eurocode 3 lecture notes
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Steel Design to Eurocode 3
Introduction
Development of Eurocode 3
Aim: to create a common structural language
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
National Annex.
Nationally Determined Parameters
(NDPs)
The 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 during
construction and service.
Structure of Eurocode 3
Eurocode 3 is broken into 6 parts:
EN 1993-1 Generic rules
EN 1993-2 Bridges
EN 1993-3 Towers, masts and chimneys
EN 1993-4 Silos, tanks and pipelines
EN 1993-5 Piling
EN 1993-6 Crane supporting structures
Eurocode 3 Part 1 has 12 sub-parts:
EN 1993-1-1 General Rules
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
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
e.g. Beams,
Plate Girders,
Compression members...
Sub-parts are based on structural phenomena
e.g. Tension, Compression, Bending, Shear
Sub-parts can be applied to any element
The arrangement of the sub-parts means less duplication of rules
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Different Axes
BS 5950 Eurocode 3
Along the member
X
Major Axis X Y
Minor Axis Y Z
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
Verification check
Resistance capacity
Different Symbols
BS
5950 EC3
BS
5950 EC3
BS
5950 EC3
A A P N py fy
Z Wel Mx My pb LTfy
S Wpl V V pc fy
Ix Iy H Iw r i
Iy Iz J It
Informative subscripts
Ed means design effect
Rd means design resistance
Therefore:
NEd is an design axial force
NRd is the design resistance to the axial force
Gamma Factors
Partial factor
M
UK NA
value Application
M0 1.00 Cross-sections
M1 1.00 Member
Buckling
M2 1.25 Fracture
Omissions
Notable omissions:
Effective lengths
Use BS 5950 effective lengths
Formulae for Mcr
Use SN003 NCCI Document
Deflection limits
Refer to National Annex
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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)
ULS Checks
EQU: static equilibrium
STR: strength/buckling etc
GEO: Failure of excessive deformation of
ground
FAT: fatigue failure
Actions
Permanent actions , G (Dead loads)
Variable actions , Q (Live loads)
Qk Characteristic value ( = 1.0)
0Qk Combination value
1Qk Frequent value
2Qk Quasi-permanent value
Partial Factors
Unfavourable Favourable
G 1.35 1.0
Q 1.5 0
Table 1: Partial Factor values from the UK NA
Combination Factors
Action 0 1 2
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
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
Snow (sites up to 1000m) 0.5 0.2 0
Wind 0.5 0.2 0
Table 2: Extract from Table NA.A1.1
Combinations of Actions
Can use either:
Equation 6.10
Less favourable of 6.10a and 6.10b
Method: Get the factors from Tables 1 and 2
and substitute them into the equation you are
using, check for a range of different loading
combinations and take the least favourable
result.
Equation 6.10
G,jGk,j + PP + Q,1Qk,1 + Q,i0,iQk,i
Equation 6.10a
G,jGk,j + PP + Q,10,iQk,1 + Q,i0,iQk,i
Equation 6.10b
jG,jGk,j + PP + Q,1Qk,1 + Q,i0,iQk,i
j is 0.925 (From NA 2.2.3.2)
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Steel Design to Eurocode 3
Structural Analysis
Analysis Types
There are four types of global analysis:
Analysis Type In
itia
l G
eo
metr
y
Defo
rme
d G
eo
me
try
Lin
ea
r m
ate
rial
be
ha
vio
ur
Non
-lin
ea
r m
ate
rial
be
ha
vio
ur
First-order elastic
Second-order elastic
First-order plastic
Second-order elastic
Table 1: Summary of Analysis Types
Figure 1: Load-Deformation graph for different analysis types
(Source: Designers Guide to EN 1993-1-1 Page 21)
Joints
Clause 5.1.2 deals with joint modelling
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.
Figure 2: Joint stiffness effects (Source: SCI CPD Course
Material)
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
second-order effects might lead to an unsafe
approach due to underestimation of some of
the internal forces and moments.
Clause 5.2.1(2) states that second order effects
shall be considered:
if they increase the action effects significantly
or modify significantly the structural behaviour
First-Order Analysis
A first-order analysis may be used if the
following criteria is satisfied:
cr 10 for elastic analysis
cr 15 for plastic analysis
cr =Fcr/FEd
cr is the factor by which the design loading
would have to increased to cause elastic
instability in a global mode (cr 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.
For portal frames (with shallow roof slopes less than 26) and beam and column plane frames:
HEd is the horizontal reaction at the bottom of the
storey
VEd is the total vertical load at the bottom of the
storey
H,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.
h is the storey height.
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Amplifier
If 10 > cr 3.0
Increase all lateral loads by the amplifier:
Limits on cr Action
cr >10 First order Analysis
10>cr >3 First order analysis plus amplification or effective length method
cr 3 Second order analysis
Table 2: Actions to be taken once cr has been
calculated
Imperfections
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 = x Vertical Forces
= 0hm
0 = 1/200 = 0.005
h is the reduction factor for height:
(h is the height of the structure in metres)
m is the reduction factor for columns
(m is the number of columns contributing to the effect on the bracing system)
Summary
1) Model the Frame
2) Put all the loads on the frame
(Including the EHFs)
3) Calculate cr
4) Check to see if second-order effects are
significant
5) If necessary use the amplifier
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Steel Design to Eurocode 3
Brittle Fracture
Steel sub-grade selection
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 +20C
S275 J0 - Charpy value of 27J can be obtained at 0C
S275 J2 - Charpy value of 27J can be obtained at -20C
EN 1993-1-10
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. it 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.
fy(t)
fy(t) = fy,nom 0.25 (t/t0)
but t0 = 1mm, so fy(t) = fy,nom 0.25 (t) .
PD 6695-1-10
Published Document is much Simpler to use
Internal Tmd is -5C (Table 2)
External is Tmd -15C (Table 3)
NOTE: Can only use this document for design in the
UK
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PD 6695-1-10 Tables
Table 2 Maximum thicknesses for internal steelwork in buildings for Tmd = -5C
Table 3 Maximum thicknesses for external steelwork in buildings for Tmd = -15C
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Steel Design to Eurocode 3
Local Buckling and Cross-
Section 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
BS 5950 EC3
Plastic Class 1
Compact Class 2
Semi-compact Class 3
Slender 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 material yield strength, fy
c/t ratio
Eurocode 3 defines the classes in Clause 5.5.2:
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.
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 local 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 factor which is calculated using fy, the yield strength of the steel.
Factor
BS 5950 EC3
= (275/py)0.5 = (235/fy)
0.5
Values of are given at the bottom of Table 5.2:
fy 235 275 355 420 460
1.00 0.92 0.81 0.75 0.71
EN 1993-1-1 Table 5.2
fy Yield Strength
The UK National Annex says that material properties should be taken from the product standards.
Extract from EN 10025-2 - fy (yield strength) values for hot rolled steel:
Steel Grade
fy (N/mm2)
nominal thickness of element, t (mm)
t1
6
16
< t
4
0
40
< t
63
63
< t
80
S 275 275 265 255 245
S 355 355 345 335 325
EN 10025-2 (Table 7)
Class 1
Class 2
Class 3
Class 4
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c/t Width-to-Thickness Ratio
The width-to-thickness ratios differ in EC3 differs from
BS 5950:
Appropriate values of c and t are defined at the top of Table 5.2 for different types of sections.
Table 5.2
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:
Sheet 1 Internal Compression Parts
Sheet 2 Outstand Flanges
Sheet 3 Angles and Tubular Sections
Cross-section Classification
Class 1: Plastic
BS (Table 11) EC3 (Table 5.2)
Lim
its
Flange outstand
b/T = < 9 c/tf = < 9
Web in bending
d/t = < 80 d/tw = < 72
Web in compression
d/tw = < 33
Class 2: Compact
BS (Table 11) EC3 (Table 5.2)
Lim
its
Flange outstand
b/T = < 10 c/tf = < 10
Web in bending
d/t = < 100 d/tw = < 83
Web in compression
d/tw = < 38
Class 3: Semi-compact
BS (Table 11) EC3 (Table 5.2)
Lim
its
Flange outstand
b/T = < 15 c/tf = < 14
Web in bending
d/t = < 120 d/tw = < 142
Web in compression
d/tw = < 42
Class 4: Slender
An element that doesnt 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:
Section Type Reference
Cold-formed sections EN 1993-1-3
Hot-rolled and fabricated section
EN 1993-1-5
CHS EN 1993-1-6
Overall Cross-Section Classification
Clause 5.5.2(6) states that a cross-section is classified according to the highest (least favourable) class of its compression parts.
Summary
1. Determine fy (UK NA recommends you use the product standards)
2. Determine from Table 5.2
3. Substitute the value of 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/tw
Class 1 9 72
Class 2 10 83
Class 3 14 124
Class 4 If it does not meet Class 3
requirements, the section is classified as Class 4
4. Take the least favourable class from the flange
and web results
BS 5950
EC3
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
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Steel Design to Eurocode 3
Restrained Beams
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 o Clause 6.2.5 deals with the cross-
sectional resistance for bending.
EN 1993-1-1 Clause 6.2.4 Equation 6.12 states
that the design moment (MEd) must be less than
the design cross-sectional moment resistance
(Mc,Rd)
The equation to calculate Mc,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 = Wplfy/M0
For Class 3 cross-sections:
Mc,Rd = Mel,Rd = Wel,minfy/M0
For Class 4 cross sections:
Mc,Rd = Weff,minfy/M0
M0 =1.0
Section Modulus, W
Subscripts are used to identify whether or not the
section modulus is plastic or elastic and the axis
about which it acts.
BS
5950 EC3
Elastic modulus about the major axis Zxx Wel,y
Elastic modulus about the minor axis Zyy Wel,z
Plastic modulus about the major axis Sxx Wpl,y
Plastic modulus about the minor axis Syy Wpl,z
Table 1.0 Section modulus terminology comparison
between BS 5950 and EC3
Cross-section Classification
Summary
1. Get fy from Table 3.1 2. Get from Table 5.2 3. Substitute the value of 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 o Clause 6.2.6 deals with the cross-
sectional resistance for shear.
EN 1993-1-1 Clause 6.2.6 Equation 6.17 states
that the design shear force (VEd) must be less than
the design plastic shear resistance of the cross-
section (Vpl,Rd)
(6.12)
(6.13)
(6.14)
(6.15) (6.17)
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M0 =1.0
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.
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 hwtw
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/
Plates and solid bars Av =A
Table 2.0: Shear area formulas
Term Definition
A Cross-sectional area
b Overall breadth
h Overall depth
hw Depth of web
r Root radius
tf Flange thickness
tw Web thickness (taken as the minimum value is the web is not of constant thickness)
Constant which may be conservatively taken as 1.0
Table 3.0: Shear area parameter descriptions
Shear Resistance Summary
1. Calculate the shear area, Av
2. Substitute the value of Av into equation 6.18
to get the design plastic shear resistance
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 2: Standard case deflections and corresponding
maximum deflection equations
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
Cantilever Length/180
Beams carrying plaster of other brittle finish
Span/360
Other beams (except purlins and sheeting rails)
Span/200
Purlins and sheeting rails To suit the characteristics of particular cladding
Table 4.0: Vertical Deflection Limits from NA 2.23
Clause 7.2.1(1) B
(6.18)
Figure 1: Visual definition
of the parameters used in
the shear area
calculation. (Source:
Blue Book)
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Steel Design to Eurocode 3
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.
Lateral torsional buckling can be discounted when:
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. Primary method (Clauses 6.3.2.2 and Clauses
6.3.2.3)
2. Simplified assessment method (Clause
6.3.2.4)
3. 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.
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
EN 1993-1-1 Clause 6.3.2.1 Equation 6.54 states that
the design moment (MEd) must be less than the design
buckling resistance moment (Mb,Rd)
where M1 =1.0 (from UK NA)
Section Modulus Wy
For Class 1 and 2 cross-sections:
Wy = Wpl,y
For Class 3 cross-sections:
Wy = Wel,y
For Class 4 cross-sections:
Wy = Weff,y
Yield Strength, fy
The UK National Annex says that we should obtain the
value of the yield strength from the product standards.
Extract from EN 10025-2 - fy (yield strength) values for hot rolled steel:
Steel Grade
fy (N/mm2)
nominal thickness of element, t (mm)
t1
6
16
< t
4
0
40
< t
63
63
< t
80
S 275 275 265 255 245
S 355 355 345 335 325
Extract from EN 10025-2 (Table 7)
Reduction Factor, LT
General Case:
where
(6.54)
(6.55)
(6.56)
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To get LT, 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 LT
Cross-section
Limits Buckling Curve
Rolled I sections
h/b 2 h/b >2
a s
Welded I sections
h/b 2 h/b >2
c d
Other - d
EN 1993-1-1 Table 6.4
Buckling curve
a b c d
LT 0.21 0.34 0.49 0.76 EN 1993-1-1 Table 6.3
Special Case (for rolled sections):
where
UK NA sets = 0.75 and = 0.4
To get LT, 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 LT
Cross-section Limits Buckling Curve
Rolled bi-symmetric I and H sections and hot-finished hollow sections
h/b 2 2.0 < h/b 3.1
b c
Angles (for moments in the major principal plane) and other hot-rolled sections
d
Welded bi-symmetric sections and cold-formed hollow sections
h/b 2 h/b > 2
c d
Table from NA.2.17 Clause 6.3.2.3(1)
Buckling curve
a b c d
LT 0.21 0.34 0.49 0.76 EN 1993-1-1 Table 6.3
You can use a modified value of LT in the special
case to give some extra resistance:
f= 1- 0.5(1 - kc)[1-2.0( - 0.8)2]
but f 1.0
kc can be obtained from Table 6.6 in the
Eurocodes:
EN 1993-1-1 Table 6.6
You will need the value of for both the general and
special cases.
Mcr
Refer to SN003 document (NCCI) for detailed
description of how to get Mcr
where
L is the distance between points of lateral restraint (Lcr)
E is the Youngs Modulus = 210000 N/mm2
G is the shear modulus = 80770 N/mm2
Iz is the second moment of area about the weak axis
It is the torsion constant
Iw is the warping constant
(6.58)
(6.57)
(6.56)
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k is an effective length factor (usually 1.0)
kw is an effective length factor (usually 1.0)
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 1 (from SN003 document)
C1 and C2 are coefficients.
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:
Table 3.1 from SN003 (Values of C1 for members with
end moments)
where
Figure 3.1 from SN003
Summary
1. Draw the bending moment diagram to obtain the
value of the maximum bending moment, MEd
2. Determine fy (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, Lcr
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 LT
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 LT
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 LT
7. Work out LT
a. For the general case use expression 6.56
b. For the special case, use expression 6.57
8. Work out LT
a. For the general case use expression 6.56
b. For the special case, use expression 6.57
9. Calculate the design buckling resistance Mc,Rd using equation 6.55.
10. Carry out the buckling resistance check in
expression 6.54.
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Steel Design to Eurocode 3 Compression Members 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
Figure 1 Behaviour of columns is determined by their
slenderness
Stocky Columns are not affected by buckling and the strength is related to the material yield stress fy.
Nmax = Npl = Aeff fy
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 slender 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 (NEd)
must be less than the design cross-sectional resistance of the sections to uniform compression force (Nc,Rd)
Cross-section resistance in compression depends on cross-section classification. For Classes 1, 2 and 3:
For Class 4 sections:
M0 =1.0
Cross-section Classification Summary 1. Get fy from Product Standards
2. Get from Table 5.2
3. Substitute the value of 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
For a more detailed description of cross-section classification, please refer to the Cross-section Classification handout.
Cross-section Resistance Check Summary
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.
(6.9)
(6.10)
(6.11)
hussaik1Text Box8
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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 (Nb,Rd)
Similarly to cross-section resistance, buckling resistance is dependent on the cross-section classification. For sections with Classes 1, 2 and 3:
For Class 4 sections:
M1 =1.0
Buckling Curves Buckling curve selection is dependent on the section geometry. Table 6.2 in EN 1993-1-1 provides guidance on a range of sections.
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.
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
Non-dimensional Slenderness
For sections with Classes 1, 2 and 3:
For Class 4 sections:
where
Imperfection Factor,
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 :
Buckling Curve a0 a b c d
Imperfection Factor
0.13 0.21 0.34 0.49 0.76
EN 1993-1-1 Table 6.1
Reduction Factor,
where Alternatively, may be read from Figure 6.4 in the Eurocodes by using and the required buckling curve.
Buckling Resistance Check Summary 1. Determine the design axial load, NEd 2. Choose a section and determine the class 3. Calculate the effective length Lcr 4. Calculate Ncr using the effective length Lcr, and
E and I which are section properties
5. Calculate 6. Determine by first determining the required
buckling curve from Table 6.2 and then reading off the required value of from Table 6.1.
7. Calculate by substituting in the values of and
8. Calculate by substituting in the values of
and 9. Determine the design buckling resistance of
the member by using equation 6.47 or 6.48 and substituting in the value of
10. Make sure that the conditions of equation 6.46 are satisfied.
(6.46)
(6.47)
(6.48)
(6.50)
(6.51)
or
or
(6.49)
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Steel Design to Eurocode 3
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.
Tensile Resistance
EN 1993-1-1 Clause 6.2.3(1) Equation 6.5 states
that the design tensile force (Nt,Ed) must be less
than the design tensile resistance moment (Nt,Rd)
The tensile resistance is limited by the lesser of:
Design Plastic Resistance Npl,Rd
Design Ultimate Resistance Nu,Rd
Design Plastic Resistance, Npl,Rd
Npl,Rd is the plastic design resistance, and is
concerned with the yielding of the gross cross-
section.
Equation 6.6 gives the expression used to
calculate Npl,Rd:
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:
Partial Factors M
M UK N.A. Value
M0 Resistance of cross-sections 1.0
M2 Resistance of cross-sections in tension to fracture
1.25
Characteristic Strengths fy and fu
The UK National Annex says you should get the
values of fy and fu from the product standards. For
hot-rolled sections you can use the table below.
Steel grade
fy (N/mm2)
fu (N/mm
2)
t 1
6
16
< t
4
0
40
< t
6
3
63
< t
8
0
t
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