σ γG, j Gk,j + γQ,1ψ0,1 Qk,1 + ΣγQ,I ψ0,I Qk,I Exp. (6.10 a)
Σ ξ⋅ γG, j Gk,j + γQ,1 Qk,1 + Σ γQ,I ψ0,I Qk,I Exp. (6.10 b)
ξ = 0.925 (UK NA)
Eurocode
For one variable action: 1.25 Gk + 1.5 QkProvided:
1. Permanent actions < 4.5 x variable actions2. Excludes storage loads
Eurocode
Design values of actions, ultimate limit state – persistent and transient design situations (Table A1.2(B) Eurocode)
Permanent actions Accompanying variable actions
Unfavourable Favourable Main(if any) Others
Eqn (6.10) γG,j,sup Gk,j,sup γG,j,inf Gk,j,inf γQ,1 Qk,1 γQ,i Ψ0,i Qk,i
Eqn (6.10a) γG,j,sup Gk,j,sup γG,j,inf Gk,j,inf γQ,1Ψ0,1Qk,1 γQ,i Ψ0,i Qk,i
Eqn (6.10b) ξ γG,j,supGk,j,sup γG,j,inf Gk,j,inf γQ,1 Qk,1 γQ,i Ψ0,i Qk,i
Leading variable action
Comb’tionexpression reference
1.5 Ψ0,i Qk,i1.5 Qk,11.0 Gk0.925x1.35GkEqn (6.10b)
1.5 Ψ0,i Qk,i1.5 Ψ0,1 Qk1.0 Gk1.35 GkEqn (6.10a)
1.5 Ψ0,i Qk,i1.5 Qk,11.0 Gk1.35 GkEqn (6.10)
Eurocode
Serviceability Limit States
Characteristic combination (Normally used for irreversible limit states)
Gk,j + Qk,1 + Σψ0,I⋅Qk,I
Frequent combination (Normally used for reversible limit states)
Gk,j + ψ1,1⋅Qk,1 + Σ ψ2,I⋅Qk,I
Quasi-permanent combination (Normally used for long term effects and appearance of the structure)
Gk,j + Σ ψ2,I⋅Qk,I
Eurocode
Eurocode: Annex A
Action ψ0 ψ1 ψ2
Category A: domestic, residential areas 0.7 0.5 0.3Category B: office areas 0.7 0.5 0.3Category C: congregation areas 0.7 0.7 0.6Category D: shopping areas 0.7 0.7 0.6Category E: storage areas 1.0 0.9 0.8Category F: traffic area(vehicle weight < 30 kN)
0.7 0.7 0.6
Category G: traffic area(30 kN < vehicle weight < 160 kN)
0.7 0.5 0.3
Category H: roofs 0.7 0 0Snow (For sites located at altitude H <1000 m asl)
0.5 0.2 0
Wind loads on buildings (BS EN 1991-1-4) 0.5 0.2 0
Eurocode: Annex A
Psi (ψ) factors for Traffic loads will be in Annex A2
• Annex A2 has been drafted
• Revised National Annex due August 2006
Nomenclature
Subscript Definition
A Accidental situationc Concrete
d DesignE Effect of actionfi Firek CharacteristicR Resistancew Shear reinforcementy Yield strength
• Introduction to the Eurocodes
• Eurocode
• Eurocode 1
• Eurocode 2· Materials· Cover· Flexure· Shear· Deflection· Axial
• Further Information
Eurocode 1 has ten parts:
• 1991-1-1 Densities, self-weight and imposed loads• 1991-1-2 Actions on structures exposed to fire• 1991-1-3 Snow loads• 1991-1-4 Wind actions• 1991-1-5 Thermal actions• 1991-1-6 Actions during execution• 1991-1-7 Accidental actions due to impact and explosions• 1991-2 Traffic loads on bridges• 1991-3 Actions induced by cranes and machinery • 1991-4 Actions in silos and tanks
Eurocode 1
Eurocode 1 Part 1-1: Densities, self-weight and imposed loads
• Bulk density of reinforced concrete is 25 kN/m3
• The draft UK NA proposes the same loads as BS 6399
• Plant loading not given
Eurocode 1
• Introduction to the Eurocodes
• Eurocode
• Eurocode 1
• Eurocode 2· Materials· Cover· Flexure· Shear· Deflection· Axial
• Further Information
Eurocode 2 has 4 parts:
EN1992-1-1 Common rules for buildings and civil engineering structures
EN1992-1-2 Structural fire design
EN1992-2 Bridges
EN1992-3 Liquid retaining and containment structures
Eurocode 2
BS EN 1990 BASIS OF STRUCTURAL
DESIGN
BS EN 1991 ACTIONS ON STRUCTURES
BS EN 1992DESIGN OF CONCRETE
STRUCTURESPart 1-1: General Rules for
StructuresPart 1-2: Structural Fire Design
BS EN 1992Part 2:
Bridges
BS EN 1992Part 3:
Liquid Ret. Structures
BS EN 1994Design of
Comp. Struct.
BS EN 13369Pre-cast Concrete
BS EN 1997GEOTECHNICAL
DESIGN
BS EN 1998SEISMIC DESIGN
BS EN 13670Execution of Structures
BS 8500Specifying Concrete
BS 4449Reinforcing
Steels
BS EN 10080Reinforcing
Steels
Eurocode 2 Relationships
• Code deals with phenomenon, rather than element types
• Design is based on characteristic cylinder strength
• Does not contain derived formulae (e.g. only the details of the stress block is given, not the flexural design formulae)
• Unit of stress in MPa
• One thousandth is represented by %o
• Partial factor for steel is 1.15
• Plain or mild steel not covered
• Notional horizontal loads considered in addition to lateral loads
• High strength, up to C90/105 covered
Eurocode 2/BS 8110 Compared
Materials
Strength classes for concretefck (MPa) 12 16 20 25 30 35 40 45 50 55 60 70 80 90
fck,cube(MPa)
15 20 25 30 37 45 50 55 60 67 75 85 95 105
fcm(MPa)
20 24 28 33 38 43 48 53 58 63 68 78 88 98
fctm(MPa)
1,6 1,9 2,2 2,6 2,9 3,2 3,5 3,8 4,1 4,2 4,4 4,6 4,8 5,0
fctk,0,05(MPa)
1 1 1,3 1,5 1,8 2,0 2,2 2,5 2,7 2,9 3,0 3,1 3,2 3,4 3,5
fctk,0,95(MPa)
2,0 2,5 2,9 3,3 3,8 4,2 4,6 4,9 5,3 5,5 5,7 6,0 6,3 6,6
Ecm(Gpa)
27 29 30 31 32 34 35 36 37 38 39 41 42 44
εc1 (‰) 1,8 1,9 2,0 2,1 2,2 2,25 2,3 2,4 2,45 2,5 2,6 2,7 2,8 2,8εcu1 (‰) 3,5 3,2 3,0 2,8 2,8 2,8εc2 (‰) 2,0 2,2 2,3 2,4 2,5 2,6εcu2 (‰) 3,5 3,1 2,9 2,7 2,6 2,6
n 2,0 1,75 1,6 1,45 1,4 1,4εc3 (‰) 1,75 1,8 1,9 2,0 2,2 2,3εcu3 (‰) 3,5 3,1 2,9 2,7 2,6 2,6
Concrete properties (Table 3.1)
• BS 8500 includes C28/35 & C32/40• For shear design, max shear strength as for C50/60
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, εuk (%)
≥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
• In UK NA max. char yield strength, fyk, = 600 MPa• BS 4449 and 4483 have adopted 500 MPa
Reinforcement properties (Annex C)
Extract from BS 8666
A
B
C
Cover
BS EN 1992-1-1 & Cover
Nominal cover, cnom
Minimum cover, cmin
cmin = max {cmin,b; cmin,dur ; 10 mm}
Axis distance, aFire protection
Allowance for deviation, ∆cdev
BS EN 1992-1-1 & Cover
Minimum cover, cmin
cmin = max {cmin,b; cmin,dur ;10 mm} cmin,b = min cover due to bond (φ)
a AxisDistance
Reinforcement cover
Axis distance, a, to
centre of bar
a = c + φm/2 + φl
BS EN 1992-1-2 Structural fire design
Scope
Part 1-2 Structural fire design gives several methods for fire
engineering
Tabulated data for various elements is given in section 5
Provides design solutions fire exposure up to 4 hours
− The tables have been developed on an empirical basis confirmed by experience and theoretical evaluation of tests
− Values are given for normal weight concrete made with siliceous aggregates
− No further checks are required for shear, torsion or anchorage
− No further checks are required for spalling up to an axis distance of 70 mm
− For HSC (> C50/60) other rules apply
Section 5. Tabulated data
µfi = NEd,fi/ NRd or conservatively 0.7
Columns: Method A
Standard fire
resistanceMinimum dimensions (mm)
Possible combinations of a and bminwhere a is the average axis
distance and bmin is the width of be am
Web thickness bw
R 30
R 60
R 90
R 120
R 180
R 240
bmin= 80a = 15*
bmin= 120a = 25
bmin= 150a = 35
bmin= 200a = 45
bmin= 240a = 60
bmin= 280a = 75
16012*
20012*
25025
30035
40050
50060
45035
55050
65060
50030
60040
70050
80
100
110
130
150
170
Continuous Beams
Flexure
For grades of concrete up to C50/60, εcu= 0.0035η = 1λ = 0.8fcd = αcc fck/ γc = 0.85 fck/1.5 = 0.57 fck
fyd = fyk/1.15 = 435 MPa
Simplified Stress Block
Design flowchart
The following flowchart outlines the design procedure for rectangular beams with concrete classes up to C50/60 and class 500 reinforcement
Determine K and K’ from:
Note: δ =1.0 means no redistribution and δ = 0.8 means 20% moment redistribution.
Beam is under-reinforced - no compression steel needed
Is K ≤ K’ ?
Beam is over-reinforced -compression steel needed
No Yes
ck2 fdb
MK = 21.018.06.0'& 2 −−= δδK
Carry out analysis to determine design moments (M)
Flow chart for over-reinforced beam
Calculate lever arm Z from: [ ]'53.3112
Kdz −+=
Calculate excess moment from: ( )'22 KKfbdM ck −=
Calculate compression steel required from:
Note fsc has been introduced to limit depth to neutral axis for members with high redistribution
( )2sc
22s ddf
MA−
=
Calculate tension steel required from:yd
sc2s
yds
'ffA
zfMMA +
−=
Check max reinforcement provided As,max ≤ 0.04Ac (Cl. 9.2.1.1)Check min spacing between bars > φbar > 20 > Agg + 5
yd2sc ))/-700(( where fxdxf ≤=
Flow chart for under-reinforced beam
Calculate lever arm Z from: [ ] dKdz 95.053.3112
≤−+=
Check minimum reinforcement requirements:
dbf
dbfA tyk
tctmmin,s 013.026.0
≥≥
Check max reinforcement provided As,max ≤ 0.04Ac (Cl. 9.2.1.1)Check min reinf. provided As,min ≥ 0.26 fctm bt d /fyk ≥ 0.13% bt dCheck min spacing between bars > φbar > 20 > Agg + 5Check max spacing between bars
Calculate tension steel required from: zfMAyd
s =
Shear
Eurocode 2/BS 8110 Compared
Strut inclination method
θcotswsRd, ywdfz
sAV =
θθναtancot
1maxRd, +
= cdwcw fzbV
21.8° < θ < 45°
We can manipulate the Expression for the concrete strut:
When cot θ = 2.5 (θ = 21.8°)VRd,max = 0.138 bw z fck (1 - fck/250)
Or in terms of stress:vRd = 0.138 fck (1 - fck/250)
where vRd = VRd/(bz) = VRd/(0.9 bd)
When vRd > vEd cot θ = 2.5 (θ = 21.8°)
When vRd < vEd we can rearrange the concrete strut expression:θ = 0,5 sin-1[vRd /(0.20 fck(1 - fck/250))]
We can also manipulate the reinforcement expression to give:Asw/s = vEd bw/(fywd cot θ)
fck
vRd (whencot θ = 2.5)
20 2.5425 3.1028 3.4330 3.6432 3.8435 4.1540 4.6345 5.0850 5.51
Shear
Design flow chart for shear
Yes (cot θ = 2.5)
Determine the concrete strut capacity vRd when cot θ = 2.5vRd = 0.138fck(1-fck/250)
Calculate area of shear reinforcement:Asw/s = vEd bw/(fywd cot θ)
Determine vEd where:vEd = shear stress at d from face of support [vEd = VEd/(0.9bwd)]
Determine θ from:θ = 0.5 sin-1[(vEd/(0.20fck(1-fck/250))]Is vRD > vEd?
No
Check maximum spacing of shear reinforcement :s⎮,max = 0.75 dFor vertical shear reinforcement
Deflection
Deflection
The deflection limits are:• Span/250 under quasi-permanent loads to avoid impairment of
appearance and general utility• Span/500 after construction under the quasi-permanent loads to
avoid damage to adjacent parts of the structure.
Deflection requirements can be satisfied by the following methods:• Direct calculation (Eurocode 2 methods considered to be an
improvement on BS 8110) .• Limiting span-to-effective-depth ratios
EC2 Span/effective depth ratios
l/d is the span/depth ratioK is the factor to take into account the different structural
systemsρ0 is the reference reinforcement ratio = √fck 10-3
ρ is the required tension reinforcement ratio at mid-span to resist the moment due to the design loads (at support for cantilevers)
ρ’ is the required compression reinforcement ratio at mid-span to resist the moment due to design loads (at support for cantilevers)
⎥⎥
⎦
⎤
⎢⎢
⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛−++=
23
0ck
0ck 12,35,111
ρρ
ρρ ffK
dl if ρ ≤ ρ0 (7.16.a)
⎥⎦
⎤⎢⎣
⎡+
−+=
0ck
0ck
'121
'5,111
ρρ
ρρρ
ffKdl if ρ > ρ0 (7.16.b)
EC2 Span/effective depth ratios
EC2 Span/effective depth ratios
12
14
16
18
20
22
24
26
28
30
32
34
36
0.30% 0.80% 1.30% 1.80%Percentage of tension reinforcement (As/bd)
Span
to d
epth
ratio
(l/d
fck = 20fck = 25fck = 28fck = 30fck = 32fck = 35fck = 40fck = 45fck = 50
Flow Chart
Is basic l/d x F1 x F2 x F3 >Actual l/d?
Yes
No
Factor F3 accounts for stress in the reinforcementF3 = 310/σs
where σs is tensile stress under quasi-permanent loadNote: As,prov ≤ 1.5 As,req’d (UK NA)
Check complete
Determine basic l/d
Factor F2 for spans supporting brittle partitions > 7mF2 = 7/leff
Factor F1 for ribbed and waffle slabs onlyF1 = 1 – 0.1 ((bf/bw) – 1) ≥ 0.8
Increase As,prov
No
Axial
Second Order Effects with Axial Load
• Second order effects may be ignored if they are less than 10% of the corresponding first order effects
• Second order effects may be ignored if the slenderness, λ < λlim
• Slenderness λ = l0/i where i = √(I/A)hence for a rectangular section λ = 3.46 l0 / h
for a circular section λ = 4 l0 / h
• With biaxial bending the slenderness should be checked separately for each direction and only need be considered in the directions where λlim is exceeded
λlim = 20⋅A⋅B⋅C/√n
Slenderness Limit
where:A = 1 / (1+0,2ϕef) ϕef is the effective creep ratio;
(if ϕef is not known, A = 0,7 may be used)
B = √(1 + 2ω) ω = Asfyd / (Acfcd)(if ω is not known, B = 1,1 may be used)
C = 1.7 - rm rm = M01/M02 M01, M02 are first order end moments,
⏐M02⏐ ≥ ⏐M01⏐(if rm is not known, C = 0.7 may be used)
n = NEd / (Acfcd)
Factor C
105 kNM 105 kNM 105 kNM
-105 kNM 105 kNM
rm = M01/ M02
= 0 / 105 = 0
C = 1.7 – 0= 1.7
rm = M01/ M02
= 105 / -105 = -1
C = 1.7 + 1= 2.7
rm = M01/ M02
= 105 / 105= 1
C = 1.7 – 1= 0.7
Effective length
l
θM
θ
l0 = l l0 = 2l l0 = 0.7l l0 = l / 2 l0 = l l /2 <l0< l l0 > 2l
⎟⎟⎠
⎞⎜⎜⎝
⎛+
+⋅⎟⎟⎠
⎞⎜⎜⎝
⎛+
+2
2
1
1
45,01
45,01
kk
kk
l0 = 0,5l⋅Braced members
Unbraced⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
⎟⎟⎠
⎞⎜⎜⎝
⎛+
+⋅⎟⎟⎠
⎞⎜⎜⎝
⎛+
++⋅
⋅+ k
kk
k kkkk 2
21
1
21
21
11
11;101maxl0 = l⋅
λ = l0/i
k = (θ / M)⋅ (EΙ / l) (from Eurocode)
Effective length (2)
⎟⎟⎠
⎞⎜⎜⎝
⎛+
+⋅⎟⎟⎠
⎞⎜⎜⎝
⎛+
+2
2
1
1
45,01
45,01
kk
kk
l0 = 0,5l⋅Braced members
Unbraced⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
⎟⎟⎠
⎞⎜⎜⎝
⎛+
+⋅⎟⎟⎠
⎞⎜⎜⎝
⎛+
++⋅
⋅+ k
kk
k kkkk 2
21
1
21
21
11
11;101maxl0 = l⋅
1.02 ≥∑
=
blEl
E
kb
c
c
I
I
(From PD 6687: Background paper to UK NA)
Where:
Ib,Ic are the beam and column uncracked second moments of area
lb,lc are the beam and column lengths
Effective length (3)
lo = Fl
Design moment
The design moment MEd is as follows:
MEd = max {M02, Moe,+ M2 , Mo1,+ 0.5M2}where:
M01 = Min {|Mtop|,|Mbottom|} + ei Ned
M02 = Max {|Mtop|,|Mbottom|} + ei Ned
ei = Max {Io/400, h/30, 20}
Moe = 0.6 M02+ 0.4 M01 ≥ 0.4 M02
M2 = Ned e2
Column design
For axial load:AsN/2 = (N - αccηfckbdc) / [(σsc - σst)γc]AsN = Total area of reinforcement
required to resist axial loadwhere
N = Axial loadαcc = 0.85 η = 1 for ≤ C50/60 b = breadth of sectionσsc = stress in compression
reinforcement σst = stress in tension reinforcement dc = effective depth of concrete in
compression = lx ≤ hwhere l = 0.8 for ≤ C50/60
x = depth to neutral axis h = height of section
For momentAsM/2 = [M - αcchfckbdc(h/2 - dc/2)] / [(h/2-d2).(σsc+σst)γc]AsM = Total area of reinforcement required to resist moment
This can be solved by iterating x such that AsN = AsM or by using charts or a spreadsheet
• Introduction to the Eurocodes
• Eurocode
• Eurocode 1
• Eurocode 2· Materials· Cover· Flexure· Shear· Deflection· Axial
• Further Information
When?
Publication dateReference TitleCode NA
Eurocode Basis of Structural Design Jul 02 Dec 04
Eurocode 1 Pt 1–1 Densities, self-weight and imposed loads Jul 02 Dec 05a
Eurocode 1 Pt 1–2 Actions on structures exposed to fire Nov 02 Mar 06a
Eurocode 1 Pt 1–3 Snow loads Jul 03 Dec 05a
Eurocode 1 Pt 1–4 Wind actions Apr 05 Aug 06a
Eurocode 1 Pt 1–5 Thermal actions Mar 03 Nov 06a
Eurocode 1 Pt 1–6 Actions during execution Dec 05 Oct 06a
Eurocode 1 Pt 1–7 Accidental actions due to impact and explosions Jun 06a Mar 07a
Eurocode 1 Pt 2 Traffic loads on bridges Oct 03 Nov 06a
Eurocode 1 Pt 3 Actions induced by cranes and machinery Jun 06a Mar 07a
Eurocode 1 Pt 4 Actions in silos and tanks Feb 06a Sep 06a
Eurocode 2 Pt 1–1 General rules and rules for buildings Dec 04 Dec 05Eurocode 2 Pt 1–2 General rules- structural fire design Dec 04 Dec 05Eurocode 2 Pt 2 Bridges Dec 05 Oct 06a
Eurocode 2 Pt 3 Liquid-retaining and containment structures Apr 06a Nov 06a
Eurocode 7 Pt 1 General rules Dec 04 Oct 06a
Eurocode 7 Pt 2 Ground investigation and testing Nov 06a May 07a
a British Standard Institution planned publication date Source: BSI
Highways Agency Documents
Topic DMRB part
Proposed Completion
% Completed
Basis of Design BD15** Sep-06 80Loading /Actions BD37** Dec-06 65
Bearings BD20** May-06 50Expansion Joints BD33** May-06 20
Concrete Bridge Design BD24** Dec-06 20Composite Bridge Design BD16** Sep-06 90
Integral Bridges BA42** Jun-06 80Backfilled RW & Bridge
AbutmentsBD30** Jun-06 80
Embedded RW & Bridge Abutments
BD42** Jun-06 50
Buried Concrete Box Structures BD31** Dec-06 60
Foundations BD74** Jun-06 80
When will British Standards be withdrawn?
There will be a period of co-existence between our current codes and the Eurocodes:
• For a maximum of 3 years after the final part of a
‘package’ is made available to BSI by CEN or:
• Until 2010 at the latest.
• BS 8110 withdrawn in 2008?
The following ‘How to…’ guides are being developed:
• Introduction to Eurocodes
• Getting started
• Slabs
• Beams
• Columns
• Foundations
• Flat slabs
• Deflections
And others are being planned:Ret Walls and Detailing
How to…
Spreadsheets
Version 3 - June 2006
To EC2
June 2006
Early 2007
www.eurocode2.info
Updated Detailing Manual
April 2006
Updated Green Book
May 2006