1

Practical Design to Eurocode 2

Columns

Steel (B600)

Steel (B500)

Stress

Strain

Concrete (C30/37)

εc1 = 0.0022 εcu1 = 0.0035

Strain compatibility

2

dh

As2

Ap

As1

∆εp

udεs ,ε pε εc

0 c2ε(ε )

c3

cu2ε(ε ) cu3

A

B

C

(1- εc2/εcu2)hor

(1- εc3/εcu3)h

εp(0)

εy

reinforcing steel tension strain limit

concrete compression strain limit

concrete pure compression strain limitC

B

A

Minimum eccentricity: e0 = h/30 but ≥ 20 mm

Bending with/without Axial LoadEC2 Figure 6.1

Concise Figure 6.3

Column Design Chart- Figure 15.5

3

Column Design Chart- Figure 15.5

Geometric ImperfectionsCl. 5.2 5.5

Deviations in cross-section dimensions are normally taken into account in the material factors and should not be included in structural analysis

Imperfections need not be considered for SLS

Out-of-plumb is represented by an inclination, θlθl = θ0 αh αm where θ0 = 1/200

αh = 2/√l; 2/3 ≤ αh ≤ 1αm = √(0.5(1+1/m))

l is the length or height (m) (see 5.2(6))m is the number of vert. members

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ei

N

Hi

N

l = l0 / 2

ei

N

l = l0Hi

N

ei = θi l0/2 for walls and isolated columns ei = l0/400

Hi = θiN for unbraced membersHi = 2θiN for braced members

or

UnbracedBraced

Isolated MembersFigure 5.1a 5.5.2

Na

Nb

Hi

l

iθ

iθNa

Nb

Hi

/2iθ

/2iθ

Bracing System Floor Diaphragm Roof

Hi = θi (Nb-Na) Hi = θi (Nb+Na)/2 Hi = θi Na

StructuresFigure 5.1b Figure 5.5

5

• Second order effects may be ignored if they are less than 10% of the corresponding first order effects

• Global second order effects in buildings may be ignored if:

Fv,Ed ≤ 0.26ns/(ns + 1.6) ⋅ ΣEcmIc/L2

where

Fv,Ed is the total vertical load (on braced and bracing members)

ns is the number of storeys

L is the total height of the building above level of moment restraint

Ecm is the secant value of the modulus of elasticity of concrete

Ic is the second moment of area of bracing members

Second Order Effects with Axial Load (1) (5.8.2, 5.8.3.1 and 5.8.3.3)

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

6

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 is less than λlim where

λlim = 20 A B C √(Acfcd/NEd)

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

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

Second Order Effects with Axial Load

Cl. 5.8.2, 5.8.3.15.6.1

llim = 20⋅A⋅B⋅C/√n (5.13N)

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) w = 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, including the effect of imperfections

M02 ≥ M01(if rm is not known, C = 0.7 may be used)

M02 = Max{|Mtop|;|Mbot|}+ei NEd ≥e0NEd

M01 = Min {|Mtop|;|Mbot|}

n = NEd / (Acfcd)

Slenderness Limit (5.8.3.1)Cl. 5.8.3.1 5.6.1.4

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Factor C

100 kNm 100 kNm 100 kNm

-100 kNm 100 kNm

rm = M01/ M02

= 0 / 100

= 0

C = 1.7 – 0

= 1.7

rm = M01/ M02

= -100 / 100

= -1

C = 1.7 + 1

= 2.7

rm = M01/ M02

= 100 / 100

= 1

C = 1.7 – 1

= 0.7

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

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⋅

λ = l0/i

k = (θ / M)⋅ (EΙ / l)

Different Column End RestraintsFigure 5.7, 5.8.3.2 Figure 5.6, 5.6.1.2

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Failing column

Non failing

column

End 1

End 2

Non failing

column From PD 6687

The contribution of ‘non failing’columns to the joint stiffness may be ignored

For beams θ/M may be taken as l/2EI(allowing for cracking in the beams)

Assuming that the beams are symmetrical about the column and their sizes are the same in the two storeys shown, then:

k1 = k2 = [EI`/`l]col / [Σ2EI/ l]beams

= [EI/`l]col / [2 x 2EI/ l]beams

Although not stated effective lengths can be used

Typical Column EffectivePD 6687 Cl.2.10 -

lo = Fl

Typical Column Effective Length- -

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MEd = M0Ed+ M2

M0Ed = Equivalent first order moment including the effect of imperfections [At about mid-height]

= M0E

= (0.6 M02 + 0.4 M01) ≥ 0.4M02

HOWEVER, this is only the mid-height moment - the two end moments should be considered too. PD 6687 advises for braced structures:

MEd = MAX{M0Ed+M2; M02; M01+0.5M02}

M02 = Max{|Mtop|;|Mbot|}+ei NEd ≥e0NEd

M01 = Min {|Mtop|;|Mbot|}

Nominal Curvature MethodCl. 5.8.8.2 5.6.2.2

Typical unbraced columnTypical braced column

1st Order

moments

2nd Order

moments Combination

of moments

1st Order

moments

2nd Order

moments

Combination

of moments

Moments in Slender Columns

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Nominal Curvature Method- Figure 5.10

M2 = NEd e2

e2 = (1/r)l02/π2

1/r = KrKϕ/r0 where 1/r0 = εyd /(0.45d)

Kr = (nu –n)/(nu-nbal) ≤ 1

Kϕ = 1 + βϕef ≥ 1

β = 0.35 + fck /200 – l /150

Second order momentCl. 5.8.8 5.6.2.2

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Biaxial BendingCl. 5.8.9 5.6.3

aaMM

M M

EdyEdz

Rdz Rdy

1,0

+ ≤

For rectangular cross-sections

NEd/NRd 0.1 0.7 1.0a 1.0 1.5 2.0

where NRd = Acfcd + Asfyd

For circular cross-sections a = 2.0

NRd

NEd

MEdz

MEdy

a = 2

a = 1

a = 1.5

Biaxial bending for rectangular column

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• h ≤ 4b

• φmin ≥ 12

• As,min = 0,10NEd/fyd but ≥ 0,002 Ac

• As,max = 0.04 Ac (0,08Ac at laps)

• Minimum number of bars in a circular column is 4.

• Where direction of longitudinal bars changes more than 1:12 the spacing of transverse reinforcement should be calculated.

Columns (1)(9.5.2)

• scl,tmax = 20 × φmin; b; 400mm

≤ 150mm

≤ 150mm

scl,tmax

• scl,tmax should be reduced by a factor 0,6:

– in sections within h above or below a beam or slab

– near lapped joints where φ > 14. A minimum of 3 bars is rqd. in lap length

Columns (2)(9.5.3)

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Worked Example

The structural grid is 7.5 m in each direction

Solution – effective length

Using PD 6687 method

14.0

7500

1225037502

3750

12300

2 3

4

=××

==∑

b

b

c

c

L

EI

L

EI

k

From Table 4 of How to…Columns

F = 0.79 ∴∴∴∴ lo = 0.79 x 3.750 = 2.96 m

Check slenderness:

λλλλ = 3.46 lo/h = 3.46 x 2.96 / 0.3 = 34.1

Say half bay width for flat slab

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Solution - slenderness

Solution – design moments

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Solution – determine As

Interaction Chart

Asfyk/bhfck

0.23

16

Interaction Chart

Asfyk/bhfck

0.25

Solution – determine As

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Workshop Problem

Assume the following:•Axial load: 8933 kN•Moment: 95.7 kNm•Nominal cover: 35mm•Pinned base•Assume bay width is 6.0 m

Practical Design to Eurocode 2

Fire Design

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Fire

a Axis

Distance

Reinforcement cover

Axis distance, a, to centre of bar

a = c + φφφφm/2 + φφφφl

Scope

Part 1-2 Structural fire design gives several methods for fire engineering

Tabulated data for various elements is given in section 5

Structural Fire DesignPart 1-2, Fig 5.2 Figure 4.2

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• High strength concrete

• Basis of fire design

• Material properties

• Tabulated data

• Design procedures

– Simplified and advanced calculation methods

– Shear and torsion

– Spalling

– Joints

– Protective layers

• Annexes A, B, C, D and E

• General

Eurocode 2: Part 1.2 Structural Fire Design

100 Pages

• Requirements: – Criteria considered are:

“R” Mechanical resistance (load bearing) “E” Integrity (compartment separation)“I” Insulation (where required)

“M” Impact resistance (where required)

• Actions - from BS EN 1991-1-2– Nominal and Parametric Fire Curves

Chapter 2: Basis of Fire Design

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• Verification methods Ed,fi ≤ Rd,fi(t)

• Member Analysis Ed,fi = ηfi Ed

Ed is the design value for normal temperature design

ηfi is the reduction factor for the fire situation

ηfi = (Gk + ψfi Qk.1)/(γGGk + γQ.1Qk.1) ψfi is taken as ψψψψ1 or ψψψψ2 (= ψψψψ1 - NA)

Chapter 2: Basis of Fire Design

• Tabulated data (Chapter 5)

• Simplified calculation methods

• Advanced calculation method

Design Procedures

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Which method?

Provides design solutions for the standard fire exposure up to 4hours

• 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

• For calcareous or lightweight aggregates minimum dimension may be reduced by 10%

• 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) the minimum cross section dimension should be increased

Section 5. Tabulated DataCl. 5.1 -

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Elements

• Approach for Beams and Slabs very similar

– Separate tables for continuous members

– One way, two way spanning and flat slabs treated separately

• Walls depend on exposure conditions

• Columns depend on load and slenderness

Standard fire

resistanceMinimum dimensions (mm)

Possible combinations ofa and bmin

wherea is the average axis

distance and bmin is the width

of beam

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 Beams1992-1-2 Table 5.6 Table 4.6

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Flat Slabs

Table 4.81992-1-2 Table 5.9

Columns Tabular Approach

Columns more Tricky!

• Two approaches

• Only for braced structures

• Unbraced structures – columns can be considered braced if there are columns outside the fire zone

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µfi = NEd,fi/ NRd

where NEd,fi is the design axial load in the fire condition

NRd is the design axial resistance at normal temperature

The minimum dimensions are larger thanBS 8110

Columns: Method ATable 5.2a Table 4.4A

Limitations to Table 5.2a

Limitations to Method A:

• Effective length of the column under fire conditions l0,fi<= 3m.

• First order eccentricity under fire conditions: e = M0Ed,fi / N0Ed,fi <= emax = 0.15 h

• Amount of reinforcement: As < 0.04 Ac

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Columns: Method B

Limitations to Table 5.2b

• l/h (or l/b) ≤ 17.3 for rectangular column (λfi ≤ 30)

• First order eccentricity under fire conditions: e/b = M0Ed,fi /b N0Ed,fi ≤ 0.25 with emax= 100 mm

• Amount of reinforcement, ω = As fyd / Ac fcd ≤ 1

For other values of these parameters see Annex C

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• EC2 distinguishes between explosive spalling that can occur in concrete under compressive conditions. such as in columns and the concrete falling off the soffit in the tension zones of beams and slabs.

• Explosive spalling occurs early on in the fire exposure and is mainly caused by the expansion of the water/steam particles trapped in the matrix of the concrete. The denser the concrete. the greater the explosive force.

− Unlikely if moisture content is less than 3% (NDP) by weight

− Tabular data OK for axis distance up to 70 mm

• Falling off of concrete occurs in the latter stage of fire exposure

Spalling

Minimum cross section should be increased:

• For walls and slabs exposed on one side only by: For Class 1: 0.1a for C55/67 to C60/75

For Class 2: 0.3a for C70/85 to C80/95

• For all other structural members by: For Class 1: 0.2a for C55/67 to C60/75 For Class 2: 0.6a for C70/85 to C80/95

Axis distance, a, increased by factor:For Class 1: 1.1 for C55/67 to C60/75For Class 2: 1.3 for C70/85 to C80/95

High Strength Concrete -Tabulated Data

27

• For C 55/67 to C 80/95 the rules for normal strength concrete apply. provided that the maximum content of silica fume is less than 6% by weight.

• For C 80/95 to C 90/105 there is a risk of spalling and at least one of the following should be provided (NA):

Method A: A reinforcement mesh

Method B: A type of concrete which resists spalling

Method C: Protective layers which prevent spalling

Method D: monofilament polypropylene fibres.

High Strength Concrete -Spalling

Other Methods

• Simplified calculation method for beams, slabs and columns

• Full Non-linear temperature dependent ……..

• But all of these must have the caveat that they are unproven for shear and torsion.

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Rd1,fi

M Rd,fi,Span

M Rd2,fi

M

M = w l / 8Ed,fi Ed,fi eff

1

1 - Free moment diagram for UDL under fire conditions

• MRd,fi,Support = (γs /γs,fi ) MEd (As,prov /As,req) (d-a)/dWhere a is the required bottom axis distance given in Section 5

As,prov /As,req should not be taken greater than 1.3

• MRd,fi,Span = (γs /γs,fi ) ks(θ) MEd (As,prov /As,req)

Annex E: Simplified Calculation Method for Beams and Slabs

500°C Isotherm Method

Ignore concrete > 500°C

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Zone Method

Divide concrete into zones and work out average

temperature of each zone, to calculate strength

Worked Example

• NEd=1824kN

• Myy,Ed=78.5kNm

• Mzz,Ed=76.8kNm

• 2 hour fire resistance required

• External, but no de-icing salts

• fck = 30MPa

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Worked Example

Cover:

cmin,b = diameter of bar (assume 25mm bars with 8mm links)

cmin,dur = (XC3/XC4) 25mm

say ∆cdev = 10mm

cnom (to main bars) = max{(25+10),(25+8+10)} = 43mm

Use cnom = 35mm to links

Worked Example

Check fire resistance of R120 to Method A

eccentricity e < 0.15b

e = MEd/NEd = 78.5x103/1824 = 43mm

0.15 x 350 = 52.5mm ∴ OK

Assume 8 bars ∴ OK

l0,fi = 0.7l = 2.8m < 3m ∴ OK

From Table 5.2a: min dimensions = 350/57

Column is 350mm, axis distance = 57mm

Check cover – 35mm + 8 (link) + φ/2 = 55.5mm

∴ Increase nominal cover to 40mm.