lecture 3 columns
TRANSCRIPT
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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
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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
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• 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
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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
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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
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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
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• 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.