Transcript
EC3 Notes Source: http://www.qub.ac.uk/structural_eu http://www.qub.ac.uk/structural_eurocodes/eurocode3 rocodes/eurocode3.html .html
1
Steel Design to Eurocode 3
•
EN 1993-5 Piling
•
EN 1993-6 Crane supporting structures
Introduction Eurocode 3 Part 1 has 12 sub-parts: sub-parts:
Development of Eurocode 3
•
EN 1993-1-1 General Rules
Aim: Aim: to create a common structural language
•
EN 1993-1-2 Fire
•
EN 1993-1-3 Cold-formed thin gauge
•
EN 1993-1-4 Stainless steel
•
EN 1993-1-5 Plated elements
•
EN 1993-1-6 Shells
•
EN 1993-1-7 Plates transversely loaded
•
EN 1993-1-8 Joints
•
EN 1993-1-9 Fatigue
•
EN 1993-1-10 Fracture Toughness
•
EN 1993-1-11 Cables
•
EN 1993-1-12 High strength steels
– And make allowances for National Choice through the use of a National Annex
National Annex •
•
Eurocode 3 allows some parameters and design methods to be determined at a national level. Where a national choice is allowed, this is indicated in the Eurocodes under the relevant clause. – values or methods to be used in a particular country are given in the t he National Annex.
Nationally Determined Parameters (NDPs) •
•
The recommended recommended values of the parameters and design methods are collectively referred to as Nationally Determined Parameters (NDPs). NDPs determine various aspects of design but perhaps most importantly the level of safety of structures str uctures during construction and service.
Key Differences between EC3 and BS 5950 There are several differences between EC3 and BS 5950: BS 5950 Structure
EC3 Structure
Separate sections for different elements types
Sub-parts are based on structural phenomena
Structure of Eurocode 3
e.g. Beams,
Eurocode 3 is broken into 6 parts: parts:
Plate Girders,
e.g. Tension, Compression, Bending, Shear
•
EN 1993-1 Generic rules
•
EN 1993-2 Bridges
•
EN 1993-3 Towers, masts and chimneys
•
EN 1993-4 Silos, tanks and pipelines
Compression members...
Sub-parts can be applied to any element The arrangement of the sub-parts means less duplication of rules
Different Axes
Informative subscripts BS 5950
Eurocode 3
Along the member
‘Ed’ means design effect
X
‘Rd’ means design resistance
Major Axis
X
Y
Therefore:
Minor Axis
Y
Z
NEd is an design axial force NRd is the design resistance to the axial force
Gamma Factors Partial factor γM
UK NA value
Application
γM0
1.00
Cross-sections
γM1
1.00
Member Buckling
γM2
1.25
Fracture
Figure 1 (Source: Arya (2009) Design of Structural Elements Pg.377)
Different Wording ‘Action’ – force or imposed displacement
Permanent action (Dead Load) Variable action (Live Load)
‘Effect’ – internal force or moment, deflections
Omissions Notable omissions:
‘Verification’ –check
•
‘Resistance’ – capacity
Effective lengths –
•
Different Symbols
Formulae for Mcr –
BS 5950
EC3
BS 5950
EC3
BS 5950
EC3
A
A
P
N
py
f y
Z
Wel
Mx
My
pb
χLTf y
S
Wpl
V
V
pc
χf y
Ix
Iy
H
Iw
r
i
Iy
Iz
J
It
•
Use BS 5950 effective lengths
Use SN003 NCCI Document
Deflection limits –
Refer to National Annex
2 Combination Factors ψ
Loading Introduction to EN 1990
Covers the „Basis of Structural Design‟
Use with the other Eurocodes
Gives safety factors needed for ULS and SLS verifications
partial factors (see Table 1)
combination factors (See Table 2)
Action Imposed loads in buildings, Category A : domestic/residential areas Category B : office areas Category C : congregation areas Category D : shopping areas Category E : storage areas Category F : traffic area, < 30kN Category G : traffic area, 30 – 160 kN Category H : roofs Snow (sites up to 1000m)
ψ0
ψ1
ψ2
0.7 0.7 0.7 0.7 1.0 0.7 0.7 0.7
0.5 0.5 0.7 0.7 0.9 0.7 0.5 0
0.3 0.3 0.6 0.6 0.8 0.6 0.3 0
0.5
0.2
0
Wind
0.5
0.2
0
Table 2: Extract from Table NA.A1.1
ULS Checks EQU: static equilibrium
Combinations of Actions
STR: strength/buckling etc
Can use either:
Equation 6.10
Less favourable of 6.10a and 6.10b
GEO: Failure of excessive deformation of ground
Method: Get the factors from Tables 1 and 2 FAT: fatigue failure
and substitute them into the equation you are using, check for a range of different loading combinations and take the least favourable
Actions
result.
Permanent actions , G (Dead loads)
Equation 6.10 Variable actions , Q (Live loads)
ΣγG,jGk,j “+” γPP “+” γQ,1Qk,1 “+” ΣγQ,iψ0,iQk,i
Qk
Characteristic value (ψ = 1.0)
ψ0Qk Combination value
ψ1Qk Frequent value
ψ2Qk Quasi-permanent value
Equation 6.10a Partial Factors
ΣγG,jGk,j “+” γPP “+” γQ,1ψ0,iQk,1 “+” ΣγQ,iψ0,iQk,i
Unfavourable
Favourable
γG
1.35
1.0
γQ
1.5
0
Table 1: Partial Factor values from the UK NA
Equation 6.10b Σξ jγG,jGk,j “+” γPP “+” γQ,1Qk,1 “+” ΣγQ,iψ0,iQk,i ξ j is 0.925 (From NA 2.2.3.2)
3
Steel Design to Eurocode 3 Structural Analysis
The choice between a first- and second- order analysis should be based on:
the flexibility of the structure
in particular, the extent to which ignoring
Analysis Types
second-order effects might lead to an unsafe
There are four types of global analysis:
approach due to underestimation of some of the internal forces and moments.
Analysis Type First-order elastic
y r t e m y r o t e e m G o d e e G m l r a f o i t i e n D I
First-order plastic Second-order elastic
Clause 5.2.1(2) states that second order effects shall be considered:
Second-order elastic
l a i r e t a r u m o r i a v a e h n i e L b
l a i r e t a m r r a u e o n i i v l - a n h o e N b
•
if they increase the action effects significantly
•
or modify significantly the structural behaviour
Table 1: Summary of Analysis Types
First-Order Analysis A first-order analysis may be used if the following criteria is satisfied: αcr ≥ 10 for elastic analysis αcr ≥ 15 for plastic analysis αcr =Fcr /FEd
Figure 1: Load-Deformation graph for different analysis types (Source: Designer’s Guide to EN 1993-1-1 Page 21)
αcr
is the factor by which the design loading would have to increased to cause elastic instability in a global mode ( λcr in BS 5950-1)
FEd
is the design loading on the structure
Fcr
is the elastic critical buckling load for global instability based on initial elastic stiffness.
Joints Clause 5.1.2 deals with joint modelling
For portal frames (with shallow roof slopes less than 26°) and beam and column plane frames:
Eurocode 3 recognises the same three types of joint, in terms of their effect on the behaviour of the frame structure, as BS 5950: Part 1.
HEd
is the horizontal reaction at the bottom of the storey
VEd
is the total vertical load at the bottom of the storey
δH,Ed is the horizontal deflection at the top of the storey under consideration relative to the bottom of the storey, with all horizontal loads applied to the structure. Figure 2: Joint stiffness effects (Source: SCI CPD Course Material)
h
is the storey height.
Amplifier
(h is the height of the structure in metres)
If 10 > αcr ≥ 3.0
α m is the reduction factor for columns
Increase all lateral loads by the amplifier: Limits on
αcr
Action
αcr >10
First order Analysis First order analysis plus amplification 10>αcr >3 or effective length method αcr ≤ 3 Second order analysis Table 2: Actions to be taken once α cr has been calculated
(m is the number of columns contributing to the effect on the bracing system)
Summary 1) Model the Frame
Imperfections
2) Put all the loads on the frame (Including the EHFs) 3) Calculate αcr 4) Check to see if second-order effects are significant
5) If necessary use the amplifier
Figure 2: Typical Imperfections that will be present when designing a structure
Frame imperfections appear in (almost) every load case. We can represent initial sway imperfections by using Equivalent Horizontal Forces (EHFs) which are based on 1/200 of the factored vertical load, with reduction factors.
Figure3:Replacing initial sway imperfections with equivalent horizontal forces
EHF = φ x Vertical Forces φ
= φ0α hα m
φ0
= 1/200 = 0.005
α h is the reduction factor for height:
4
Steel Design to Eurocode 3
S275 J0
- Charpy value of 27J can be
•
obtained at 0°C
Brittle Fracture
S275 J2
- Charpy value of 27J can be
•
obtained at -20°C
Steel sub-grade selection
EN 1993-1-10 Brittle failure is most likely to occur at very low temperatures. It should be considered where there are tensile stresses. It can be avoided by choosing a steel with sufficient fracture toughness Failure mainly dependent on: Steel strength grade
•
•
Thickness
•
Lowest service temperature
•
Material toughness
•
Tensile Stress
•
Notches or defects in the element
Steel toughness Steel toughness is measured by Charpy V-notch value. The Charpy test measures how much energy is absorbed by a steel sample, at a given temperature.
•
S275 JR
- Charpy value of 27 J can be
obtained at +20°C
The method given in the Eurocodes can be quite complex to use, it is recommended that you use Published Document PD 6695 instead. The service temperature is lowered i.e. i t becomes a reference temperature. Refer to table 2.1 of the Eurocodes so determine the steel sub grade, below is an extract from that table.
f y(t) f y(t) = f y,nom 0.25 (t/t0) –
but t0 = 1mm, so f y(t) = f y,nom 0.25 (t) . –
PD 6695-1-10 •
Published Document is much Simpler to use –
Internal Tmd is -5°C (Table 2)
–
External is Tmd -15°C (Table 3)
NOTE: Can only use this document for design in the UK
PD 6695-1-10 Tables Table 2 Maximum thicknesses for internal steelwork in buildings for T md = -5°C
Table 3 Maximum thicknesses for external steelwork in buildings for Tmd = -15°C
5
Steel Design to Eurocode 3 Local Buckling and CrossSection Classification In Eurocode 3 you will need to refer to the following clauses when classifying a section and determining the cross-sectional resistance: •
Clause 5.5 covers the cross section classification
•
Clauses 6.1 and 6.2 covers the cross-sectional resistance
Sections with slender webs or flanges will be more susceptible to local buckling, where the element will fail before the design strength is reached. Eurocode 3 takes into account the effects of local through the process of cross section classification.
Classes
Class 2 cross-sections are those which can develop their plastic moment resistance, but have limited rotation capacity because of local buckling. Class 3 cross-sections are those in which the stress in the extreme compression fibre of the steel member assuming an elastic distribution of stresses can reach the yield strength, but l ocal buckling is liable to prevent development of the plastic moment resistance. Class 4 cross-sections are those in which local buckling will occur before the attainment of yield stress in one or more parts of the cross-section.
Limits The limits between the classes depend on the ε factor which is calculated using f y, the yield strength of the steel.
ε Factor
BS 5950
EC3
Plastic
Class 1
Compact
Class 2
Semi-compact
Class 3
Slender
Class 4
BS 5950
EC3
ε = (275/py)0.5
ε = (235/f y)0.5
Values of ε are given at the bottom of Table 5.2: f y
235
275
355
420
460
ε
1.00
0.92
0.81
0.75
0.71
EN 1993-1-1 Table 5.2
Class 1 Class 2 Class 3
f y Yield Strength Class 4
Image Source: http://www.steel-insdag.org/new/pdfs/Chapter8.pdf
Similarly to BS 5950, cross sections will be placed into one of four behaviour classes. Class 1 is the least susceptible to local buckling and class 4 is the most susceptible. The classification of a section will depend mainly on:
The UK National Annex says that material properties should be taken from the product standards. Extract from EN 10025-2 - f y (yield strength) values for hot rolled steel: 2
f y (N/mm ) nominal thickness of element, t (mm) Steel Grade 6 1 ≤ t
The material yield strength, f y c/t ratio
3 6
0 8
≤ t
≤ t
≤ t
< 6 1
<
<
0 4
3 6
0 4
Eurocode 3 defines the classes in Clause 5.5.2:
S 275
275
265
255
245
Class 1 cross-sections are those which can form a plastic hinge with the rotation capacity required from plastic analysis without reduction of the resistance.
S 355
355
345
335
325
EN 10025-2 (Table 7)
Class 3: Semi-compact
c/t Width-to-Thickness Ratio The width-to-thickness ratios differ in EC3 differs from BS 5950: BS 5950
EC3
s t i m i L
Flange outstand Web in bending Web in compression
BS (Table 11)
EC3 (Table 5.2)
b/T = < 15 ε
c/tf = < 14 ε
d/t = < 120 ε
d/tw = < 142 ε d/tw = < 42 ε
Class 4: Slender
Outstand Flange
b = B/2
c = (b – tw – 2 r)/2
Internal Compressio n Part
d= D – 2 T – 2 r
c= h – 2 tf – 2 r
An element that doesn’t meet the class 3 limits should be taken as a class 4 section. Effective widths are assigned to Class 4 compression elements to make allowance for the reduction in resistance as a result of local buckling To calculate the effective width of a Class 4 section, refer to the relevant section in the Eurocodes:
Appropriate values of c and t are defined at the top of Table 5.2 for different types of sections.
Table 5.2
Section Type
Reference
Cold-formed sections
EN 1993-1-3
Hot-rolled and fabricated section
EN 1993-1-5
CHS
EN 1993-1-6
Internal compression parts and outstand flanges are assessed against the limiting width to thickness ratios for each class. The limits are provided in table 5.2. Table 5.2 is made up of three sheets:
Overall Cross-Section Classification
Sheet 1 – Internal Compression Parts Sheet 2 – Outstand Flanges
Clause 5.5.2(6) states that a cross-section is classified according to the highest (least favourable) class of its compression parts.
Sheet 3 – Angles and Tubular Sections
Summary
Cross-section Classification
1. Class 1: Plastic
s t i m i L
Flange outstand Web in bending Web in compression
BS (Table 11)
EC3 (Table 5.2)
b/T = < 9 ε
c/tf = < 9 ε
d/t = < 80 ε
d/tw = < 72 ε
2. Determine ε from Table 5.2 3.
s t i m i L
Substitute the value of ε into the class limits in Table 5.2 to work out the class of the flange and web Flange outstand limiting value, c/tf
Web in bending limiting value, d/t w
Class 1
9ε
72 ε
Class 2
10 ε
83 ε
Class 3
14 ε
124 ε
d/tw = < 33 ε
Class 2: Compact
Flange outstand Web in bending Web in compression
Determine f y (UK NA recommends you use the product standards)
BS (Table 11)
EC3 (Table 5.2)
b/T = < 10 ε
c/tf = < 10 ε
d/t = < 100 ε
d/tw = < 83 ε d/tw = < 38 ε
Class 4
4.
If it does not meet Class 3 requirements, the section is classified as Class 4
Take the least favourable class from the flange and web results
6
Steel Design to Eurocode 3
Section Modulus, W
Restrained Beams
Subscripts are used to identify whether or not the section modulus is plastic or elastic and the axis about which it acts.
Elastic modulus about the major axis
BS 5950 Z xx
Wel,y
Elastic modulus about the minor axis
Z yy
Wel,z
Plastic modulus about the major axis
S xx
Wpl,y
A beam is considered restrained if:
The section is bent about its minor axis Full lateral restraint is provided Closely spaced bracing is provided making the slenderness of the weak axis low The compressive flange is restrained again torsion The section has a high torsional and lateral bending stiffness
There are a number of factors to consider when designing a beam, and they all must be satisfied for the beam design to be adopted: Bending Moment Resistance Shear Resistance Combined Bending and Shear Serviceability
Bending Moment Resistance In Eurocode 3: Clause 6.2 covers the cross-sectional resistance Clause 6.2.5 deals with the crosso sectional resistance for bending.
EN 1993-1-1 Clause 6.2.4 Equation 6.12 states that the design moment (M Ed) must be less than the design cross-sectional moment resistance (Mc,Rd) (6.12)
The equation to calculate M c,Rd is dependent on the class of the section. A detailed assessment of cross-section classification can be found in the ‘Local Buckling and Cross-Section Classification’ handout. For Class 1 and 2 cross-sections: Mc,Rd = Mpl,Rd = Wplf y/ɣM0(6.13) For Class 3 cross-sections: Mc,Rd = Mel,Rd = Wel,minf y/ɣM0 (6.14) For Class 4 cross sections: Mc,Rd = Weff,min f y/ɣM0 γM0 =1.0
(6.15)
EC3
Plastic modulus about the minor axis S yy Wpl,z Table 1.0 Section modulus terminology comparison between BS 5950 and EC3
Cross-section Classification Summary 1. Get f y from Table 3.1 2. Get ε from Table 5.2 3. Substitute the value of ε into the class limits in Table 5.2 to work out the class of the flange and web 4. Take the least favourable class from the flange outstand, web in bending and web in compression results to get the overall section class
Bending Moment Resistance Summary 1. Determine the design moment, MEd 2. Choose a section and determine the section classification 3. Determine Mc,Rd, using equation 6.13 for Class 1 and 2 cross-sections, equation 6.14 for Class 3 cross-sections, and equation 6.15 for Class 4 sections. Ensure that the correct value of W, the section modulus is used. 4. Carry out the cross-sectional moment resistance check by ensuring equation 6.12 is satisfied.
Shear Resistance In Eurocode 3: Clause 6.2 covers the cross-sectional resistance Clause 6.2.6 deals with the crosso sectional resistance for shear.
EN 1993-1-1 Clause 6.2.6 Equation 6.17 states that the design shear force (V Ed) must be less than the design plastic shear resistance of the crosssection (Vpl,Rd) (6.17)
(6.18)
Shear Resistance Summary 1. Calculate the shear area, Av
γM0 =1.0
2. Substitute the value of Av into equation 6.18 to get the design plastic shear resistance
Shear Area, Av EC3 should provide a slightly larger shear area compared to BS 5950 meaning that the overall resistance will be larger as shown in Figure 1.
3. Carry out the cross-sectional plastic shear resistance check by ensuring equation 6.17 is satisfied.
Serviceability Deflection checks should be made against unfactored permanent actions and unfactored variable actions.
Figure 1: Differences in shear area calculated using BS 5950 and EC3 Type of member
Shear Area, Av
Rolled I and H sections (load parallel to web)
Av = A – 2btf + (tw + 2r)tf but ≥ ηhwtw
Rolled Channel sections (load parallel to web)
Av = A – 2btf + (tw + r)tf
Rolled PHS of uniform thickness (load parallel to depth)
Av =Ah/(b+h)
CHS and tubes of uniform thickness
Av =2A/π
Plates and solid bars
A v =A
Table 2.0: Shear area formulas Term A
Figure 2: Standard case deflections and corresponding maximum deflection equations
Definition Cross-sectional area
b
Overall breadth
h
Overall depth
hw
Depth of web
The maximum deflection calculated must not exceed the deflection limit. The deflection limits are not given directly in Eurocode 3, instead, reference must be made to the National Annex. Design Situation
Deflection limit Length/180
r
Root radius
Cantilever
tf
Flange thickness
tw
Web thickness (taken as the minimum value is the web is not of constant thickness)
Beams carrying plaster of other brittle finish Other beams (except purlins and sheeting rails)
η
Constant which may be conservatively taken as 1.0 Table 3.0: Shear area parameter descriptions
Figure 1: Visual definition of the parameters used in the shear area calculation. (Source: Blue Book)
Span/360 Span/200
To suit the characteristics of particular cladding Table 4.0: Vertical Deflection Limits from NA 2.23 Clause 7.2.1(1) B
Purlins and sheeting rails
7 Steel Design to Eurocode 3
(6.54)
Unrestrained Beams Beams without continuous lateral restraint are prone to buckling about their major axis, this mode of buckling is called lateral torsional buckling (LTB). This handout is a continuation of the ’Restrained Beams’ one and covers the design of unrestrained beams that are prone to lateral torsional buckling.
(6.55)
where γM1 =1.0 (from UK NA)
Section Modulus Wy For Class 1 and 2 cross-sections: Wy = Wpl,y
Lateral torsional buckling can be discounted when:
For Class 3 cross-sections:
The section is bent about its minor axis Full lateral restraint is provided Closely spaced bracing is provided making the slenderness of the weak axis low The compressive flange is restrained again torsion The section has a high torsional and lateral bending stiffness The non-dimensional slenderness, < 0.2
Eurocode 3 Approach There are three methods for calculating the LTB resistance of a member in Eurocode 3: 1. 2. 3.
Primary method (Clauses 6.3.2.2 and Clauses 6.3.2.3) Simplified assessment method (Clause 6.3.2.4) General method (Clause 6.3.4)
Note: This handout will only deal with the primary method.
General and Special Cases When using the primary method, there are two cases which are available for you to use. The first case is the ‘General Case’ which can be used for all sections, and the second case is the ‘Special Case’ which is specifically for rolled sections of standard dimensions.
Wy = Wel,y For Class 4 cross-sections: W y = Weff,y
Yield Strength, f y The UK National Annex says that we should obtain the value of the yield strength from the product standards. Extract from EN 10025-2 - f y (yield strength) values for hot rolled steel: 2
f y (N/mm ) nominal thickness of element, t (mm) Steel Grade 6 1 ≤ t
EN 1993-1-1 Clause 6.3.2.1 Equation 6.54 states that the design moment (M Ed) must be less than the design buckling resistance moment (M b,Rd)
0 8
≤ t
≤ t
≤ t
< 6 1
<
<
0 4
3 6
S 275
275
265
255
245
S 355
355
345
335
325
Extract from EN 10025-2 (Table 7)
Reduction Factor, χLT General Case: (6.56)
The methods for both cases are very similar with the addition of a few extra parameters in the Special Case. This small amount of extra work for the Special Case is worthwhile as it provides greater resistance of the section.
LTB Resistance
3 6
0 4
where
To get αLT, determine the buckling curve that you need to use from table 6.4 and then refer to table 6.3 to get the corresponding value of αLT
(6.58)
2
f= 1- 0.5(1 - k c)[1-2.0( - 0.8) ] Crosssection Rolled I sections Welded I sections Other
Buckling curve αLT
Limits
Buckling Curve h/b ≤ 2 a h/b >2 s c h/b ≤ 2 h/b >2 d d EN 1993-1-1 Table 6.4 a
b
but f ≤1.0 kc can be obtained from Table 6.6 in the Eurocodes:
c
d
0.21 0.34 0.49 0.76 EN 1993-1-1 Table 6.3
Special Case (for rolled sections): (6.57)
where
EN 1993-1-1 Table 6.6
UK NA sets β = 0.75 and = 0.4
You will need the value of for both the general and special cases.
To get αLT, determine the buckling curve that you need to use from the table from the National Annex NA.2.17 Clause 6.3.2.3(1) and then refer to table 6.3 to get the corresponding value of αLT Cross-section Rolled bi-symmetric I and H sections and hotfinished hollow sections Angles (for moments in the major principal plane) and other hotrolled sections Welded bi-symmetric sections and coldformed hollow sections
Limits
Buckling Curve
h/b ≤ 2 2.0 < h/b ≤ 3.1
b c
(6.56)
Mcr Refer to SN003 document (NCCI) for detailed description of how to get Mcr
d
c d
h/b ≤ 2 h/b > 2
Table from NA.2.17 Clause 6.3.2.3(1)
where L is the distance between points of lateral restraint (L cr ) 2
E is the Young’s Modulus = 210000 N/mm Buckling curve αLT
a
b
c
d G is the shear modulus = 80770 N/mm
0.21 0.34 0.49 0.76 EN 1993-1-1 Table 6.3
You can use a modified value of χLT in the special case to give some extra resistance:
2
Iz is the second moment of area about the weak axis It is the torsion constant Iw is the warping constant
k is an effective length factor (usually 1.0)
Table 3.1 from SN003 (Values of C1 for members with end moments)
kw is an effective length factor (usually 1.0) where zg is the distance between the point of load application and the shear centre. The value will be positive or negative depending on where the load is applied as shown in figure 1.
Figure 3.1 from SN003
Summary Figure 1 (from SN003 document) C1 and C2 are coefficients.
1.
Draw the bending moment diagram to obtain the value of the maximum bending moment, M Ed
2.
Determine f y (UK NA recommends you use the product standards) and calculate the class of the section. Once you know the class of the section then you will know which value of the section modulus you will need to use in the equation 6.55.
3.
Work out the effective length, L cr
4.
Refer to SN003 document and work out the value of Mcr , the critical moment
5.
Work out using expression 6.56.
6.
Determine the values of αLT a. For the general case use Table 6.4 to work out the buckling curve and then refer to Table 6.3 to get a value of αLT b. For the special case, refer to the table in the National Annex (NA.2.17 Clause 6.3.2.3(1)) to get the buckling curve and then refer to Table 6.3 to get the value of αLT
7.
Work out ΦLT a. For the general case use expression 6.56 b. For the special case, use expression 6.57 Work out χLT a. For the general case use expression 6.56 b. For the special case, use expression 6.57
For transverse loading, C1 and C2 are obtained from Table 5.2 in SN003:
Table 5.2 from SN003 (C1 and C2 values for transverse loading)
For members with end moments, the value of C1 is obtained from Table 3.1 in SN003:
8.
9.
Calculate the design buckling resistance Mc,Rd using equation 6.55.
10. Carry out the buckling resistance check in expression 6.54.
8
Steel Design to Eurocode 3 Compression Members
must be less than the design cross-sectional resistance of the sections to uniform compression force (Nc,Rd) (6.9)
Columns are vertical members used to carry axial compression loads and due to their slender nature, they are prone to buckling. The behaviour of a column will depend on its slenderness as shown in Figure 1
Cross-section resistance in compression depends on cross-section classification. For Classes 1, 2 and 3: (6.10)
For Class 4 sections: (6.11)
γM0 =1.0 Figure 1 Behaviour of columns is determined by their slenderness
Cross-section Classification Summary 1. Get f y from Product Standards
Stocky Columns are not affected by buckling and the strength is related to the material yield stress f y. Nmax = Npl = Aeff f y
2. Get ε from Table 5.2 3. Substitute the value of ε into the class limits in Table 5.2 to work out the class of the flange and web
4. Take the least favourable class from the flange outstand, web in bending and web in compression results to get the overall section class
Figure 2: Resistance of columns depends on different factors
Eurocode 3 Approach To take into account the various imperfections which the Euler formula does not allow for, the Eurocode uses the Perry-Robertson approach. This is approach is the similar to that used in BS 5950. Table 1 shows the checks required for both s lender and stocky columns: Slender column > 0.2
Stocky Column < 0.2
Cross-section Resistance check, Nc,Rd
Buckling Resistance Check, Nb,Rd
Table 1.0 Resistance checks required for slender and stocky columns
Cross-Section Resistance EN 1993-1-1 Clause 6.2.4 Equation 6.9 states that the design value of the Compression force (N Ed)
For a more detailed description of cross-section classification, please refer to the ‘Cross -section Classification’ handout.
Cross-section Summary
Resistance
Check
1. Determine the design compression force 2. Choose a section and determine the section classification
3. Determine Nc,Rd, using equation 6.10 for Class 1,2 and 3 sections, and equation 6.11 for Class 4 sections.
4. Carry out the cross-sectional resistance check by ensuring equation 6.9 is satisfied.
Effective Area Aeff The effective area of the cross-section used for design of compression members with Class 1, 2 or 3 cross-sections, is calculated on the basis of the gross cross-section using the specified dimensions. Holes, if they are used with fasteners in connections, need not be deducted.
Member Buckling Resistance EN 1993-1-1 Clause 6.3.1 Equation 6.46 states that the design values of the Compression force (NEd) must be less than the buckling resistance of the compression member (N b,Rd) (6.46)
Non-dimensional Slenderness
For sections with Classes 1, 2 and 3: (6.50)
or
For Class 4 sections: (6.51)
or
Similarly to cross-section resistance, buckling resistance is dependent on the cross-section classification. For sections with Classes 1, 2 and 3: where
(6.47)
For Class 4 sections:
Imperfection Factor, (6.48)
is an imperfection factor, first you will need to determine the required buckling curve from Table 6.2 and refer to Table 6.1 to get the value of :
γM1 =1.0
Buckling Curves
Buckling Curve Imperfection Factor
Buckling curve selection is dependent on the section geometry. Table 6.2 in EN 1993-1-1 provides guidance on a range of sections.
a0
a
b
c
d
0.13
0.21
0.34
0.49
0.76
EN 1993 -1-1 Table 6.1
Reduction Factor,
Effective Buckling Lengths The effective length of a member will depend on its end conditions. EC3 gives no direct guidance on calculating the buckling length, therefore it is acceptable to use those given in BS 5950 Table 13. Some typical effective lengths are given in Figure 3.
χ
(6.49)
where Alternatively, χ may be read from Figure 6.4 in the and the required buckling Eurocodes by using curve.
Buckling Resistance Check Summary
Pinned Pinned
Fixed - Fixed
Fixed - Pinned
Figure 3: Effective Lengths for three types of end conditions
Elastic Critical Buckling Load Ncr is the elastic critical buckling load for the relevant buckling mode based on the gross properties of the cross section
1. 2. 3. 4.
Determine the design axial load, N Ed Choose a section and determine the class Calculate the effective length L cr Calculate Ncr using the effective length L cr , and E and I which are section properties 5. Calculate 6. Determine α by first determining the required buckling curve from Table 6.2 and then reading off the required value of α from Table 6.1. 7. Calculate Φ by substituting in the values of α and 8. Calculate χ by substituting in the values of Φ and 9. Determine the design buckling resistance of the member by using equation 6.47 or 6.48 and substituting in the value of χ 10. Make sure that the conditions of equation 6.46 are satisfied.
9
Steel Design to Eurocode 3
Partial Factors γM UK N.A. Value
γM
Tension Members As the tensile force increases on a member it will straighten out as the load is increased. For a member that is purely in tension, we do not need to worry about the section classification since it will not buckle locally. A tension member fails when it reached the ultimate stress and the failure load is independent of the length of the member. Tension members are generally designed using rolled section, bars or flats.
γM0
Resistance of cross-sections
1.0
γM2
Resistance of cross-sections in tension to fracture
1.25
Characteristic Strengths f y and f u The UK National Annex says you should get the values of f y and f u from the product standards. For hot-rolled sections you can use the table below. 2
Steel grade
Tensile Resistance
6 1
EN 1993-1-1 Clause 6.2.3(1) Equation 6.5 states that the design tensile force (N t,Ed) must be less than the design tensile resistance moment (N t,Rd)
≤ t
Design Plastic Resistance N pl,Rd Design Ultimate Resistance N u,Rd
f u (N/mm )
0 4
3 6
0 8
≤ t
≤ t
≤ t
<
<
<
3
6 1
0 4
3 6
< t
S 275
275
265
255
245
S 355
355
345
335
325
0 0 1
430580 510680
≤ t < 3
410560 470630
Extract from Table 7 of EN 10025-2
(6.5)
The tensile resistance is limited by the lesser of:
2
f y (N/mm )
Anet for Non staggered fasteners Anet = A – Σd0t
Design Plastic Resistance, Npl,Rd Npl,Rd is the plastic design resistance, and is concerned with the yielding of the gross crosssection. Equation 6.6 gives the expression used to calculate Npl,Rd:
Anet for Staggered Fasteners:
(6.6)
Design Ultimate Resistance, Nu,Rd Nu,Rd is the design ultimate resistance of the net cross-section, and is concerns with the ultimate fracture of the net cross-section, which will normally occur at fastener holes. Equation 6.7 gives the expression used to calculate Nu,Rd:
The total area to be deducted should be taken as the greater of: a) The maximum sum of the sectional areas of the holes on any line perpendicular to the member axis
(6.7)
b)
where:
Tension Member Design Steps Summary
t is the thickness of the plate
1. Determine the design axial load N Ed
p is the spacing of the centres of the same two holes measured perpendicular to the member axis
2. Choose a section
s is the staggered pitch of the two consecutive holes
4. Get the gross area A and the net area Anet
n is the number of holes extending in any diagonal or zig-zag line progressively across the section
3. Find f y and f u from the product standards
5. Substitute the values into the equations to work out Npl,Rd and Nu,Rd (6.6)
d0 is the diameter of the hole
Angles with welded end connections (6.7)
Clause 4.13(2) of EN 1993-1-8 states that for an equal angle, or unequal angle welded along its larger leg, the effective area = gross area.
For angles connected by a single row of bolts, use the required equation to work out N u,Rd
Angles Connected by a single row of bolts
from EN 1993-1-8 which will depend on the
Refer to EN 1993-1-8.
number of bolts. For 1 bolt: (3.11)
For 2 bolts:
For 1 bolt:
(3.12)
(3.11)
For 3 or more bolts:
For 2 bolts: (3.12)
(3.13)
For 3 or more bolts: (3.13)
6. The design tensile Resistance is the lesser of the values of Npl,Rd and Nu,Rd 7. Carry out the tension check:
Values of reduction factors β 2 and β3 can be found in Table 3.8: Pitch p1
≤ 2.5 d0
≥ 5.0 d0
β2 (for 2 bolts)
0.4
0.7
β3 (for 3 or 0.5 0.7 more bolts) Note: For intermediate values of pitch p 1 values of β may be determined by linear interpolation. EN 1993-1-8Table 3.8
(6.5)
10
Steel Design to Eurocode 3 Combined axial compression and bending
Clause 6.3.3(4) Members which are subjected to combined bending and axial compression should satisfy both:
Equation 6.61 Uniform members in bending and axial compression demonstrate complex structural behaviour
Interaction Method
Equation 6.62
When using the interaction method you will need to refer to Clause 6.3.3 of EN 1993-1-1, and you will also need to refer to Annex A or B depending on the specific method being used. where:
Clause 6.3.3(1) When checking uniform members in bending and axial compression, a distinction is made for:
members not susceptible to torsional deformation (e.g. SHS, CHS, fully restrained members)
NEd
design values of the compression force and the maximum moments about the y-y
My,Ed
and z-z axes along the member,
Mz,Ed
respectively
ΔMy,Rd
moments due to the shift of the centroidal
members susceptible to torsional deformation ΔMz,Rk
axis according to 6.2.9.3
Clause 6.3.3(2) The resistance of the cross-sections at each end of the member should also satisfy the requirements given in Clause 6.2
Χy χz
Clause 6.3.3(3) For members of structural systems the resistance check may be carried out on the basis of the individual single span members regarded as cut out of the system. Second-order effects of the sway system (P- Δ effects) have to be taken into account, whether by the end moments of the member or by means of appropriate buckling lengths respectively.
reduction factors due to flexural buckling from clause 6.3.1
reduction factor due to lateral torsional χLT
kyy, kyz, kzy, kzz
buckling from clause 6.3.2
interaction factors k ij.
Table 6.7 – Values for N Rk, Mi,Rk and ΔMi,Ed NRk = f y Ai
Mi,Rk = f yWi
Class
1
2
3
4
AI
A
A
A
Aeff
Wy
Wpl,y
Wpl,y
Wel,y
W eff,y
Wz
Wpl,z
Wpl,z
Wel,z
W eff,z
ΔMY,Ed
0
0
0
eN,yNEd
ΔMz,Ed
0
0
0
eN,zNEd
NOTE1 : For members not susceptible to torsional deformation χLT would be χLT = 1.0 NOTE 2: Nb,Rd = χ NRk/γM1
Interaction Factors k ij Interaction factors are obtained from one of two methods:
Method 1 (given in Annex A) Method 2 (given in Annex B)
Annex A (Method 1) •
Use Table A.1 of EN 1993-1-1
Equivalent uniform moment factors C mi,0 depend on the shape of the bending moments diagram and these factors are determined from Table A.2 of EN 1993-1-1
Annex B (Method 2) • •
Use Table B.1 of EN 1993-1-1 for members not susceptible to LTB Use Table B.2 of EN 1993-1-1for members that are susceptible to LTB.
Determine the equivalent uniform moment factors from Table B.3 of EN 1993-1-1.
Annex A Table A.1 Interaction factors kij for interaction formula in clause 6.3.3 (4)
Table A.2
– Equivalent
uniform moment factors Cmi,0
Annex B Table B.1 - Interaction factors kij for members not susceptible to torsional deformations
Table B.2 - Interaction factors kij for members susceptible to torsional deformations
Table B.3 Equivalent uniform factors Cm in Tables B.1 and B.2
11
Steel Design to Eurocode 3
EN 1993-1-8 Clause 2.2 •
Joints
Partial safety factors, γ M for joints are given in Table 2.1 of EC 3-8.
•
Eurocode 3 Part: 1-8
Refer to NA to get the required values of the different partial safety factors
•
Resistance of bolts and welds, γM2 = 1.25
Joint Types CL 5.2.2.2 ‘Nominally pinned’ joints are capable of transmitting internal forces without developing significant moments, and capable of accepting the resulting rotations under the design loads. CL 5.2.2.3 ‘Rigid and full strength’ joints have sufficient rotational stiffness to justify analysis based on full continuity.
Bolted Joints – Table 3.4 Table 3.4 of EN 1993-1-8 gives the different checks required for individual fasteners subjected to shear and/or tension. Checks need to be carried out for a number of possible failure modes:
‘Semi-rigid’ joints lie somewhere between ‘nominally pinned’ and ‘rigid’.
Eurocode 3 •
Principles mostly the same as BS 5950
•
Results are similar although EC3 results are slightly more conservative and this is due to the larger
•
Shear resistance per shear plate
•
Bearing Resistance
•
Tension Resistance
•
Combined shear and tension
Bolted Joints – Tension Tension resistance for ordinary bolts:
partial safety factor (γM2=1.25)
Bolt Strength These values should be adopted as characteristic values in design calculations : Bolt classes 2
f yb (N/mm ) 2
f ub (N/mm )
4.6
5.6
8.8
10.9
240
300
640
900
400
500
800
1000
EN 1993-1-8 Table 3.1 - Nominal values of f yb and f ub for bolts
Steel Strength 2
2
f y (N/mm ) Steel grade
S 275
S 355
f u (N/mm )
Nominal thickness of element t (mm) t ≤ 16
16 < t ≤ 40
40 < t ≤ 63
63 < t ≤ 80
275
265
255
245
355
345
335
325
Nominal thickness of element t (mm) 3≤t t<3 ≤ 100 430 to 580 510 to 680
Extract from Table 7 of EN 10025-2
410 to 560 470 to 630
where: •
As is the tensile stress area of the bolt
•
γM2 = 1.25
•
f ub is the ultimate tensile strength of the bolt
•
k2 = 0.63 for countersunk bolt, otherwise k 2 = 0.9
Bolted Joints – Shear Shear resistance per shear plane for ordinary bolts where the shear plane passes through the threaded portion of the bolt:
where: •
As is the tensile stress area of the bolt
•
f ub is the ultimate tensile strength of the bolt
•
γM2 = 1.25
Bolt classes
4.6
4.8
5.6
5.8
6.8
8.8
10.9
αv
0.6
0.5
0.6
0.5
0.5
0.6
0.5
