TTT Handout LG Nov2008 Lecture Note on EC3 Design

1

1L. Gardner

Introduction

Overview

Objectives

Design of steel structures to Eurocode 3

Dr Leroy GardnerSenior Lecturer in Structural Engineering

Eurocode 3: Design of steel structures

2L. Gardner

Introduction

Overview

Objectives Introduction



Session 1

2

3L. Gardner

Introduction

Overview

Objectives

• Session 1: General Introduction

• Session 2: Introduction to EN 1990 & EN 1991

• Session 3: Overview of Eurocode 3

• Session 4: Structural analysis

• Session 5: Design of tension members

• Session 6: Local buckling and cross-section classification

Outline:

Overview of Course

4L. Gardner

Introduction

Overview

Objectives

• Session 7: Design of columns

• Session 8: Design of beams

• Session 9: Design of beam-columns

• Session 10: Design of joints

Outline (continued):

Overview of Course

3

5L. Gardner

Introduction

Overview

Objectives

• Senior Lecturer in Structural Engineering

• Research into stability and design of steel structures

• Specialist advisory work

• Development and assessment of Eurocode 3

• Author of TTT guide to Eurocode 3

• [email protected]

Dr Leroy GardnerDr Leroy GardnerBEng MSc PhD DIC CEng MICE MIStructE

6L. Gardner

Introduction

Overview

Objectives

• Introduce yourselves

• Organisation

• Your experience with steel design/ Eurocodes

• Any particular interests/concerns

Your experience with Eurocodes

4

7L. Gardner

Introduction

Overview

Objectives

• Clear training requirements

• Textbooks and design guides

• Background information

• Most of the Eurocodes are now published

• Conflicting British Standards to be withdrawn

• Designers need to be prepared

Motivation for course

8L. Gardner

Introduction

Overview

Objectives

Introduction of Eurocodes:

Designers

• Biggest change since limit states

• Designers unfamiliar with format

• Resistance to uptake

• Supporting material and training

• Basis for other National design codes

5

9L. Gardner

Introduction

Overview

Objectives

Designers’ Guide

• Covers Eurocode 3: Part 1.1

• Also Parts 1.3, 1.5 and 1.8

• EN 1990 and EN 1991

• Sections aligned with code

Designers’ Guide to EN 1993-1-1:

10L. Gardner

Introduction

Overview

Objectives

Textbook

• Structural phenomena

• Theoretical background

• Code implementation

• Worked examples

The behaviour and design of steel structures to EC3:Trahair, Bradford, Nethercot & Gardner (2008)

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11L. Gardner

Introduction

Overview

Objectives • Idea of Eurocodes dates back to 1974

• Family of design codes

• Harmonisation of treatment

• Removal of barriers to trade

• Framework for development

Historical developments

Historical development of Eurocodes:

12L. Gardner

Introduction

Overview

Objectives

• EN 1990 – Basis of structural design

• EN 1991 – Actions on structures

Scope of Eurocodes

The first 2 codes are material independent:

• A total of 10 codes (comprising 58 documents)

Scope of structural Eurocodes:

7

13L. Gardner

Introduction

Overview

Objectives

• EN 1992 – Design of concrete structures

• EN 1993 – Design of steel structures

• EN 1994 – Design of composite structures

• EN 1995 – Design of timber structures

• EN 1996 – Design of masonry structures

• EN 1997 – Geotechnical design

• EN 1998 – Design of structures for earthquakes

• EN 1999 – Design of aluminium structures

Scope of EurocodesRemaining 8 codes focus on materials:

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Introduction

Overview

Objectives

• Codes published by CEN

• Comité Europeén de Normalisation

• European Committee for Standardisation

• National standards bodies adopt (BSI)

• Two years to produce National Annex

• Three year co-existance period

• Conflicting existing standards withdrawn

Timetable for introduction

Timetable for introduction of codes:

8

15L. Gardner

Introduction

Overview

Objectives

• Codes will be published by CEN in 3 languages:

• English

• French

• German

• All codes originally developed in English, and then ‘exactly’ translated

• Other participating counties will either use 1 of 3 language versions available, or translate at own cost.

Eurocodes

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Introduction

Overview

Objectives• Familiarity with layout, notation, philosophy of Eurocodes

• Understanding of background and design procedure for principal structural components

Objectives

Course objectives:

9

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Introduction

Overview

Objectives



Session 1

Introduction

1

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EN 1990

EN 1991 Introduction to EN 1990 & EN 1991



Session 2

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EN 1990

EN 1991

• Introduction to EN 1990

• Introduction to EN 1991

• Conclusions

Outline:

Overview

2

3L. Gardner

EN 1990

EN 1991

• EN 1990 – Basis of structural design

• UK National Annex published

• ‘Should read at least once’….

EN 1990 (2002)

EN 1990 (2002):

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EN 1990

EN 1991

• Structural resistance

• Serviceability

• Durability

• Fire resistance

• Robustness

Basic requirements

EN 1990 states that a structure shall have adequate:

3

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EN 1990

EN 1991• Persistent design situations: normal use

• Transient design situations: temporary conditions, e.g. during construction or repair

• Accidental design situations: exceptional conditions such as fire, explosion or impact

• Seismic design situations: where the structure is subjected to seismic events.

Design situations

All relevant design situations must be examined:

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EN 1990

EN 1991

Actions and Effects

Action (F):• Direct actions – applied loads

• Indirect actions – imposed deformations or accelerations e.g. by temperature changes, vibrations etc

• Both essentially produce same effect

Effect of action (E):• On structural members and whole structure

• For example internal forces and moments, deflections ..

CAUSE

EFFECT

4

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EN 1990

EN 1991

• Permanent, G

• Variable, Q (leading and non-leading)

• Accidental, A

Types of actions

Types of actions:

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EN 1990

EN 1991

Load combinations

Fundamental combinations of actions may be determined from EN 1990 using either of:

• Equation 6.10

• Less favourable of Equation 6.10a and 6.10b

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EN 1990

EN 1991

Load combinations

Equation 6.10:

∑∑1>i

i,ki,0i,Q1,k1,QPj,k1≥j

j,G Q"+"Q"+"P"+"G ψγγγγ

1.35 x Permanent actions

‘to be combined with’ Actions due to

prestressing

1.5 x Leading variable action

1.5 x combination factor x Other variable actions

Load factors 1.35 and 1.5 are applied when actions are ‘unfavourable’.

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EN 1990

EN 1991

• In Equation 6.10, the full value of the leading variable action is applied γQ,1Qk,1 (i.e. 1.5 x characteristic imposed load)

• The leading variable action is the one that leads to the most unfavourable effect (i.e. the critical combination)

• To generate the various load combinations, each variable action should be considered in turn as the leading one, (and consideration should be given to whether loading is favourable or unfavourable.)

Leading variable actions Qk,1

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EN 1990

EN 1991

Combination factor ψ0

The combination factor ψ0 is intended specifically to take account of the reduced probability of the simultaneous occurrence of two or more variable actions.

0.5*Wind loading

0.7Imposed loading

Combination factor ψ0

Loading

* 0.5 is UK NA value, 0.6 is the unmodified EC value

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EN 1990

EN 1991

Loads may be considered as ‘unfavourable’or ‘favourable’ in any given combination, depending on whether they increase or reduce the effects (bending moments, axial forces etc) in the structural members.

For unfavourable dead loads: γG = 1.35

For favourable dead loads: γG = 1.00

For unfavourable variable loads: γQ = 1.5

For favourable variable loads: γQ = 0

Unfavourable and favourable loading

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EN 1990

EN 1991

Equivalent horizontal forces:

Equivalent horizontal forces (EHFs), previously known as notional horizontal loads (NHL), are required to account for imperfections that exist in all structural frames.

EHFs should be included in all load combinations, and since their value is related to the level of vertical loading, they will generally be different for each load combination (and will already be factored).

Equivalent horizontal forces

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EN 1990

EN 1991

Load combinations for a typical structure from Equation 6.10:

Exercise solution – Equation 6.10

1.0

1.0

1.0

1.0

EHF

Dead + Wind (uplift)

D + I + W(wind leading)

D + I + W(imposed leading)

Dead + Imposed

Combination WindImposedDead

Note EHF are always present and already based on factored loads

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EN 1990

EN 1991

Load combinations

∑∑1 1i

i,ki,0i,Q1,k1,01,Qj

Pj,kj,G Q""Q""P""G>≥

ψγ+ψγ+γ+γ

∑∑1j 1i

i,ki,0i,Q1,k1,QPj,kj,Gj Q""Q""P""G>≥

ψγ+γ+γ+γξ

Equations 6.10a and 6.10b – use less favourable result:

Unfavourable dead load reduction factor (i.e. not applied when γG = 1)ξ = 0.925 in UK NA(0.85 is the unmodified EC value)

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EN 1990

EN 1991

Load combinations from Eqs 6.10a and 6.10b – All combinations except last one are from Eq. 6.10b.

Exercise solution – Eqs 6.10a and 6.10b

1.0D + I + W (6.10a)*

1.0

1.0

1.0

1.0

EHF

Dead + Wind (uplift) (6.10b)

D + I + W (6.10b)(wind leading)

D + I + W (6.10b)(imposed leading)

Dead + Imposed (6.10b)

Combination WindImposedDead

* Unlikely to govern unless Dead >> Imposed

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17L. Gardner

EN 1990

EN 1991 For checking sliding or overturning of the structure as a rigid body, only Eq. 6.10 may be used. Dead loads are factored by 0.9 when favourable and 1.1 when unfavourable.

The critical case will generally arise when wind load is unfavourable and the leading variable action, and dead load is favourable, resulting in:

Equilibrium check (EQU)

Equilibrium check (EQU):

0.9Gk + 1.5Wk + EHF

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EN 1990

EN 1991

Equilibrium check (EQU)Favourable and unfavourable loading:

1.5 Wk 0.9 Gk

Wind load unfavourable, dead load favourable, imposed load favourable

Overturning point

1.5 Wk 0.9 Gk

Wind load unfavourable, part of dead load favourable, part unfavourable, part of imposed unfavourable

1.1 Gk + 1.05 Qk

Overturning point

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EN 1990

EN 1991

SLS load combinations

The UK National Annex to EN 1993-1-1 states that deflections may be checked using the SLS characteristic combination, ignoring dead load and with some specified deflection limits.

1.0Qk + 0.5Wk + EHF (Vertical deflections)

1.0Wk + 0.7Qk + EHF (Horizontal deflections)

Deflection limits are as given in BS 5950

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EN 1990

EN 1991

• EN 1991-1: General actions

• EN 1991-2: Traffic loads on bridges

• EN 1991-3: Actions from cranes and machinery

• EN 1991-4: Actions in silos and tanks

Parts of EN 1991

EN 1991 contains the following parts:

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21L. Gardner

EN 1990

EN 1991 • EN 1991-1-1: Densities, self-weight, imposed loads

• EN 1991-1-2: Fire

• EN 1991-1-3: Snow loads

• EN 1991-1-4: Wind actions

• EN 1991-1-5: Thermal actions

• EN 1991-1-6: Actions during execution

• EN 1991-1-7: Impact and explosions

Sub-parts of EN 1991-1

EN 1991-1 contains the following sub-parts:

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EN 1990

EN 1991

Concluding comments:

Conclusions

• Presentation of load combinations unfamiliar

• Idea of leading variable actions and combination factors etc is new

• Other than format and notation, loading codes are similar to existing BS

• Using Eq. 6.10a and 6.10b (with 6.10 for EQU), four basic load combinations arise (ignoring those unlikely to govern).

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EN 1990

EN 1991



Session 2

Introduction to EN 1990 & EN 1991

1

Background

Overview

1L. Gardner 1L. Gardner

Overview of Eurocode 3



Session 3

Background

Overview


• Development of Eurocode 3

• Introduction to design to Eurocode 3

• Conclusions

Outline:

Overview

2

Background

Overview


• Work began back in 1975

• Eurocode 3 contains a number of parts

• … and sub-parts

• The first 5 parts were published in 2005

EN 1993: Eurocode 3

Eurocode 3:

Background

Overview


EN 1993: Eurocode 3

Eurocode 3 contains six parts:

• EN 1993-1 Generic rules

• EN 1993-2 Bridges

• EN 1993-3 Towers, masts & chimneys

• EN 1993-4 Silos, tanks & pipelines

• EN 1993-5 Piling

• EN 1993-6 Crane supporting structures

3

Background

Overview


EN 1993-1

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

Background

Overview


EN 1993-1

• 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

4

Background

Overview


National Annexes

National Annexes:

• Every Eurocode will contain a National Annex

• National choice

• Non Conflicting Complementary Information

• Timescale

Background

Overview


Axes convention

Different axes convention:

ZYMinor axis

YXMajor axis

XAlong the member

Eurocode 3BS 5950

5

Background

Overview


Labelling convention

Labelling convention:

b

h d

tw

tf

r

y y

z

z

t

b

r

h y y

z

z

Background

Overview


SubscriptsExtensive use of sub-scripts – generally helpful:

• ‘Ed’ means design effect (i.e. factored member force or moment)

• ‘Rd’ means design resistance

So,

• NEd is an axial force

• NRd is the resistance to axial force

Sometimes tedious e.g. Ac,eff,loc

6

Background

Overview


Different symbols

ItJIzIy

irIwHIyIx

pcVVWplS

Z

A

BS5950

Wel

A

EC3

Mx

P

BS5950

My

N

EC3

pb

fypy

EC3BS5950

yLTfχ

For example:

yfχ

Background

Overview


Gamma factors γ

Gamma factors γ:

• Appear everywhere

• Partial safety factors

• γF for actions (loading)

• γM for resistance

7

Background

Overview


Gamma factors γM

Gamma factors γM account for material and modelling uncertainties:

1.25 (1.10)

1.00 (1.00)

1.00 (1.00)

EC 3 value (UK NA value)

Fracture

Member buckling

Cross-sections

Application

γM2

γM1

γM0

Partial factor γM

Background

Overview


Material propertiesMaterial properties are taken from product standards (generally EN 10025-2). The Young’s modulus of steel should be taken as 210000 N/mm2.

550430450S450

470345355S355

410265275S275

360235235S235

Ultimate strength fu (N/mm2)

3 ≤ t ≤ 100 mm

Yield strength fy(N/mm2)

16 < t ≤ 40 mm

Yield strength fy(N/mm2)

t ≤ 16mm

Steel grade

8

Background

Overview


Structural design

• Reference to EN 1990 and EN 1991

• Identify clauses open to National choice

• Materials, reference to material standards

• Durability

Early sections (1-4) of EN 1993-1-1:

Background

Overview


Structural design

• Section 5 – Structural analysis

• Global analysis

• Cross-section classification

• Requirements for plastic analysis

• Section 6 – ULS

• General

• Resistance of cross-sections

Subsequent sections of EN 1993-1-1:

9

Background

Overview


Structural design

• Buckling resistance of members

• Built-up members

• SLS

• Annexes A, B, AB and BB

Background

Overview


Omissions

• Effective lengths

• Formulae for Mcr

• Deflection limits

• National Annex and NCCIs to resolve

Notable omissions:

10

Background

Overview


Sources of further information

• http://www.eurocodes.co.uk/

• Latest news and developments

• http://www.steel-sci.org/publications/

• Design guides

• http://www.access-steel.com/

• NCCIs

• Worked examples

Background

Overview




Session 3

Overview of Eurocode 3

1

Introduction

Deformed geometry

Imperfections

Actions


Structural analysis



Session 4

Introduction

Deformed geometry

Imperfections

Actions


Outline:

• Introduction• Analysis types• Second order effects• Imperfections

Overview

2

Introduction

Deformed geometry

Imperfections

Actions


Analysis types:

• First order elastic• Second order elastic• First order plastic• Second order plastic

Analysis types

Introduction

Deformed geometry

Imperfections

Actions


General approach

General approach:

• Choose an appropriate analysis• Make an appropriate model• Apply all actions (loads) and combinations

of actions• Check cross-sections, members and joints

3

Introduction

Deformed geometry

Imperfections

Actions


Frame Stability is assured by checking:• Cross-sections• Members• Joints

But will be unsafe unless:• Frame model• Loads on frame• Analysisare appropriate.

Frame stability

Introduction

Deformed geometry

Imperfections

Actions


Effects of deformed geometry

• if they increase the action effects significantly

• or modify significantly the structural behaviour

EN 1993-1-1 Clause 5.2.1(2) states that deformed geometry (second order effects) shall be considered:

4

Introduction

Deformed geometry

Imperfections

Actions


For elastic analysis:

where

αcr is the factor by which the design loading would have to be 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.

10FF

Ed

crcr ≥=α

Limits for ignoring deformed geometry

Introduction

Deformed geometry

Imperfections

Actions


For plastic analysis: 15FF

Ed

crcr ≥=α

Limits for ignoring deformed geometry

Stricter limit for plastic analysis due to loss of stiffness associated with material yielding.

So, for αcr ≥ 10 (or 15), the effects of deformed geometry may be ignored and a first order analysis will suffice

5

Introduction

Deformed geometry

Imperfections

Actions


• Portals with shallow roof slopes

• Beam and column frames (each storey)

where

HEd horizontal reaction at bottom of the storey

VEd total vertical load at bottom of the storey

δH,Ed storey sway when loaded with horizontal loads (eg wind, equivalent horizontal forces)

⎟⎟⎠

⎞⎜⎜⎝

⎛

δ⎟⎟⎠

⎞⎜⎜⎝

⎛=α

Ed,HEd

Edcr

hVH

Simple estimate for αcr

Simple estimate for αcr may be applied to:

Introduction

Deformed geometry

Imperfections

Actions


Limit on portal rafter slope for (Clause 5.2)• not steeper than 1:2 (26 degrees)

Limit on axial compression in beams or rafters for (Clause 5.2):

where NEd is the design value of compression in the beam or rafter and Ncr is its elastic buckling resistance.

Limits on use of simple estimate

09.0NN

cr

Ed ≤

6

Introduction

Deformed geometry

Imperfections

Actions


Distinguish between:

• Analysis method (1st or 2nd order)

• Analysis achievement i.e. can achieve 2nd

order by:

1) 2nd order analysis

2) 1st and amplified sway

3) 1st and increased effective length.

Analysis method and achievement

Introduction

Deformed geometry

Imperfections

Actions


Limits for treatment of second order effects depend on αcr: Ed

crcr F

F=α

Frame stability

Second order effects more accurately

Second order effects by approximate means

First order only

Achievement

Second order analysisαcr<3

First order analysis plus amplification or effective length method

10>αcr>3

First order analysisαcr>10

ActionLimits on αcr

7

Introduction

Deformed geometry

Imperfections

Actions


Global initial sway imperfections:

Global imperfections for frames

mh0 ααφ=φ

factorsreductionareand

200/1valuebasictheiswhere

mh

0

αα

=φ

Introduction

Deformed geometry

Imperfections

Actions


Global imperfections for frames

• Much easier to apply as equivalent horizontal forces φNEd, where NEd is the design compressive force in the column

• Saves changing the model for opposite direction in asymmetric buildings

• Many buildings have such complicated arrangements that it will be best to ignore the αh and αm reductions and use 1/200

• Don’t forget them.

8

Introduction

Deformed geometry

Imperfections

Actions


• EN 1991-1-1: Densities, self-weight, imposed loads

• EN 1991-1-2: Fire

• EN 1991-1-3: Snow loads

• EN 1991-1-4: Wind actions

• EN 1991-1-5: Thermal actions

• EN 1991-1-6: Actions during execution

• EN 1991-1-7: Impact and explosions

Actions to be specified

Actions to be specified:

Introduction

Deformed geometry

Imperfections

Actions


Other Actions

• Equivalent horizontal forces

- unless using initial imperfection model

• Derived from imperfections

• Applied in ALL combinations

(only in gravity combinations in BS 5950)

9

Introduction

Deformed geometry

Imperfections

Actions


• Analyse structure

• Classify sections using clause 5.5- for plastic global analysis, check clause 5.6

• Check cross-sectional resistance to clause 6.2

• Check buckling resistance to clause 6.3- check built-up members to clause 6.4

Checks

Introduction

Deformed geometry

Imperfections

Actions


Structural analysis



Session 4

1


Introduction

Design

Example

Exercise

Design of tension members



Session 5


Introduction

Design

Example

Exercise

Overview

Outline:

• Introduction• Tension member design• Example

2


Introduction

Design

Example

Exercise

Eurocode 3

Eurocode 3 states that tensile resistance should be verified as follows:

Rd,tEd,t NN ≤ Tension check

Nt,Ed is the tensile design effect

Nt,Rd is the design tensile resistance


Introduction

Design

Example

Exercise

Design tensile resistance Nt,Rd

Design tensile resistance Nt,Rd is limited either by:

• Yielding of the gross cross-section Npl,Rd

• or ultimate failure (fracture) of the net cross-section (at holes for fasteners) Nu,Rd

whichever is the lesser.

3


Introduction

Design

Example

Exercise

Yielding of gross cross-section

0M

yRd,pl

AfN

γ=

The Eurocode 3 design expression for yielding of the gross cross-section (plastic resistance) given as:

This criterion is applied to prevent excessive deformation of the member.


Introduction

Design

Example

Exercise

Ultimate resistance of net section

M2

unetf0.9Aγ

= Rdu,N

And for the ultimate resistance of the net cross-section (defined in clause 6.2.2.2), the Eurocode 3 design expression is:

Anet is the reduced cross-sectional area to account for bolt holes

4


Introduction

Design

Example

Exercise

Partial factors γM

)NAUKin1.1(25.1and0.1 2M0M =γ=γ

Plastic resistance of the gross cross-section Npl,Rd utliises γM0, whilst ultimate fracture of the net cross-section Nu,Rd utilises γM2.

The larger safety factor associated with fracture reflects the undesirable nature of the failure mode.


Introduction

Design

Example

Exercise

For a non-staggered arrangement of fasteners, the total area to be deducted should be taken as the sum of the sectional areas of the holes on any line (A-A) perpendicular to the member axis that passes through the centreline of the holes.

Non-staggered arrangement of fasteners

p

s s

A

A

Non-staggered fasteners

5


Introduction

Design

Example

Exercise

Net area at bolts holes Anet on any line (A-A) perpendicular to the member axis:

Anet = A - nd0t

Non-staggered fasteners

A = gross cross-sectional arean = number of bolt holesd0 = diameter of bolt holest = material thickness


Introduction

Design

Example

Exercise

For a staggered arrangement of fasteners, the total area to be deducted should be taken as the greater of:

1. the maximum sum of the sectional areas of the holes on any line (A-A) perpendicular to the member axis

2. ⎟⎟⎠

⎞⎜⎜⎝

⎛−∑ p4

sndt2

0

where

s is the staggered pitch of two consecutive holesp is the spacing of the centres of the same two holes measured perpendicular to the member axis

Staggered fasteners

6


Introduction

Design

Example

Exercise

Staggered arrangement of fasteners

p

s

A

A

s

B

Staggered fastenersn is the number of holes extending in any diagonal or zig-zag line progressively across the section

Σ relates to the number of diagonal paths


Introduction

Design

Example

Exercise

Single angles in tension connected by a single row of bolts through one leg, may be treated as concentrically loaded, but with an effective net section, to give the design ultimate tensile resistance as below.

2M

uetn3Rd,u

2M

uetn2Rd,u

2M

u02Rd,u

fAN:boltsmoreor3With

fAN:bolts2With

tf)d5.0e(0.2N:bolt1With

γβ

=

γβ

=

γ−

=

Angles connected by a single row of bolts

7


Introduction

Design

Example

Exercise

where β2 and β3 are reduction factors dependent upon the bolt spacing (pitch) p1.

Anet is the net area of the angle. For an unequal angle connected by its smaller leg, Anet should be taken as the net section of an equivalent equal angle of leg length equal to the smaller leg of the unequal angle. Other symbols are defined below:

Definitions for e1, e2, p1 and d0


e1 p1 p1

e2 d0


Introduction

Design

Example

Exercise

Reduction factors β2 and β3

Note: For intermediate values of pitch p1 values of β may be determined by linear interpolation. d0 is the bolt hole diameter.

0.70.5β3 (for 3 or more bolts)

0.70.4β2 (for 2 bolts)

≥ 5.0d0≤ 2.5d0Pitch p1


8


Introduction

Design

Example

Exercise

Angles with welded end connections

In the case of welded end connections:

For an equal angle, or an unequal angle connected by its larger leg, the eccentricity may be neglected, and the effective area may be taken as equal to the gross area (clause 4.13(2) of EN 1993-1-8).


Introduction

Design

Example

Exercise

A

B

Tension member AB in truss

Example: Tension member design

Design a single angle tie, using grade S355 steel, for the member AB shown below. Consider a bolted and a welded arrangement.

NEd = 541 kN

9


Introduction

Design

Example

Exercise


Cross-section resistance in tension is covered in clause 6.2.3 of EN 1993-1-1, with reference to clause 6.2.2 for the calculation of cross-section properties.


Introduction

Design

Example

Exercise

Gusset plate

125×75×10 unequal angle

Welded connection

Try a 125×75×10 unequal angle, welded by the longer leg.

For an unequal angle connected (welded) by its larger leg, the effective area may be taken as equal to the gross area (clause 4.13(2) of EN 1993-1-8)

125×75×10 unequal angle welded by longer leg


10


Introduction

Design

Example

Exercise

For a nominal material thickness t of 10 mm, yield strength fy = 355 N/mm2 and ultimate tensile strength fu = 470 N/mm2 (from EN 10025-2).

Partial factors from UK National Annex are γM0 = 1.00 and γM2 = 1.10.

Gross area of cross-section, A = 1920 mm2

(from Section Tables).



Introduction

Design

Example

Exercise

kN682N10x6821.0

3551920AfN 3

M0

yRd,pl ==

×=

γ=

For yielding of the gross cross-section, plastic resistance is given as:=

And for the ultimate resistance of the net cross-section, concentrically loaded (defined in clause 6.2.2.2), the Eurocode 3 design expression is:

kN738N1073810.1

47019209.0fA9.0N 3

2M

unetRd,u =×=

××=

γ=


11


Introduction

Design

Example

Exercise

The tensile resistance Nt,Rd is taken as the lesser of these two values, and is therefore 682 kN.

682 kN > 541 kN (i.e. Nt,Rd > NEd)

Unequal angle 125×75×10 in grade S355 steel, connected by the longer leg is therefore acceptable. For efficiency, a smaller angle may be checked.



Introduction

Design

Example

Exercise

Bolted connection

Try a 150×75×10 unequal angle, bolted (with a line of four 22 mm HSFG bolts, at 125 mm centres) through the longer leg. Material properties and partial factors are as for the welded case.

150×75×10 unequal angle bolted by longer leg


150×75×10 unequal angle

Gusset plate

24 mm diameter holes for 22 mm HSFG bolts

12


Introduction

Design

Example

Exercise kN770N104.7700.1

3552170AfN 3

0M

yRd,pl =×=

×=

γ=


Gross area of cross-section, A = 2170 mm2 (from Section Tables).

For yielding of the gross cross-section, plastic resistance is given as:

The net cross-sectional area Anet:

Anet = A – allowance for bolt holes = 2170 –(24×10) = 1930 mm2


Introduction

Design

Example

Exercise

kN577N1057710.1

47019307.0fAN 3

2M

unet3Rd,u =×=

××=

γβ

=


The tensile resistance Nt,Rd is taken as the lesser of these two values, and is therefore 577 kN.

577 kN > 541 kN (i.e. Nt,Rd > NEd)

Unequal angle 150×75×10 in grade S355 steel, connected by the longer leg (using four 22 mm diameter HSFG bolts) is therefore acceptable.

From Table, β3 = 0.7 (since the pitch p1 > 5d0).

13


Introduction

Design

Example

Exercise

Tension member design exercise

A flat bar 200 mm wide × 25 mm thick is to be used as a tie (tension member). Erection conditions require that the bar be constructed from two lengths connected together with a lap splice using six M20 bolts as shown below. Assume 22 mm diameter bolt holes. Calculate the tensile strength of the bar assuming grade S275 steel.

T

50 mm

50 mm

100 mm T

100 mm

A

A

T

T

100 mm

25 mm thick plates


Introduction

Design

Example

Exercise



Session 5

Design of tension members

1

1L. Gardner

Introduction

Local buckling

Classification

Class 4

Exercise

Local buckling and cross-section

classificationDr Leroy GardnerSenior Lecturer in Structural Engineering


Session 6

2L. Gardner

Introduction

Local buckling

Classification

Class 4

Exercise

Overview

Outline:

• Introduction• Local buckling• Cross-section classification• Class 4 – effective widths

2

3L. Gardner

Introduction

Local buckling

Classification

Class 4

Exercise

Background

Background:

• For efficiency, structural members are generally composed of relatively thin elements (i.e. thicknesses substantially less than other cross-sectional dimensions)

• Although favourable in terms of overall structural efficiency, the slender nature of these thin elements results in susceptibility to local instabilities (buckling) under compressive stress, which must be considered in design.

4L. Gardner

Introduction

Local buckling

Classification

Class 4

Exercise

Local buckling

Local buckling in structural components

Local buckling

3

5L. Gardner

Introduction

Local buckling

Classification

Class 4

Exercise

Cross-section classification

• Whether in the elastic or inelastic material range, cross-sectional resistance and rotation capacity are limited by the effects of local buckling.

• Eurocode 3 (and BS 5950) account for the effects of local buckling through cross-section classification.

• The classifications from BS 5950 of plastic, compact, semi-compact and slender are replaced in Eurocode 3 with Class 1, Class 2, Class 3 and Class 4, respectively.

6L. Gardner

Introduction

Local buckling

Classification

Class 4

Exercise

The factors that affect local buckling (and therefore the cross-section classification) are:

• Width/thickness ratios of plate components

• Element support conditions

• Material strength, fy

• Fabrication process

• Applied stress system

Factors affecting local buckling

4

7L. Gardner

Introduction

Local buckling

Classification

Class 4

Exercise

Classification is made by comparing actual width-to-thickness ratios of the plate elements with a set of limiting values, given in Table 5.2 of EN 1993-1-1).

A plate element is Class 4 (slender) if it fails to meet the limiting values for a class 3 element.

The classification of the overall cross-section is taken as the least favourable of the constituent elements (for example, a cross-section with a class 3 flange and class 1 web has an overall classification of Class 3).

Cross-section classification

8L. Gardner

Introduction

Local buckling

Classification

Class 4

Exercise

Definition of 4 classes

Deformation

Moment

Mel

Mpl

Class 1

Class 2

Class 4

Class 3

Eurocode 3 defines four classes of cross-section:

5

9L. Gardner

Introduction

Local buckling

Classification

Class 4

Exercise

Compression

0M

yRd,c

AfN

γ=

Cross-section resistance in compression Nc,Rd:

Class 1, 2 and 3:

Class 4:0M

yeffRd,c

fAN

γ=

10L. Gardner

Introduction

Local buckling

Classification

Class 4

Exercise

Bending

• Class 1 & 2 cross-sections:

0M

yplplRd,c

fWMM

γ==

0M

yelelRd,c

fWMM

γ==

• Class 3 cross-sections:

6

11L. Gardner

Introduction

Local buckling

Classification

Class 4

Exercise

Bending


0M

yeffRd,c

fWM

γ=

12L. Gardner

Introduction

Local buckling

Classification

Class 4

Exercise

Definition of compressed widths – flat widths:

Compressed widths c

c

(a) Outstand flanges (b) Internal compression parts

c

c

Rolled

Welded c

Rolled

Welded

yf/235=ε

Limits on slenderness e.g. c/t ≤ 9ε

7

13L. Gardner

Introduction

Local buckling

Classification

Class 4

Exercise

Internal compression parts

14L. Gardner

Introduction

Local buckling

Classification

Class 4

Exercise

Outstand flanges

8

15L. Gardner

Introduction

Local buckling

Classification

Class 4

Exercise

Angles and tubular sections

16L. Gardner

Introduction

Local buckling

Classification

Class 4

Exercise

Class 4 cross-sections

Class 4 (slender) cross-sections

• For class 4 (slender) cross-sections, reduced (effective) cross-section properties must be calculated to account explicitly for the occurrence of local buckling prior to yielding.

• Effective width formulae for individual elements are provided in Eurocode 3 Part 1.5 (EN 1993-1-5).

9

17L. Gardner

Introduction

Local buckling

Classification

Class 4

Exercise

Class 4 – effective width concept

18L. Gardner

Introduction

Local buckling

Classification

Class 4

Exercise

Cross-section classification exercise

b

h d

tw

tf

r

y y

z

z

h = 254.1 mm

b = 254.6 mm

tw = 8.6 mm

tf = 14.2 mm

r = 12.7 mm

A = 9310 mm2

Section properties for 254 x 254 x 73 UC

Determine the classification and resistance Nc,Rd for a 254 x 254 x 73 UC in pure compression, assuming grade S 355 steel.

10

19L. Gardner

Introduction

Local buckling

Classification

Class 4

Exercise

Summary

Local buckling and cross-section classification:

• Local buckling accounted for through cross-section classification

• 4 Classes of cross-section

• Classification influences resistance

• Effective widths for Class 4 sections

20L. Gardner

Introduction

Local buckling

Classification

Class 4

Exercise

Local buckling and cross-section

classificationDr Leroy GardnerSenior Lecturer in Structural Engineering


Session 6

1

1L. Gardner

Background

Cross-section

Buckling

Example

Exercise

Compression members



Session 7

2L. Gardner

Background

Cross-section

Buckling

Example

Exercise

Outline

Overview:

• Background

• Cross-section resistance Nc,Rd

• Member buckling resistance Nb,Rd

2

3L. Gardner

Background

Cross-section

Buckling

Example

Exercise

Elastic buckling theoryN

w

N

x

L

N

N

(a) Unloaded member

(b) Loaded member (straight)

(c) Loaded member (displaced)

4L. Gardner

Background

Cross-section

Buckling

Example

Exercise

Elastic buckling theory

From stability theory, the elastic buckling load of a perfect pin-ended column is given by:

Other boundary conditions may be accounted for through the effective (critical) length concept.

2

2

cr LEIN π

=

3

5L. Gardner

Background

Cross-section

Buckling

Example

Exercise

Elastic buckling theory

Two bounds: Yielding and buckling

Afy

Non-dimensional slenderness

Material yielding (squashing)

Euler (critical) buckling Ncr

NEd

NEd

Lcr

Load

6L. Gardner

Background

Cross-section

Buckling

Example

Exercise

Imperfections

• Geometric imperfections• Eccentricity of loading• Residual stresses• Non-homogeneity of material

properties• End restraint• etc

Forms of imperfection:

4

7L. Gardner

Background

Cross-section

Buckling

Example

Exercise

Residual stresses

Welding

Hot-rolling

8L. Gardner

Background

Cross-section

Buckling

Example

Exercise

Behaviour of imperfect columns

d,0

cr

Edmax e

NN1

1w−

=

1MwN

NN

Rd

maxEd

Rd

Ed ≤+

1M

eN

NN1

1NN

Rd

d,0Ed

cr

EdRd

Ed ≤−

+

wmax = (w0+w) at mid-height

e0,d is the magnitude of the initial imperfection w0

NEd

NEd

x

w0

w0 = Initial imperfection

ww = additional deflection

5

9L. Gardner

Background

Cross-section

Buckling

Example

Exercise

Perry-Robertson

Perry observed:

• All columns contain imperfections and will deflect laterally from the onset of loading

• The maximum stress along the column length will occur at mid-height and on the inner surface

• The maximum stress will comprise 2 components – axial stress and bending stress

• Failure may be assumed when the maximum stress reaches yield

10L. Gardner

Background

Cross-section

Buckling

Example

Exercise

Perry-Robertson

Robertson contribution:

• The bending stress component is a function of the lateral deflection, which is, in turn, an amplification of the initial imperfection e0,d

• Robertson determined suitable values for these initial imperfections for a range of structural cross-sections

• Eurocode 3 uses the Perry-Robertson concept

• Five different imperfection amplitudes are included (through the imperfection factor α), giving five buckling curves

6

11L. Gardner

Background

Cross-section

Buckling

Example

Exercise

Buckling curves

0.0

0.2

0.4

0.6

0.8

1.0

1.2

0 0.5 1 1.5 2 2.5

Curve a0 Curve a Curve b Curve c Curve d

Curve a0

Red

uctio

n fa

ctor

χ

Non-dimensional slenderness λ

12L. Gardner

Background

Cross-section

Buckling

Example

Exercise

Eurocode 3

Eurocode 3 states, as with BS 5950, that both cross-sectional and member resistance must be verified:

Rd,bEd NN ≤

Rd,cEd NN ≤ Cross-section check

Member buckling check

7

13L. Gardner

Background

Cross-section

Buckling

Example

Exercise

Cross-section resistance

sections4ClassforfA

N

ectionss3or21,ClassforAf

N

0M

yeffRd,c

0M

yRd,c

γ=

γ=

• Cross-section resistance in compression Nc,Rd depends on cross-section classification:

γM0 is specified as 1.0 in EN 1993

This value will also be adopted in the UK

14L. Gardner

Background

Cross-section

Buckling

Example

Exercise

Compression buckling resistance Nb,Rd:

4Class)symmetric(forfA

N

3and2,1ClassforfA

N

1M

yeffRd,b

1M

yRd,b

γ

χ=

γ

χ=

Member buckling

8

15L. Gardner

Background

Cross-section

Buckling

Example

Exercise

Compression buckling resistance:

Equivalence to BS 5950

Pc BS 5950

Eurocode 3Nb,Rd

pc A=

=Aχ fy

γM1

16L. Gardner

Background

Cross-section

Buckling

Example

Exercise

Member buckling

Calculate non-dimensional slenderness λ

3and2,1ClassforN

fA

cr

y=λ

4ClassforN

fA

cr

yeff=λ

Ncr is the elastic critical buckling load for the relevant buckling mode based on the gross properties of the cross-section

9

17L. Gardner

Background

Cross-section

Buckling

Example

Exercise


gyrationofradiusisiandi/Lwhere =λ

2

2

cr LEIN π

=

2

2

2

2

2

2

crE

)i/L(E

ALEIf

λπ

=π

=π

=

The theoretical slenderness boundary λ1between material yielding and elastic member buckling may be found by setting fcr = fy:

y12

1

2

y fEEf π=λ⇒

λπ

=

18L. Gardner

Background

Cross-section

Buckling

Example

Exercise


The non-dimensional slenderness used in EC3 is defined as:

cr

y

cr

y

y

cr

y

cr

1 NAf

ff

f1f1

fEfE

===π

π=

λλ

=λ

Non-dimensionalising in terms of the material as well as the geometry makes it easier to compare the buckling behaviour of columns of different strength material.

10

19L. Gardner

Background

Cross-section

Buckling

Example

Exercise


1.0

N

Afy

Material yielding (in-plane bending)

Elastic member buckling (LTB)

NEd

NEd

Lcr


20L. Gardner

Background

Cross-section

Buckling

Example

Exercise

• Calculate reduction factor, χ

α is the imperfection factor

1)(

15.022 ≤

λ−ϕ+ϕ=χ

))2.0(1(5.0 2λ+−λα+=ϕ

Member buckling

11

21L. Gardner

Background

Cross-section

Buckling

Example

Exercise

Imperfection factor α

0.760.490.340.210.13Imperfection factor α

dcbaa0Buckling curve

Imperfection factors α for 5 buckling curves:

22L. Gardner

Background

Cross-section

Buckling

Example

Exercise

cd

cd

y – yz - ztf > 40 mm

bc

bc

y – yz - ztf ≤ 40 mmWelded

I-sections

cc

dd

y – yz - ztf > 100 mm

aa

bc

y – yz - ztf ≤ 100 mm

h/b ≤ 1.2

aa

bc

y – yz - z

40 mm < tf≤ 100 mm

a0a0

ab

y – yz - ztf ≤ 40 mm

h/b > 1.2

Rolled I-sections

S460

S235S275S355S420

Buckling curve

Buckling about axis

LimitsCross-section

b

tw

tfr

y y

z

z

h

tf

y y

z

ztf

y y

z

z

Buckling curve selection

12

23L. Gardner

Background

Cross-section

Buckling

Example

Exercise

bbanyL-sections

ccanyU-, T- and

solid sections

ccanythick welds: a > 0.5tf

b/tf < 30h/tw < 30

bbanygenerally (except as below)

Welded box sections

ccanycold formed

a0aanyhot finishedHollow

sections

tf

y y

z

zb

h

tw


24L. Gardner

Background

Cross-section

Buckling

Example

Exercise

End restraint (in the plane under consideration) Buckling length Lcr

Effectively restrained in direction at both ends 0.7 L

Partially restrained in direction at both ends 0.85 L

Restrained in direction at one end 0.85 L

Effectively held in position at both ends

Not restrained in direction at either end 1.0 L

One end Other end

Effectively restrained in direction 1.2 L

Partially restrained in direction 1.5 L

Effectively held in position and restrained in direction

Not held in position

Not restrained in direction 2.0 L

Effective (buckling) lengths Lcr

13

25L. Gardner

Background

Cross-section

Buckling

Example

Exercise

Effective (buckling) lengths Lcr

Non-sway Sway

26L. Gardner

Background

Cross-section

Buckling

Example

Exercise

Column buckling design procedure

Design procedure for column buckling:

1. Determine design axial load NEd

2. Select section and determine geometry

3. Classify cross-section (if Class 1-3, no account need be made for local buckling)

4. Determine effective (buckling) length Lcr

5. Calculate Ncr and Afy

14

27L. Gardner

Background

Cross-section

Buckling

Example

Exercise

Column buckling design procedure

6. Non-dimensional slenderness

7. Determine imperfection factor α

8. Calculate buckling reduction factor χ

9. Design buckling resistance

10. Check

cr

y

NfA

=λ

1M

yRd,b

fAN

γ

χ=

0.1NN

Rd,b

Ed ≤

28L. Gardner

Background

Cross-section

Buckling

Example

Exercise

Member buckling resistance example

A circular hollow section member is to be used as an internal column in a multi-storey building. The column has pinned boundary conditions at each end, and the inter-storey height is 4 m.

4.0 m

NEd = 2110 kN

The critical combination of actions results in a design axial force of 2110 kN.

15

29L. Gardner

Background

Cross-section

Buckling

Example

Exercise


d = 244.5 mm

t = 10.0 mm

A = 7370 mm2

Wel,y = 415000 mm3

Wpl,y = 550000 mm3

I = 50730000 mm4

t

d

Assess the suitability of a hot-rolled 244.5×10 CHS in grade S 355 steel for this application.

30L. Gardner

Background

Cross-section

Buckling

Example

Exercise


For a nominal material thickness (t = 10.0 mm) of less than or equal to 16 mm the nominal values of yield strength fy for grade S 355 steel is 355 N/mm2 (from EN 10210-1).

From clause 3.2.6: E = 210000 N/mm2

16

31L. Gardner

Background

Cross-section

Buckling

Example

Exercise


Cross-section classification (clause 5.5.2):

Tubular sections (Table 5.2, sheet 3)

d/t = 244.5/10.0 = 24.5

Limit for Class 1 section = 50 ε2 = 40.7 > 24.5

∴ Cross-section is Class 1

81.0355/235f/235 y ===ε

32L. Gardner

Background

Cross-section

Buckling

Example

Exercise


OKisresistanceectionsCrosskN21102616

kN2616N10261600.1

3557370N

sections-cross 3 or21,ClassforAf

N

3Rd,c

0M

yRd,c

−∴>

=×=×

=∴

γ=

Cross-section compression resistance (clause 6.2.4):

17

33L. Gardner

Background

Cross-section

Buckling

Example

Exercise


( )[ ]

sections-cross 3 & 21, Class forNAf

and

0.2-10.5 where

1.0but1

sections- cross 3 & 21, Class forfA

N

cr

y

2

22

1M

yRd,b

=λ

λ+λα+=Φ

≤χλ−Φ+Φ

=χ

γ

χ=

Member buckling resistance in compression (clause 6.3.1):

34L. Gardner

Background

Cross-section

Buckling

Example

Exercise


63.01065713557370

kN 65714000

50730000210000L

EIN

3

2

2

2cr

2

cr

=××

=λ∴

=××π

=π

=

Elastic critical force and non-dimensional slenderness for flexural buckling Ncr

From Table 6.2 of EN 1993-1-1:

For a hot-rolled CHS, use buckling curve a

18

35L. Gardner

Background

Cross-section

Buckling

Example

Exerciseccanycold formed

a0aanyhot finishedHollow

sections

S460

S235S275S355S420

Buckling curve

Buckling about axis

LimitsCross-section


Extract from Table 6.2 of EN 1993-1-1:

For a hot-rolled CHS, use buckling curve a

36L. Gardner

Background

Cross-section

Buckling

Example

Exercise

Graphical approach

0.0

0.2

0.4

0.6

0.8

1.0

1.2

0 0.5 1 1.5 2 2.5

Curve a0 Curve a Curve b Curve c Curve d

Curve a0

Red

uctio

n fa

ctor

χ


0.63

≈0.88

19

37L. Gardner

Background

Cross-section

Buckling

Example

Exercise

From Table 6.1 of EN 1993-1-1, for buckling curve a, α = 0.21

2297 > 2110 kN ∴Buckling resistance is OK.

The chosen cross-section, 244.5x10 CHS, in grade S 355 steel is acceptable.

kN2297N1022970.1

355737088.0N

88.063.074.074.0

174.0]63.0)2.063.0(21.01[5.0

3Rd,b

22

2

=×=××

=∴

=−+

=χ

=+−+=Φ


38L. Gardner

Background

Cross-section

Buckling

Example

Exercise

Member buckling resistance exercise

A UC section member is to be used as an internal column in a multi-storey building. The column has pinned boundary conditions at each end, and the inter-storey height is 4.5 m.

4.5 m

NEd = 305.6 kN

The critical combination of actions results in a design axial force of 305.6 kN.

20

39L. Gardner

Background

Cross-section

Buckling

Example

Exercise

b

h d

tw

tf

r

y y

z

z

h = 157.6 mmb = 152.9 mmtw = 6.5 mmtf = 9.4 mmr = 7.6 mmA = 3830 mm2

Iy = 17480000 mm4

Iz = 5600000 mm4

Section properties for 152x152x30 UC

Try a 152x152x30 UC in grade S 275 steel.

Member buckling resistance exercise

40L. Gardner

Background

Cross-section

Buckling

Example

Exercise



Session 7

Compression members

1

1L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

Beams



Session 8

2L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

Outline

Overview:

• Background• In-plane bending• Shear• Deflections• Lateral torsional buckling

2

3L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

Eurocode 3

Eurocode 3 states, as with BS 5950, that both cross-sectional and member bending resistance must be verified:

Rd,bEd MM ≤

Rd,cEd MM ≤ Cross-section check(In-plane bending)

Member buckling check

4L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises


1.0

Beam behaviour analogous to yielding/buckling of columns.

M

Wyfy

Material yielding (in-plane bending)

Elastic member buckling Mcr

Lcr

MEd MEd

Non-dimensional slenderness LTλ

3

5L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

Cross-sections in bending

• Class 1 & 2 cross-sections:

0M

yplplRd,c

fWMM

γ==

0M

yelelRd,c

fWMM

γ==


6L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

Cross-sections in bending


0M

yeffRd,c

fWM

γ=

4

7L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

Section moduli W

z

Subscripts are used to differentiate between the plastic, elastic or effective section modulus

Plastic modulus Wpl (S in BS 5950)

Elastic modulus Wel (Z in BS 5950)

Effective modulus Weff (Zeff in BS 5950)

The partial factor γM0 is applied to all cross-section bending resistances, and equal 1.0.

8L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

Shear resistance

The design shear force is denoted by VEd(shear force design effect).

The design shear resistance of a cross-section is denoted by Vc,Rd and may be calculated based on a plastic (Vpl,Rd) or an elastic distribution of shear stress.

0.1VV

Rd,c

Ed ≤ Shear check

5

9L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

Plastic shear resistance Vpl,Rd

The usual approach is to use the plastic shear resistance Vpl,Rd

The plastic shear resistance is essentially defined as the yield strength in shear multiplied by a shear area Av:

0M

yvRd,pl

)3/f(AV

γ=

10L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

Shear area Av

The shear area Av is in effect the area of the cross-section that can be mobilised to resist the applied shear force with a moderate allowance for plastic redistribution

For sections where the load is applied parallel to the web, this is essentially the area of the web (with some allowance for the root radii in rolled sections).

6

11L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

Shear areas Av

Shear areas Av are given in clause 6.2.6(3).

• 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 RHS of uniform thickness, load parallel to depth:

Av = Ah/(b+h)

• CHS and tubes of uniform thickness:

Av = 2A/π

12L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

Definition of terms

A is the cross-sectional area

b is the overall section breadth

h is the overall section depth

hw is the overall web depth (measured between flanges)

r is the root radius

tf is the flange thickness

tw is the web thickness (taken as the minimum value if the web is not of constant thickness)

η = 1.0 (from UK NA)

7

13L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

Shear buckling

The resistance of the web to shear buckling should also be checked, though this is unlikely to affect cross-sections of standard hot-rolled proportions.

Shear buckling need not be considered provided:

)NAUKfrom(0.1;f

235where

websdunstiffene for 72th

y

w

w

=η=ε

ηε

≤

14L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises


b

h

tw

tf

r

y y

z

z

Determine the shear resistance of a rolled channel section 229x89 in grade S 275 steel loaded parallel to the web.

Section properties for 229x89 rolled channel section

Shear resistance example

8

15L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

Shear resistance is determined according to clause 6.2.6

0M

yvRd,pl

)3/f(AV

γ=

Shear resistance example

For a nominal material thickness (tf=13.3 mm and tw = 8.6 mm) of less than or equal to 16 mm the nominal values of yield strength fyfor grade S 275 steel (to EN 10025-2) is found from Table 3.1 to be 275 N/mm2.

16L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

Shear resistance exampleShear area Av

For a rolled channel section, loaded parallel to the web, the shear area is given by:

Av = A – 2btf + (tw + r)tf

= 4160 – (2×88.9×13.3) + (8.6+13.7)×13.3

= 2092 mm2

kN332N33200000.1

)3/275(2092V Rd,pl ==×

=

For the same cross-section BS 5950 (2000) gives a shear resistance of 324 kN.

9

17L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

Serviceability

Excessive serviceability deflections may impair the function of a structure, for example, leading to cracking of plaster, misalignments of crane rails, causing difficulty in opening doors, etc.

Deflection checks should therefore be performed against suitable limiting values.

From the UK National Annex, deflection checks should be made under unfactored variable actions Qk.

18L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

Serviceability

Vertical deflection limits

Horizontal deflection limits

Design situation Deflection limit

Cantilevers Length/180

Beams carrying plaster or other brittle finish Span/360

Other beams (except purlins and sheeting rails) Span/200

Purlins and sheeting rails To suit cladding

Design situation Deflection limit

Tops of columns in single storey buildings, except portal frames Height/300

Columns in portal frame buildings, not supporting crane runways To suit cladding

In each storey of a building with more than one storey Height of storey/300

10

19L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

Lateral torsional buckling


Lateral torsional buckling is the member buckling mode associated with slender beams loaded about their major axis, without continuous lateral restraint.

If continuous lateral restraint is provided to the beam, then lateral torsional buckling will be prevented and failure will occur in another mode, generally in-plane bending (and/or shear).

20L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises


Can be discounted when:

• Minor axis bending

• CHS, SHS, circular or square bar

• Fully laterally restrained beams

• < 0.2 (or 0.4 in some cases)LTλ

11

21L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

Checks should be carried out on all unrestrained segments of beams (between the points where lateral restraint exists).

Lateral torsional buckling resistance

Beam on plan

Lateral restraint

Lcr = 1.0 L

Lateral restraint

Lateral restraint

22L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

Eurocode 3Three methods to check LTB in EC3:

• The primary method adopts the lateral torsional buckling curves given by equations 6.56 and 6.57, and is set out in clause 6.3.2.2 (general case) and clause 6.3.2.3 (for rolled sections and equivalent welded sections).

• The second is a simplified assessment method for beams with restraints in buildings, and is set out in clause 6.3.2.4.

• The third is a general method for lateral and lateral torsional buckling of structural components, given in clause 6.3.4.

12

23L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises


Eurocode 3 design approach for lateral torsional buckling is analogous to the column buckling treatment.

The design buckling resistance Mb,Rd of a laterally unrestrained beam (or segment of beam) should be taken as:

1M

yyLTRd,b

fWM

γχ=

Reduction factor for LTB

24L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

Lateral torsional buckling resistance:

Equivalence to BS 5950

Mb

Eurocode 3

BS 5950

Mb,Rd

pb Sx (or Zx)=

=WyχLT fy

γM1

Wy will be Wpl,y or Wel,y

13

25L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

Buckling curves – general case

Lateral torsional buckling curves for the general case are given below:

0.1but1LT2

LT2LTLT

LT ≤χλ−Φ+Φ

=χ

])2.0(1[5.0 2LTLTLTLT λ+−λα+=Φ

Plateau length

Imperfection factor from Table 6.3

26L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises


For the general case, refer to Table 6.4:

d-Other cross-sections

dh/b > 2

ch/b ≤ 2Welded I-sections

bh/b > 2

ah/b ≤ 2Rolled I-sections

Buckling curveLimitsCross-section

14

27L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

Imperfection factor αLT

0.760.490.340.21Imperfection factor αLT

dcbaBuckling curve

Imperfection factors αLT for 4 buckling curves:

28L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

LTB curves

4 buckling curves for LTB (a, b, c and d)

0.0

0.2

0.4

0.6

0.8

1.0

1.2

0 0.5 1 1.5 2 2.5

Curve a

Curve b

Curve c

Curve d

Red

uctio

n fa

ctor

χLT

Non-dimensional slenderness LTλ0.2

15

29L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

Buckling curves – rolled or equivalent welded sections case

LTB curves for the rolled or equivalent welded sections case are given below; Table 6.5 is used to select buckling curve:

])(1[5.0 2LT0,LTLTLTLT λβ+λ−λα+=Φ

Plateau lengthRecommended value = 0.4

⎪⎩

⎪⎨⎧

λ≤χ

≤χ

λβ−Φ+Φ=χ

LTLT

LT

2LT

2LTLT

LT 10.1

but1

β factor Recommended value = 0.75

30L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

LTB curvesComparison between general curves and curves for rolled and equivalent welded sections (I-sections – h/b>2)

Red

uctio

n fa

ctor

χLT

Non-dimensional slenderness LTλ

0.00

0.20

0.40

0.60

0.80

1.00

1.20

0 0.5 1 1.5 2 2.5

General (h/b>2)

Rolled (h/b>2)

16

31L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

cr

yyLT M

fW=λ


• Buckling curves as for compression (except curve a0)

• Wy depends on section classification• Mcr is the elastic critical LTB moment

• Calculate lateral torsional buckling slenderness:

32L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

Designers familiar with BS 5950 will be accustomed to simplified calculations, where determination of the elastic critical moment for lateral torsional buckling Mcr is aided, for example, by inclusion of the geometric quantities ‘u’ and ‘v’ in section tables.

Such simplifications do not appear in the primary Eurocode method.

Elastic critical buckling moment Mcr

17

33L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

Mcr under uniform moment

G is the shear modulusIT is the torsion constantIw is the warping constantIz is the minor axis second moment of areaLcr is the buckling length of the beam

5.0

z2

T2

cr

z

w2

cr

z2

0,cr EIGIL

II

LEIM ⎥

⎦

⎤⎢⎣

⎡

π+

π=

For typical end conditions, and under uniform moment the elastic critical lateral torsional buckling moment Mcr is:

34L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

Mcr under non-uniform moment

Numerical solutions have been calculated for a number of other loading conditions. For uniform doubly-symmetric cross-sections, loaded through the shear centre at the level of the centroidal axis, and with the standard conditions of restraint described, Mcr may be calculated by:

5.0

z2

T2

cr

z

w2

cr

z2

1cr EIGIL

II

LEICM ⎥

⎦

⎤⎢⎣

⎡π

+π

=

18

35L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

C1 factor – end moments

For end moment loading C1 may be approximated by the equation below, though other approximations also exist.

C1= 1.88 – 1.40ψ + 0.52ψ2 but C1 ≤ 2.70

where ψ is the ratio of the end moments (defined in the following table).

36L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

C1 factor – transverse loading

C1 values for transverse loading

Loading and support conditions Bending moment diagram Value of C1

1.132

1.285

1.365

1.565

1.046

w

w

F

F

F F

= = = =

C L

19

37L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

For hot-rolled doubly symmetric I and H sections without destabilising loads, may be conservatively simplified to:

LTλ

LTofassessmentSimplified λ

1

z

1z

1LT 9.0

C19.0

C1

λλ

=λ=λ

y1zz f

E;i/L π=λ=λ

As a further simplification, C1 may also be conservatively taken = 1.0.

38L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

Substituting in numerical values for , the following simplified expressions result.

S355S275S235

96i/L

C1 z

1LT =λ

LTofassessmentSimplified λ

104i/L

C1 z

1LT =λ 85

i/LC1 z

1LT =λ

C1 may be conservatively taken = 1.0, though the level of conservatism increases the more the actual bending moment diagram differs from uniform moment.

1λ

20

39L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

Design procedure for LTB

Design procedure for LTB:

1. Determine BMD and SFD from design loads

2. Select section and determine geometry

3. Classify cross-section (Class 1, 2, 3 or 4)

4. Determine effective (buckling) length Lcr –depends on boundary conditions and load level

5. Calculate Mcr and Wyfy

40L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

Design procedure for LTB

6. Non-dimensional slenderness

7. Determine imperfection factor αLT

8. Calculate buckling reduction factor χLT

9. Design buckling resistance

10. Check for each unrestrained portion

cr

yyLT M

fW=λ

1M

yyLTRd,b

fWM

γχ=

0.1MM

Rd,b

Ed ≤

21

41L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

Simplified method (Cl. 6.3.2.4)Simplified method for beams with restraints in buildings (Clause 6.3.2.4)

This method treats the compression flange of the beam and part of the web as a strut:

b

h

Beam

Strut

Compression

Tension

b

Compression flange + 1/3 of the compressed

area of web

42L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

General method (Cl. 6.3.4)

General method for lateral and lateral torsional buckling of structural components

• May be applied to single members, plane frames etc.

• Requires determination of plastic and elastic (buckling) resistance of structure, which subsequently defines global slenderness

• Generally requires FE

22

43L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

Restrained beam exercise

The simply supported 610×229×125 UB of S275 steel shown below has a span of 6.0 m. Check moment resistance, shear and deflections.

6.0 m

Dead load = 60 kN/mImposed load = 70 kN/m

Beam is fully laterally restrained

44L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

Restrained beam exercise

b

h d

tw

tf

r

y y

z

z


Wy,pl = 3676×103 mm3

Iy = 986.1×106 mm4

610×229×125 UB

23

45L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

LTB Example

Description

A simply-supported primary beam is required to span 10.8 m and to support two secondary beams as shown below. The secondary beams are connected through fin plates to the web of the primary beam, and full lateral restraint may be assumed at these points. Select a suitable member for the primary beam assuming grade S 275 steel.

46L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

LTB Example

General arrangement

24

47L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

LTB Example

Design loading is as follows:

Loading

AB C

D

2.5 m 3.2 m 5.1 m

425.1 kN 319.6 kN

48L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

LTB Example

Shear force diagram

A D

477.6 kN

267.1 kN

B

C

SF

52.5 kN

A D

1194 kNm 1362 kNm

B C

BM

Bending moment diagram

25

49L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

LTB Example

For the purposes of this example, lateral torsional buckling curves for the general case will be utilised. Lateral torsional buckling checks to be carried out on segments BC and CD. By inspection, segment AB is not critical.

Try 762×267×173 UB in grade S 275 steel.

50L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

LTB Example

b

h d

tw

tf

r

y y

z

z


Wy,pl = 6198×103 mm3

Iz = 68.50×106 mm4

It = 2670×103 mm4

Iw = 9390×109 mm6

26

51L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

LTB Example

For a nominal material thickness (tf = 21.6 mm and tw = 14.3 mm) of between 16 mm and 40 mm the nominal values of yield strength fy for grade S 275 steel (to EN 10025-2) is 265 N/mm2.

From clause 3.2.6: E = 210000 N/mm2 and G ≈81000 N/mm2.

52L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

LTB Example

Cross-section classification (clause 5.5.2):

Outstand flanges (Table 5.2, sheet 2)

cf = (b – tw – 2r) / 2 = 109.7 mm

cf / tf = 109.7 / 21.6 = 5.08

Limit for Class 1 flange = 9ε = 8.48 > 5.08

∴ Flange is Class 1

94.0265/235f/235 y ===ε

27

53L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

LTB Example

Web – internal part in bending (Table 5.2, sheet 1)

cw = h – 2tf – 2r = 686.0 mm

cw / tw= 686.0 / 14.3 = 48.0

Limit for Class 1 web = 72 ε = 67.8 > 48.0

∴ Web is Class 1

Overall cross-section classification is therefore Class 1.

54L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

LTB Example

Bending resistance of cross-section (clause 6.2.5):

kNm1362kNm1642

Nmm1016420.1

265106198

tionssec2and1ClassforfW

M

63

0M

yy,plRd,y,c

>=

×=××

=

γ=

∴ Cross-section resistance in bending is OK.

28

55L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

LTB ExampleLateral torsional buckling check (clause 6.3.2.2) – Segment BC:

1M

yyLTRd,b

Ed

fWM

kNm1362M

γχ=

=

where Wy = Wpl,y for Class 1 and 2 sections

Determine Mcr for segment BC (Lcr = 3200 mm)

5.0

z2

T2

cr

z

w2

cr

z2

1cr EIGIL

II

LEICM ⎥

⎦

⎤⎢⎣

⎡

π+

π=

56L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

LTB Example

For end moment loading C1 may be approximated from:

C1 = 1.88 – 1.40ψ + 0.52ψ2 but C1 ≤ 2.70

05.1C

88.013621194momentsendtheofratiotheis

1 =⇒

==ψ

5.0

62

32

6

9

2

62

cr 105.68210000102670810003200

105.68109390

3200105.6821000005.1M ⎥

⎦

⎤⎢⎣

⎡×××π

×××+

×××××π

×=

= 5699x106 Nmm = 5699 kNm

29

57L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

LTB Example

Non-dimensional lateral torsional slenderness for segment BC:

54.0105699

265106198M

fW6

3

cr

yyLT =

×××

==λ

Select buckling curve and imperfection factor αLT:

From Table 6.4: h/b = 762.2/266.7 = 2.85

For a rolled I-section with h/b > 2, use buckling curve b

58L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

LTB Example

From Table 6.3 of EN 1993-1-1:

For buckling curve b, αLT = 0.34

Calculate reduction factor for lateral torsional buckling, χLT – Segment BC:

0.1but1LT2

LT2LTLT

LT ≤χλ−Φ+Φ

=χ

])2.0(1[5.0where 2LTLTLTLT λ+−λα+=Φ

30

59L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

LTB Example

ΦLT = 0.5[1+0.34(0.54-0.2) + 0.542] = 0.70

87.054.070.070.0

122LT =

−+=χ∴

kNm1425Nmm101425

0.126510619887.0

fWM

6

3

1M

yyLTRd,b

=×=

×××=γ

χ=

Lateral torsional buckling resistance Mb,Rd –Segment BC:

60L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

LTB Example

OKisBCSegment0.196.014251362

MM

Rd,b

Ed ∴≤==

Lateral torsional buckling check (clause 6.3.2.2) – Segment CD:

1M

yyLTRd,b

Ed

fWM

kNm1362M

γχ=

=

where Wy = Wpl,y for Class 1 and 2 sections

Determine Mcr for segment CD (Lcr = 5100 mm)

31

61L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

LTB Example5.0

z2

T2

cr

z

w2

cr

z2

1cr EIGIL

II

LEICM ⎥

⎦

⎤⎢⎣

⎡

π+

π=

Determine ψ from Table:

88.1C

01362

0momentsendtheofratiotheis

1 =⇒

==ψ

5.0

62

32

6

9

2

62

cr 105.68210000102670810005100

105.68109390

5100105.6821000088.1M ⎥

⎦

⎤⎢⎣

⎡×××π

×××+

×××××π

=

= 4311×106 Nmm = 4311 kNm

62L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

LTB Example

Non-dimensional lateral torsional slenderness for segment CD:

62.0104311

265106198M

fW6

3

cr

yyLT =

×××

==λ

The buckling curve and imperfection factor αLTare as for segment BC.

32

63L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

LTB Example

0.1but1LT2

LT2LTLT

LT ≤χλ−Φ+Φ

=χ

])2.0(1[5.0where 2LTLTLTLT λ+−λα+=Φ

83.062.076.076.0

122LT =

−+=χ∴

= 0.5[1+0.34(0.62-0.2) + 0.622] = 0.76

Calculate reduction factor for lateral torsional buckling, χLT – Segment CD:

64L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises

LTB Example

kNm1360Nmm101360

0.126510619883.0

fWM

6

3

1M

yyLTRd,b

=×=

×××=γ

χ=

Lateral torsional buckling resistance Mb,Rd –Segment CD:

00.113601362

MM

Rd,b

Ed ==

Segment CD is critical and marginally fails LTB check.

33

65L. Gardner

Background

In-plane bending

Shear

Serviceability

LTB

Exercises



Session 8

Beams

1

1L. Gardner

Introduction

Cross-section

Members

Annex A & B

Simple construction

Beam-columns



Session 9

2L. Gardner

Introduction

Cross-section

Members

Annex A & B

Simple construction

Beam-columns:

• Cross-section check

• Member buckling check

Introduction

2

3L. Gardner

Introduction

Cross-section

Members

Annex A & B

Simple construction

1MM

MM

NN

Rdz,

Edz,

Rdy,

Edy,

Rd

Ed ≤++

Cross-section checks similar to BS 5950, including a simplified linear interaction, as below:

Cross-section checks

More sophisticated expressions are also provided for Class 1 and 2 for greater efficiency.

4L. Gardner

Introduction

Cross-section

Members

Annex A & B

Simple construction

Two philosophies:

• Interaction method - Clause 6.3.3

• Interaction ‘k’ factors from Annex A or B.

• General method - Clause 6.3.4

• Not for hand calculations (requires FE or similar)

Beam-columns – member checks

3

5L. Gardner

Introduction

Cross-section

Members

Annex A & B

Simple construction

1MM

kMM

kNN

1MM

kMM

kNN

Rd,z,c

Ed,zzz

Rd,b

Ed,yzy

Rd,z,b

Ed

Rd,z,c

Ed,zyz

Rd,b

Ed,yyy

Rd,y,b

Ed

≤++

≤++

Simple construction

In general, both Eqs. 6.61 and 6.62 must be examined and satisfied:

Eq. 6.61

Eq. 6.62

6L. Gardner

Introduction

Cross-section

Members

Annex A & B

Simple construction

Annex A (Method 1) – French-Belgian

• Derived the necessary coefficients explicitly- so far as it is possible

• Correct by calibration for plasticity etc.- with FE and test results

Annex B (Method 2) – German-Austrian

• Derived all coefficients from FE- Calibrated with test results

Interaction factors kij

4

7L. Gardner

Introduction

Cross-section

Members

Annex A & B

Simple construction

Good news - simple construction

‘Simple construction’ is commonly used for the design of multi-storey buildings (particularly in the UK).

• Beams are designed as simply supported

• Columns are designed for nominal moments arising from the eccentricity at the beam-to-column connection.

8L. Gardner

Introduction

Cross-section

Members

Annex A & B

Simple construction

Multi-storey frame:

Simple construction

Gkr & QkrWk & NHF

Wk & NHF

Wk & NHF

Wk & NHF

Gkf & Qkf

Gkf & Qkf

Gkf & Qkf

5

9L. Gardner

Introduction

Cross-section

Members

Annex A & B

Simple construction

1MM

kMM

kNN

1MM

kMM

kNN

Rd,z,c

Ed,zzz

Rd,b

Ed,yzy

Rd,z,b

Ed

Rd,z,c

Ed,zyz

Rd,b

Ed,yyy

Rd,y,b

Ed

≤++

≤++

Simple construction

In general, both Eqs. 6.61 and 6.62 must be examined and satisfied:

Eq. 6.61

Eq. 6.62

10L. Gardner

Introduction

Cross-section

Members

Annex A & B

Simple construction

1MM

kMM

kNN

Rd,z,c

Ed,zzz

Rd,b

Ed,yzy

Rd,z,b

Ed ≤++

NCCI Simplification

For columns in simple construction, the first term (i.e. the axial load) of both expressions (Eq. 6.61 and 6.62) dominates.

For UC sections, Iy > Iz (usually around 3 times greater), so Nb,y,Rd > Nb,z,Rd (greater difference for higher slenderness).

Therefore, for practical simple construction situations and UC sections, Eq. 6.62 will always govern.

Eq. 6.62

6

11L. Gardner

Introduction

Cross-section

Members

Annex A & B

Simple construction

1MM

5.1MM

0.1NN

Rd,z,c

Ed,z

Rd,b

Ed,y

Rd,z,b

Ed ≤++

NCCI Simplification

Given that the moment components are small for simple construction, the interaction factors can be conservatively simplified without any significant overall loss of efficiency, resulting in:

kzy = 1.0 and kzz = 1.5

Eq. 6.62

12L. Gardner

Introduction

Cross-section

Members

Annex A & B

Simple construction

Recommendations:

• For pencil and paper calculations, use:

- Clause 6.3.3 with Annex B

• Use NCCI simplification for columns in simple construction

• Make spreadsheets to check calculations

• Full worked examples in Designers’ Guide and Trahair et al textbook.

Recommendations

7

13L. Gardner

Introduction

Cross-section

Members

Annex A & B

Simple construction

Beam-columns



Session 9

1

1L. Gardner

Introduction

Bolted joints

Welded joints Joints



Session 10

2L. Gardner

Introduction

Bolted joints

Welded joints

Overview:

• Introduction• Bolted joints• Welded joints

Outline

2

3L. Gardner

Introduction

Bolted joints

Welded joints

Part 1.8 of Eurocode 3 is some 50% longer than the general Part 1.1.

It provides a much more extensive treatment of the whole subject area of connections than a UK designer would expect to find in a code.

EN 1993-1-8

4L. Gardner

Introduction

Bolted joints

Welded joints

EN 1993-1-8

Essentially, the coverage of Part 1.8 focuses on 4 topics:

1. Fasteners (Sections 3 and 4 of Part 1.8) covering the basic strength of bolts in shear, the resistance of fillet welds etc.

2. The role of connections in overall frame design (Section 5 of Part 1.8), covering the various possible approaches to joint classification and global frame analysis.

3

5L. Gardner

Introduction

Bolted joints

Welded joints

EN 1993-1-8

3. Joints between I-sections (Section 6 of Part 1.8), being more akin to the BCSA/SCI Green Books treatment than to the current content of BS 5950 Part 1.

4. Joints between structural hollow sections (Section 7 of Part 1.8), being very similar to several existing CIDECT guides.

6L. Gardner

Introduction

Bolted joints

Welded joints

Bolted joints

Bolted joints:

• Shear resistance Fv,Rd

• Bearing resistance Fb,Rd

• Tension resistance Ft,Rd

• Combined shear and tension

• Bolt spacing

4

7L. Gardner

Introduction

Bolted joints

Welded joints

Bolt shear resistance per shear plane for ordinary bolts:

where:

αv = 0.6 for classes 4.6, 5.6 and 8.8 where the shear plane passes through the threaded portion of the bolt, and for all classes where the shear plane passes through the unthreaded portion of the bolt

= 0.5 for classes 4.8, 5.8, 6.8 and 10.9 where the shear plane passes through the threaded portion of the bolt

2M

ubvRd,v

AfF

γα

=

Bolted joints

8L. Gardner

Introduction

Bolted joints

Welded joints

fub is the ultimate tensile strength of the bolt

A is the tensile stress area As (i.e. area at threads) when the shear plane passes through the threaded portion of the bolt or the gross cross-sectional area when the shear plane passes through the unthreaded (shank) portion of the bolt.

γM2 may be taken as 1.25

Bolted joints

5

9L. Gardner

Introduction

Bolted joints

Welded joints

Bearing resistance Fb,Rd

Bearing resistance is governed by the projected contact area between a bolt and connected parts, the ultimate material strength (of the bolt or the connected parts), and may be limited by bolt spacing and edge and end distances.

From EN 1993-1-8, bearing resistance is given by:

2M

ub1Rd,b

dtfkFγα

=

Bolted joints

10L. Gardner

Introduction

Bolted joints

Welded joints

Definitions of terms:

αb is the smallest of: αd; fub/fu or 1.0, and accounts for various failure modes

d is the bolt diameter

t is the minimum thickness of the connected parts


fu is the ultimate tensile strength of the connected parts

αd and k1 relate to bolt spacing and edge and end distances.

Bolted joints

6

11L. Gardner

Introduction

Bolted joints

Welded joints

Combined tension and shear

In some situations, bolts may experience tension and shear in combination. In general, bolt capacities would be expected to reduce when high values of shear and tension are coexistent. EN 1993-1-8 provides the following interaction expression to deal with such cases:

Bolted joints

0.1F4.1

FFF

Rd,t

Ed,t

Rd,v

Ed,v ≤+

12L. Gardner

Introduction

Bolted joints

Welded joints

Bolted joints

Spacing requirements

Minimum bolt spacings and edge and end distances are as below, where d0 is the fastener (bolt) hole diameter. These values are defined in Table 3.3 of EN 1993-1-8.

• Minimum spacing of bolts in the direction of load transfer p1 = 2.2d0

• Minimum end distance in the direction of load transfer e1 = 1.2d0

7

13L. Gardner

Introduction

Bolted joints

Welded joints

Bolted joints

• Minimum spacing of bolts perpendicular to the direction of load transfer p2 = 2.4d0

• Minimum edge distance perpendicular to the direction of load transfer e2 = 1.2d0

14L. Gardner

Introduction

Bolted joints

Welded joints

Bolted joint example

Description

Calculate the strength of the bolts in the lap splice shown below assuming the use of M20 Grade 4.6 bolts in 22 mm clearance holes and Grade S275 plate.

8

15L. Gardner

Introduction

Bolted joints

Welded joints

Shear resistance:

Bolts are in single shear, and it is assumed that the shear plane passes through the threaded portion of the bolts:

αv = 0.6, fub = 400 N/mm2, A = As = 245 mm2, γM2= 1.25

Shear resistance per bolt Fv,Rd:

kN0.47N4704025.1

2454006.0AfF2M

ubvRd,v ==

××=

γα

=


16L. Gardner

Introduction

Bolted joints

Welded joints

Bearing resistance:Bearing resistance per bolt Fb,Rd:

From geometry: p1 = 60 mm, e1 = 40 mm, e2 = 40 mm, d0 = 22 mm.

From EN 10025-2, fu of plate (Grade S275, t > 3 mm) = 410 N/mm2.

2M

ub1Rd,b

dtfkFγα

=


9

17L. Gardner

Introduction

Bolted joints

Welded joints

For end bolts, αd = = (40/66) = 0.61

For inner bolts, αd = = (60/66 – 0.25) = 0.66

For edge bolts, k1 is the smaller of or 2.5

(2.8×(40/22) – 1.7) = 3.4. ∴ k1 = 2.5

fub/fu = 400/410 = 0.98

αb is the smaller of: αd, fu/fub or 1.0

For end bolts αb = 0.61, and for inner bolts αb = 0.66

0

1

d3e

0

1

d3p

)7.1de8.2(

0

2 −


18L. Gardner

Introduction

Bolted joints

Welded joints

Therefore, for end bolts,

And, for inner bolt,

Clearly the resistance of the joint is controlled by the strength in shear. Therefore, the resistance of the tension splice as governed by the shear resistance of the bolts = 3 × 47.0 = 141 kN.

kN1.16025.1

162041061.05.2dtfkF2M

ub1Rd,b =

××××=

γα

=

kN2.17325.1

162041066.05.2dtfkF2M

ub1Rd,b =

××××=

γα

=


10

19L. Gardner

Introduction

Bolted joints

Welded joints

Design of welded joints:

• Butt welds

• Fillet welds

- directional method

- simplified method

Welded joints

20L. Gardner

Introduction

Bolted joints

Welded joints

Butt welds

Strength of butt weld taken as that of parent metal (i.e. fy in tension or compression or fy/ in shear) provided that suitable electrodes are used.

Throat thickness taken as minimum depth of penetration, reduced by 3 mm for most partial-penetration butt welds.

3

Welded joints

11

21L. Gardner

Introduction

Bolted joints

Welded joints

Two methods are permitted for the design of fillet welds:

• the directional method, in which the forces transmitted by a unit length of weld are resolved into parallel and perpendicular components.

• the simplified method, in which only longitudinal shear is considered.

These approaches broadly mirror those used in BS5950: Part 1.

Welded joints

22L. Gardner

Introduction

Bolted joints

Welded joints

Simplified approach

Check Fw,Ed ≤ Fw,Rd

Fw,Ed is the design value of the weld force per unit length

Fw,Rd is the design resistance of the weld per unit length

The design resistance of the weld per unit length may be calculated as follows:

Welded joints

afF d,vwRd,w =

12

23L. Gardner

Introduction

Bolted joints

Welded joints

fvw,d is the design shear strength of the weld

a is the throat thickness of the weld

3f

f2Mw

ud,vw γβ=

Welded joints

fu is the minimum ultimate tensile strength of the connected parts

βw is a correlation factor that depends on the material grade


24L. Gardner

Introduction

Bolted joints

Welded joints

Values for correlation factor βw

Welded joints

4703 < t ≤ 1000.90

510t ≤ 3S355

4103 < t ≤ 1000.85

430t ≤ 3S275

3603 < t ≤ 1000.80

360t ≤ 3S235

Correlation factor βw


Thickness range (mm)

Steel grade

13

25L. Gardner

Introduction

Bolted joints

Welded joints

Description

A 150×20 mm tie in Grade S275 steel carrying 400 kN is spliced using a single-sided cover plate 100×20 mm as shown in the figure below. Design a suitable fillet weld to carry the applied load.

Welded joint example

26L. Gardner

Introduction

Bolted joints

Welded joints

Try 8 mm fillet welds:

Throat thickness a = 0.7 s = 0.7×8 = 5.6 mm


Design shear strength of weld:

From Table, fu = 410 N/mm2 and βw = 0.85

2d,vw mm/N223

325.185.0410f =

××=

14

27L. Gardner

Introduction

Bolted joints

Welded joints

The design resistance of the weld per unit length (i.e. per mm run) Fvw,d:

Fvw,d = fvw,d a = 223×5.6 = 1248 N/mm = 1.25 kN/mm

Total resistance of weld = 1.25×350

= 437 kN (> 400 kN)

Above arrangement, using 8 mm fillet welds, with a 350 mm weld length is acceptable.


28L. Gardner

Introduction

Bolted joints

Welded joints Joints



Session 10

15

29L. Gardner

Introduction

Bolted joints

Welded joints

Conclusions

• ‘The construction industry has not previously faced the challenge of implementing a complete suite of new codes encompassing all the major materials and loading requirements

• This burden will not be eased by the format and terminology of the Eurocodes both of which are different from British Standards’.

National Strategy for Implementation of the Structural Eurocodes (IStructE, 2004)

30L. Gardner

Introduction

Bolted joints

Welded joints

Conclusions

• ‘The Eurocodes will become the Europe wide means of designing Civil and Structural engineering works and so, they are of vital importance to both the design and construction sectors of the Civil and Building Industries’.

European Union website

16

31L. Gardner

Introduction

Bolted joints

Welded joints

Conclusions:

Conclusions

• Advanced design codes

• Greater in scope

• Biggest change since limit states

• Unfamiliar format/ resistance to uptake

• Guidance material and training emerging

• Basis for other National design codes

32L. Gardner

Introduction

Bolted joints

Welded joints Thank youDr Leroy GardnerSenior Lecturer in Structural Engineering


Eurocode 3

1

Table 1: Values for yield strength fy and ultimate strength fu (from EN 10025-2)

Steel grade Yield strength

fy (N/mm2)

t ≤ 16 mm

Yield strength fy (N/mm2)

16 < t ≤ 40 mm


3 < t ≤ 100 mm

S235 235 235 360

S275 275 265 410

S355 355 345 470

S450 450 430 550

2

Table 2 (sheet 1): Maximum width-to-thickness ratios for compression parts (Table 5.2 of EN

1993-1-1)

3

Table 2 (sheet 2): Maximum width-to-thickness ratios for compression parts (Table 5.2 of EN

1993-1-1)

4

Table 3: Selection of buckling curve for a cross-section (Table 6.2 of EN 1993-1-1)

Buckling curve

Cross-section Limits Buckling

about axis

S 235 S 275 S 355 S 420

S 460

tf ≤ 40 mm y – y z - z

a b

a0 a0

h/b

> 1.

2

40 mm < tf ≤ 100 mm y – y z - z

b c

a a

tf ≤ 100 mm y – y z - z

b c

a a R

olle

d I-s

ectio

ns

h/b ≤

1.2

tf > 100 mm y – y z - z

d d

c c

tf ≤ 40 mm y – y z - z

b c

b c

Wel

ded

I-se

ctio

ns

tf > 40 mm y – y z - z

c d

c d

hot finished any a a0

Hol

low

sec

tions

cold formed any c c

generally (except as below) any b b

Wel

ded

box

sect

ions

thick welds: a > 0.5tf b/tf < 30 h/tw < 30

any c c

U-,

T- a

nd

solid

sec

tions

any c c

L-se

ctio

ns

any b b

b

tw

tf r

y y

z

z

h

tf

y y

z

z

tf

y y

z

z

tf

y y

z

z

b

h

tw

5

Table 4: Imperfection factors for buckling curves (Table 6.1 of EN 1993-1-1)

Buckling curve a0 a b c d

Imperfection factor α 0.13 0.21 0.34 0.49 0.76

Figure 1: Eurocode 3 Part 1.1 buckling curves (Figure 6.4 of EN 1993-1-1)

0.0

0.2

0.4

0.6

0.8

1.0

1.2

0 0.5 1 1.5 2 2.5

Curve a0

Curve a

Curve b

Curve c

Curve d

a0

Red

uctio

n fa

ctor

χ


TTT Handout LG Nov2008 Lecture Note on EC3 Design

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Transcript of TTT Handout LG Nov2008 Lecture Note on EC3 Design