Eurocode 3 Lecture Notes

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

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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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


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


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


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


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


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)


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


Wy = Wel,y


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)

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.

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

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:


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:


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)


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

Eurocode 3 Lecture Notes

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Transcript of Eurocode 3 Lecture Notes