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SPATA Training 4 Oct 2012 - Eurocode 2 1
Introduction to Eurocode 2
SPATA Training4 October 2012
Charles Goodchild
BSc CEng MCIOB MIStructE
The Concrete Centre
2
•Setting the scene for the Eurocodes,
• their format,
• their hierarchy,
• how they interact.
• An overview of Eurocode 2,
• highlighting changes from and
• comparing it to BS8110
• How it all fits together.
Outline
3
Setting the scene
Eurocodes are being/ will be used in:
• EU countries
• EFTA Countries
• Malaysia
• Singapore
• Vietnam
• Sri Lanka
• Others?
CEN National Members
Austria Belgium
Cyprus Czech Republic
Denmark Estonia Finland
France Germany Greece
Hungary Iceland Ireland
Italy Latvia Lithuania
Luxembourg Malta The
Netherlands Norway
Poland Portugal Romania
Slovakia Slovenia Spain
Sweden Switzerland
United Kingdom
4
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
5
• 58 Parts to Eurocodes plus National Annexes
• Culture shock / steep learning curve
• New symbols and terminology
• Affects all materials
• Confusion over timescales
• Costs:
◦ Training
◦ Resources
Challenges of the Eurocodes
6
BS 8110 and all old structural design British Standards have now been ‘withdrawn’. There will be a period of co-existence between our current codes and the Eurocodes.
DCLG letter: “Building Control will continue to consider the appropriate use of relevant standards on a case by case basis….. [The ‘traditional’] British Standards may not necessarily be suitable ….. in the medium and long term.”
DCLG 2012 Consultation document – Eurocodes only in AD A by 2013?
Insurers? Large projects? International projects?
Scottish Technical Handbook: ‘The structural design and construction of a building should be carried out in accordance with the following Structural Eurocodes’.
Eurocodes: Timescales
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Eurocodes: Timescales
Highways:
HA IAN 124/11 July 2011
3 Implementation
“Unless otherwise agreed with HA Project Sponsors/Project
Managers and the Technical Approval Authority (TAA),
Eurocodes must be used for the design of new and
modification of existing highway structures (including
geotechnical works), . . . .”
8
• Most of Europe using the same basic design codes:◦ Increased market for UK consultants◦ Increased market for UK manufacturers◦ Reduced costs when working in several European
markets◦ Greater transferability of highly skilled staff◦ Greater understanding of research, proprietary products
etc. ◦ Reduce software development costs
• Technically advanced codes
• Logical, organised to avoid conflicts between codes
Opportunities
9
Each Eurocode Contains:
a. National front cover
(e.g. Eurocode 2)Format of the Eurocodes
10
Each Eurocode Contains:
a. National front cover
b. National forward
Format of the Eurocodes
11
Each Eurocode Contains:
a. National front cover
b. National forward
c. CEN front cover
Format of the Eurocodes
12
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)
Format of the Eurocodes
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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
Format of the Eurocodes
National Annex(NA).
Format of the Eurocodes
15
• Values of Nationally Determined Parameters (NDPs)
(NDPs have been allowed for reasons of safety, economy and durability)
• Example: Min diameter for longitudinal steel in columns
min = 8 mm in text min = 12 mm in N.A.
• The decision where main text allows alternatives
• Example: Load arrangements in Cl. 5.1.3 (1) P
• The choice to adopt informative annexes
• Example: Annexes E [Strength class for durability] and J [particular detailing rules] are not used in the UK
• Non-contradictory complementary information (NCCI)
• TR 43: Post-tensioned concrete floors – design handbook
The National Annex provides:
16
+ PDs
+ NA + NA
+ NAs
+ NA
+ NAEN 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
These
affect
concrete
design
17
• 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
EurocodeBasis of structural design
EN 1990 provides comprehensive information and guidance for all the Eurocodes, on the principles and requirements for safety and serviceability.
It gives the safety factors for actions and combinations of action for the verification of both ultimate andserviceability limit states.
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Published 27 July 2002
Says that 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: BS EN 1990 (EC0):Basis of design
Eurocode – EC0Representative value of an action
Design value of an action = Fd
= F Frep
= F ( FK )where
FK = the characteristic value of actionFrep = FK - is the representative value = Four values, namely, 1.0 or 0 or 1 or 2
Qk = Characteristic Value (of a variable action)0 Qk = Combination Value1 Qk = Frequent Value2 Qk =Quasi-permanent Value
Greek Alphabet
The ULS is divided into the following categories:
EQU Loss of equilibrium of the structure.Ed,dst ≤ Ed,stb
STR Internal failure or excessive deformation of thestructure or structural member.
Ed Rd;
GEO Failure due to excessive deformation of the ground.
FAT Fatigue failure of the structure or structural members.
Eurocode – EC0Ultimate Limit State – Categories
23
Generally for one variable action: 1.25 Gk + 1.5 Qk
Provided:1. Permanent actions < 4.5 x variable actions2. Excludes storage loads
Eurocode: ULS Actions
Design values of actions, ultimate limit state – persistent and transient design situations (Table A1.2(B) Eurocode)
Comb’tion
expression
reference
Permanent actions Leading
variable
action
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
Eqn (6.10) 1.35 Gk 1.0 Gk 1.5 Qk,1 1.5 Ψ0,i Qk,i
Eqn (6.10a) 1.35 Gk 1.0 Gk 1.5 Ψ0,1 Qk 1.5 Ψ0,i Qk,i
Eqn (6.10b) 0.925x1.35Gk 1.0 Gk 1.5 Qk,1 1.5 Ψ0,i Qk,i
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Load arrangements to EC2
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Load arrangements to EC2alternative to UK NA
26
Characteristic combination (Normally used for irreversible limit states)
Gk,j + Qk,1 + 0,IQk,I
Frequent combination (Normally used for reversible limit states)
Gk,j + 1,1Qk,1 + 2,IQk,I
Quasi-permanent combination (Normally used for long term effects and appearance of the structure)
Gk,j + 2,IQk,I
Eurocode: SLS Actions
27
EurocodeEurocode: SLS Actions -
28
Eurocode: Annex A
Action 0 1 2
Category A: domestic, residential areas 0.7 0.5 0.3
Category B: office areas 0.7 0.5 0.3
Category C: congregation areas 0.7 0.7 0.6
Category D: shopping areas 0.7 0.7 0.6
Category E: storage areas 1.0 0.9 0.8
Category 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 0
Snow (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
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• 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
30
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: Actions
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Eurocode 1 Part 1-1: Densities, self-weight and imposed loads
• Bulk density of reinforced concrete is 25 kN/m3
• The UK NA uses the same loads as BS 6399
• Plant loading not given
Eurocode 1
32
• 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
33
Date UK CEB/fib Eurocode 2
1968 CP114 (CP110 draft) Blue Book (Limit state design)
1972 CP110 (Limit state design) Red Book
1975 Treaty of Rome
1978 Model code
1985 BS8110 Eurocode 2 (EC)
1990 Model Code
1993 EC2: Part 1-1(ENV) (CEN)
2004 EC2: Part 1-1 (EN)
2005 UK Nat. Annex.
2006 BS110/EC2 PD 6687
2010 EC2 Model Code 2010
Eurocode 2 is more extensive than old codes
Eurocode 2 is less restrictive than old codes
Eurocode 2 can give more economic structures [?]
Eurocode 2: Context
34
• Code deals with phenomenon, rather than element types so Bending, Shear, Torsion, Punching, Crack control, Deflection control (not beams, slabs, columns)
• Design is based on characteristic cylinder strength
• No derived formulae (e.g. only the details of the stress block is given, not the flexural design formulae)
• No ‘tips’ (e.g. concentrated loads, column loads, )
• Unit of stress in MPa
• Plain or mild steel not covered
• Notional horizontal loads considered in addition to lateral loads
• High strength, up to C90/105 covered
• No materials and workmanship
• Part of the Eurocode system
Eurocode 2 & BS 8110 Compared
35
Concrete properties (Table 3.1)
• BS 8500 includes C28/35 & C32/40
• For shear design, max shear strength as for C50/60
Strength classes for concrete
fck (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
Ecm (GPa) 27 29 30 31 33 34 35 36 37 38 39 41 42 44
fck = Concrete cylinder strength fck,cube = Concrete cube strength fcm = Mean concrete strength fctm = Mean concrete tensile strength Ecm = Mean value of elastic modulus
Eurocode 2
36
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, uk (%)
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
• In UK NA max. char yield strength, fyk, = 600 MPa• BS 4449 and 4483 have adopted 500 MPa
Reinforcement properties (Annex C)
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Extract BS 8666
38
Nominal cover, cnom
Minimum cover, cmin
cmin = max {cmin,dur; cmin,b ; 10 mm}
Axis distance, aFire protection
Allowance for deviation, ∆cdev
bond ≡durability as per BS 8500
10 mm
Tables in Section 5 of part 1-2
Eurocode 2 - Cover
39
BS EN 1992-1-1 & Cover
Minimum cover, cmin = max {cmin,b; cmin,dur ;10 mm}
cmin,b = min cover due to bond (= )
cmin,dur = min cover due to exposure – see BS 8500 Tables A3, A4, A5 etc
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
BS EN 1992-1-2 Structural Fire Design
EC2 - Cover
41
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
Part 1-2 Fire: Section 5.
42fi = NEd,fi/ NRd or conservatively 0.7
Part 1-2 Fire Section 5. Tabulated data
Columns: Method A
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Standard fire
resistanceMinimum dimensions (mm)
Possible combinations of a and bmin
where 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
Part 1-2 Fire Section 5. Tabulated data
Continuous Beams
44
For grades of concrete up to C50/60,
εcu= 0.0035; = 1 ; = 0.8 ;
fcd = cc fck/ c = 0.85 fck/1.5 = 0.57 fck fyd = fyk/1.15 = 435 MPa
Derived formulae include:
z/d = (1 + (1 + 3.529K)0.5] / 2 (where K = M/bd2fck)
As = MEd/(1.15 fykz )
K’ = 0.207 ( = 1. But UK best practice limits x/d to 0.45 max
which in turn limits K’ to 0.167)
Eurocode 2 - Flexure
The following flowchart outlines the design procedure for rectangularbeams with concrete classes up to C50/60 and grade 500 reinforcement
Determine K and K’ from:
Note: =1.0 means no redistribution and = 0.8 means 20% moment redistribution.
Beam doubly reinforced –compression steel needed
Is K ≤ K’ ?
Beam singly reinforced
Yes No
ck
2 fdbM
K 21.018.06.0'& 2 K
Carry out analysis to determine design moments (M)
It is often recommended in the UK that K’ is limited to 0.168 to ensure ductile failure
K’
1.00 0.208
0.95 0.195
0.90 0.182
0.85 0.168
0.80 0.153
0.75 0.137
0.70 0.120
EC2 - FlexureDesign Flowchart
Calculate lever arm z from:
* A limit of 0.95d is considered good practice, it is not a requirement of Eurocode 2.
*95.053.3112
dKd
z
Check minimum reinforcement requirements:
dbf
dbfA t
yk
tctmmin,s 0013.0
26.0
Check max reinforcement provided As,max 0.04Ac (Cl. 9.2.1.1)
Check min spacing between bars > bar > 20 > Agg + 5Check max spacing between bars
Calculate tension steel required from:zf
MA
yd
s
EC2 - FlexureFlow Chart for singly reinforced section
*.. dKd
z 950533112
EC2 - Flexureessential design by hand
435 MPa = 500/1.15 =
where K = M/bd2fck
z = d x z/d
As = MEd/fydz
Check min reinforcement provided As,min > 0.26(fctm/fyk)btd (Cl. 9.2.1.1)
Check max reinforcement provided As,max 0.04Ac (Cl. 9.2.1.1)Check min spacing between bars > bar > 20 > Agg + 5
Check max spacing between bars
48
Strut inclination method
cotswsRd, ywdfz
s
AV
21.8 < < 45
Eurocode 2 – Beam shear
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Eurocode 2 vs BS8110: Shear
Shear reinforcement
density
Asfyd/s
Shear Strength, VR
BS8110: VR = VC + VS
Test results VR
Eurocode 2:
VRmax
Minimum links
Less links!(but more critical)
Safer!
EC2 - ShearDesign Flow Chart for Shear
Yes (cot = 2.5)
Determine the concrete strut capacity vRd when cot = 2.5vRd = 0.138fck(1-fck/250)
Calculate area of shear reinforcement:Asw/s = vEd bw/(fywd cot )
Determine vEd where:vEd = design shear stress [vEd = VEd/(bwz) = VEd/(bw 0.9d)]
Determine from: = 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
51
We can manipulate the Expressions for concrete struts so that
when
vEd < vRd,cot =2.5,
then
cot = 2.5 ( = 21.8°)
and
Asw/s = vEd bw/(fywd.2.5)
fck
MPa
vRd cot = 2.5
MPa
20 2.5425 3.1028 3.4330 3.6432 3.8435 4.1540 4.6345 5.0850 5.51
ShearEurocode 2 – Beam shearessential design by hand
52
The deflection limits stated to be:
• 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
Eurocode 2 – Deflection
53
Is basic l/d x F1 x F2 x F3 >Actual l/d?
Yes
No
Factor F3 accounts for stress in the reinforcementF3 = 310/s ≤ 1.5
where s is tensile stress under characteristic load orAs,prov/As,req’d
Check complete
Determine basic l/d including K for structural system
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
or fck
No
Eurocode 2 – Flow chart for L/d
54
Basic span/effective depth ratios
20.5
Percentage of tension reinforcement (As,req’d/bd)
Sp
an t
o d
epth
rat
io (
l/d)
Structural system
K
Simply supported
1.0
End span 1.3
Internal span 1.5
Flat slab 1.2
Cantilever 0.4
fck = 30,
= 0.50%
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EC2 Columns: Design moments
1st order moments:
M01 = Min {|Mtop|,|Mbottom|} + ei Ned
M02 = Max {|Mtop|,|Mbottom|} + ei Ned
where
ei = Max {Io/400, h/30, 20}
(20 mm usually critical)
For stocky columns:
Design moment, MEd = M02
56
For Slender columns,
MEd = Max[M02, M 0e +M2,M01 + M 2/2]
Where
M2 = nominal 2nd order momentM2 = NEd e2 where e2 = fn(deflection)
There are alternative methods for calculating
eccentricity, e2, for slender columns
Actions
Effective length, l0
First order moments
Slenderness,
Slenderness limit, lim
Is lim?Yes
No
Design Moments MEd
Slen-
der
Calculate As
Detailing
M0e M0e + M2
EC2 Columns: Slenderness (7)& 2nd order moments
57
Slenderness = l0/i
where
l0 = Effective length,
= Fl
. . . . . of which more later (or use BS8110 factors!}
Actions
Effective length, l0
First order moments
Slenderness,
Slenderness limit, lim
Is lim?Yes
No
Design Moments, MEd
Slen-
der
Calculate As
Detailing
EC2 Columns: Slenderness & 2nd order moments: Slenderness
i = radius of gyration= (I/A)
For a rectangular section, = 3.46 l0 / h
For a circular section, = 4 l0 / h
58
Actions
Effective length, l0
First order moments
Slenderness,
Slenderness limit, lim
Is lim?Yes
No
Design Moments, MEd
Slen-
der
Calculate As
Detailing
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
k
k
k
kF = 0,5
Braced members:
Unbraced members:
k
k
k
k
kk
kk 2
21
1
21
21
11
11;101maxF =
M
EC2 Columns: Slenderness (2)& 2nd order moments: Effective length & F
F
59
1.02
bl
El
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
k = relative stiffness= ( / M) (E / l)
Actions
Effective length, l0
First order moments
Slenderness,
Slenderness limit, lim
Is lim?Yes
No
Design Moments, MEd
Slen-
der
Calculate As
Detailing
EC2 Columns: Slenderness (3)& 2nd order moments: Effective length & F
F: working out k (each end)
(From Eurocode 2)
Alternatively...
60
Slenderness = l0/i
Actions
Effective length, l0
First order moments
Slenderness,
Slenderness limit, lim
Is lim?Yes
No
Design Moments, MEd
Slen-
der
Calculate As
Detailing
EC2 Columns: Slenderness (4) & 2nd order moments: Effective length : F from k
1.02
bl
El
E
kb
c
c
I
I
ki = relative stiffnesseach end
F
l0 = Fl
And
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Allowable Slenderness
lim = 20ABC/n
where:A = 1 / (1+0,2ef)
ef is the effective creep ratio;
(if ef is not known, A = 0,7 may be used)
B = (1 + 2) = Asfyd / (Acfcd)
(if 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)
Actions
Effective length, l0
First order moments
Slenderness,
Slenderness limit, lim
Is lim?Yes
No
Design Moments, MEd
Slen-
der
Calculate As
Detailing
EC2 Columns: Slenderness (5)& 2nd order moments: Allowable Slenderness
62
Actions
Effective length, l0
First order moments
Slenderness,
Slenderness limit, lim
Is lim?Yes
No
Design Moments, MEd
Slen-
der
Calculate As
Detailing
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
lim = 20ABC/n
EC2 Columns: Slenderness (6)& 2nd order moments: Allowable Slenderness & C
63
If Slenderness > Allowable slendernessThen include nominal 2nd order moment, M2
M2 = NEd e2 where e2 = fn(deflection)There are alternative methods for calculating
eccentricity, e2, for slender columns
Actions
Effective length, l0
First order moments
Slenderness,
Slenderness limit, lim
Is lim?Yes
No
Design Moments MEd
Slen-
der
Calculate As
Detailing
M0e M0e + M2
EC2 Columns: Slenderness (7)& 2nd order moments
64
Eurocode 2: Column design
So we haveNEd and MEd !!!!
If using column charts we want:NEd/bhfck and MEd/bh2fck
from which we get: Asfyk/bhfck
65
Eurocode 2: Column design
Asfyk/bhfck = 1 ≡ As/bd = 6%
for C30/37 concrete and B500 steel
The design value of the ultimate bond stress, fbd = 2.25 12fctdwhere fctd should be limited to C60/75
1 =1 for ‘good’ and 0.7 for ‘poor’ bond conditions2 = 1 for 32, otherwise (132- )/100
a) 45º 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 conditionshatched zone - ‘poor’ bond conditions
Direction of concreting
250
Direction of concreting
EC2 – Detailing: Ultimate bond stress
08/10/2012
SPATA Training 4 Oct 2012 - Eurocode 2 12
lbd = α1 α2 α3 α4 α5 lb,rqd lb,min
However:
(α2 α3 α5) 0.7
lb,min > max(0.3lb; 15, 100mm)
EC2 – Detailing:
Design Anchorage Length, lbd
EC2 – Detailing: Alpha values
• For members without shear reinforcement this is satisfied with al = d
a l
Ftd
a l
Envelope of (M Ed /z +NEd)
Acting tensile force
Resisting tensile force
lbd
lbd
lbd
lbd
lbd lbd
lbd
lbd
Ftd
“Shift rule”
• For members with shear reinforcement: al = (MEd/z) + 0.5VEd Cot
But it is always conservative to use al = 1.125d
EC2 – Detailing Curtailment of reinforcement
70
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 –
BS EN 206Concrete
NSCS
DMRB?
NBS?
Rail?
CESWI?
BS EN 10138Prestressing
Steels
71
Specifications
BS EN 13670
72
BS EN 13670 & NSCS
New Types of Finish
Hierarchy of Tolerances
Includes NA
Types of Finish as BS EN 13670
Hierarchy of Tolerances
Green Issues
BS EN 13670
08/10/2012
SPATA Training 4 Oct 2012 - Eurocode 2 13
73
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 –
BS EN 206Concrete
NSCS
DMRB?
NBS?
Rail?
CESWI?
BS EN 10138Prestressing
Steels
74
Eurocode 2 & the UK – what does it mean?
A paper by Moss and Webster (BS8110 vs EC2, TSE 16/03/04)
concluded:·
• big impact
• learning curve
• not wildly different from BS8110 in terms of the design approach.
• similar answers
• marginally more economic.
• less prescriptive and more extensive than BS8110
• gives designers the opportunity to derive benefit from the
considerable advances in concrete technology over recent years
• believe that after an initial acclimatisation period, EC2 will be
generally regarded as a very good code.
75
Flat slabs: Economic depths
150
200
250
300
350
400
450
500
4.0 5.0 6.0 7.0 8.0 9.0 10.0 11.0 12.0
SPAN, m
SLAB D
EPT
H,
mm
IL = 5 kN/m2
To BS8110 incl 1.5 SDL IL = 2.5 kN/m2
To BS8110 incl 1.5 SDL
To EC2
EC2: up to 25 mm shallower @ 9 m
EC2: up to 15 mm shallower @ 6 m
Rev’d 12 May 10
To BS8110
5 to 7 % savings?
76
Concise Eurocode 2RC Spreadsheets
‘How to’
compendium
www. eurocode2.info
ECFE – scheme sizing
Worked Examples
Properties
of concrete
Technical publications (CCIP)
Scheme design
Precast Design Manual
Precast Worked Examples
Concise Eurocode 2 for Bridges
77
Concise Eurocode 2
Clarity
Clear references
Comment
Design aids
78
‘How to’ compendium
08/10/2012
SPATA Training 4 Oct 2012 - Eurocode 2 14
79
Spreadsheets to BS EN 1992-1-1 (and UK NA) TCC11 Element designTCC12 Bending and Axial ForceTCC13 Punching ShearTCC14 Crack WidthTCC21 Subframe analysisTCC31 One-way Solid Slabs (A & D)TCC31R Rigorous* One-way Solid SlabTCC32 Ribbed slabs (A & D)TCC33 Flat Slabs (A & D) (single bay)TCC33X Flat Slabs. Xls (whole floor)TCC41 Continuous beams (A & D)TCC41R Rigorous* Continuous BeamsTCC42 (β) Post-tensioned Slabs & Beams (A & D)TCC43 Wide Beams (A & D)
Spreadsheets
TCC51 Column Load Take-down & DesignTCC52 Column Chart generationTCC53 Column DesignTCC54 Circular Column DesignTCC55 Axial Column ShorteningTCC71 Stair Flight & Landing – SingleTCC81 Foundation PadsTCC82 Pilecap Design
80
Design GuidanceNew Concrete Industry Design Guidance is written for Eurocode 2
• TR 64 Flat Slab
• TR43 PT
• TR58 Deflections
Text books
81
Introduction to Eurocode 2
Charles Goodchild,
BSc CEng MCIOB MIStructE
The Concrete Centre
www.concretecentre.com
www.eurocode2.info