Introduction to Eurocode 2

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Introduction to Eurocode 2

SPATA Training

4 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 1990 Basis of Design EN 1991 Actions on Structures EN 1992 Concrete EN 1993 Steel EN 1994 Composite EN 1995 Timber EN 1996 Masonry EN 1999 Aluminium EN 1997 Geotechnical Design EN 1998 Seismic 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6 BS 8110 and all old structural design British Standards have now been ‘withdrawn’.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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7

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

a.National front cover b.National forward

Format of the Eurocodes

a.National front cover b.National forward c.CEN front cover

Format of the Eurocodes

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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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 diameterfor 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 + NA EN 1990 Basis of Design EN 1991 Actions on Structures EN 1992 Concrete EN 1993 Steel EN 1994 Composite EN 1995 Timber EN 1996 Masonry EN 1999 Aluminium EN 1997 Geotechnical Design EN 1998 Seismic 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

Eurocode

Basis 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

and

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19

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 – EC0

Representative value of an action

Design value of an action = F

d

=



_F



F

_rep

=



F



(





F

K

)

where

FK = the characteristic value of action

Frep = 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 Value

1Qk= Frequent Value

2 Qk =Quasi-permanent Value

Greek Alphabet

The ULS is divided into the following categories: EQU Loss of equilibrium of the structure.

E

d,dst

≤

E

d,stb

STR Internal failure or excessive deformation of the structure or structural member.

E

d



R

d

;

GEO Failure due to excessive deformation of the ground.

FAT Fatigue failure of the structure or structural members.

Eurocode – EC0

Ultimate Limit State – Categories

23 Generally for one variable action:

1.25 G

_k

+ 1.5 Q

_k

Provided:

1. Permanent actions < 4.5 x variable actions 2. 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,1Qk,1 γQ,iΨ0,iQk,i

Eqn (6.10a) γG,j,supGk,j,sup γG,j,inf Gk,j,inf γQ,1Ψ0,1Qk,1 γQ,iΨ0,iQk,i

Eqn (6.10b) ξ γG,j,supGk,j,sup γG,j,infGk,j,inf γQ,1Qk,1 γQ,iΨ0,iQk,i

Eqn (6.10) 1.35 Gk 1.0 Gk 1.5 Qk,1 1.5Ψ0,iQk,i

Eqn (6.10a) 1.35 Gk 1.0 Gk 1.5Ψ0,1Qk 1.5Ψ0,iQk,i

Eqn (6.10b) 0.925x1.35G_k 1.0 G_k 1.5 Q_k,1 1.5Ψ_0,iQ_k,i

24

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25

Load arrangements to

EC2

alternative to

UK NA

26 Characteristiccombination (Normally used for irreversible limit states)

Gk,j + Qk,1+ 0,IQk,I

Frequentcombination (Normally used for reversible limit states)

Gk,j + 1,1Qk,1 + 2,IQk,I

Quasi-permanentcombination (Normally used for long term effects and appearance of the structure)

Gk,j + 2,IQk,I

Eurocode: SLS Actions

27

Eurocode

Eurocode: 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

29

• BS EN 1990 (EC0): Basis of structural design

• BS EN 1991 (EC1): Actions on Structures

• BS EN 1992 (EC2): Design of concrete 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

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31

Eurocode 1 Part 1-1: Densities, self-weight and

imposed loads

• Bulk density of

reinforced

concrete is

25 kN/m

3

• 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

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 additionto 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 <1,35 1,15 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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37

Extract BS 8666

38

Nominal cover,

c

nom

Minimum cover,

c

min

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

Axis distance,

a

Fire protection

Allowance for deviation,

∆

c

dev

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, c

min = 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

Axis Distance 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.

42

fi= NEd,fi/ NRd or conservatively 0.7

Part 1-2 Fire Section 5. Tabulated data

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43

Standard fire

resistance Minimum dimensions (mm) Possible combinations ofaand bmin

whereais the average axis distance and bminis the width of

be am Web thickness bw R 30 R 60 R 90 R 120 R 180 R 240 bmin= 80 a= 15* bmin= 120 a= 25 bmin= 150 a= 35 bmin= 200 a= 45 bmin= 240 a= 60 bmin= 280 a= 75 160 12* 200 12* 250 25 300 35 400 50 500 60 450 35 550 50 650 60 500 30 600 40 700 50 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= ccfck/ c= 0.85fck/1.5 = 0.57fck fyd= fyk/1.15 = 435 MPa

Derived formulae include:

z/d = (1 + (1 + 3.529K)0.5_{] / 2}_{(where K = M/bd}2_f ck)

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 rectangular beams 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_f d b M K & _K'_0.6__0.18_2_0.21 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 - Flexure

Design Flowchart

Calculate lever arm zfrom:

*A limit of 0.95dis considered good practice, it is not a requirement of Eurocode 2.



1 1 3.53



0.95 *

2 K d

d

z   

Check minimum reinforcement requirements: d b f d b f A _t yk t ctm min , s 0.26 0.0013

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

Calculate tension steel required from: z f M A yd s

EC2 - Flexure

Flow Chart for singly reinforced section



. K



. d* d z 1 1 353 095 2    

EC2 - Flexure

essential design by hand

435 MPa

= 500/1.15 =

where K = M/bd

2

_f

ck

z = d x z/d

A

s

= M

Ed

/f

yd

z

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

 cot sw s Rd, _s zfywd A V  21.8< < 45

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49

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 - Shear

Design Flow Chart for Shear

Yes (cot = 2.5)

Determine the concrete strut capacity vRdwhen cot = 2.5 vRd = 0.138fck(1-fck/250)

Calculate area of shear reinforcement: Asw/s= vEdbw/(fywdcot )

DeterminevEdwhere:

vEd= design shear stress [vEd= VEd/(bwz) = VEd/(bw 0.9d)]

Determine  from:

= 0.5 sin-1_[(_v

Ed/(0.20fck(1-fck/250))] Is vRd >vEd? No

Check maximum spacing of shear reinforcement : s,max= 0.75 d

For 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

A

_sw

/

s

=

v

_Ed

b

_w

/(

f

_ywd

.2.5)

f

ck MPa

v

Rdcot = 2.5 MPa

20

2.54

25

3.10

28

3.43

30

3.64

32

3.84

35

4.15

40

4.63

45

5.08

50

5.51 Shear

Eurocode 2 – Beam shear

essential design by hand

52

The deflection limits stated to be:

• Span/250 under quasi-permanentloads 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/dx F1 x F2 x F3 >Actual l/d?

Yes No

Factor F3 accounts for stress in the reinforcement F3 = 310/s ≤1.5

where sis tensile stress under characteristic load or

As,prov/As,req’d

Check complete Determine basic l/d including K for structural system

Factor F2 for spans supporting brittle partitions > 7m F2= 7/leff

Factor F1 for ribbed and waffle slabs only F1= 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)

Span to depth ratio ( 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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55

EC2 Columns: Design moments

1st order moments: M01 = Min {|Mtop|,|Mbottom|} + eiNed M02 = Max {|Mtop|,|Mbottom|} + eiNed where

ei = Max {Io/400, h/30, 20}

(20 mm usually critical)

For stocky columns: Design moment, MEd= M02

56 For Slender columns,

M

Ed

=

Max

[M

02

, M

0e

+ M

2

, M

01

+ M

2

/2]

Where

M2

=

nominal 2ndorder moment

M2 = NEde2 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, 

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

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 , 0 1 45 , 0 1 k k k k F= 0,5 Braced members: Unbraced members:                              k k k k k k k k 2 2 1 1 2 1 2 1 1 1 1 1 ; 10 1 max F =  M 

EC2 Columns: Slenderness (2)

& 2nd_{order moments: Effective length &}_F

F

59 1 . 0 2    b l E l E k b c c I I _{(From PD 6687:} Background paper to UK NA) Where:

Ib,Icare the beam and column uncracked second moments of area

lb,lcare the beam and column lengths k= relative stiffness

= (/ M)(E/ l)

Actions

Effective length, l0 First order moments Slenderness, 

Yes No

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, 

Yes No

Calculate As

Detailing

EC2 Columns: Slenderness (4)

& 2nd_{order moments: Effective length :}_F_from_k

1 . 0 2    b l E l E k b c c I I ki= relative stiffness each end

F

l₀= Fl And

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61

Allowable Slenderness

lim= 20ABC/n where:

A= 1 / (1+0,2ef)

ef is the effective creep ratio; (if efis 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, M02are first order end moments, M02  M01

(if rmis not known, C= 0.7 may be used)

n = NEd/ (Acfcd)

Actions

Yes No

Calculate As

Detailing

EC2 Columns: Slenderness (5)

& 2nd_{order moments: Allowable Slenderness}

62 Actions

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= 20ABC/n

EC2 Columns: Slenderness (6)

& 2nd_{order moments:}_{Allowable Slenderness &}_C

63 If

Slenderness > Allowable slenderness

Then include nominal 2nd_{order moment,}_M 2

M2 = NEde2 where e2= fn(deflection)

There are alternative methods for calculating eccentricity, e2, for slender columns

Actions

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 have

N

Ed

and M

Ed

!!!!

If using column charts we want:

N

_Ed

/bhf

_ck

and M

_Ed

/bh

2

_f

ck

from which we get:

A

s

f

yk

/bhf

ck

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 12fctd

where fctdshould be limited to C60/75

1 =1 for ‘good’ and 0.7 for ‘poor’ bond conditions

2= 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 conditions hatched zone - ‘poor’ bond conditions

 Direction of concreting

250

Direction of concreting

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l

bd

=

α

1

α

2

α

3

α

4

α

5

l

b,rqd



l

b,min

However:

(

α

2

α

3

α

5

) 

0.7 l

b,min

> max(0.3

l

b

; 15



, 100mm)

EC2 – Detailing:

Design Anchorage Length, l

bd

EC2 – Detailing:

Alpha values

•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”

•For members with shear reinforcement: al = (MEd/z) + 0.5VEdCot

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 1992 DESIGN OF CONCRETE STRUCTURES Part 1-1: General Rules for

Structures Part 1-2: Structural Fire Design

BS EN 1992 Part 2: Bridges BS EN 1992 Part 3: Liquid Ret. Structures BS EN 1994 Design of Comp. Struct. BS EN 13369 Pre-cast Concrete BS EN 1997 GEOTECHNICAL DESIGN BS EN 1998 SEISMIC DESIGN BS EN 13670 Execution of Structures BS 8500 Specifying Concrete BS 4449 Reinforcing Steels BS EN 10080 Reinforcing Steels

Eurocode 2: relationships –

BS EN 206 Concrete NSCS DMRB? NBS? Rail? CESWI? BS EN 10138 Prestressing 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

(13)

73 BS EN 1990 BASIS OF STRUCTURAL DESIGN BS EN 1991 ACTIONS ON STRUCTURES BS EN 1992 DESIGN OF CONCRETE STRUCTURES Part 1-1: General Rules for

Structures Part 1-2: Structural Fire Design

BS EN 1992 Part 2: Bridges BS EN 1992 Part 3: Liquid Ret. Structures BS EN 1994 Design of Comp. Struct. BS EN 13369 Pre-cast Concrete BS EN 1997 GEOTECHNICAL DESIGN BS EN 1998 SEISMIC DESIGN BS EN 13670 Execution of Structures BS 8500 Specifying Concrete BS 4449 Reinforcing Steels BS EN 10080 Reinforcing Steels

Eurocode 2: relationships –

BS EN 206 Concrete NSCS DMRB? NBS? Rail? CESWI? BS EN 10138 Prestressing 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 differentfrom BS8110 in terms of the design approach. •similar answers

• marginally more economic.

•less prescriptiveand more extensivethan 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 SL A B D E P T H , m m IL = 5 kN/m2 To BS8110 incl 1.5 SDL _{IL = 2.5 kN/m}2 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 2 _{RC 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

(14)

79 Spreadsheets to BS EN 1992-1-1

(and UK NA) TCC11 Element design TCC12 Bending and Axial Force TCC13 Punching Shear TCC14 Crack Width TCC21 Subframe analysis TCC31 One-way Solid Slabs (A & D) TCC31R Rigorous* One-way Solid Slab TCC32 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 Beams TCC42 (β) Post-tensioned Slabs & Beams (A & D) TCC43 Wide Beams (A & D)

Spreadsheets

TCC51 Column Load Take-down & Design TCC52 Column Chart generation TCC53 Column Design TCC54 Circular Column Design TCC55 Axial Column Shortening TCC71 Stair Flight & Landing – Single TCC81 Foundation Pads TCC82 Pilecap Design