Practical Design to Eurocode 2

Columns

Steel (B600)

Steel (B500)

Stress

Strain

Concrete (C30/37)

c1 = 0.0022 cu1 = 0.0035

Strain compatibility

dh

As2

Ap

As1

p

uds , p c0 c2c3cu2cu3

A

B

C

(1- c2/cu2)hor

(1- c3/cu3)h

p(0)

yreinforcing 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 Load

EC2 Figure 6.1

Concise Figure 6.3

Column Design Chart- Figure 15.5b

Column Design Chart- Figure 15.5e

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, ii = 0 h m where 0 = 1/200h = 2/l; 2/3 h 1m = (0.5(1+1/m))l is the length or height (m) (see 5.2(6))m is the number of vert. members

For isolated columns in braced systems, m = 1 and h may be taken as 1 ie i = 0 = 1/200

Effect of ImperfectionsCl. 5.2 (7) & (9) 5.6.2.1

For isolated members

The effect of imperfections may be taken into account in two ways:

a) as an eccentricity, ei = i l0/2For isolated columns in braced systems, ei = l0/400 may be used.

b) as a transverse force, Hi

Hi = i N for unbraced membersHi = 2i N for braced members = N/100

Minimum eccentricity: e0 = h/30 but 20 mm cl 6.1(4)

EC 2: Concise:

ei = i l0/2 For walls and isolated columns ei = l0/400Hi = iN for unbraced membersHi = 2iN for braced members = N/100

or

Unbraced Braced

Isolated MembersFigure 5.1a 5.5.2

Na

Nb

Hi

l

i

iNa

Nb

Hi

/2i

/2i

Bracing System Floor Diaphragm Roof

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

StructuresFigure 5.1b Figure 5.5

Column Design Process

Determine the actions on the column

Determine the effective length, l0

Determine the first order moments

Determine slenderness, Determine slenderness limit, lim

Is lim? YesNo

Column is not slender, MEd = Max (M02, NEde0)

Column is slenderSecond order effects

Calculate As (eg using column chart)

Check detailing requirements

EC 2: Concise:

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, is less than lim where

lim = 20 A B C/nWith biaxial bending the slenderness should be checked separately for each direction and only need be considered in the directions wherelim is exceeded

Slenderness = l0/i where i = (I/A)hence for a rectangular section = 3.46 l0 / h

for a circular section = 4 l0 / h

Second Order Effects with Axial Load Cl. 5.8.2, 5.8.3.1 5.6.1

EC 2: Concise:

lim = 20ABC/n (5.13N)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

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

M01, M02 are first order end moments, including the effect of imperfections, M02 M01

M02 = Max{|Mtop|;|Mbot|} +ei NEdM01 = Min {|Mtop|;|Mbot|} +ei NEd

n = NEd / (Acfcd)

Slenderness Limit (5.8.3.1)Cl. 5.8.3.1 5.6.1.4

C = 1.7 - rm rm = M01/M02

Note:

In the following cases, rm should be taken as 1.0 (i.e. C= 0.7)

for braced members in which the first order moments arise only from or predominantly due to imperfections or transverse loading

For unbraced members in general

Slenderness Limit (5.8.3.1)Cl. 5.8.3.1 5.6.1.4

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

EC 2: Concise:

l

M

l0 = l l0 = 2l l0 = 0.7l l0 = l / 2 l0 = l l /2 2l

22

1

1

45,01

45,01

kk

kk

l0 = 0.5lBraced members:Fig f)

Unbraced members:Fig g)

kk

kk

kkkk 2

21

1

21

21

11

11;101maxl0 = l

= l0/i

k = ( / M) (E / l)

Different Column End RestraintsFigure 5.7, 5.8.3.2 Figure 5.6, 5.6.1.2

f) g)

EC 2: Concise:

Failing column

Non failing column

End 1

End 2

Non failing column From PD 6687

The contribution of non failingcolumns to the joint stiffness may be ignored

For beams /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`/`lcol / 2EI/ lbeams

= EI/`lcol / 2 x 2EI/ lbeams 0.1Although not stated effective lengths can be used?

Typical Column EffectivePD 6687 Cl.2.10 -

EC 2: Concise:

lo = Fl

Typical Column Effective Length- -

l is clear height

Conservative to use the method in BS 8110 Table 3.19

EC 2: Concise:

2nd order effects Slender columnsMethods of analysis

The methods of analysis include a general method. Based on non-linear second order analysis and the following two simplified methods:

Method based on nominal stiffness

Method based on nominal curvature

This method is primarily suitable for isolated members with constant normal force and defined effective length. The method gives a nominal second order moment based on a deflection, which in turn is based on the effective length and an estimated maximum curvature.

Cl 5.8.5, cl 5.8.8 5.6.2.1

EC 2: Concise:

MEd = M0Ed+ M2M0Ed = Equivalent first order moment including the effect

of imperfections [At about mid-height]. May be taken as = M0e

M0e = (0.6 M02 + 0.4 M01) 0.4M02HOWEVER, this is only the mid-height moment - the two

end moments should be considered too. PD 6687 advises for braced structures:

MEd = MAX{M0e+M2; M02; M01+0.5M2} e0NEd

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

M01 = Min {|Mtop|;|Mbot|} + ei NEdM2 = nominal 2nd order moment in slender columns = NEde2

Nominal Curvature MethodCl. 5.8.8.2 5.6.2.2

Typical unbraced columnTypical braced column

1st Ordermoments

2nd Ordermoments Combination

of moments

1st Ordermoments

2nd Ordermoments

Combinationof moments

Moments in Slender Columns

Nominal Curvature Method- Figure 5.10

M2 = NEd e2

e2 = (1/r)l02/2

1/r = KrK/r0 where 1/r0 = yd /(0.45d)Kr = (nu n)/(nu-nbal) 1K = 1 + ef 1

= 0.35 + fck /200 /150

Second order momentCl. 5.8.8 5.6.2.2

Biaxial BendingCl. 5.8.9 5.6.3

aa MMM 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

MEdzMEdy

a = 2

a = 1

a = 1.5

Biaxial bending for rectangular column

h 4b min 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 = min {20 min; 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 > 14.

A min of 3 bars is required in lap length

scl,tmax = min {12 min; 0.6b ; 240mm}

Columns (2)(9.5.3)

Worked Example

The structural grid is 7.5 m in each direction

Worked Examples to EC2 - Example 5.1

38.5 kN.m

38.5 kN.m

Solution effective length

Using PD 6687 methodClear span is 4000 250 = 3750 mm

14.0

75001225037502

375012300

2 3

4

b

b

c

c

LEILEI

k

From Table 4 of How toColumns

F = 0.62 lo = 0.62 x 3.750 = 2.33 mCheck slenderness:

= 3.46 lo/h = 3.46 x 2.33 / 0.3 = 26.8

Take beam width as, say, half the bay width

Column moments

First order moments

M02 = M + eiNEd

ei = l0/400 = 2300/400 = 5.8 mm

M02 = 38.5 + 0.0058 x 1620

= 47.9 kNm

M01 = 38.5 + 0.0058 x 1620

= 47.9 kNm

Slenderness

A = 0.7 (use default value)

B = 1.1 (use default value)

C = 1.7 rm = 1.7 M01/M02 = 1.7 (-47.9/47.9) = 2.70

n = NEd/Acfcd = 1620 x 1000/(3002 x 0.85 x 30/1.5) = 1.06

lim = 20 ABC/n

= 20 x 0.7 x 1.1 x 2.7/1.06 = 40.4lim > (26.8) ...column is not slender M2 = 0

Column moments

Design moments

MEd = Max { M02, e0NEd}

e0 = Max[h/30,20mm] = Max[300/30,20mm] = 20 mm

MEd = Max { 47.9, 0.02 x 1620} = Max { 47.9 , 32.4}

MEd = 47.9 kN.m

Solution determine Asd2 = cnom + link + /2 = 25 + 8 +16 = 49 mm

d2/h = 49/300 = 0.163

Column design charts interpolate between d2/h = 0.15 and 0.20

MEd/(bh2fck) = 47.9 x 106/(3003 x 30) = 0.059

NEd/(bhfck) = 1620 x 1000/(3002 x 30) = 0.60

Charts are for symmetrically reinforced columnswhere bars are in the corners.

See concise 15.9.3 for method where bars are not concentrated in the corners

Interaction Chart

Asfyk/bhfck

0.22

Interaction Chart

Asfyk/bhfck

0.24

Interpolating:

Asfyk/(bhfck) = 0.23

As = 0.23 x 3002 x 30/500 = 1242 mm2

Try 4 H25 (1964 mm2) (note 4 H20 is 1260 mm2)

Solution determine As

Workshop Problem

Design column C2 between ground and first floors for bending about axis parallel to line 2.

Assume the following:Axial load: 7146kNTop Moment: 95.7kN.mNominal cover:35mmPinned baseBay width is 6.0 mElastic modulus is the same for column and slab

Take beamwidth as, say, half the bay width

Solution - effective length

41.0

96001230030002

860012300300024200

12500

2 33

4

b

b

c

c

LEILEI

k

From Table 4 of How toColumns

F = 0.86 lo = 0.86 x 4.2 = 3.612 mCheck slenderness:

= 3.46 lo/h = 3.46 x 3.612 /0.5 = 25.0

Using PD 6687 methodClear span is 4500 300 = 4200 mm

Column moments

First order moments

M02 = M + eiNEd

ei = l0/400 =3612/400 = 9.0 mm

M02 = 95.7 + 0.0090 x 7146

= 160.0 142.9 kNm

M01 = 0 kNm (pinned base)

Slenderness

A = 0.7 (use default value)

B = 1.1 (use default value)

C = 1.7 rm = 1.7 M01/M02 = 1.7 (0/160.0) = 1.7

n = NEd/Acfcd = 7146 x 1000/(5002 x 0.85 x 50/1.5) = 1.01

lim = 20 ABC/n

= 20 x 0.7 x 1.1 x 1.7/1.01 = 26.1lim > (25.0) ...column is not slender M2 = 0

Column moments

Design moments

MEd = Max { M02, e0NEd}

e0 = Max[h/30,20mm] = Max[500/30,20mm] = 20 mm

MEd = Max { 160.0, 0.02 x 7146} = Max { 160.0 , 142.9}

MEd = 160.0 kN.m

d2 = cnom + link + /2 = 35 + 8 + 16 = 59 mm

d2/h = 59/500 = 0.118

MEd/(bh2fck) = 160.0 x 106/(5003 x 50) = 0.026

NEd/(bhfck) = 7146 x 1000/(5002 x 50) = 0.57

Solution determine As

Interaction Chart

A sf yk

/bhf ck

0.09

Solution determine As

Asfyk/bhfck = 0.09

As = 0.09 x 5002 x 50 / 500 = 2250 mm2

Use 8 H20 (2513 mm2)

Practical Design to Eurocode 2

Fire Design

Fire

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

Structural Fire DesignPart 1-2, Fig 5.2 Figure 4.2

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

Verification methods Ed,fi Rd,fi(t) Member Analysis Ed,fi = fi Ed

Ed is the design value for normal temperature design fi is the reduction factor for the fire situation

fi = (Gk + fi Qk.1)/(GGk + Q.1Qk.1) fi is taken as 1 or 2 (= 1 - NA)

Chapter 2: Basis of Fire Design

Tabulated data (Chapter 5) Simplified calculation methods Advanced calculation method

Design Procedures

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 -

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

EC 2: Concise:

Continuous BeamsTable 5.6 Table 4.6

Flat SlabsTable 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

EC 2: Concise:

fi = NEd,fi/ NRd = Gk + 1,1 Qk,1/(1.35Gk + 1.5 Qk) Conservatively 0.7where 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

Columns: Method B

Limitations to Table 5.2b

l/h (or l/b) 17.3 for rectangular column (fi 30)

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

Amount of reinforcement, = As fyd / Ac fcd 1

For other values of these parameters see Annex C

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

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.

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 = (s /s,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 = (s /s,fi ) ks() MEd (As,prov /As,req)

Annex E: Simplified Calculation Method for Beams and Slabs

500C Isotherm Method

Ignore concrete > 500C

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

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 OKAssume 8 bars OKl0,fi = 0.7l = 2.8m < 3m OKFrom Table 5.2a: min dimensions = 350/57

Column is 350mm, axis distance = 57mm

Check cover 35mm + 8 (link) + /2 = 55.5mm Increase nominal cover to 40mm.