Transcript
Introduction Overview Objectives
Design of steel structures to Eurocode 3 Dr Leroy Gardner Senior Lecturer in Structural Engineering
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Eurocode 3: Design of steel structures
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Session 1 Introduction Overview Objectives
Introduction Dr Leroy Gardner Senior Lecturer in Structural Engineering
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Eurocode 3: Design of steel structures
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Overview of Course Introduction Overview Objectives
Outline: • 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
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Overview of Course Introduction Overview Objectives
Outline (continued): • Session 7: Design of columns • Session 8: Design of beams • Session 9: Design of beam-columns • Session 10: Design of joints
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Dr Leroy Gardner Introduction
Dr Leroy Gardner BEng MSc PhD DIC CEng MICE MIStructE
Overview
• Senior Lecturer in Structural Engineering Objectives
• 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]
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Your experience with Eurocodes Introduction Overview
• Introduce yourselves
Objectives
• Organisation • Your experience with steel design/ Eurocodes • Any particular interests/concerns
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Motivation for course Introduction Overview
• Most of the Eurocodes are now published • Conflicting British Standards to be withdrawn
Objectives
• Designers need to be prepared • Clear training requirements • Textbooks and design guides • Background information
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Designers Introduction
Introduction of Eurocodes: Overview Objectives
• Biggest change since limit states • Designers unfamiliar with format • Resistance to uptake • Supporting material and training • Basis for other National design codes
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Designers’ Guide Introduction
Designers’ Guide to EN 1993-1-1:
Overview Objectives
• Covers Eurocode 3: Part 1.1 • Also Parts 1.3, 1.5 and 1.8 • EN 1990 and EN 1991 • Sections aligned with code
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Textbook Introduction Overview
The behaviour and design of steel structures to EC3: Trahair, Bradford, Nethercot & Gardner (2008)
Objectives
• Structural phenomena • Theoretical background • Code implementation • Worked examples
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Historical developments Introduction Overview Objectives
Historical development of Eurocodes: • Idea of Eurocodes dates back to 1974 • Family of design codes • Harmonisation of treatment • Removal of barriers to trade • Framework for development
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Scope of Eurocodes Introduction
Scope of structural Eurocodes: Overview
• A total of 10 codes (comprising 58 documents) Objectives
The first 2 codes are material independent: • EN 1990 – Basis of structural design • EN 1991 – Actions on structures
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Scope of Eurocodes Introduction Overview Objectives
Remaining 8 codes focus on materials: • 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
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• EN 1999 – Design of aluminium structures 13
Timetable for introduction Introduction Overview Objectives
Timetable for introduction of codes: • 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
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Eurocodes Introduction Overview
• Codes will be published by CEN in 3 languages: • English • French
Objectives
• 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.
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Objectives Introduction
Course objectives:
Overview Objectives
• Familiarity with layout, notation, philosophy of Eurocodes • Understanding of background and design procedure for principal structural components
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Session 1 Introduction Overview Objectives
Introduction Dr Leroy Gardner Senior Lecturer in Structural Engineering
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Eurocode 3: Design of steel structures
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Session 2 EN 1990 EN 1991
Introduction to EN 1990 & EN 1991 Dr Leroy Gardner Senior Lecturer in Structural Engineering
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Eurocode 3: Design of steel structures
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Overview EN 1990
Outline: EN 1991
• Introduction to EN 1990 • Introduction to EN 1991 • Conclusions
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EN 1990 (2002) EN 1990 EN 1991
EN 1990 (2002): • EN 1990 – Basis of structural design • UK National Annex published • ‘Should read at least once’….
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Basic requirements EN 1990 EN 1991
EN 1990 states that a structure shall have adequate: • Structural resistance • Serviceability • Durability • Fire resistance • Robustness
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Design situations EN 1990 EN 1991
All relevant design situations must be examined: • 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.
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Actions and Effects EN 1990 EN 1991
Action (F): • Direct actions – applied loads
CAUSE
• Indirect actions – imposed deformations or accelerations e.g. by temperature changes, vibrations etc • Both essentially produce same effect
Effect of action (E):
EFFECT
• On structural members and whole structure • For example internal forces and moments, deflections ..
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Types of actions EN 1990 EN 1991
Types of actions: • Permanent, G • Variable, Q (leading and non-leading) • Accidental, A
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Load combinations EN 1990 EN 1991
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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Load combinations EN 1990
Equation 6.10:
EN 1991
‘to be combined with’
∑γ
G, j
1.5 x combination factor x Other variable actions Actions due to prestressing
G k , j "+" γ PP "+" γ Q ,1Q k ,1 "+"
j ≥1
1.35 x Permanent actions
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∑γ
Q ,i
ψ 0 ,i Q k , i
i >1
1.5 x Leading variable action
Load factors 1.35 and 1.5 are applied when actions are ‘unfavourable’.
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Leading variable actions Qk,1 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.)
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Combination factor ψ0 EN 1990 EN 1991
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The combination factor ψ0 is intended specifically to take account of the reduced probability of the simultaneous occurrence of two or more variable actions. Loading
Combination factor ψ0
Imposed loading
0.7
Wind loading
0.5*
* 0.5 is UK NA value, 0.6 is the unmodified EC value
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Unfavourable and favourable loading 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
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Equivalent horizontal forces 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).
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Exercise solution – Equation 6.10 EN 1990
Load combinations for a typical structure from Equation 6.10:
EN 1991
Combination
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Dead
Imposed
Wind
EHF
Dead + Imposed
1.0
Dead + Wind (uplift)
1.0
D+I+W (imposed leading)
1.0
D+I+W (wind leading)
1.0
Note EHF are always present and already based on factored loads
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Load combinations EN 1990 EN 1991
Equations 6.10a and 6.10b – use less favourable result:
∑γ
G, jGk , j
"+ " γ PP "+ " γ Q ,1ψ 0,1Q k ,1 "+ "
j≥1
∑γ
Q ,i ψ 0 ,iQ k ,i
i> 1
∑ξ γ
j G, jGk , j
"+ " γ PP "+ " γ Q ,1Q k ,1 "+ "
j≥1
∑γ
Q ,i ψ 0 ,i Q k ,i
i> 1
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) L. Gardner
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Exercise solution – Eqs 6.10a and 6.10b EN 1990
Load combinations from Eqs 6.10a and 6.10b – All combinations except last one are from Eq. 6.10b.
EN 1991
Combination
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Dead
Imposed
Wind
EHF
Dead + Imposed (6.10b)
1.0
Dead + Wind (uplift) (6.10b)
1.0
D + I + W (6.10b) (imposed leading)
1.0
D + I + W (6.10b) (wind leading)
1.0
D + I + W (6.10a)*
1.0
* Unlikely to govern unless Dead >> Imposed
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Equilibrium check (EQU) EN 1990
Equilibrium check (EQU):
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: 0.9Gk + 1.5Wk + EHF
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Equilibrium check (EQU) EN 1990 EN 1991
Favourable and unfavourable loading: Wind load unfavourable, dead load favourable, imposed load favourable
Wind load unfavourable, part of dead load favourable, part unfavourable, part of imposed unfavourable 1.1 Gk + 1.05 Qk
1.5 Wk
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0.9 Gk
Overturning point
1.5 Wk
0.9 Gk
Overturning point
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SLS load combinations EN 1990 EN 1991
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 L. Gardner
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Parts of EN 1991 EN 1990
EN 1991 contains the following parts: 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
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Sub-parts of EN 1991-1 EN 1990 EN 1991
EN 1991-1 contains the following sub-parts: • 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
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Conclusions EN 1990
Concluding comments:
EN 1991
• 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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Session 2 EN 1990 EN 1991
Introduction to EN 1990 & EN 1991 Dr Leroy Gardner Senior Lecturer in Structural Engineering
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Eurocode 3: Design of steel structures
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Session 3 Background Overview
Overview of Eurocode 3 Dr Leroy Gardner Senior Lecturer in Structural Engineering
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Eurocode 3: Design of steel structures
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Overview Background
Outline: Overview
• Development of Eurocode 3 • Introduction to design to Eurocode 3 • Conclusions
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EN 1993: Eurocode 3 Background
Eurocode 3:
Overview
• Work began back in 1975 • Eurocode 3 contains a number of parts • … and sub-parts • The first 5 parts were published in 2005
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EN 1993: Eurocode 3 Background Overview
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
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EN 1993-1 Background
Eurocode 3: Part 1 has 12 sub-parts: Overview
• 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
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EN 1993-1 Background Overview
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• 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
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National Annexes Background Overview
National Annexes: • Every Eurocode will contain a National Annex • National choice • Non Conflicting Complementary Information • Timescale
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Axes convention Background
Different axes convention:
Overview
BS 5950 Along the member
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Eurocode 3 X
Major axis
X
Y
Minor axis
Y
Z 8
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Labelling convention Background
Labelling convention: b
Overview
z z r
tw h
d
y
h
y
y
t y
z
r tf
b
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Subscripts Background Overview
Extensive 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
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Different symbols Background Overview
For example: BS5950
EC3
BS5950
EC3
BS5950
EC3
A
A
P
N
py
fy
Z
Wel
Mx
My
pb
χLT fy
S
Wpl
V
V
pc
χf y
Ix
Iy
H
Iw
r
i
Iy
Iz
J
It
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Gamma factors γ Background Overview
Gamma factors γ: • Appear everywhere • Partial safety factors • γF for actions (loading) • γM for resistance
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Gamma factors γM Background Overview
Gamma factors γM account for material and modelling uncertainties:
Partial factor γM
EC 3 value (UK NA value)
Application
γM0
1.00 (1.00)
Cross-sections
γM1
1.00 (1.00)
Member buckling
γM2
1.25 (1.10)
Fracture
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Material properties Background Overview
Material properties are taken from product standards (generally EN 10025-2). The Young’s modulus of steel should be taken as 210000 N/mm2. Yield strength fy (N/mm2)
Yield strength fy (N/mm2)
Ultimate strength fu (N/mm2)
t ≤ 16mm
16 < t ≤ 40 mm
3 ≤ t ≤ 100 mm
S235
235
235
360
S275
275
265
410
S355
355
345
470
S450
450
430
550
Steel grade
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Structural design Background Overview
Early sections (1-4) of EN 1993-1-1: • Reference to EN 1990 and EN 1991 • Identify clauses open to National choice • Materials, reference to material standards • Durability
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Structural design Background Overview
Subsequent sections of EN 1993-1-1: • Section 5 – Structural analysis • Global analysis • Cross-section classification • Requirements for plastic analysis • Section 6 – ULS • General • Resistance of cross-sections
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Structural design Background
• Buckling resistance of members Overview
• Built-up members • SLS • Annexes A, B, AB and BB
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Omissions Background Overview
Notable omissions: • Effective lengths • Formulae for Mcr • Deflection limits • National Annex and NCCIs to resolve
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Sources of further information Background Overview
• 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
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Session 3 Background Overview
Overview of Eurocode 3 Dr Leroy Gardner Senior Lecturer in Structural Engineering
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Eurocode 3: Design of steel structures
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Session 4 Introduction Deformed geometry
Structural analysis
Imperfections Actions
Dr Leroy Gardner Senior Lecturer in Structural Engineering L. Gardner
Eurocode 3: Design of steel structures
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Overview Introduction
Outline:
Deformed geometry Imperfections Actions
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• • • •
Introduction Analysis types Second order effects Imperfections
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Analysis types Introduction
Analysis types:
Deformed geometry Imperfections Actions
• • • •
First order elastic Second order elastic First order plastic Second order plastic
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General approach Introduction
General approach: Deformed geometry Imperfections Actions
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• Choose an appropriate analysis • Make an appropriate model • Apply all actions (loads) and combinations of actions • Check cross-sections, members and joints
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Frame stability 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 • Analysis are appropriate.
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Effects of deformed geometry Introduction Deformed geometry
EN 1993-1-1 Clause 5.2.1(2) states that deformed geometry (second order effects) shall be considered:
Imperfections Actions
• if they increase the action effects significantly • or modify significantly the structural behaviour
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Limits for ignoring deformed geometry Introduction Deformed geometry Imperfections Actions
For elastic analysis:
α cr =
Fcr ≥ 10 FEd
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.
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Limits for ignoring deformed geometry Introduction Deformed geometry
For plastic analysis:
α cr =
Fcr ≥ 15 FEd
Imperfections Actions
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
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Simple estimate for αcr Introduction Deformed geometry
Simple estimate for αcr may be applied to: • Portals with shallow roof slopes • Beam and column frames (each storey)
Imperfections Actions
where
⎛ H ⎞⎛ h ⎞ ⎟ α cr = ⎜⎜ Ed ⎟⎟⎜⎜ ⎟ δ V ⎝ Ed ⎠⎝ H,Ed ⎠
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) L. Gardner
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Limits on use of simple estimate 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):
NEd ≤ 0.09 Ncr where NEd is the design value of compression in the beam or rafter and Ncr is its elastic buckling resistance.
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Analysis method and achievement Introduction
Distinguish between:
Deformed geometry
• Analysis method (1st or 2nd order)
Imperfections
• Analysis achievement i.e. can achieve 2nd order by:
Actions
1) 2nd order analysis 2) 1st and amplified sway 3) 1st and increased effective length.
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Frame stability Introduction Deformed geometry Imperfections
Limits for treatment of second order effects depend on αcr:
=
Fcr FEd
Limits on αcr Action
Achievement
αcr>10
First order analysis
First order only
10>αcr>3
Second order First order analysis effects by plus amplification or approximate effective length method means
αcr<3
Second order analysis
Actions
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α cr
Second order effects more accurately 12
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Global imperfections for frames Introduction
Global initial sway imperfections:
Deformed geometry
φ = φ 0 αhα m
Imperfections
where φ0 is the basic value = 1/ 200 αh and α m are reduction factors
Actions
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Global imperfections for frames Introduction Deformed geometry Imperfections
• 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
Actions
• 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. L. Gardner
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Actions to be specified Introduction
Actions to be specified:
Deformed geometry
• EN 1991-1-1: Densities, self-weight, imposed loads
Imperfections
• EN 1991-1-2: Fire • EN 1991-1-3: Snow loads
Actions
• 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
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Other Actions Introduction
• Equivalent horizontal forces Deformed geometry Imperfections Actions
- unless using initial imperfection model • Derived from imperfections • Applied in ALL combinations (only in gravity combinations in BS 5950)
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Checks Introduction
• Analyse structure Deformed geometry
• Classify sections using clause 5.5 - for plastic global analysis, check clause 5.6
Imperfections
• Check cross-sectional resistance to clause 6.2 Actions
• Check buckling resistance to clause 6.3 - check built-up members to clause 6.4
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Session 4 Introduction Deformed geometry Imperfections Actions
Structural analysis Dr Leroy Gardner Senior Lecturer in Structural Engineering
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Eurocode 3: Design of steel structures
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Session 5 Introduction Design Example Exercise
Design of tension members Dr Leroy Gardner Senior Lecturer in Structural Engineering
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Eurocode 3: Design of steel structures
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Overview Introduction
Outline:
Design Example Exercise
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• Introduction • Tension member design • Example
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Eurocode 3 Introduction Design
Eurocode 3 states that tensile resistance should be verified as follows:
Example Exercise
Nt ,Ed ≤ Nt ,Rd
Tension check
Nt,Ed is the tensile design effect Nt,Rd is the design tensile resistance L. Gardner
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Design tensile resistance Nt,Rd Introduction Design
Design tensile resistance Nt,Rd is limited either by:
Example
• Yielding of the gross cross-section Npl,Rd
Exercise
• or ultimate failure (fracture) of the net crosssection (at holes for fasteners) Nu,Rd whichever is the lesser.
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Yielding of gross cross-section Introduction Design
The Eurocode 3 design expression for yielding of the gross cross-section (plastic resistance) given as:
Example Exercise
Npl,Rd =
Afy γ M0
This criterion is applied to prevent excessive deformation of the member. L. Gardner
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Ultimate resistance of net section Introduction Design
And for the ultimate resistance of the net cross-section (defined in clause 6.2.2.2), the Eurocode 3 design expression is:
Example Exercise
Nu,Rd
=
0.9A net fu γ M2
Anet is the reduced cross-sectional area to account for bolt holes L. Gardner
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Partial factors γM Introduction Design
Plastic resistance of the gross cross-section Npl,Rd utliises γM0, whilst ultimate fracture of the net cross-section Nu,Rd utilises γM2.
Example Exercise
γ M 0 = 1 .0
and
γ M2 = 1.25 (1.1 in UK NA)
The larger safety factor associated with fracture reflects the undesirable nature of the failure mode. L. Gardner
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Non-staggered fasteners Introduction Design
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.
Example
A
Exercise
p
A s L. Gardner
s
Non-staggered arrangement of fasteners
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Non-staggered fasteners Net area at bolts holes Anet on any line (AA) perpendicular to the member axis:
Introduction Design
Anet = A - nd0t
Example Exercise
A = n = d0 = t =
gross cross-sectional area number of bolt holes diameter of bolt holes material thickness
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Staggered fasteners Introduction Design
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
Example
2. Exercise
⎛ s2 ⎞ ⎟⎟ ⎜ t⎜ nd0 − ∑ 4 p ⎠ ⎝
where s is the staggered pitch of two consecutive holes p is the spacing of the centres of the same two holes measured perpendicular to the member axis L. Gardner
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Staggered fasteners Introduction Design
n 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 A
Example Exercise
p
B s L. Gardner
A
s
Staggered arrangement of fasteners
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Angles connected by a single row of bolts
Introduction Design Example Exercise
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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. With 1 bolt :
Nu,Rd =
2.0 (e 2 − 0.5d0 )tfu γ M2
With 2 bolts :
Nu,Rd =
β 2 Anet fu γ M2
With 3 or more bolts :
Nu,Rd =
β 3 A net fu γ M2 12
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Angles connected by a single row of bolts
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: e1
p1
p1
e2
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d0
Definitions for e1, e2, p1 and d0
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Angles connected by a single row of bolts
Introduction
Reduction factors β2 and β3
Design
≤ 2.5d0
≥ 5.0d0
β2 (for 2 bolts)
0.4
0.7
β3 (for 3 or more bolts)
0.5
0.7
Pitch p1 Example Exercise
Note: For intermediate values of pitch p1 values of β may be determined by linear interpolation. d0 is the bolt hole diameter.
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Angles with welded end connections Introduction Design Example Exercise
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).
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Example: Tension member design Introduction Design
Design a single angle tie, using grade S355 steel, for the member AB shown below. Consider a bolted and a welded arrangement.
Example Exercise
B
NEd = 541 kN
A
Tension member AB in truss L. Gardner
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Example: Tension member design Introduction Design
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.
Example Exercise
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Example: Tension member design Welded connection Introduction Design Example Exercise
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
Gusset plate
125×75×10 unequal angle welded by longer leg L. Gardner
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Example: Tension member design 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).
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Example: Tension member design Introduction Design
For yielding of the gross cross-section, plastic resistance = is given as: Npl,Rd =
Example Exercise
γ M0
=
1920 × 355 = 682x103 N = 682 kN 1.0
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: Nu,Rd =
L. Gardner
Afy
0.9 A net fu 0.9 × 1920 × 470 = = 738 × 10 3 N = 738 kN γ M2 1.10 20
10
Example: Tension member design Introduction Design Example
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)
Exercise
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.
L. Gardner
21
Example: Tension member design Bolted connection Introduction Design Example
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.
Exercise 150×75×10 unequal angle 24 mm diameter holes for 22 mm HSFG bolts Gusset plate
150×75×10 unequal angle bolted by longer leg L. Gardner
22
11
Example: Tension member design Introduction Design
Gross area of cross-section, A = 2170 mm2 (from Section Tables). For yielding of the gross cross-section, plastic resistance is given as:
Example Exercise
Npl,Rd =
Afy γ M0
=
2170 × 355 = 770.4 × 10 3 N = 770 kN 1 .0
The net cross-sectional area Anet: Anet = A – allowance for bolt holes = 2170 – (24×10) = 1930 mm2 L. Gardner
23
Example: Tension member design Introduction Design
From Table, β3 = 0.7 (since the pitch p1 > 5d0). Nu,Rd =
β 3 A net fu 0.7 × 1930 × 470 = = 577 × 10 3 N = 577 kN γ M2 1.10
Example Exercise
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.
L. Gardner
24
12
Tension member design exercise Introduction Design Example
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. A
Exercise
50 mm
T
100 mm
T
50 mm A T
T
25 mm thick plates L. Gardner
100 mm 100 mm
25
Session 5 Introduction Design Example Exercise
Design of tension members Dr Leroy Gardner Senior Lecturer in Structural Engineering
L. Gardner
Eurocode 3: Design of steel structures
26
13
Session 6 Introduction Local buckling Classification Class 4 Exercise
Local buckling and cross-section classification Dr Leroy Gardner Senior Lecturer in Structural Engineering
L. Gardner
Eurocode 3: Design of steel structures
1
Overview Introduction
Outline:
Local buckling Classification Class 4 Exercise
L. Gardner
• Introduction • Local buckling • Cross-section classification • Class 4 – effective widths
2
1
Background Introduction
Background:
Local buckling
• For efficiency, structural members are generally composed of relatively thin elements (i.e. thicknesses substantially less than other cross-sectional dimensions)
Classification Class 4 Exercise
• 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.
L. Gardner
3
Local buckling Introduction Local buckling Classification
Local buckling
Class 4 Exercise
L. Gardner
Local buckling in structural components
4
2
Cross-section classification Introduction Local buckling
• Whether in the elastic or inelastic material range, cross-sectional resistance and rotation capacity are limited by the effects of local buckling.
Classification Class 4
• Eurocode 3 (and BS 5950) account for the effects of local buckling through cross-section classification.
Exercise
• 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. L. Gardner
5
Factors affecting local buckling Introduction Local buckling Classification
The factors that affect local buckling (and therefore the cross-section classification) are: • Width/thickness ratios of plate components
Class 4
• Element support conditions
Exercise
• Material strength, fy • Fabrication process • Applied stress system
L. Gardner
6
3
Cross-section classification Introduction Local buckling Classification Class 4
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.
Exercise
The classification of the overall cross-section is taken as the least favourable of the constituent elements (for example, a crosssection with a class 3 flange and class 1 web has an overall classification of Class 3). L. Gardner
7
Definition of 4 classes Introduction
Eurocode 3 defines four classes of cross-section:
Local buckling
Moment Classification Class 4 Exercise
Class 1 Mpl Mel
Class 2 Class 3
Class 4 Deformation L. Gardner
8
4
Compression Introduction Local buckling
Cross-section resistance in compression Nc,Rd:
Classification Class 4
Class 1, 2 and 3:
Nc ,Rd =
Exercise
Class 4:
Nc ,Rd =
Afy γ M0
A eff fy γ M0
L. Gardner
9
Bending Introduction
• Class 1 & 2 cross-sections:
Local buckling Classification
Mc ,Rd = Mpl =
Class 4 Exercise
γ M0
• Class 3 cross-sections:
Mc ,Rd = Mel = L. Gardner
Wpl fy
Wel fy γ M0 10
5
Bending Introduction Local buckling
• Class 4 cross-sections:
Classification Class 4
Mc ,Rd =
Exercise
Weff fy γ M0
L. Gardner
11
Compressed widths c Introduction
Definition of compressed widths – flat widths: c
Local buckling Rolled
c
Rolled
Welded
c
Welded
Classification Class 4
c
Exercise
(a) Outstand flanges
(b) Internal compression parts
Limits on slenderness e.g. c/t ≤ 9ε L. Gardner
ε =
235 / fy
12
6
Internal compression parts Introduction Local buckling Classification Class 4 Exercise
L. Gardner
13
Outstand flanges Introduction Local buckling Classification Class 4 Exercise
L. Gardner
14
7
Angles and tubular sections Introduction Local buckling Classification Class 4 Exercise
L. Gardner
15
Class 4 cross-sections Introduction
Class 4 (slender) cross-sections
Local buckling Classification Class 4 Exercise
• 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).
L. Gardner
16
8
Class 4 – effective width concept Introduction Local buckling Classification Class 4 Exercise
L. Gardner
17
Cross-section classification exercise Introduction Local buckling
Determine the classification and resistance Nc,Rd for a 254 x 254 x 73 UC in pure compression, assuming grade S 355 steel.
Classification
b z
Class 4 Exercise
h = 254.1 mm b = 254.6 mm tw = 8.6 mm
tw h
d
y
tf = 14.2 mm
y r z
tf
r = 12.7 mm A = 9310 mm2
Section properties for 254 x 254 x 73 UC L. Gardner
18
9
Summary Introduction Local buckling Classification Class 4 Exercise
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
L. Gardner
19
Session 6 Introduction Local buckling Classification Class 4 Exercise
Local buckling and cross-section classification Dr Leroy Gardner Senior Lecturer in Structural Engineering
L. Gardner
Eurocode 3: Design of steel structures
20
10
Session 7 Background Cross-section Buckling Example
Compression members
Exercise
Dr Leroy Gardner Senior Lecturer in Structural Engineering L. Gardner
Eurocode 3: Design of steel structures
1
Outline Background
Overview:
Cross-section Buckling
• Background
Example
• Cross-section resistance Nc,Rd
Exercise
L. Gardner
• Member buckling resistance Nb,Rd
2
1
Elastic buckling theory N
N
Background Cross-section Buckling
w
L Example Exercise
x N (a) Unloaded member
(b) Loaded member (straight)
N (c) Loaded member (displaced)
L. Gardner
3
Elastic buckling theory Background Cross-section
From stability theory, the elastic buckling load of a perfect pin-ended column is given by:
Buckling Example
Ncr
=
π 2EI L2
Exercise
Other boundary conditions may be accounted for through the effective (critical) length concept. L. Gardner
4
2
Elastic buckling theory Background
Two bounds: Yielding and buckling
Cross-section
Load
Material yielding (squashing)
Buckling
NEd
Afy Example
Lcr Euler (critical) buckling Ncr
Exercise
NEd
Non-dimensional slenderness
L. Gardner
5
Imperfections Background Cross-section Buckling Example Exercise
L. Gardner
Forms of imperfection: • • • •
Geometric imperfections Eccentricity of loading Residual stresses Non-homogeneity of material properties • End restraint • etc
6
3
Residual stresses Background Cross-section Buckling Example Exercise
Welding
Hot-rolling L. Gardner
7
Behaviour of imperfect columns Background
wmax = (w0+w) at mid-height
NEd
e0,d is the magnitude of the initial imperfection w0
Cross-section Buckling
w0 = Initial imperfection
Example Exercise
w0
=
w max
w w = additional deflection
NEd NRd
+
NEd NRd
+
1 e NEd 0,d 1− Ncr
NEd w max MRd
≤
1
x
NEd L. Gardner
1 NEde 0,d N 1 − Ed MRd Ncr
≤
1
8
4
Perry-Robertson Background Cross-section
Perry observed: • All columns contain imperfections and will deflect laterally from the onset of loading
Buckling Example Exercise
• 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
L. Gardner
9
Perry-Robertson Background Cross-section Buckling Example Exercise
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
L. Gardner
10
5
Buckling curves 1.2
Cross-section Buckling Example Exercise
Reduction factor χ
Background
1.0
Curve aa0 0
Curve a
0.8
Curve b Curve c Curve d
0.6 0.4 0.2 0.0 0
0.5
1
1.5
Non-dimensional slenderness
2
2.5
λ
L. Gardner
11
Eurocode 3 Background Cross-section
Eurocode 3 states, as with BS 5950, that both cross-sectional and member resistance must be verified:
Buckling Example
NEd ≤ Nc ,Rd
Cross-section check
NEd ≤ Nb,Rd
Member buckling check
Exercise
L. Gardner
12
6
Cross-section resistance Background
• Cross-section resistance in compression Nc,Rd depends on cross-section classification:
Cross-section Buckling
Nc ,Rd =
Afy
for Class 1, 2 or 3 sections
γ M0
Example Exercise
Nc ,Rd =
A eff fy γ M0
for Class 4 sections
γM0 is specified as 1.0 in EN 1993 This value will also be adopted in the UK L. Gardner
13
Member buckling Background
Compression buckling resistance Nb,Rd:
Cross-section Buckling Example
Nb,Rd
=
Nb,Rd
=
Exercise
L. Gardner
χ A fy γ M1 χ A eff fy γ M1
for Class 1, 2 and 3
for (symmetric) Class 4
14
7
Equivalence to BS 5950 Background
Compression buckling resistance:
Cross-section Buckling
χ A fy
Nb,Rd =
γM1
Example
Eurocode 3
Exercise
Pc = pc A
BS 5950
L. Gardner
15
Member buckling Background
Calculate non-dimensional slenderness
Cross-section Buckling
λ=
A fy Ncr
λ
for Class 1, 2 and 3
Example Exercise
λ=
A eff fy Ncr
for Class 4
Ncr is the elastic critical buckling load for the relevant buckling mode based on the gross properties of the cross-section L. Gardner
16
8
Non-dimensional slenderness Background Cross-section
=
Ncr
π 2EI AL2
=
fcr
Buckling Example Exercise
π 2EI L2
π 2E (L / i)2
=
where λ = L / i
=
π 2E λ2
and i is radius of gyration
The theoretical slenderness boundary λ1 between material yielding and elastic member buckling may be found by setting fcr = fy: fy
L. Gardner
=
π 2E λ 12
⇒
λ1
=
π
E fy 17
Non-dimensional slenderness Background
The non-dimensional slenderness used in EC3 is defined as:
Cross-section Buckling Example Exercise
λ =
λ = λ1
π π
E fcr E fy
=
1 fcr 1 fy
=
fy fcr
=
Afy Ncr
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. L. Gardner
18
9
Non-dimensional slenderness Background
N
NEd
Material yielding (in-plane bending)
Cross-section
Afy Buckling
Lcr Elastic member buckling (LTB)
Example
NEd
Exercise
1.0
Non-dimensional slenderness
λ
L. Gardner
19
Member buckling Background
•
Calculate reduction factor, χ
Cross-section Buckling
χ=
1 ϕ + ( ϕ 2 − λ 2 ) 0 .5
≤
1
Example Exercise
ϕ = 0.5 (1 + α(λ − 0.2) + λ 2 ) α is the imperfection factor
L. Gardner
20
10
Imperfection factor α Background
Imperfection factors α for 5 buckling curves:
Cross-section Buckling Example Exercise
Buckling curve Imperfection factor α
a0
a
b
c
d
0.13
0.21
0.34
0.49
0.76
L. Gardner
21
Buckling curve selection Background
Cross-section
Limits
Buckling about axis
Cross-section
Example
z
Rolled Isections
a0 a0
40 mm < tf ≤ 100 mm
y–y z-z
b c
a a
tf ≤ 100 mm
y–y z-z
b c
a a
tf > 100 mm
y–y z-z
d d
c c
tf ≤ 40 mm
y–y z-z
b c
b c
tf > 40 mm
y–y z-z
c d
c d
y
Exercise
r
h/b ≤ 1.2
tf
z z
z
Welded Isections
y
y
z
L. Gardner
a b
h/b > 1.2
y
y
S460
y–y z-z
tw h
S235 S275 S355 S420
tf ≤ 40 mm
b
Buckling
Buckling curve
y
tf
tf z
22
11
Buckling curve selection Background
hot finished
any
a
a0
cold formed
any
c
c
generally (except as below)
any
b
b
thick welds: a > 0.5tf b/tf < 30 h/tw < 30
any
c
c
U-, T- and solid sections
any
c
c
L-sections
any
b
b
Hollow sections
Cross-section z
Buckling
Welded box sections
h
y
y tw b
z
Example Exercise
tf
L. Gardner
23
Effective (buckling) lengths Lcr Background
End restraint (in the plane under consideration)
Cross-section Buckling
Effectively held in position at both ends
Example Exercise
One end
Effectively held in position and restrained in direction
L. Gardner
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
Not restrained in direction at either end
1.0 L
Other end
Not held in position
Effectively restrained in direction
1.2 L
Partially restrained in direction
1.5 L
Not restrained in direction
2.0 L
24
12
Effective (buckling) lengths Lcr Background Cross-section Buckling Example Exercise
Non-sway
Sway
L. Gardner
25
Column buckling design procedure Background
Design procedure for column buckling:
Cross-section
1. Determine design axial load NEd
Buckling
2. Select section and determine geometry
Example Exercise
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
L. Gardner
26
13
Column buckling design procedure Background Cross-section Buckling
6. Non-dimensional slenderness λ =
Ncr
7. Determine imperfection factor α 8. Calculate buckling reduction factor χ
Example Exercise
A fy
9. Design buckling resistance Nb,Rd = 10. Check
χ A fy γ M1
NEd ≤ 1.0 Nb,Rd
L. Gardner
27
Member buckling resistance example Background Cross-section
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.
Buckling Example
NEd = 2110 kN
Exercise
4.0 m
L. Gardner
The critical combination of actions results in a design axial force of 2110 kN.
28
14
Member buckling resistance example Background
Assess the suitability of a hot-rolled 244.5×10 CHS in grade S 355 steel for this application.
Cross-section Buckling
t
Example Exercise
d
= 244.5 mm
t
= 10.0 mm
A
= 7370 mm2
Wel,y = 415000 mm3 d
Wpl,y = 550000 mm3 I
= 50730000 mm4
L. Gardner
29
Member buckling resistance example Background Cross-section Buckling Example
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
Exercise
L. Gardner
30
15
Member buckling resistance example Cross-section classification (clause 5.5.2): Background Cross-section
ε=
235 / fy = 235 / 355 = 0.81
Buckling
Tubular sections (Table 5.2, sheet 3) Example Exercise
d/t = 244.5/10.0 = 24.5 Limit for Class 1 section = 50 ε2 = 40.7 > 24.5 ∴ Cross-section is Class 1
L. Gardner
31
Member buckling resistance example Background Cross-section Buckling
Cross-section compression resistance (clause 6.2.4): Nc ,Rd =
Afy γ M0
for Class 1, 2 or 3 cross - sections
Example Exercise
∴ Nc ,Rd =
7370 × 355 = 2616 × 10 3 N = 2616 kN 1.00
2616 > 2110 kN ∴ Cross − section resistance is OK L. Gardner
32
16
Member buckling resistance example Background
Member buckling resistance in compression (clause 6.3.1):
Cross-section Buckling
χ=
Example Exercise
=
Nb ,Rd
χ A fy γ M1
for Class 1, 2 & 3 cross - sections
1 Φ + Φ 2 − λ2
[
but χ ≤ 1.0
where Φ = 0.5 1 + α (λ - 0.2 ) + λ2
and λ =
Afy Ncr
]
for Class 1, 2 & 3 cross - sections
L. Gardner
33
Member buckling resistance example Background
Elastic critical force and non-dimensional slenderness for flexural buckling Ncr
Cross-section Buckling
Ncr =
π 2EI π 2 × 210000 × 50730000 = = 6571 kN 2 4000 2 L cr
Example Exercise
∴λ =
7370 × 355 = 0.63 6571 × 10 3
From Table 6.2 of EN 1993-1-1: For a hot-rolled CHS, use buckling curve a L. Gardner
34
17
Buckling curve selection Buckling curve
Background
Cross-section
Cross-section
Buckling about axis
Limits
Buckling Example
S235 S275 S355 S420
S460
hot finished
any
a
a0
cold formed
any
c
c
Hollow sections
Exercise
Extract from Table 6.2 of EN 1993-1-1: For a hot-rolled CHS, use buckling curve a L. Gardner
35
Graphical approach 1.2
Cross-section Buckling Example Exercise
Reduction factor χ
Background
1.0
Curve aa0 Curve 0 Curve a
≈0.88
0.8
Curve b Curve c Curve d
0.6 0.4 0.2 0.0 0
0.5
0.63
1
1.5
Non-dimensional slenderness L. Gardner
2
2.5
λ 36
18
Member buckling resistance example Background
From Table 6.1 of EN 1993-1-1, for buckling curve a, α = 0.21
Cross-section
Φ
Buckling
χ
= 0.5[1 + 0.21(0.63 − 0.2) + 0.63 2 ] = 0.74 1 = = 0.88 0.74 + 0.74 2 − 0.63 2
Example Exercise
∴ Nb ,Rd =
0.88 × 7370 × 355 = 2297 × 10 3 N = 2297 kN 1 .0
2297 > 2110 kN
L. Gardner
∴Buckling resistance is OK.
The chosen cross-section, 244.5x10 CHS, in grade S 355 steel is acceptable.
37
Member buckling resistance exercise Background Cross-section
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.
Buckling Example
NEd = 305.6 kN
Exercise
4.5 m
L. Gardner
The critical combination of actions results in a design axial force of 305.6 kN.
38
19
Member buckling resistance exercise Background
Try a 152x152x30 UC in grade S 275 steel.
Cross-section
b z
Buckling
tw
Example
h
d
y
y r
Exercise
tf
z
h b tw tf r A Iy Iz
= 157.6 mm = 152.9 mm = 6.5 mm = 9.4 mm = 7.6 mm = 3830 mm2 = 17480000 mm4 = 5600000 mm4
Section properties for 152x152x30 UC L. Gardner
39
Session 7 Background Cross-section Buckling Example
Compression members
Exercise
Dr Leroy Gardner Senior Lecturer in Structural Engineering L. Gardner
Eurocode 3: Design of steel structures
40
20
Session 8 Background In-plane bending
Beams
Shear Serviceability LTB Exercises
Dr Leroy Gardner Senior Lecturer in Structural Engineering L. Gardner
Eurocode 3: Design of steel structures
1
Outline Background In-plane bending Shear Serviceability LTB Exercises
L. Gardner
Overview: • Background • In-plane bending • Shear • Deflections • Lateral torsional buckling
2
1
Eurocode 3 Background In-plane bending
Eurocode 3 states, as with BS 5950, that both cross-sectional and member bending resistance must be verified:
Shear Serviceability
M Ed ≤ M c ,Rd
Cross-section check (In-plane bending)
MEd ≤ Mb ,Rd
Member buckling check
LTB Exercises
L. Gardner
3
Non-dimensional slenderness Background In-plane bending Shear
Beam behaviour analogous to yielding/buckling of columns. M
Material yielding (in-plane bending)
MEd
MEd
Wyfy
Serviceability
Elastic member buckling Mcr
LTB
Lcr
Exercises
1.0 L. Gardner
Non-dimensional slenderness
λ LT 4
2
Cross-sections in bending Background
• Class 1 & 2 cross-sections:
In-plane bending Shear
Mc ,Rd = Mpl =
Wpl fy γ M0
Serviceability LTB
• Class 3 cross-sections:
Exercises
Mc ,Rd = M el =
Wel fy γ M0
L. Gardner
5
Cross-sections in bending Background In-plane bending
• Class 4 cross-sections:
Shear Serviceability LTB
Mc ,Rd =
Weff fy γ M0
Exercises
L. Gardner
6
3
Section moduli W Background In-plane bending
Subscripts are used to differentiate between the plastic, elastic or effective section modulus
Shear Serviceability LTB
Plastic modulus
Wpl
(S in BS 5950)
Elastic modulus
Wel
(Z in BS 5950)
Effective modulus Weff z
(Zeff in BS 5950)
Exercises
The partial factor γM0 is applied to all crosssection bending resistances, and equal 1.0. L. Gardner
7
Shear resistance Background In-plane bending Shear Serviceability
The design shear force is denoted by VEd (shear force design effect). The design shear resistance of a crosssection is denoted by Vc,Rd and may be calculated based on a plastic (Vpl,Rd) or an elastic distribution of shear stress.
LTB Exercises
L. Gardner
VEd ≤ 1.0 Vc ,Rd
Shear check
8
4
Plastic shear resistance Vpl,Rd Background In-plane bending Shear Serviceability
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:
LTB Exercises
Vpl,Rd =
A v (fy / 3 ) γ M0
L. Gardner
9
Shear area Av Background In-plane bending Shear Serviceability LTB Exercises
L. Gardner
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).
10
5
Shear areas Av Background In-plane bending Shear Serviceability
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
LTB
• Rolled RHS of uniform thickness, load parallel to depth: Exercises
Av = Ah/(b+h) • CHS and tubes of uniform thickness:
L. Gardner
Av = 2A/π
11
Definition of terms Background
A is the cross-sectional area
In-plane bending
b is the overall section breadth
Shear
hw is the overall web depth (measured between flanges)
Serviceability
h is the overall section depth r is the root radius tf is the flange thickness
LTB Exercises
tw is the web thickness (taken as the minimum value if the web is not of constant thickness) η = 1.0 (from UK NA)
L. Gardner
12
6
Shear buckling Background In-plane bending
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 Serviceability
Shear buckling need not be considered provided:
LTB
ε hw ≤ 72 η tw
Exercises
for unstiffened webs
where ε =
235 ; η = 1.0 (from UK NA) fy
L. Gardner
13
Shear resistance example Background In-plane bending
Determine the shear resistance of a rolled channel section 229x89 in grade S 275 steel loaded parallel to the web. b
Shear
z Serviceability LTB
tw h
y
y r
Exercises
z L. Gardner
tf
h = 228.6 mm b = 88.9 mm tw = 8.6 mm tf = 13.3 mm r = 13.7 mm A = 4160 mm2
Section properties for 229x89 rolled channel section
14
7
Shear resistance example Background In-plane bending Shear
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 fy for grade S 275 steel (to EN 10025-2) is found from Table 3.1 to be 275 N/mm2.
Serviceability LTB
Shear resistance is determined according to clause 6.2.6
Exercises
Vpl,Rd =
A v (fy / 3 ) γ M0
L. Gardner
15
Shear resistance example Background
Shear area Av
In-plane bending
For a rolled channel section, loaded parallel to the web, the shear area is given by:
Shear
Av =
A – 2btf + (tw + r)tf
Serviceability
= 4160 – (2×88.9×13.3) + (8.6+13.7)×13.3
LTB
= 2092 mm2
Exercises
L. Gardner
Vpl,Rd =
2092 × (275 / 3 ) = 332000 N = 332 kN 1.00
For the same cross-section BS 5950 (2000) gives a shear resistance of 324 kN.
16
8
Serviceability Background In-plane bending Shear Serviceability LTB Exercises
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.
L. Gardner
17
Serviceability Background In-plane bending
Vertical deflection limits Design situation Cantilevers Beams carrying plaster or other brittle finish
Shear
Other beams (except purlins and sheeting rails) Purlins and sheeting rails
Deflection limit Length/180 Span/360 Span/200 To suit cladding
Serviceability LTB Exercises
Horizontal deflection limits Design situation Tops of columns in single storey buildings, except portal frames Columns in portal frame buildings, not supporting crane runways In each storey of a building with more than one storey
L. Gardner
Deflection limit Height/300 To suit cladding Height of storey/300
18
9
Lateral torsional buckling 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).
L. Gardner
19
Lateral torsional buckling Background In-plane bending Shear Serviceability
Can be discounted when: • Minor axis bending • CHS, SHS, circular or square bar
LTB
• Fully laterally restrained beams Exercises
• λ LT < 0.2 (or 0.4 in some cases) L. Gardner
20
10
Lateral torsional buckling resistance Background In-plane bending
Checks should be carried out on all unrestrained segments of beams (between the points where lateral restraint exists).
Shear Serviceability LTB Exercises
Lateral restraint
Lateral restraint
Lateral restraint Lcr = 1.0 L
Beam on plan L. Gardner
21
Eurocode 3 Background In-plane bending Shear Serviceability LTB Exercises
Three 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.
L. Gardner
22
11
Lateral torsional buckling Background In-plane bending Shear Serviceability
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:
LTB Exercises
Mb,Rd = χLT Wy
fy γ M1
Reduction factor for LTB
L. Gardner
23
Equivalence to BS 5950 Background
Lateral torsional buckling resistance:
In-plane bending Shear Serviceability
Mb,Rd =
χLT Wy fy γM1
Eurocode 3
LTB Exercises
L. Gardner
Mb = pb Sx (or Zx) Wy will be Wpl,y or Wel,y
BS 5950
24
12
Buckling curves – general case Background In-plane bending Shear Serviceability LTB
Lateral torsional buckling curves for the general case are given below:
χLT
=
1 ΦLT +
2 ΦLT
−
λ2LT
but χLT ≤ 1.0
Φ LT = 0.5 [ 1 + αLT ( λ LT − 0.2) + λ2LT ]
Exercises
Plateau length Imperfection factor from Table 6.3 L. Gardner
25
Buckling curve selection Background
For the general case, refer to Table 6.4: In-plane bending Shear
Cross-section
Serviceability
Rolled I-sections
LTB Exercises
Welded Isections Other crosssections
L. Gardner
Limits
Buckling curve
h/b ≤ 2
a
h/b > 2
b
h/b ≤ 2
c
h/b > 2
d
-
d
26
13
Imperfection factor αLT Background In-plane bending
Imperfection factors αLT for 4 buckling curves:
Shear
Buckling curve
Serviceability
a
b
c
d
Imperfection 0.21 0.34 0.49 factor αLT
LTB Exercises
0.76
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27
LTB curves Background
4 buckling curves for LTB (a, b, c and d)
In-plane bending
Serviceability LTB Exercises
Reduction factor χLT
Shear
1.2 1.0
Curve a Curve b
0.8
Curve c Curve d
0.6 0.4 0.2 0.0 0
0.5
0.2 L. Gardner
1
1.5
Non-dimensional slenderness
2
2.5
λ LT
28
14
Buckling curves – rolled or equivalent welded sections case
Background In-plane bending Shear
LTB curves for the rolled or equivalent welded sections case are given below; Table 6.5 is used to select buckling curve:
χLT
=
Serviceability LTB
1 2 Φ LT + Φ LT − β λ2LT
⎧ χLT ≤ 1.0 ⎪ 1 but ⎨ χLT ≤ ⎪⎩ λ LT
Φ LT = 0.5 [ 1 + αLT ( λ LT − λ LT ,0 ) + β λ2LT ]
Exercises
Plateau length
β factor
Recommended value = 0.4 29
Recommended value = 0.75 L. Gardner
LTB curves Background
Comparison between general curves and curves for rolled and equivalent welded sections (I-sections – h/b>2)
In-plane bending
Serviceability LTB Exercises
Reduction factor χLT
Shear
1.20 1.00
General (h/b>2) Rolled (h/b>2)
0.80 0.60 0.40 0.20 0.00 0
L. Gardner
0.5
1
1.5
2
Non-dimensional slenderness
2.5
λ LT
30
15
Non-dimensional slenderness Background In-plane bending Shear Serviceability LTB Exercises
L. Gardner
• Calculate lateral torsional buckling slenderness:
λ LT
=
Wy f y Mcr
• Buckling curves as for compression (except curve a0) • Wy depends on section classification • Mcr is the elastic critical LTB moment 31
Elastic critical buckling moment Mcr Background In-plane bending Shear Serviceability
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.
LTB Exercises
L. Gardner
Such simplifications do not appear in the primary Eurocode method.
32
16
Mcr under uniform moment Background In-plane bending
For typical end conditions, and under uniform moment the elastic critical lateral torsional buckling moment Mcr is:
Shear
Mcr ,0
Serviceability LTB Exercises
L. Gardner
G IT Iw Iz Lcr
π 2EIz = 2 L cr
⎡ Iw L cr 2GIT ⎤ ⎢ + 2 ⎥ π I EI z z ⎣ ⎦
0 .5
is the shear modulus is the torsion constant is the warping constant is the minor axis second moment of area is the buckling length of the beam
33
Mcr under non-uniform moment Background In-plane bending Shear Serviceability
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:
LTB Exercises
L. Gardner
π 2EIz Mcr = C1 2 L cr
⎡ Iw L cr 2GIT ⎤ ⎢ + 2 ⎥ π EIz ⎦ ⎣ Iz
0 .5
34
17
C1 factor – end moments Background In-plane bending Shear Serviceability
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
LTB Exercises
where ψ is the ratio of the end moments (defined in the following table).
L. Gardner
35
C1 factor – transverse loading C1 values for transverse loading
Background In-plane bending
Loading and support conditions
Bending moment diagram
Value of C1
w 1.132
Shear w
1.285
Serviceability F
1.365
LTB F
Exercises
1.565
F
CL
F 1.046
=
L. Gardner
=
=
=
36
18
Simplified assessment of λ LT Background In-plane bending Shear
For hot-rolled doubly symmetric I and H sections without destabilising loads, λLTmay be conservatively simplified to:
λ LT =
Serviceability LTB Exercises
L. Gardner
λ 1 0 .9 z λ1 C1
1 0 .9 λ z = C1 E fy
λ z = L / iz ; λ 1 = π
As a further simplification, C1 may also be conservatively taken = 1.0.
37
Simplified assessment of λ LT Background In-plane bending Shear
Substituting in numerical values for λ1 , the following simplified expressions result. S235
Serviceability LTB Exercises
L. Gardner
λLT =
1 L / iz C1 104
S275 λLT =
1 L / iz C1 96
S355 λ LT =
1 L / iz C1 85
C1 may be conservatively taken = 1.0, though the level of conservatism increases the more the actual bending moment diagram differs from uniform moment.
38
19
Design procedure for LTB Background In-plane bending Shear
Design procedure for LTB: 1. Determine BMD and SFD from design loads 2. Select section and determine geometry
Serviceability
3. Classify cross-section (Class 1, 2, 3 or 4) LTB Exercises
4. Determine effective (buckling) length Lcr – depends on boundary conditions and load level 5. Calculate Mcr and Wyfy
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39
Design procedure for LTB Background In-plane bending Shear
6. Non-dimensional slenderness λ LT =
Wy fy Mcr
7. Determine imperfection factor αLT 8. Calculate buckling reduction factor χLT
Serviceability LTB Exercises
L. Gardner
9. Design buckling resistance Mb,Rd = χLT
Wy fy γ M1
M
Ed 10. Check ≤ 1.0 for each unrestrained M b ,Rd portion
40
20
Simplified method (Cl. 6.3.2.4) Background In-plane bending Shear
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
Serviceability LTB Exercises
b
Compression h Tension
L. Gardner
Compression flange + 1/3 of the compressed area of web
Strut
Beam
41
General method (Cl. 6.3.4) Background In-plane bending Shear
General method for lateral and lateral torsional buckling of structural components • May be applied to single members, plane frames etc.
Serviceability LTB Exercises
• Requires determination of plastic and elastic (buckling) resistance of structure, which subsequently defines global slenderness • Generally requires FE
L. Gardner
42
21
Restrained beam exercise Background In-plane bending
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.
Shear
Dead load = 60 kN/m Imposed load = 70 kN/m
Serviceability LTB Exercises
6.0 m Beam is fully laterally restrained L. Gardner
43
Restrained beam exercise Background
b z
In-plane bending Shear
tw
Serviceability
h
d
y
y
LTB
r Exercises
tf
h b tw tf r A Wy,pl Iy
= 612.2 mm = 229.0 mm = 11.9 mm = 19.6 mm = 12.7 mm = 15900 mm2 = 3676×103 mm3 = 986.1×106 mm4
z
610×229×125 UB L. Gardner
44
22
LTB Example Background
Description In-plane bending Shear Serviceability LTB Exercises
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.
L. Gardner
45
LTB Example Background In-plane bending Shear Serviceability LTB Exercises
General arrangement L. Gardner
46
23
LTB Example Background
Design loading is as follows:
In-plane bending
425.1 kN
Shear
319.6 kN
A B
Serviceability LTB
D
C
2.5 m
3.2 m
Exercises
5.1 m
Loading
L. Gardner
47
LTB Example 267.1 kN
Background
A
B
In-plane bending
D 52.5 kN
SF
C
477.6 kN
Shear
Shear force diagram
Serviceability LTB
A
B
C D
Exercises
BM
L. Gardner
1194 kNm
1362 kNm
Bending moment diagram
48
24
LTB Example Background In-plane bending Shear Serviceability LTB
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.
Exercises
L. Gardner
49
LTB Example Background
b
In-plane bending
z
Shear
tw Serviceability LTB
h
d
y
y
r
Exercises
z
L. Gardner
tf
h b tw tf r A Wy,pl Iz It Iw
= 762.2 mm = 266.7 mm = 14.3 mm = 21.6 mm = 16.5 mm = 22000 mm2 = 6198×103 mm3 = 68.50×106 mm4 = 2670×103 mm4 = 9390×109 mm6 50
25
LTB Example Background In-plane bending Shear
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.
Serviceability LTB
From clause 3.2.6: E = 210000 N/mm2 and G ≈ 81000 N/mm2.
Exercises
L. Gardner
51
LTB Example Background In-plane bending Shear
Cross-section classification (clause 5.5.2):
ε = 235 / fy = 235 / 265 = 0.94 Outstand flanges (Table 5.2, sheet 2)
Serviceability
cf = (b – tw – 2r) / 2 = 109.7 mm LTB
cf / tf
= 109.7 / 21.6 = 5.08
Exercises
Limit for Class 1 flange = 9ε = 8.48 > 5.08 ∴ Flange is Class 1 L. Gardner
52
26
LTB Example Background In-plane bending
Web – internal part in bending (Table 5.2, sheet 1) cw = h – 2tf – 2r = 686.0 mm
Shear
cw / tw= 686.0 / 14.3 = 48.0
Serviceability
Limit for Class 1 web = 72 ε = 67.8 > 48.0
LTB
∴ Web is Class 1
Exercises
Overall cross-section classification is therefore Class 1.
L. Gardner
53
LTB Example Background In-plane bending Shear
Bending resistance of cross-section (clause 6.2.5):
Mc ,y,Rd =
Serviceability LTB Exercises
=
Wpl,y fy γ M0
for Class 1 and 2 sec tions
6198 × 10 3 × 265 = 1642 × 10 6 Nmm 1. 0
= 1642 kNm > 1362 kNm ∴ Cross-section resistance in bending is OK.
L. Gardner
54
27
LTB Example Background
Lateral torsional buckling check (clause 6.3.2.2) – Segment BC:
In-plane bending
MEd = 1362 kNm
Shear
Mb ,Rd = χ LT Wy
Serviceability LTB Exercises
L. Gardner
fy γ M1
where Wy = Wpl,y for Class 1 and 2 sections Determine Mcr for segment BC (Lcr = 3200 mm)
π 2EIz Mcr = C1 2 L cr
⎡ Iw L cr 2GIT ⎤ ⎥ ⎢ + 2 π EIz ⎦ ⎣ Iz
0 .5
55
LTB Example Background In-plane bending
For end moment loading C1 may be approximated from: C1 = 1.88 – 1.40ψ + 0.52ψ2
but C1 ≤ 2.70
Shear
ψ is the ratio of the end moments = Serviceability LTB Exercises
1194 = 0.88 1362
⇒ C1 = 1.05 Mcr = 1.05 ×
π 2 × 210000 × 68.5 × 10 6 3200 2
⎡ 9390 × 10 9 3200 2 × 81000 × 2670 × 10 3 ⎤ + ⎢ ⎥ 6 π 2 × 210000 × 68.5 × 10 6 ⎦ ⎣ 68.5 × 10
0 .5
= 5699x106 Nmm = 5699 kNm L. Gardner
56
28
LTB Example Background In-plane bending
Non-dimensional lateral torsional slenderness for segment BC: λLT =
Shear
Wy fy Mcr
=
6198 × 103 × 265 = 0.54 5699 × 106
Serviceability LTB
Select buckling curve and imperfection factor αLT:
Exercises
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 L. Gardner
57
LTB Example Background
From Table 6.3 of EN 1993-1-1:
In-plane bending
For buckling curve b, αLT = 0.34
Shear
Calculate reduction factor for lateral torsional
Serviceability LTB Exercises
buckling, χLT – Segment BC:
χ LT
=
1 2 Φ LT + Φ LT − λ2LT
but χ LT ≤ 1.0
where Φ LT = 0.5 [ 1 + α LT ( λ LT − 0.2) + λ2LT ] L. Gardner
58
29
LTB Example Background In-plane bending
ΦLT = 0.5[1+0.34(0.54-0.2) + 0.542] = 0.70
∴ χLT =
Shear Serviceability LTB Exercises
1 0.70 + 0.702 − 0.54 2
= 0.87
Lateral torsional buckling resistance Mb,Rd – Segment BC: Mb,Rd
= χ LT Wy
fy γ M1
= 0.87 × 6198 × 10 3 ×
265 1 .0
= 1425 × 10 6 Nmm = 1425 kNm L. Gardner
59
LTB Example Background In-plane bending Shear Serviceability LTB Exercises
MEd 1362 = = 0.96 ≤ 1.0 ∴ Segment BC is OK Mb ,Rd 1425
Lateral torsional buckling check (clause 6.3.2.2) – Segment CD: MEd = 1362 kNm Mb ,Rd = χ LT Wy
fy γ M1
where Wy = Wpl,y for Class 1 and 2 sections Determine Mcr for segment CD (Lcr = 5100 mm) L. Gardner
60
30
LTB Example π 2EIz Mcr = C1 2 L cr
Background In-plane bending Shear Serviceability LTB
⎡ Iw L cr 2GIT ⎤ ⎥ ⎢ + 2 π EIz ⎦ ⎣ Iz
0 .5
Determine ψ from Table: ψ is the ratio of the end moments =
0 =0 1362
⇒ C1 = 1.88
Exercises
Mcr = 1.88
L. Gardner
π 2 × 210000 × 68.5 × 10 6 ⎡ 9390 × 10 9 5100 2 × 81000 × 2670 × 10 3 ⎤ + ⎢ ⎥ 6 5100 2 π 2 × 210000 × 68.5 × 10 6 ⎦ ⎣ 68.5 × 10
= 4311×106 Nmm = 4311 kNm
0. 5
61
LTB Example Background In-plane bending Shear Serviceability LTB Exercises
L. Gardner
Non-dimensional lateral torsional slenderness for segment CD:
λLT =
Wy fy Mcr
=
6198 × 103 × 265 = 0.62 4311× 106
The buckling curve and imperfection factor αLT are as for segment BC.
62
31
LTB Example Background In-plane bending Shear
Calculate reduction factor for lateral torsional buckling, χLT – Segment CD:
χ LT
1
=
2 Φ LT + Φ LT − λ2LT
but χ LT ≤ 1.0
Serviceability LTB Exercises
where Φ LT = 0.5 [ 1 + α LT ( λ LT − 0.2) + λ2LT ] = 0.5[1+0.34(0.62-0.2) + 0.622] = 0.76 ∴ χ LT
=
1 0.76 + 0.76 2 − 0.62 2
= 0.83
L. Gardner
63
LTB Example Background In-plane bending Shear Serviceability LTB Exercises
Lateral torsional buckling resistance Mb,Rd – Segment CD: Mb,Rd
= χ LT Wy
fy γ M1
= 0.83 × 6198 × 10 3 ×
265 1 .0
= 1360 × 10 6 Nmm = 1360 kNm MEd 1362 = = 1.00 Mb ,Rd 1360
Segment CD is critical and marginally fails LTB check. L. Gardner
64
32
Session 8 Background In-plane bending Shear Serviceability
Beams
LTB Exercises
Dr Leroy Gardner Senior Lecturer in Structural Engineering L. Gardner
Eurocode 3: Design of steel structures
65
33
Session 9 Introduction Cross-section Members Annex A & B
Beam-columns
Simple construction
Dr Leroy Gardner Senior Lecturer in Structural Engineering L. Gardner
Eurocode 3: Design of steel structures
1
Introduction Introduction Cross-section
Beam-columns:
Members
• Cross-section check
Annex A & B
• Member buckling check
Simple construction
L. Gardner
2
1
Cross-section checks Introduction Cross-section Members Annex A & B
Cross-section checks similar to BS 5950, including a simplified linear interaction, as below:
My,Ed Mz,Ed NEd + + ≤ 1 NRd My,Rd Mz,Rd
Simple construction
More sophisticated expressions are also provided for Class 1 and 2 for greater efficiency. L. Gardner
3
Beam-columns – member checks Introduction
Two philosophies:
Cross-section
• Interaction method - Clause 6.3.3 Members Annex A & B Simple construction
• Interaction ‘k’ factors from Annex A or B. • General method - Clause 6.3.4 • Not for hand calculations (requires FE or similar)
L. Gardner
4
2
Simple construction Introduction Cross-section Members Annex A & B Simple construction
In general, both Eqs. 6.61 and 6.62 must be examined and satisfied:
My,Ed M NEd + k yy + k yz z ,Ed ≤ 1 Nb ,y,Rd Mb ,Rd Mc ,z ,Rd
Eq. 6.61
My,Ed M NEd + k zy + k zz z ,Ed ≤ 1 Nb ,z ,Rd Mb ,Rd Mc ,z ,Rd
Eq. 6.62
L. Gardner
5
Interaction factors kij 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
L. Gardner
6
3
Good news - simple construction Introduction Cross-section
‘Simple construction’ is commonly used for the design of multi-storey buildings (particularly in the UK).
Members Annex A & B Simple construction
• Beams are designed as simply supported • Columns are designed for nominal moments arising from the eccentricity at the beam-tocolumn connection.
L. Gardner
7
Simple construction Introduction Cross-section Members
Multi-storey frame: Wk & NHF Wk & NHF
Gkr & Qkr Gkf & Qkf
Annex A & B Simple construction
Wk & NHF Wk & NHF
L. Gardner
Gkf & Qkf Gkf & Qkf
8
4
Simple construction Introduction Cross-section Members Annex A & B Simple construction
In general, both Eqs. 6.61 and 6.62 must be examined and satisfied:
My,Ed M NEd + k yy + k yz z ,Ed ≤ 1 Nb ,y,Rd Mb ,Rd Mc ,z ,Rd
Eq. 6.61
My,Ed M NEd + k zy + k zz z ,Ed ≤ 1 Nb ,z ,Rd Mb ,Rd Mc ,z ,Rd
Eq. 6.62
L. Gardner
9
NCCI Simplification Introduction Cross-section Members Annex A & B Simple construction
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.
My,Ed M NEd + k zy + k zz z ,Ed ≤ 1 Nb ,z ,Rd Mb ,Rd Mc ,z ,Rd L. Gardner
Eq. 6.62 10
5
NCCI Simplification Introduction Cross-section Members
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:
Annex A & B Simple construction
kzy = 1.0 and kzz = 1.5
My,Ed M NEd + 1 .0 + 1.5 z ,Ed ≤ 1 Nb ,z ,Rd Mb ,Rd Mc ,z ,Rd
Eq. 6.62
L. Gardner
11
Recommendations Introduction Cross-section Members
Recommendations: • For pencil and paper calculations, use: - Clause 6.3.3 with Annex B
Annex A & B
• Use NCCI simplification for columns in simple construction
Simple construction
• Make spreadsheets to check calculations • Full worked examples in Designers’ Guide and Trahair et al textbook.
L. Gardner
12
6
Session 9 Introduction Cross-section Members Annex A & B
Beam-columns
Simple construction
Dr Leroy Gardner Senior Lecturer in Structural Engineering L. Gardner
Eurocode 3: Design of steel structures
13
7
Session 10 Introduction Bolted joints Welded joints
Joints Dr Leroy Gardner Senior Lecturer in Structural Engineering
L. Gardner
Eurocode 3: Design of steel structures
1
Outline Introduction Bolted joints
Overview:
Welded joints
• Introduction • Bolted joints • Welded joints
L. Gardner
2
1
EN 1993-1-8 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.
L. Gardner
3
EN 1993-1-8 Introduction
Essentially, the coverage of Part 1.8 focuses on 4 topics:
Bolted joints Welded joints
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.
L. Gardner
4
2
EN 1993-1-8 Introduction Bolted joints Welded joints
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.
L. Gardner
5
Bolted joints Bolted joints: Introduction Bolted joints Welded joints
• Shear resistance Fv,Rd • Bearing resistance Fb,Rd • Tension resistance Ft,Rd • Combined shear and tension • Bolt spacing
L. Gardner
6
3
Bolted joints Bolt shear resistance per shear plane for ordinary bolts: Introduction
Fv,Rd =
Bolted joints Welded joints
where:
αvfubA γM2
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
L. Gardner
7
Bolted joints Introduction
fub is the ultimate tensile strength of the bolt
Bolted joints
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.
Welded joints
γM2 may be taken as 1.25
L. Gardner
8
4
Bolted joints Bearing resistance Fb,Rd Introduction Bolted joints Welded joints
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:
Fb ,Rd =
k 1αb fudt γ M2
L. Gardner
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Bolted joints Definitions of terms: Introduction Bolted joints
αb is the smallest of: αd; fub/fu or 1.0, and accounts for various failure modes
Welded joints
d is the bolt diameter t is the minimum thickness of the connected parts γM2 may be taken as 1.25 fu is the ultimate tensile strength of the connected parts αd and k1 relate to bolt spacing and edge and end distances.
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Bolted joints Combined tension and shear Introduction Bolted joints Welded joints
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:
Fv,Ed Fv,Rd
+
Ft ,Ed 1.4 Ft ,Rd
≤ 1.0
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Bolted joints Introduction Bolted joints Welded 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
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Bolted joints Introduction Bolted joints Welded 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
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Bolted joint example Introduction Bolted joints Welded joints
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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.
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Bolted joint example Shear resistance: Introduction Bolted joints
Bolts are in single shear, and it is assumed that the shear plane passes through the threaded portion of the bolts:
Welded joints
αv = 0.6, fub = 400 N/mm2, A = As = 245 mm2, γM2= 1.25 Shear resistance per bolt Fv,Rd:
Fv ,Rd =
α v fub A 0.6 × 400 × 245 = = 47040 N = 47.0 kN γ M2 1.25
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Bolted joint example Bearing resistance: Introduction
Bearing resistance per bolt Fb,Rd:
Bolted joints Welded joints
Fb ,Rd =
k 1α b fu dt γ M2
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.
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Bolted joint example Introduction Bolted joints Welded joints
For end bolts, αd =
e1 = (40/66) = 0.61 3d0
p1 = (60/66 – 0.25) = 0.66 3d0 e For edge bolts, k1 is the smaller of (2.8 2 − 1.7) or 2.5 d0
For inner bolts, αd =
(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 L. Gardner
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Bolted joint example Therefore, for end bolts, Introduction Bolted joints Welded joints
Fb ,Rd =
k 1α b fudt 2.5 × 0.61 × 410 × 20 × 16 = = 160 .1 kN γ M2 1.25
And, for inner bolt, Fb ,Rd =
k 1α b fudt 2.5 × 0.66 × 410 × 20 × 16 = = 173 .2 kN γ M2 1.25
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. L. Gardner
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Welded joints Introduction
Design of welded joints:
Bolted joints
• Butt welds
Welded joints
• Fillet welds - directional method - simplified method
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Welded joints 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/ 3 in shear) provided that suitable electrodes are used. Throat thickness taken as minimum depth of penetration, reduced by 3 mm for most partialpenetration butt welds.
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Welded joints 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.
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Welded joints Simplified approach Introduction
Check Fw,Ed ≤ Fw,Rd Bolted joints Welded joints
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:
Fw ,Rd = fvw ,d a L. Gardner
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Welded joints fvw,d is the design shear strength of the weld Introduction
a is the throat thickness of the weld
Bolted joints
fvw ,d =
Welded joints
fu β w γ M2 3
fu is the minimum ultimate tensile strength of the connected parts βw is a correlation factor that depends on the material grade γM2 may be taken as 1.25 L. Gardner
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Welded joints Introduction
Values for correlation factor βw
Bolted joints
Steel grade
Thickness range (mm)
Ultimate strength fu (N/mm2)
Welded joints
S235
t≤3
360
3 < t ≤ 100
360
t≤3
430
3 < t ≤ 100
410
t≤3
510
3 < t ≤ 100
470
S275
S355
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Correlation factor βw 0.80
0.85
0.90
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Welded joint example Description Introduction Bolted joints Welded joints
A 150×20 mm tie in Grade S275 steel carrying 400 kN is spliced using a singlesided cover plate 100×20 mm as shown in the figure below. Design a suitable fillet weld to carry the applied load.
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Welded joint example Introduction Bolted joints
Try 8 mm fillet welds: Throat thickness a = 0.7 s = 0.7×8 = 5.6 mm
Welded joints
Design shear strength of weld: From Table, fu = 410 N/mm2 and βw = 0.85 fvw ,d =
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410 = 223 N / mm2 0.85 × 1.25 × 3 26
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Welded joint example Introduction
The design resistance of the weld per unit length (i.e. per mm run) Fvw,d:
Bolted joints Welded joints
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.
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Session 10 Introduction Bolted joints Welded joints
Joints Dr Leroy Gardner Senior Lecturer in Structural Engineering
L. Gardner
Eurocode 3: Design of steel structures
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Conclusions Introduction Bolted joints Welded joints
• ‘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)
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Conclusions Introduction Bolted joints Welded joints
• ‘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
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Conclusions Introduction
Conclusions:
Bolted joints
• Advanced design codes
Welded joints
• Greater in scope • Biggest change since limit states • Unfamiliar format/ resistance to uptake • Guidance material and training emerging • Basis for other National design codes
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Eurocode 3 Introduction Bolted joints Welded joints
Thank you Dr Leroy Gardner Senior Lecturer in Structural Engineering
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Eurocode 3: Design of steel structures
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Table 1: Values for yield strength fy and ultimate strength fu (from EN 10025-2) Yield strength 2 fy (N/mm )
Yield strength fy (N/mm2)
Ultimate strength fu (N/mm2)
t ≤ 16 mm
16 < t ≤ 40 mm
3 < t ≤ 100 mm
S235
235
235
360
S275
275
265
410
S355
355
345
470
S450
450
430
550
Steel grade
1
Table 2 (sheet 1): Maximum width-to-thickness ratios for compression parts (Table 5.2 of EN 1993-1-1)
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Table 2 (sheet 2): Maximum width-to-thickness ratios for compression parts (Table 5.2 of EN 1993-1-1)
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Table 3: Selection of buckling curve for a cross-section (Table 6.2 of EN 1993-1-1)
Buckling curve Cross-section
Limits
S 460
a b
a0 a0
40 mm < tf ≤ 100 mm
y–y z-z
b c
a a
tf ≤ 100 mm
y–y z-z
b c
a a
tf > 100 mm
y–y z-z
d d
c c
tf ≤ 40 mm
y–y z-z
b c
b c
tf > 40 mm
y–y z-z
c d
c d
hot finished
any
a
a0
cold formed
any
c
c
generally (except as below)
any
b
b
thick welds: a > 0.5tf b/tf < 30 h/tw < 30
any
c
c
any
c
c
any
b
b
h/b > 1.2
y–y z-z
Rolled I-sections
tf ≤ 40 mm
U-, T- and solid sections
S 235 S 275 S 355 S 420
L-sections
b z
Buckling about axis
tw y
y
r
h/b ≤ 1.2
h
tf
z z
Welded Isections
z y
y
y
tf
tf z
Hollow sections
z
y
Welded box sections
z h
y
tf y tw
b z
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Table 4: Imperfection factors for buckling curves (Table 6.1 of EN 1993-1-1)
Buckling curve Imperfection factor α
a0
a
b
c
d
0.13
0.21
0.34
0.49
0.76
1.2
1.0
Curve aa0 0
Reduction factor χ
Curve a Curve b
0.8
Curve c Curve d
0.6
0.4
0.2
0.0 0
0.5
1
1.5
2
2.5
Non-dimensional slenderness λ
Figure 1: Eurocode 3 Part 1.1 buckling curves (Figure 6.4 of EN 1993-1-1)
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