Practical Design to Eurocode 2 Columns
Strain compatibility Steel (B600) Stress
Steel (B500)
Concrete (C30/37)
ε
c1
= 0.0022
ε
cu1
= 0.0035
Strain
1
Bending with/without Axial Concise Figure 6.3 Load c / εcu2 c u2)h (1- ε c2 2/ ε
EC2 Figure 6.1
or
c / εcu3 c u3)h (1- ε c3 3/ ε
B
As2 h
d
C Ap
∆ε p
A
ε p(0) p(0)
As1
ε s , ε p
ε y
ε ud
0
A reinforcing steel tension strain limit
ε c2 (ε c3 )
ε cu2 (ε cu3 )
ε c
B concrete compression strain limit C concrete pure compression strain limit
Minimum eccentricity: e0 = h/30 but
≥ 20 mm
Column Design Chart -
Figure 15.5
2
Column Design Chart -
Figure 15.5
Geometric Imperfections Cl. 5.2
5.5
Deviations in cross-section dimensions are normally taken into account in the material factors factors and should not be included in structural analysis Imperfections need not be considered for SLS Out-of-plumb is represented by an inclination, θl θ l = θ 0 α h α m where θ 0 = 1/200 α h = 2/√l; 2/3 ≤ α h ≤ 1 α m = √(0.5(1+1/m)) l is the length or height (m) (see 5.2(6)) m is the number of vert. members
3
Isolated Members Figure 5.1a
5.5.2 e i
e i N
N
N
N H i
Braced
Unbraced
H i
l = l 0 / 2
l = l 0
ei = θ i l0/2 for walls and isolated columns ei = l0/400 or H i = θ iN for unbraced members H i = 2θ iN for braced members
Structures Figure 5.1b
θ i
H i
Figure 5.5
N a N b
l
θ i /2 H i
θ i /2
θ i N a N b
Bracing System
Floor Diaphragm
H i = θ i (N b-N a)
H i = θ i (N b+N a)/2
Roof H i = θ i N a
4
Second Order Effects with Axial Load (1) (5.8.2, 5.8.3.1 and 5.8.3.3) • 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)
⋅ Σ EcmIc/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
Column Design Process Determine the actions on the column Determine the effective length,
l0
Determine the first order moments Determine slenderness, λ λ Determine slenderness limit, λ λlim Is λ λ
≥ λ λlim ?
Yes
Column is slender
No Column is not slender, MEd = M02
Calculate As (eg using column chart) Check detailing requirements
5
Second Order Effects with Axial Load 5.6.1
Cl. 5.8.2, 5.8.3.1 Second order effects may be ignored if they are less than 10% of the corresponding first order effects
Slenderness λ = l0/i where i = √(I/A) hence for a rectangular section for a circular section
λ = 3.46 l0 / h λ = 4 l0 / h
Second order effects may be ignored if the slenderness, λ is less than λ lim where λ lim = 20 A B C √( Ac f cd/N Ed) With biaxial bending the slenderness should be checked separately for each direction and only need be considered in the d irections where λlim is exceeded
Slenderness Limit (5.8.3.1) Cl. 5.8.3.1 llim
5.6.1.4
= 20 ⋅A⋅B⋅C/√n
where: A = 1 / (1+0,2 ϕ ef )
(5.13N) ϕ ef is the effective creep ratio; (if ϕ ef is not known, A = 0.7 may be used)
B = √(1 + 2ω)
w = A f s yd / ( Ac f cd) (if ω is not known, B = 1.1 may be used)
C = 1.7 - r m
r m = M01/M02 M01, M02 are first order end moments, including the effect of imperfections M02 ≥ M01 (if r m is not known, C = 0.7 may be used)
M02 = Max{|Mtop|;|Mbot|}+ei N Ed ≥e0N Ed M01 = Min {|Mtop|;|Mbot|} n = N Ed / ( Ac f cd)
6
Factor C 100 kNm
100 kNm
100 kNm
100 kNm
-100 kNm r m
C
r m
= M01/ M02 = 0 / 100 =0 = 1.7 – 0 = 1.7
C
= M01/ M02 = -100 / 100 = -1 = 1.7 + 1 = 2.7
r m
C
= M01/ M02 = 100 / 100 =1 = 1.7 – 1 = 0.7
Different Column End Restraints Figure 5.7, 5.8.3.2
Figure 5.6, 5.6.1.2
λ = l0/i θ
θ
l0
=
l
2l
Braced members:
l0
= 2l
l0
= 0.7l
l0
=
l/ 2
l0
=l
l /2
l
M
l
l0
>
k 1 k 2 1 + ⋅ 1 + 0,45 + k 1 0,45 + k 2
l0 = 0.5l⋅
k ⋅ k k 1 k 2 ⋅ 1 + 1 + 10 ⋅ 1 2 ; 1 + k 1 + k 2 1 + k 1 1 + k 2
Unbraced members: l0 = l⋅max
k = (θ / M)⋅ (E Ι / l)
7
Typical Column Effective PD 6687 Cl.2.10
-
Non failing column
End 1 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)
End 2 Non failing column
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 Effective Length
-
lo
=
Fl
8
Nominal Curvature Method Cl. 5.8.8.2
5.6.2.2
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 N Ed ≥e0N Ed M01 = Min {|Mtop|;|Mbot|}
Moments in Slender Columns 2nd Order moments
Combination of moments
1st Order moments
Typical braced column
Combination of moments
2nd Order moments 1st Order moments
Typical unbraced column
9
Nominal Curvature Method -
Figure 5.10
Second order moment Cl. 5.8.8
5.6.2.2
M2 = N Ed e2 e2 = (1/r )l02/π 2
1/r = K rK ϕ/r 0
where 1/r 0 = ε yd /(0.45d )
K r = (nu –n)/(nu-nbal) K ϕ = 1 + βϕ ef
≤1
≥1
β = 0.35 + f ck /200 – l /150
10
Biaxial Bending Cl. 5.8.9
5.6.3
M Edz MRdz
a
M Edy + M Rdy
a
≤ 1,0
For rectangular cross-sections N Ed/N Rd 0.1 0.7 1.0 1.0 1.5 2.0 a where N Rd = Ac f cd + A f s y d For circular cross-sections
a = 2.0
Biaxial bending for rectangular column N Rd a =
2
N Ed a =
1.5
a =
M Edy
1
M Edz
11
Columns (1) (9.5.2) • h ≤ 4b • φ min ≥ 12 • As,min = 0,10N Ed /f yd 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 (2) (9.5.3) ≤ 150mm
s cl,tmax
• s cl,tmax = 20 × φ min ; b; 400mm
≤ 150mm
• s cl,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
12
Worked Example
The structural grid is 7.5 m in each direction
Solution – effective length Say half bay width for flat slab
Using PD 6687 method EI c k
=
L c
∑
2EI b
L b
300
=
4
12
3750 2
× 3750 × 250
3
12
= 0.14
7500
From Table 4 of How to…Columns
F = 0.79 ∴ lo = 0.79 x 3.750 = 2.96 m Check slenderness: λ λ = 3.46 lo/h = 3.46 x 2.96 / 0.3 = 34.1
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Solution - slenderness
Solution – design moments
14
Solution – determine As
Interaction Chart
Asf yk/bhf ck
0.23
15
Interaction Chart
Asf yk/bhf ck
0.25
Solution – determine As
16
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
17
Fire
Structural Fire Design Part 1-2, Fig 5.2
Figure 4.2
Scope
Part 1-2 Structural fire design gives several methods for fire engineering Tabulated data for various elements is given in section 5 Reinforcement cover Axis distance, a, to centre of bar a = c + φ φm /2 + φ φl
a
Axis Distance
18
Eurocode 2: Part 1.2 Structural Fire Design • General • Basis of fire design • Material properties • Design procedures – Simplified and advanced calculation methods – Shear and torsion – Spalling – Joints – Protective layers
100 Pages
• Tabulated data • High strength concrete • Annexes A, B, C, D and E
Chapter 2: Basis of Fire Design • 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
19
Chapter 2: Basis of Fire Design • Verification methods E d,fi ≤ Rd,fi(t) • Member Analysis E d,fi = η fi E d E d is the design value for normal temperature design η fi is the reduction factor for the fire situation η fi = (Gk + ψ fi Q k.1)/(γ GGk + γ Q.1Q k.1) ψ fi is taken as ψ ψ1 or ψ ψ 2 (= ψ ψ1 - NA)
Design Procedures • • •
Tabulated data (Chapter 5) Simplified calculation methods Advanced calculation method
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Which method?
Section 5. Tabulated Data Cl. 5.1
-
Provides design solutions for the standard fire exposure up to 4 hours
• 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
21
Elements • Approach for Beams and Slabs very similar – Separate tables for continuous members – One way, two way spanning and flat slabs treated separately
• Columns depend on load and slenderness • Walls depend on exposure conditions
Continuous Beams 1992-1-2 Table 5.6
Table 4.6
Standard fire resistance
Minimum dimensions (mm) Possible combinations of a andb min wherea is the average axis distance and b min is the width beam of
Web thickness
b w
R 30
b min= 80 a = 15*
160 12*
80
R 60
b min= 120 a = 25
200 12*
100
R 90
b min= 150 a = 35
250 25
110
R 120
b min= 200 a = 45
300 35
450 35
500 30
130
R 180
b min= 240 a = 60
400 50
550 50
600 40
150
R 240
b min= 280 a = 75
500 60
650 60
700 50
170
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Flat Slabs 1992-1-2 Table 5.9
Table 4.8
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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Columns: Method A Table 5.2a
Table 4.4A
The minimum dimensions are larger than BS 8110
µ fi
= N Ed,fi/
N Rd
where
N Ed,fi N Rd
is the design axial load in the fire condition
is the design axial resistance at normal temperature
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 / N 0Ed,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 N 0Ed,fi ≤ 0.25 with emax= 100 mm • Amount of reinforcement,
ω = As f yd / Ac f cd ≤ 1
For other values of these parameters see Annex C
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Spalling • 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
High Strength Concrete Tabulated Data 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 C60/75 For Class 2: 1.3 for C70/85 to C80/95
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High Strength Concrete Spalling • 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.
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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Annex E: Simplified Calculation Method for Beams and Slabs M Rd1,fi M Rd2,fi M Ed,fi = w Ed,fil eff /
8
1
M Rd,fi,Span
1
- Free moment diagram for UDL under fire conditions
• MRd,fi,Span = (γ s /γ s,fi ) ks(θ ) MEd ( As,prov / As,req) • MRd,fi,Support = (γ s /γ s,fi ) MEd ( As,prov / As,req) (d -a)/d Where
a is the required bottom axis distance given in Section 5 As,prov / As,req should not be taken greater than 1.3
500°C Isotherm Method Ignore concrete > 500°C
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Zone Method
Divide concrete into zones and work out avera ge 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 • f ck = 30MPa
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