EC2 23rd April 2010 - ASELB 23rd April 2010.pdf · • 1991-3 Actions induced by cranes and machinery • 1991-4 Actions in silos and tanks Eurocode 1. Eurocode 1 Part 1-1: ... •

Concrete Design to Eurocode 2

Jenny BurridgeMA CEng MICE MIStructE

Head of Structural Engineering

• Introduction to the Eurocodes

• Eurocode

• Eurocode 1

• Eurocode 2· Materials· Cover· Flexure· Shear· Deflection

• Further Information

• 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

• 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

• Each Eurocode part has a National Annex which modifies the

main text of the Eurocode

Features 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)

• Example: Min diameter for longitudinal steel in columns

φφφφmin = 8 mm in text φφφφmin = 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 and J are not used in the UK

• Non-contradictory complementary information (NCCI)

• TR 43: Post-tensioned concrete floors – design handbook


• Eurocode

• Eurocode 1



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 of actions for ultimate limit state

• Persistent

• Transient

• Accidental

• Seismic

• Combinations of actions for serviceability limit state

Eurocode

For one variable action: 1.25 Gk + 1.5 Qk

Provided:1. Permanent actions < 4.5 x variable actions

2. Excludes storage loads

Eurocode

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,iQk,i

Eqn (6.10b) 0.925x1.35Gk 1.0 Gk 1.5 Qk,1 1.5 Ψ0,i Qk,i


• Eurocode

• Eurocode 1

• Eurocode 2· Materials· Cover· Flexure· Shear· Deflection· Axial


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 UK NA uses the same loads as BS 6399

• Plant loading not given

Eurocode 1


• Eurocode

• Eurocode 1



BS EN 1990

BASIS OF STRUCTURAL

DESIGN

BS EN 1991

ACTIONS ON STRUCTURES

BS EN 1992DESIGN 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

• Code deals with phenomena, 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

• 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

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

Product form Bars and de-coiled rods Wire Fabrics

Class

A

B

C

A

B

C

Characteristic yield strength fyk or f0,2k (MPa)

400 to 600

k = (ft/fy)k

≥1,05

≥1,08

≥1,15 <1,35

≥1,05

≥1,08

≥1,15 <1,35

Characteristic strain at

maximum force, εεεεuk (%)

≥2,5

≥5,0

≥7,5

≥2,5

≥5,0

≥7,5

Fatigue stress range

(N = 2 x 106) (MPa) with an upper limit of 0.6fyk

150

100

• In UK NA max. char yield strength, fyk, = 600 MPa

• BS 4449 and 4483 have adopted 500 MPa

Reinforcement properties (Annex C)

Extract BS 8666

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 Axis

Distance

Reinforcement cover

Axis distance, a, to

centre of bar

a = c + φφφφm/2 + φφφφl

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

µµµµfi = NEd,fi/ NRd or conservatively 0.7

Columns: Method A

Flexure

For grades of concrete up to C50/60,

εcu= 0.0035

ηηηη = 1

λλλλ = 0.8

fcd = ααααcc fck/ γγγγc = 0.85 fck/1.5 = 0.57 fck

fyd = fyk/1.15 = 435 MPa

Simplified Stress Block

Design flowchart

The following flowchart outlines a design procedure for rectangular beams withconcrete classes up to C50/60 and class 500 reinforcement

Determine K and K’ from:

Note: δδδδ = 0.8 means 20% moment redistribution.ck

2 fdb

MK ==== 21.018.06.0'& 2 −−−−−−−−==== δδδδδδδδK

Carry out analysis to determine design moments (M)

δδδδ 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

It is often recommended in the UK that K’ is limited to 0.168 to ensure ductile failure

Beam is over-reinforced -compression steel needed

Is K ≤ K’ ?

Beam is under-reinforced - no compression steel needed

Yes No

Flow chart for under-reinforced beam

Calculate lever arm Z from: [[[[ ]]]] dKd

z 95.053.3112

≤≤≤≤−−−−++++====

Check minimum reinforcement requirements:

dbf

dbfA

t

yk

tctm

min,s013.0

26.0≥≥≥≥≥≥≥≥

Check max reinforcement provided As,max ≤≤≤≤ 0.04Ac (Cl. 9.2.1.1)

Check min spacing between bars > φφφφbar > 20 > Agg + 5

Check max spacing between bars

Calculate tension steel required from:zf

MA

yd

s====

Flow chart for over-reinforced beam

Calculate lever arm Z from: [[[[ ]]]]'53.3112

Kd

z −−−−++++====

Calculate excess moment from: (((( ))))'2

2KKfbdM

ck−−−−====

Calculate compression steel required from:

(((( ))))2yd

22s

ddf

MA

−−−−====

Calculate tension steel required from:

yd

sc

2s

yd

s

'

f

fA

zf

MMA ++++

−−−−====

Check max reinforcement provided As,max ≤≤≤≤ 0.04Ac (Cl. 9.2.1.1)Check min spacing between bars > φφφφbar > 20 > Agg + 5

Shear

Eurocode 2/BS 8110 Compared

Strut inclination method

θθθθcotswsRd, ywdfz

s

AV ====

θθθθθθθθννννααααtancot

1maxRd, ++++

==== cdwcw fzbV

21.8°°°° < θθθθ < 45°°°°

We can manipulate the Expression for the concrete strut:

When cot θ θ θ θ = 2.5 (θ θ θ 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/(b z) = VRd/(0.9 bd)

When vRd > vEd cot θ θ θ θ = 2.5 (θθθθ = 21.8°)

When vRd < vEd we can rearrange the concrete

strut expression:

θθθθ = 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 θθθθ)

fck

vRd (whencot θθθθ = 2.5)

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

Design flow chart for shear

Yes (cot θθθθ = 2.5)

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

Calculate area of shear reinforcement:Asw/s = vEd bw/(fywd cot θθθθ)

Determine θθθθ from:

θθθθ = 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

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

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

systemsρρρρ0 is the reference reinforcement ratio = √√√√fck 10-3

ρρρρ is the required tension reinforcement ratio at mid-span to resist the moment due to the design loads (at support for cantilevers)

ρρρρ’ 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

ffKd

l if ρ ≤ ρ0 (7.16.a)

+

−+=

0

ck0

ck

'

12

1

'5,111

ρρ

ρρρ

ffKd

lif ρ > ρ0 (7.16.b)


Structural system K

Simply supported beam, one- or two-way simply

supported slab

1.0

End span of continuous beam or one-way

spanning slab continuous slab or two-way slab

over continuous over one long side

1.3

Interior span of beam or one-way or two-way

spanning slab

1.5

Slab supported without beams (flat slab) (based

on longer span)

1.2

Cantilever 0.4


18.5

Percentage of tension reinforcement (As,req’d/bd)

Sp

an

to

de

pth

ra

tio

(l/

d)

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/σσσσs

where σσσσs 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

or fck

No

Axial

Column design process

Determine the actions on the column

Determine the effective length, l0

Determine the first order moments

Determine slenderness, λλλλ

Determine slenderness limit, λλλλlim

Is λλλλ ≥≥≥≥ λλλλlim?Yes

No

Column is not slender, MEd = M02

Column is slender

Calculate As (eg using column chart)

Check detailing requirements

Effective length

Actions

Effective length, l0

First order moments

Slenderness, λλλλ

Slenderness limit, λλλλlim


No

Not slender, MEd = M02

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

kl0 = 0,5l⋅

Braced members:

Unbraced members:

++⋅

++

+⋅

⋅+ k

k

k

k

kk

kk 2

21

1

21

21

11

11;101maxl0 = l⋅

θM

θ

Effective length (2)

1.02

≥≥≥≥∑∑∑∑

====

bl

E

l

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

From Eurocode 2:

k = (θθθθ / M)⋅⋅⋅⋅ (EΙΙΙΙ / l)

Alternatively...

Actions


First order moments




No


Slen-

der

Calculate As

Detailing

Effective length (3)

lo = Fl

How to…Columns has a look up table Actions


First order moments




No


Slen-

der

Calculate As

Detailing

Design moment

The design moment MEd is as

follows:

M01 = Min {|Mtop|,|Mbottom|} + ei Ned

M02 = Max {|Mtop|,|Mbottom|} + ei Ned

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

M2 = Ned e2

For stocky columns:

MEd = M02

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

Actions


First order moments




No


Slen-

der

Calculate As

Detailing

Slenderness

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, λλλλ < λλλλlim

Slenderness λλλλ = l0/i where i = √√√√(IIII/A)

hence

for a rectangular section λλλλ = 3.46 l0 / h

for a circular section λλλλ = 4 l0 / h

With biaxial bending the slenderness should be checked separately for each direction and only need be considered in the directions where λλλλlim is exceeded

Actions


First order moments




No


Slen-

der

Calculate As

Detailing

λλλλlim = 20⋅⋅⋅⋅A⋅⋅⋅⋅B⋅⋅⋅⋅C/√√√√n

Slenderness Limit

where:A = 1 / (1+0,2ϕϕϕϕef)

ϕϕϕϕef is the effective creep ratio;

(if ϕϕϕϕef is not known, A = 0,7 may be used)

B = √√√√(1 + 2ωωωω) ωωωω = Asfyd / (Acfcd)

(if ωωωω 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


First order moments




No


Slen-

der

Calculate As

Detailing

Slenderness limit – factor C

Actions


First order moments




No


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

Column design (2)


• Eurocode

• Eurocode 1



Design aids from the UK concrete sector

Concise Eurocode 2

RC Spreadsheets‘How to’ compendiumwww.eurocode2.info

ECFE –scheme sizing

Worked Examples

Properties of concrete

TCC Courses

• Eurocode 2 half-day course for building

designers

• Background to Eurocode 2 for building designers

(one day)

• Eurocode 2 with design workshops for building

designers (one day)

• Background to Eurocode 2, including liquid

retaining structures (one day)

• Design of Concrete Bridges to Eurocodes (one

day)

• Two-day course for building designers

Other Resources

Updated

Detailing

Manual

Updated

‘Green’ bookText Books Designer’s Guides

Recent Concrete Industry

Design Guidance is

written for Eurocode 2

Design Guidance

TR 64 Flat Slab

TR43 Post-tensioned Slabs

TR58 Deflections


• Eurocode

• Eurocode 1

• Eurocode 2· Materials· Flexure· Shear· Deflection· Axial


• Worked Example

Worked Example

Cover = 40mm to each face

fck = 30

Check the beam for flexure, shear and deflection

Gk = 75 kN/m, Qk = 50 kN/m

10 m

600

1000

Determine K and K’ from:

Beam is under-reinforced - no compression steel needed

Is K ≤ K’ ?

Yes

ck

2 fdb

MK ====

21.018.06.0'& 2 −−−−−−−−==== δδδδδδδδK

Carry out analysis to determine design moments (M)

Solution - Flexure

ULS = (75 x 1.25 + 50 x 1.5)= 168.75 kN/m

Mult = 168.75 x 102/8

= 2109 kNm

d = 1000 - 40 - 10 - 16

= 934

δδδδ K’

1.00 0.205

0.95 0.193

0.90 0.180

0.85 0.166

0.80 0.151

0.75 0.135

134.030934600

1021092

6

====××××××××

××××====K

Calculate lever arm Z

[[[[ ]]]] dKd

z 95.053.3112

≤≤≤≤−−−−++++====

Check max reinforcement provided

Check min reinforcement provided

Check min spacing between bars

Check max spacing between bars

Calculate tension steel

zf

MA

yd

s====

Solution - Flexure

[[[[ ]]]]d

z

95.0806

134.0x53.3112

934

≤≤≤≤====

−−−−++++====

2

6

smm6015

806x435

10x2109========A

Provide 8 H32 (6430 mm2)

Space between bars

= 35mm > φ φ φ φ ⇒⇒⇒⇒ OK


Determine the concrete strut capacity vRd

Determine vEd where:vEd = VEd/(bwd)

Shear force: VEd = 168.75 x (10/2 - 0.934)

= 686.1 kN

Shear stress:vEd= VEd/(bwd)

= 686.1 x 103/(1000 x 600)

= 1.14 MPa

Solution - Shear

fck vRd (when cot θθθθ = 2.5)

20 2.28

25 2.79

28 3.08

30 3.27

32 3.46

35 3.73

40 4.17

45 4.58

50 4.96


Yes (cot θθθθ = 2.5)

Determine the concrete strut capacity vRd

Area of shear reinforcement:Asw/s = vEd bw/(0.9 fywd cot θθθθ)

Determine vEd where:vEd = VEd/(bwd)

Is vRD > vEd?

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

Shear force: VEd = 168.75 x (10/2 - 0.934)

= 686.1 kN

Shear stress:vEd= VEd/(bwd)

= 686.1 x 103/(1000 x 600)

= 1.14 MPa

vRd = 3.27 MPa

vRd > vEd ∴∴∴∴ cot θθθθ = 2.5

Asw/s = 1.14 x 600 /(0.9 x 435 x 2.5)

Asw/s = 0.70 mm

Try H10 links with 2 legs.

Asw = 157 mm2

s < 157 /0.70 = 224 mm

⇒⇒⇒⇒ provide H10 links at 200 mm CRS

Determine basic l/d

Solution - Deflection

Reinforcement ratio:

ρρρρ = As/bd

= 6430 x 100/(600 x 934)

= 1.15%

Basic span-to-depth ratios(for simply supported condition)

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)

Sp

an

to

dep

th r

atio

(l/d

)

fck = 20

fck = 25

fck = 28

fck = 30

fck = 32

fck = 35

fck = 40

fck = 45

fck = 50

14.9


Is actuall/d < (l/d).j1.j2?

j1 = 1.0

Yes

Determine basic l/d

NoIs bf > 3bw

Beam > 7m & support brittle partitions?

No

j2 = 1.0

Check complete

Solution - Deflection

Reinforcement ratio:

ρρρρ = As/bd

= 6430 x 100/(600 x 934)

= 1.15%

Req’d l/d = 14.9 x 1.0 = 14.9

Actual l/d = 10000/934 = 10.7

Basic l/d > Actual l/d

Documents

EC2 23rd April 2010 - ASELB 23rd April 2010.pdf · • 1991-3 Actions induced by cranes and machinery • 1991-4 Actions in silos and tanks Eurocode 1. Eurocode 1 Part 1-1: ... •