Lecture 3 Columns_nov 11_leeds

7/31/2019 Lecture 3 Columns_nov 11_leeds

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Practical Design to Eurocode 2

Columns

Steel (B600)

Steel (B500)

Stress

Strain

Concrete (C30/37)

c1 = 0.0022 cu1 = 0.0035

Strain compat ibi li t y


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dh

As2

Ap

As1

p

uds , p c

0 c2c3

cu2cu3

A

B

C

(1- c2/cu2)hor

(1- c3/cu3)h

p(0)

y

reinforcing steel tension strain limit

concrete compression strain limitconcrete pure compression strain limitC

B

A

Minimum eccentricity: e0 = h/30 but 20 mm

Bending with/wi thout AxialLoad

EC2 Figure 6.1

Concise Figure 6.3

Column Design Chart- Figure 15.5b


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Column Design Chart- Figure 15.5e

Geometric ImperfectionsCl. 5.2 5.5

Deviations in cross-section dimensions are normally taken into accountin 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 = 1/200

h = 2/l; 2/3 h 1m = (0.5(1+1/m))

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

For isolated columns in braced systems, m = 1 andh may be taken as 1

ie i = 0 = 1/200


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Effect of Imperfect ionsCl. 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 l0/2

For 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 = l0/400

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

or

UnbracedBraced

Isolated MembersFigure 5.1a 5.5.2


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Na

Nb

Hi

l

i

i

Na

Nb

Hi

/2i

/2i

Bracing System Floor DiaphragmRoof

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 fi rst order moments

Determine slenderness, Determine slenderness limit , li m

Is lim? Yes

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

Column is slenderSecond order effects

Calculate As (eg using column chart )

Check detailing requirements


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EC 2: Concise:

Second order effects may be ignored if they are less than 10% of thecorresponding first order effects

Second order effects may be ignored if the slenderness, is less thanlim where

lim = 20 A B C/n

With biaxial bending the slenderness should be checked separately foreach 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 wi th 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, includingthe effect ofimperfections, 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


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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 onlyfrom 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


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EC 2: Concise:

l

M

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

2

2

1

1

45,01

45,01

k

k

k

kl0 = 0.5lBraced members:

Fig f)

Unbraced members:Fig g)

k

k

k

k

kk

kk 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 beignored

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

Assuming that the beams are symmetr ical about t he column and thei r

sizes are t he same in t he two storeys shown, t hen:

k1 = k2 = EI/`lcol / 2EI/ lbeams= EI/`lcol / 2 x 2EI/ lbeams 0.1

Although not stated effective lengths can be used?

Typical Column EffectivePD 6687 Cl.2.10 -


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EC 2: Concise:

lo = Fl

Typical Column Effective Length- -

l is clear height

Conservative touse the method inBS 8110 Table3.19

EC 2: Concise:

2nd order effects Slender columns

Methods of analysis

The methods of analysis include a general method. Basedon non-linear second order analysis and the following twosimplified methods:

Method based on nominal stiffness

Method based on nominal curvature

This method is primarily suitable for isolated members withconstant normal force and defined effective length. The

method gives a nominal second order moment based on adeflection, which in turn is based on the effective length andan estimated maximum curvature.

Cl 5.8.5, cl 5.8.8 5.6.2.1


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EC 2: Concise:

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

of imperfections [At about mid-height]. May betaken 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 6687advises 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 2

nd order moment in slender columns = NEde2

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)l02/2

1/r = KrK/r0 where 1/r0 = yd /(0.45d)

Kr = (nu n)/(nu-nbal) 1

K = 1 + ef 1

= 0.35 + fck /200 /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

MEdzMEdy

a = 2

a = 1

a = 1.5

Biaxial bending for rectangular

column


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h4b

min 12

As,min=0,10NEd/ fyd but 0,002 Ac

As,max=0.04Ac (0,08Acat laps)

Minimum number of bars in a circular column is 4.

Where direct ion of longit udinal bars changes more t han1:12 the spacing of t ransverse reinforcement should becalculated.

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)


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Worked Example

The structural grid is 7.5 m in each directionWorked Examples to EC2 - Example 5.1

38.5 kN.m

38.5 kN.m

Solut ion effect ive length

Using PD 6687 method

Clear span is 4000 250 = 3750 mm

14.0

7500

1225037502

3750

12300

2 3

4

b

b

c

c

L

EI

L

EI

k

From Table 4 of How t oColumns

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


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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 = 40.4

lim > (26.8) ...column is not slender M2 = 0


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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

Solut ion 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 10

6/(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 notconcentrated in the corners


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Interact ion Chart

Asfyk/bhfck

0.22

Interact ion Chart

Asfyk/bhfck

0.24


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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)

Solut ion determine As

Workshop Problem

Design column C2between ground and firstfloors for bending aboutaxis parallel to line 2.

Assume the following:Axial load: 7146kNTop Moment: 95.7kNNominal cover:35mmPinned baseBay width is 6.0 mElastic modulus is thesame for column and slab


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Take beamwidth as, say,half the baywidth

Solution - effective length

41.0

9600

1230030002

8600

1230030002

4200

12500

2 33

4

b

b

c

c

L

EI

L

EI

k

From Table 4 of How t oColumns

F = 0.86 lo

= 0.86 x 4.2 = 3.612 m

Check slenderness:

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

Using PD 6687 method

Clear 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)


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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 = 26.1

lim > (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


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

d2/h = 59/500 = 0.118

MEd/(bh2fck) = 160.0 x 10

6/(5003 x 50) = 0.026

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


Interact ion Chart

A sf yk

/bhf c

k

0.09


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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


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Fire

a AxisDistance

Reinforcement cover

Axis distance, a, t o 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


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High strength concrete

Basis of fire design

Material properties

Tabulated data

Design pr ocedures

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 StructuralFire 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 = fi EdEd 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)

Simpli fi ed 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 basisconfirmed by experience and theoretical evaluation of tests

Values are given for normal weight concrete made wit h siliceousaggregates

For calcareous or lightweight aggregates minimum dimensionmay be reduced by 10%

No fur ther checks are required for shear, torsion or anchorage

No furt her checks are required for spalling up to an axisdistance of 70 mm

For HSC (> C50/60) t he minimum cross sect ion di mension shouldbe 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 slabstreated separately

Walls depend on exposure conditions

Columns depend on load and slenderness

EC 2: Concise:

Cont inuous BeamsTable 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 columnscan be considered braced ifthere are columns outside thefire zone


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EC 2: Concise:

fi = NEd,fi/NRd = Gk + 1,1 Qk,1/ (1. 35Gk + 1.5 Qk) Conservatively 0.7

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

NRd is the design axial r esistance at nor mal temperature

The minimumdimensions arelarger 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


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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


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EC2 distinguishes between explosive spall ingthat can occurin concrete under compressive conditions, such as incolumns, and the concret e fall ing off the soffit in thetension zones of beams and slabs.

Explosive spallingoccurs early on in the fire exposure and ismainly caused by the expansion of the water/steam particlestrapped in the matrix of the concrete. The denser theconcrete. the greater the explosive force.

Unlikely if moisture content is less than 3% (NDP) byweight

Tabular data OK for axis distance up to 70 mm

Fall ing off of concrete occursin the latter stage of fireexposure

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 C60/75For 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


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For C 55/67 to C 80/95 the rules for normal strength concreteapply. provided that the maximum content of silica fume is lessthan 6% by weight.

For C 80/95 to C 90/105 there is a risk of spalling and at leastone 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, slabsand 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

MRd,fi,Span

MRd2,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 ais 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: Simpli f ied Calculat ionMethod for Beams and Slabs

500C Isotherm Method

Ignore concrete > 500C


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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 resistancerequired

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 = 350/57Column is 350mm, axis distance = 57mm

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

Increase nominal cover to 40mm.

Documents

Lecture 3 Columns_nov 11_leeds