TTT Handout LG Nov2008 Lecture Note on EC3 Design

June 14, 2018 | Author: Nguyen Tran Hieu | Category: Buckling, Screw, Deformation (Engineering), Applied And Interdisciplinary Physics, Building Engineering
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Introduction Overview ObjectivesDesign of steel structures to Eurocode 3 Dr Leroy Gardner Senior Lecturer in Structural Engineering Eurocode 3: Design of steel structures L. Gardner 1 Session 1 Introduction Overview Objectives Introduction Dr Leroy Gardner Senior Lecturer in Structural Engineering L. Gardner Eurocode 3: Design of steel structures 2 1 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 L. Gardner 3 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 L. Gardner 4 2 Dr Leroy Gardner Introduction Overview Dr Leroy Gardner BEng MSc PhD DIC CEng MICE MIStructE • 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] L. Gardner 5 Your experience with Eurocodes Introduction Overview Objectives • Introduce yourselves • Organisation • Your experience with steel design/ Eurocodes • Any particular interests/concerns L. Gardner 6 3 Gardner 7 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 L.Motivation for course Introduction Overview Objectives • Most of the Eurocodes are now published • Conflicting British Standards to be withdrawn • Designers need to be prepared • Clear training requirements • Textbooks and design guides • Background information L. Gardner 8 4 . Bradford. 1.Designers’ Guide Introduction Overview Objectives Designers’ Guide to EN 1993-1-1: • Covers Eurocode 3: Part 1.5 and 1. Gardner 9 Textbook Introduction Overview Objectives The behaviour and design of steel structures to EC3: Trahair.8 • EN 1990 and EN 1991 • Sections aligned with code L.1 • Also Parts 1. Nethercot & Gardner (2008) • Structural phenomena • Theoretical background • Code implementation • Worked examples L.3. Gardner 10 5 . 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 L. Gardner 12 6 . Gardner 11 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 L. Gardner 14 7 .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 • EN 1999 – Design of aluminium structures 13 L. Gardner 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 L. Gardner 15 Objectives Introduction Overview Objectives Course objectives: • Familiarity with layout. philosophy of Eurocodes • Understanding of background and design procedure for principal structural components L. or translate at own cost. and then ‘exactly’ translated • Other participating counties will either use 1 of 3 language versions available. L.Eurocodes Introduction Overview Objectives • Codes will be published by CEN in 3 languages: • English • French • German • All codes originally developed in English. Gardner 16 8 . notation. Session 1 Introduction Overview Objectives Introduction Dr Leroy Gardner Senior Lecturer in Structural Engineering L. Gardner Eurocode 3: Design of steel structures 17 9 . Gardner 2 1 .Session 2 EN 1990 EN 1991 Introduction to EN 1990 & EN 1991 Dr Leroy Gardner Senior Lecturer in Structural Engineering L. Gardner Eurocode 3: Design of steel structures 1 Overview EN 1990 Outline: EN 1991 • Introduction to EN 1990 • Introduction to EN 1991 • Conclusions L. 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’…. Gardner 4 2 . Gardner 3 Basic requirements EN 1990 EN 1991 EN 1990 states that a structure shall have adequate: • Structural resistance • Serviceability • Durability • Fire resistance • Robustness L. L. vibrations etc • Both essentially produce same effect Effect of action (E): EFFECT • On structural members and whole structure • For example internal forces and moments. Gardner 6 3 .g.g. e. Gardner 5 Actions and Effects EN 1990 EN 1991 Action (F): • Direct actions – applied loads CAUSE • Indirect actions – imposed deformations or accelerations e. during construction or repair • Accidental design situations: exceptional conditions such as fire.. deflections . L.Design situations EN 1990 EN 1991 All relevant design situations must be examined: • Persistent design situations: normal use • Transient design situations: temporary conditions. by temperature changes. L. explosion or impact • Seismic design situations: where the structure is subjected to seismic events. Gardner 7 Load combinations EN 1990 EN 1991 Fundamental combinations of actions may be determined from EN 1990 using either of: • Equation 6.Types of actions EN 1990 EN 1991 Types of actions: • Permanent. A L. Q (leading and non-leading) • Accidental.10b L.10a and 6.10 • Less favourable of Equation 6. G • Variable. Gardner 8 4 . the full value of the leading variable action is applied γQ. j G k .e.1Qk. j "+" γ PP "+" γ Q .i ψ 0 . 1. Gardner 10 5 .i Q k .1Q k . each variable action should be considered in turn as the leading one.5 x combination factor x Other variable actions Actions due to prestressing ∑γ j ≥1 G.e.10. Gardner Load factors 1. i 1.5 are applied when actions are ‘unfavourable’.10: ‘to be combined with’ 1.1 (i.) L.Load combinations EN 1990 EN 1991 Equation 6.1 EN 1990 EN 1991 • In Equation 6.35 and 1. 9 Leading variable actions Qk.35 x Permanent actions L.5 x Leading variable action ∑γ i >1 Q . (and consideration should be given to whether loading is favourable or unfavourable. the critical combination) • To generate the various load combinations.1 "+" 1.5 x characteristic imposed load) • The leading variable action is the one that leads to the most unfavourable effect (i. 35 For favourable dead loads: γG = 1. Gardner * 0. Loading Imposed loading Wind loading Combination factor ψ0 0.Combination factor ψ0 EN 1990 EN 1991 The combination factor ψ0 is intended specifically to take account of the reduced probability of the simultaneous occurrence of two or more variable actions.00 For unfavourable variable loads: γQ = 1.5 For favourable variable loads: γQ = 0 L.5 is UK NA value. axial forces etc) in the structural members. depending on whether they increase or reduce the effects (bending moments. For unfavourable dead loads: γG = 1.6 is the unmodified EC value Unfavourable and favourable loading EN 1990 EN 1991 Loads may be considered as ‘unfavourable’ or ‘favourable’ in any given combination.5* 11 L. Gardner 12 6 . 0.7 0. 10: Combination Dead + Imposed Dead + Wind (uplift) D+I+W (imposed leading) D+I+W (wind leading) Dead Imposed Wind EHF 1. are required to account for imperfections that exist in all structural frames.0 L.Equivalent horizontal forces EN 1990 EN 1991 Equivalent horizontal forces: Equivalent horizontal forces (EHFs). previously known as notional horizontal loads (NHL).10 EN 1990 EN 1991 Load combinations for a typical structure from Equation 6. Gardner Note EHF are always present and already based on factored loads 14 7 .0 1.0 1.0 1. EHFs should be included in all load combinations. Gardner 13 Exercise solution – Equation 6. L. they will generally be different for each load combination (and will already be factored). and since their value is related to the level of vertical loading. e.0 1.85 is the unmodified EC value) L.0 1.10b) Dead + Wind (uplift) (6.1Q k .i ψ 0 .10a)* Dead Imposed Wind EHF 1.10b) (wind leading) D + I + W (6.925 in UK NA (0.10a and 6. jGk .i ψ 0 .i Q k . jGk .iQ k .10b EN 1990 EN 1991 Load combinations from Eqs 6.10b – use less favourable result: γ ∑ j≥1 G.1 "+ " γ ∑ i> 1 Q .Load combinations EN 1990 EN 1991 Equations 6.0 L.i Unfavourable dead load reduction factor (i.0 1.1Q k . not applied when γG = 1) ξ = 0.10b.10b) D + I + W (6.10b – All combinations except last one are from Eq. Gardner 15 Exercise solution – Eqs 6.1ψ 0. 6. j "+ " γ PP "+ " γ Q .i ξ ∑γ j≥1 j G. Gardner * Unlikely to govern unless Dead >> Imposed 16 8 .10a and 6. j "+ " γ PP "+ " γ Q .1 "+ " γ ∑ i> 1 Q .0 1.10a and 6.10b) (imposed leading) D + I + W (6. Combination Dead + Imposed (6. imposed load favourable Wind load unfavourable.1 when unfavourable. Dead loads are factored by 0.Equilibrium check (EQU) EN 1990 EN 1991 Equilibrium check (EQU): For checking sliding or overturning of the structure as a rigid body.9 Gk 1.10 may be used.05 Qk 1. part of imposed unfavourable 1. Gardner Overturning point Overturning point 18 9 . part of dead load favourable.1 Gk + 1. Gardner 17 Equilibrium check (EQU) EN 1990 EN 1991 Favourable and unfavourable loading: Wind load unfavourable. part unfavourable. only Eq.9 Gk L.9 when favourable and 1. dead load favourable.5Wk + EHF L. The critical case will generally arise when wind load is unfavourable and the leading variable action. 6.5 Wk 0.5 Wk 0. and dead load is favourable.9Gk + 1. resulting in: 0. ignoring dead load and with some specified deflection limits.0Wk + 0.0Qk + 0. Gardner 19 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 L. Gardner 20 10 .5Wk + EHF 1.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. 1.7Qk + EHF (Vertical deflections) (Horizontal deflections) Deflection limits are as given in BS 5950 L. 10a and 6.Sub-parts of EN 1991-1 EN 1990 EN 1991 EN 1991-1 contains the following sub-parts: • EN 1991-1-1: Densities. Gardner 21 Conclusions EN 1990 EN 1991 Concluding comments: • Presentation of load combinations unfamiliar • Idea of leading variable actions and combination factors etc is new • Other than format and notation. L. loading codes are similar to existing BS • Using Eq.10b (with 6. self-weight.10 for EQU). 6. 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 L. four basic load combinations arise (ignoring those unlikely to govern). Gardner 22 11 . Gardner Eurocode 3: Design of steel structures 23 12 .Session 2 EN 1990 EN 1991 Introduction to EN 1990 & EN 1991 Dr Leroy Gardner Senior Lecturer in Structural Engineering L. Gardner 2 1 . Gardner Eurocode 3: Design of steel structures 1 Overview Background Outline: Overview • Development of Eurocode 3 • Introduction to design to Eurocode 3 • Conclusions L.Session 3 Background Overview Overview of Eurocode 3 Dr Leroy Gardner Senior Lecturer in Structural Engineering L. EN 1993: Eurocode 3 Background Overview Eurocode 3: • Work began back in 1975 • Eurocode 3 contains a number of parts • … and sub-parts • The first 5 parts were published in 2005 L. tanks & pipelines • EN 1993-5 Piling • EN 1993-6 Crane supporting structures L. Gardner 3 EN 1993: Eurocode 3 Background Overview Eurocode 3 contains six parts: • EN 1993-1 Generic rules • EN 1993-2 Bridges • EN 1993-3 Towers. Gardner 4 2 . masts & chimneys • EN 1993-4 Silos. EN 1993-1 Background Eurocode 3: Part 1 has 12 sub-parts: Overview • EN 1993-1-1 • EN 1993-1-2 • EN 1993-1-3 • EN 1993-1-4 • EN 1993-1-5 • EN 1993-1-6 L. Gardner General rules Fire Cold-formed thin gauge Stainless steel Plated elements Shells 5 EN 1993-1 Background Overview • EN 1993-1-7 • EN 1993-1-8 • EN 1993-1-9 • EN 1993-1-10 • EN 1993-1-11 • EN 1993-1-12 Plates transversely loaded Joints Fatigue Fracture toughness Cables High strength steels L. Gardner 6 3 . Gardner 7 Axes convention Background Overview Different axes convention: BS 5950 Along the member Major axis Minor axis Eurocode 3 X X Y Y Z 8 L.National Annexes Background Overview National Annexes: • Every Eurocode will contain a National Annex • National choice • Non Conflicting Complementary Information • Timescale L. Gardner 4 . factored member force or moment) • ‘Rd’ means design resistance So. Ac. Gardner z b 9 Subscripts Background Overview Extensive use of sub-scripts – generally helpful: • ‘Ed’ means design effect (i.Labelling convention Background Overview Labelling convention: b z z tw h d y y h y r t y r tf z L.loc L. • NEd is an axial force • NRd is the resistance to axial force Sometimes tedious e.e. Gardner 10 5 .eff.g. Gardner 12 6 . Gardner 11 Gamma factors γ Background Overview Gamma factors γ: • Appear everywhere • Partial safety factors • γF for actions (loading) • γM for resistance L.Different symbols Background Overview For example: BS5950 A Z S Ix Iy EC3 A Wel Wpl Iy Iz BS5950 P Mx V H J EC3 N My V Iw It BS5950 py pb pc r EC3 fy χLT fy χf y i L. Gamma factors γM Background Overview Gamma factors γM account for material and modelling uncertainties: Partial factor γM γM0 γM1 γM2 L.00) 1.00 (1. Yield strength fy (N/mm2) t ≤ 16mm S235 S275 S355 S450 235 275 355 450 Yield strength fy (N/mm2) 16 < t ≤ 40 mm 235 265 345 430 Ultimate strength fu (N/mm2) 3 ≤ t ≤ 100 mm 360 410 470 550 Steel grade L.00 (1.25 (1. Gardner 14 7 .00) 1. Gardner EC 3 value (UK NA value) 1. The Young’s modulus of steel should be taken as 210000 N/mm2.10) Application Cross-sections Member buckling Fracture 13 Material properties Background Overview Material properties are taken from product standards (generally EN 10025-2). reference to material standards • Durability L. Gardner 15 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 L.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. Gardner 16 8 . AB and BB L. B.Structural design Background • Buckling resistance of members Overview • Built-up members • SLS • Annexes A. Gardner 17 Omissions Background Overview Notable omissions: • Effective lengths • Formulae for Mcr • Deflection limits • National Annex and NCCIs to resolve L. Gardner 18 9 . steel-sci.Sources of further information Background Overview • http://www.eurocodes.uk/ • Latest news and developments • http://www. Gardner 19 Session 3 Background Overview Overview of Eurocode 3 Dr Leroy Gardner Senior Lecturer in Structural Engineering L. Gardner Eurocode 3: Design of steel structures 20 10 .co.com/ • NCCIs • Worked examples L.org/publications/ • Design guides • http://www.access-steel. Session 4 Introduction Deformed geometry Imperfections Actions Structural analysis Dr Leroy Gardner Senior Lecturer in Structural Engineering L. Gardner 2 1 . Gardner Eurocode 3: Design of steel structures 1 Overview Introduction Deformed geometry Imperfections Actions Outline: • • • • Introduction Analysis types Second order effects Imperfections L. Analysis types Introduction Deformed geometry Imperfections Actions Analysis types: • • • • First order elastic Second order elastic First order plastic Second order plastic L. members and joints L. Gardner 3 General approach Introduction General approach: Deformed geometry Imperfections Actions • Choose an appropriate analysis • Make an appropriate model • Apply all actions (loads) and combinations of actions • Check cross-sections. Gardner 4 2 . 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. Gardner Effects of deformed geometry Introduction Deformed geometry Imperfections Actions EN 1993-1-1 Clause 5.2. 5 L. Gardner 6 3 .1(2) states that deformed geometry (second order effects) shall be considered: • if they increase the action effects significantly • or modify significantly the structural behaviour L. Limits for ignoring deformed geometry Introduction Deformed geometry Imperfections Actions For elastic analysis: where α cr = Fcr ≥ 10 FEd 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. So. Gardner 7 Limits for ignoring deformed geometry Introduction Deformed geometry Imperfections Actions For plastic analysis: α cr = Fcr ≥ 15 FEd Stricter limit for plastic analysis due to loss of stiffness associated with material yielding. Gardner 8 4 . for αcr ≥ 10 (or 15). the effects of deformed geometry may be ignored and a first order analysis will suffice L. L. 2) • not steeper than 1:2 (26 degrees) Limit on axial compression in beams or rafters for (Clause 5. L. equivalent horizontal forces) L.2): NEd ≤ 0. Gardner 9 Limits on use of simple estimate Introduction Deformed geometry Imperfections Actions Limit on portal rafter slope for (Clause 5.Simple estimate for αcr Introduction Deformed geometry Imperfections Actions Simple estimate for αcr may be applied to: • Portals with shallow roof slopes • Beam and column frames (each storey) where ⎛ H ⎞⎛ h ⎞ ⎟ α cr = ⎜ Ed ⎟⎜ ⎜ V ⎟⎜ δ ⎟ ⎝ Ed ⎠⎝ H.Ed storey sway when loaded with horizontal loads (eg wind.Ed ⎠ HEd horizontal reaction at bottom of the storey VEd total vertical load at bottom of the storey δH. Gardner 10 5 .09 Ncr where NEd is the design value of compression in the beam or rafter and Ncr is its elastic buckling resistance. Gardner 11 Frame stability Introduction Deformed geometry Imperfections Actions Limits for treatment of second order effects depend on αcr: Limits on αcr Action αcr>10 First order analysis α cr = Fcr FEd Achievement First order only 10>αcr>3 Second order First order analysis effects by plus amplification or approximate effective length method means Second order analysis Second order effects more accurately 12 αcr<3 L. L. Gardner 6 .Analysis method and achievement Introduction Deformed geometry Imperfections Actions Distinguish between: • Analysis method (1st or 2nd order) • Analysis achievement i. can achieve 2nd order by: 1) 2nd order analysis 2) 1st and amplified sway 3) 1st and increased effective length.e. Global imperfections for frames Introduction Deformed geometry Imperfections Actions Global initial sway imperfections: φ = φ 0 αhα m where φ0 is the basic value = 1/ 200 αh and α m are reduction factors L. L. Gardner 14 7 . Gardner 13 Global imperfections for frames Introduction Deformed geometry Imperfections Actions • 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. Gardner 16 8 .unless using initial imperfection model • Derived from imperfections • Applied in ALL combinations (only in gravity combinations in BS 5950) L. 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 L. self-weight.Actions to be specified Introduction Deformed geometry Imperfections Actions Actions to be specified: • EN 1991-1-1: Densities. Gardner 15 Other Actions Introduction • Equivalent horizontal forces Deformed geometry Imperfections Actions . check built-up members to clause 6. check clause 5.3 . Gardner 17 Session 4 Introduction Deformed geometry Imperfections Actions Structural analysis Dr Leroy Gardner Senior Lecturer in Structural Engineering L.Checks Introduction • Analyse structure Deformed geometry Imperfections • Classify sections using clause 5. Gardner Eurocode 3: Design of steel structures 18 9 .5 .for plastic global analysis.6 • Check cross-sectional resistance to clause 6.4 L.2 Actions • Check buckling resistance to clause 6. 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 1 Overview Introduction Design Example Exercise Outline: • Introduction • Tension member design • Example L. Gardner 2 1 . Rd Introduction Design Example Exercise Design tensile resistance Nt.Rd is limited either by: • Yielding of the gross cross-section Npl. Gardner 4 2 .Rd Tension check Nt.Eurocode 3 Introduction Design Example Exercise Eurocode 3 states that tensile resistance should be verified as follows: Nt .Rd whichever is the lesser.Rd is the design tensile resistance L.Rd • or ultimate failure (fracture) of the net crosssection (at holes for fasteners) Nu.Ed is the tensile design effect Nt. Gardner 3 Design tensile resistance Nt. L.Ed ≤ Nt . 9A net fu γ M2 Anet is the reduced cross-sectional area to account for bolt holes L. Gardner 5 Ultimate resistance of net section Introduction Design Example Exercise And for the ultimate resistance of the net cross-section (defined in clause 6. L.2). the Eurocode 3 design expression is: Nu. Gardner 6 3 .Yielding of gross cross-section Introduction Design Example Exercise The Eurocode 3 design expression for yielding of the gross cross-section (plastic resistance) given as: Npl.2.2.Rd = Afy γ M0 This criterion is applied to prevent excessive deformation of the member.Rd = 0. Rd utliises γM0. 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.25 (1. Gardner s Non-staggered arrangement of fasteners 8 4 .Rd utilises γM2. whilst ultimate fracture of the net cross-section Nu. Gardner 7 Non-staggered fasteners Introduction Design Example Exercise For a non-staggered arrangement of fasteners. γ M 0 = 1 . L.Partial factors γM Introduction Design Example Exercise Plastic resistance of the gross cross-section Npl. A p A s L.0 and γ M2 = 1.1 in UK NA) The larger safety factor associated with fracture reflects the undesirable nature of the failure mode. nd0t A = n = d0 = t = gross cross-sectional area number of bolt holes diameter of bolt holes material thickness L. the maximum sum of the sectional areas of the holes on any line (A-A) perpendicular to the member axis 2. Gardner 9 Staggered fasteners Introduction Design Example For a staggered arrangement of fasteners. Gardner 10 5 .Non-staggered fasteners Introduction Design Example Exercise Net area at bolts holes Anet on any line (AA) perpendicular to the member axis: Anet = A . the total area to be deducted should be taken as the greater of: 1. Exercise ⎛ s2 ⎞ ⎟ ⎜ nd0 − ∑ t⎜ 4p ⎟ ⎠ ⎝ 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 A s Staggered arrangement of fasteners 11 Angles connected by a single row of bolts Single angles in tension connected by a single row of bolts through one leg. but with an effective net section.Rd = Nu. Gardner Introduction Design Example Exercise Nu.Rd = 2.Rd = Nu. With 1 bolt : With 2 bolts : With 3 or more bolts : L. may be treated as concentrically loaded. to give the design ultimate tensile resistance as below.0 (e 2 − 0.Staggered fasteners Introduction Design Example Exercise 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 p B s L.5d0 )tfu γ M2 β 2 Anet fu γ M2 β 3 A net fu γ M2 12 6 . Gardner Definitions for e1. d0 is the bolt hole diameter.0d0 0.5 ≥ 5. Other symbols are defined below: e1 e2 p1 p1 d0 Introduction Design Example Exercise L.Angles connected by a single row of bolts where β2 and β3 are reduction factors dependent upon the bolt spacing (pitch) p1. 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. L. Anet is the net area of the angle.5d0 0.4 0. e2.7 Example β2 (for 2 bolts) Exercise β3 (for 3 or more bolts) Note: For intermediate values of pitch p1 values of β may be determined by linear interpolation. Gardner 14 7 . p1 and d0 13 Angles connected by a single row of bolts Introduction Design Reduction factors β2 and β3 Pitch p1 ≤ 2.7 0. For an unequal angle connected by its smaller leg. Gardner 15 Example: Tension member design Introduction Design Example Exercise Design a single angle tie. L. or an unequal angle connected by its larger leg. B A NEd = 541 kN Tension member AB in truss L. Gardner 16 8 .Angles with welded end connections Introduction Design Example Exercise In the case of welded end connections: For an equal angle. Consider a bolted and a welded arrangement. the eccentricity may be neglected. and the effective area may be taken as equal to the gross area (clause 4. for the member AB shown below. using grade S355 steel.13(2) of EN 1993-1-8). 3 of EN 1993-1-1. the effective area may be taken as equal to the gross area (clause 4. L.2.Example: Tension member design Introduction Design Example Exercise Cross-section resistance in tension is covered in clause 6. For an unequal angle connected (welded) by its larger leg. Gardner 18 9 .2.2 for the calculation of cross-section properties. with reference to clause 6.13(2) of EN 1993-1-8) 125×75×10 unequal angle Gusset plate 125×75×10 unequal angle welded by longer leg L. Gardner 17 Example: Tension member design Welded connection Introduction Design Example Exercise Try a 125×75×10 unequal angle. welded by the longer leg. concentrically loaded (defined in clause 6. Gross area of cross-section.9 × 1920 × 470 = = 738 × 10 3 N = 738 kN γ M2 1. L. Partial factors from UK National Annex are γM0 = 1. the Eurocode 3 design expression is: Nu.10 20 L. yield strength fy = 355 N/mm2 and ultimate tensile strength fu = 470 N/mm2 (from EN 10025-2).00 and γM2 = 1.Example: Tension member design Introduction Design Example Exercise For a nominal material thickness t of 10 mm.Rd = 0.2. A = 1920 mm2 (from Section Tables).Rd = Afy γ M0 = 1920 × 355 = 682x103 N = 682 kN 1. plastic resistance is given as: = Npl.9 A net fu 0.2).10. Gardner 10 . Gardner 19 Example: Tension member design Introduction Design Example Exercise For yielding of the gross cross-section.2.0 And for the ultimate resistance of the net cross-section. Rd is taken as the lesser of these two values. at 125 mm centres) through the longer leg. L.e. bolted (with a line of four 22 mm HSFG bolts. connected by the longer leg is therefore acceptable. 150×75×10 unequal angle bolted by longer leg L. and is therefore 682 kN.Rd > NEd) Unequal angle 125×75×10 in grade S355 steel. Material properties and partial factors are as for the welded case. a smaller angle may be checked.Example: Tension member design Introduction Design Example Exercise The tensile resistance Nt. For efficiency. Gardner 22 11 . Gardner 21 Example: Tension member design Bolted connection Introduction Design Example Exercise 150×75×10 unequal angle 24 mm diameter holes for 22 mm HSFG bolts Gusset plate Try a 150×75×10 unequal angle. 682 kN > 541 kN (i. Nt. 0 The net cross-sectional area Anet: Anet = A – allowance for bolt holes = 2170 – (24×10) = 1930 mm2 L. A = 2170 mm2 (from Section Tables).e. Nu. Gardner 24 12 . β3 = 0. L.10 The tensile resistance Nt.7 × 1930 × 470 = = 577 × 10 3 N = 577 kN γ M2 1.4 × 10 3 N = 770 kN 1 . connected by the longer leg (using four 22 mm diameter HSFG bolts) is therefore acceptable.Rd is taken as the lesser of these two values. Gardner 23 Example: Tension member design Introduction Design Example Exercise From Table. Nt. and is therefore 577 kN.Rd = β 3 A net fu 0. For yielding of the gross cross-section.Rd = Afy γ M0 = 2170 × 355 = 770. plastic resistance is given as: Npl.Rd > NEd) Unequal angle 150×75×10 in grade S355 steel. 577 kN > 541 kN (i.Example: Tension member design Introduction Design Example Exercise Gross area of cross-section.7 (since the pitch p1 > 5d0). 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.Tension member design exercise Introduction Design Example Exercise A flat bar 200 mm wide × 25 mm thick is to be used as a tie (tension member). Assume 22 mm diameter bolt holes. Erection conditions require that the bar be constructed from two lengths connected together with a lap splice using six M20 bolts as shown below. A T 50 mm 100 mm 50 mm A T T T 25 mm thick plates L. Calculate the tensile strength of the bar assuming grade S275 steel. 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 Eurocode 3: Design of steel structures L. Gardner 2 1 . Gardner 1 Overview Introduction Local buckling Classification Class 4 Exercise Outline: • Introduction • Local buckling • Cross-section classification • Class 4 – effective widths L. Gardner Local buckling Introduction Local buckling Classification Class 4 Exercise Local buckling L. the slender nature of these thin elements results in susceptibility to local instabilities (buckling) under compressive stress. 3 L. structural members are generally composed of relatively thin elements (i.Background Introduction Local buckling Classification Class 4 Exercise Background: • For efficiency. which must be considered in design. Gardner Local buckling in structural components 4 2 .e. thicknesses substantially less than other cross-sectional dimensions) • Although favourable in terms of overall structural efficiency. • The classifications from BS 5950 of plastic. Gardner 5 Factors affecting local buckling 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. Class 3 and Class 4. compact. cross-sectional resistance and rotation capacity are limited by the effects of local buckling. semi-compact and slender are replaced in Eurocode 3 with Class 1.Cross-section classification Introduction Local buckling Classification Class 4 Exercise • Whether in the elastic or inelastic material range. L. Class 2. Gardner 6 3 . • Eurocode 3 (and BS 5950) account for the effects of local buckling through cross-section classification. respectively. fy • Fabrication process • Applied stress system L. Gardner 8 4 .2 of EN 1993-1-1). Gardner 7 Definition of 4 classes Introduction Local buckling Eurocode 3 defines four classes of cross-section: Moment Classification Class 4 Exercise Class 1 Mpl Mel Class 2 Class 3 Class 4 Deformation L.Cross-section classification 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. a crosssection with a class 3 flange and class 1 web has an overall classification of Class 3). A plate element is Class 4 (slender) if it fails to meet the limiting values for a class 3 element. given in Table 5. The classification of the overall cross-section is taken as the least favourable of the constituent elements (for example. L. Rd: Class 1. Gardner Wel fy γ M0 10 5 .Rd = Mel = L.Compression Introduction Local buckling Classification Class 4 Exercise Cross-section resistance in compression Nc. Gardner Nc .Rd = Mpl = Wpl fy γ M0 • Class 3 cross-sections: Mc . 2 and 3: Nc .Rd = A eff fy γ M0 Afy γ M0 Class 4: L.Rd = 9 Bending Introduction Local buckling Classification Class 4 Exercise • Class 1 & 2 cross-sections: Mc . Gardner 11 Compressed widths c Introduction Local buckling Rolled Classification Class 4 Exercise c Definition of compressed widths – flat widths: c Rolled c Welded c Welded (a) Outstand flanges (b) Internal compression parts Limits on slenderness e.g.Bending Introduction Local buckling Classification Class 4 Exercise • Class 4 cross-sections: Mc . Gardner ε = 235 / fy 12 6 . c/t ≤ 9ε L.Rd = Weff fy γ M0 L. Gardner 13 Outstand flanges Introduction Local buckling Classification Class 4 Exercise L. Gardner 14 7 .Internal compression parts Introduction Local buckling Classification Class 4 Exercise L. reduced (effective) cross-section properties must be calculated to account explicitly for the occurrence of local buckling prior to yielding. Gardner 16 8 .Angles and tubular sections Introduction Local buckling Classification Class 4 Exercise L.5 (EN 1993-1-5). L. Gardner 15 Class 4 cross-sections Introduction Local buckling Classification Class 4 Exercise Class 4 (slender) cross-sections • For class 4 (slender) cross-sections. • Effective width formulae for individual elements are provided in Eurocode 3 Part 1. b z tw h d y r z y tf h = 254.6 mm tw = 8.Class 4 – effective width concept Introduction Local buckling Classification Class 4 Exercise L. Gardner 18 9 .1 mm b = 254. Gardner 17 Cross-section classification exercise Introduction Local buckling Classification Class 4 Exercise Determine the classification and resistance Nc.7 mm A = 9310 mm2 Section properties for 254 x 254 x 73 UC L.2 mm r = 12.6 mm tf = 14.Rd for a 254 x 254 x 73 UC in pure compression. assuming grade S 355 steel. 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 20 10 . 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 Eurocode 3: Design of steel structures L. Gardner Eurocode 3: Design of steel structures 1 Outline Background Cross-section Buckling Example Exercise Overview: • Background • Cross-section resistance Nc.Session 7 Background Cross-section Buckling Example Exercise Compression members Dr Leroy Gardner Senior Lecturer in Structural Engineering L.Rd • Member buckling resistance Nb. Gardner 2 1 .Rd L. L.Elastic buckling theory N Background Cross-section Buckling N L Example Exercise w x N (a) Unloaded member (b) Loaded member (straight) N (c) Loaded member (displaced) L. the elastic buckling load of a perfect pin-ended column is given by: Ncr = π 2EI L2 Other boundary conditions may be accounted for through the effective (critical) length concept. Gardner 4 2 . Gardner 3 Elastic buckling theory Background Cross-section Buckling Example Exercise From stability theory. Gardner 6 3 . Gardner Non-dimensional slenderness 5 Imperfections Background Cross-section Buckling Example Exercise Forms of imperfection: • • • • Geometric imperfections Eccentricity of loading Residual stresses Non-homogeneity of material properties • End restraint • etc L.Elastic buckling theory Background Cross-section Two bounds: Yielding and buckling Load Buckling Material yielding (squashing) NEd Afy Example Exercise Lcr Euler (critical) buckling Ncr NEd L. d N 1 − Ed MRd Ncr ≤ 8 1 4 .d is the magnitude of the initial imperfection w0 w0 = Initial imperfection w0 w w max = w = additional deflection x 1 e NEd 0.Residual stresses Background Cross-section Buckling Example Exercise Welding Hot-rolling L.d 1− Ncr NEd NRd NEd NRd + NEd w max MRd ≤ 1 NEd L. Gardner 7 Behaviour of imperfect columns Background Cross-section Buckling Example Exercise NEd wmax = (w0+w) at mid-height e0. Gardner + 1 NEde 0. 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). which is. Gardner 9 Perry-Robertson Background Cross-section Buckling Example Exercise Robertson contribution: • The bending stress component is a function of the lateral deflection. in turn. giving five buckling curves L. Gardner 10 5 .Perry-Robertson Background Cross-section Buckling Example Exercise 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 L. Gardner λ 11 Eurocode 3 Background Cross-section Buckling Example Exercise Eurocode 3 states.4 0. as with BS 5950.5 Curve a0 a0 Curve a Curve b Curve c Curve d 2 2.Buckling curves Background 1.Rd L.0 0.Rd Cross-section check NEd ≤ Nb. that both cross-sectional and member resistance must be verified: NEd ≤ Nc .2 Reduction factor χ Cross-section Buckling Example Exercise 1.0 0 0.5 Non-dimensional slenderness L.8 0.2 0.5 1 1.6 0. Gardner Member buckling check 12 6 . Rd = for Class 1.Rd = Nc . Gardner 14 7 .Rd depends on cross-section classification: Afy γ M0 A eff fy γ M0 Nc .Cross-section resistance Background Cross-section Buckling Example Exercise • Cross-section resistance in compression Nc. 2 and 3 Nb.0 in EN 1993 This value will also be adopted in the UK L.Rd = for Class 1. Gardner 13 Member buckling Background Cross-section Buckling Example Exercise Compression buckling resistance Nb.Rd = for (symmetric) Class 4 L.Rd: χ A fy γ M1 χ A eff fy γ M1 Nb. 2 or 3 sections for Class 4 sections γM0 is specified as 1. 2 and 3 λ= 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.Rd = Eurocode 3 Pc = pc A BS 5950 L. Gardner 16 8 .Equivalence to BS 5950 Background Cross-section Buckling Example Exercise Compression buckling resistance: χ A fy γM1 Nb. Gardner 15 Member buckling Background Cross-section Buckling Example Exercise Calculate non-dimensional slenderness λ λ= A fy Ncr A eff fy Ncr for Class 1. Non-dimensional slenderness Background Cross-section Buckling Example Exercise Ncr = π 2EI L2 fcr = π 2EI AL2 = π 2E (L / i)2 = π 2E λ2 where λ = L / i 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 = π 2E λ 12 ⇒ λ1 = π E fy 17 L. L. Gardner Non-dimensional slenderness Background Cross-section Buckling Example Exercise The non-dimensional slenderness used in EC3 is defined as: 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. Gardner 18 9 . χ χ= 1 ϕ + ( ϕ 2 − λ 2 ) 0 .0 L.2) + λ 2 ) α is the imperfection factor L.Non-dimensional slenderness Background N Cross-section Material yielding (in-plane bending) NEd Afy Buckling Example Exercise Lcr Elastic member buckling (LTB) NEd 1. Gardner Non-dimensional slenderness λ 19 Member buckling Background Cross-section Buckling Example Exercise • Calculate reduction factor.5 ≤ 1 ϕ = 0.5 (1 + α(λ − 0. Gardner 20 10 . Gardner 21 Buckling curve selection Background Cross-section Buckling Example Exercise z z z b z Cross-section Limits Buckling about axis Buckling curve S235 S275 S355 S420 a b b c b c d d b c c d S460 tf ≤ 40 mm y–y z-z y–y z-z y–y z-z y–y z-z y–y z-z y–y z-z a0 a0 a a a a c c b c c d h/b > 1.2 40 mm < tf ≤ 100 mm y Rolled Isections tw h y tf ≤ 100 mm r tf h/b ≤ 1.Imperfection factor α Background Cross-section Buckling Example Exercise Imperfection factors α for 5 buckling curves: Buckling curve Imperfection factor α a0 0. Gardner 22 11 .2 tf > 100 mm Welded Isections tf ≤ 40 mm y tf z y y tf y z tf > 40 mm L.49 0.21 0.34 0.76 L.13 a b c d 0. 0 L L.5 L 2. Gardner 23 Effective (buckling) lengths Lcr Background Cross-section Buckling Example Exercise End restraint (in the plane under consideration) Effectively restrained in direction at both ends Effectively held in position at both ends Partially restrained in direction at both ends Restrained in direction at one end Not restrained in direction at either end One end Other end Effectively restrained in direction Not held in position Partially restrained in direction Not restrained in direction Buckling length Lcr 0.5tf b/tf < 30 h/tw < 30 any b b Buckling Example Exercise Welded box sections h y any c c U-.0 L Effectively held in position and restrained in direction 1. Gardner 24 12 .and solid sections any c c L-sections any b b L.7 L 0.85 L 0.2 L 1. T.85 L 1.Buckling curve selection hot finished Background Cross-section z tf y tw b z any any a c Hollow sections a0 c cold formed generally (except as below) thick welds: a > 0. Effective (buckling) lengths Lcr Background Cross-section Buckling Example Exercise Non-sway Sway L. Gardner 26 13 . no account need be made for local buckling) 4. Classify cross-section (if Class 1-3. Select section and determine geometry 3. Calculate Ncr and Afy L. Determine design axial load NEd 2. Determine effective (buckling) length Lcr 5. Gardner 25 Column buckling design procedure Background Cross-section Buckling Example Exercise Design procedure for column buckling: 1. Non-dimensional slenderness λ = 7. L. The column has pinned boundary conditions at each end. Calculate buckling reduction factor χ 9.0 Nb. Design buckling resistance Nb. Gardner 28 14 .Rd χ A fy γ M1 L.Column buckling design procedure Background Cross-section Buckling Example Exercise 6. and the inter-storey height is 4 m.Rd = 10. Gardner 27 Member buckling resistance example Background Cross-section Buckling Example Exercise A circular hollow section member is to be used as an internal column in a multi-storey building. NEd = 2110 kN 4. Check NEd ≤ 1.0 m The critical combination of actions results in a design axial force of 2110 kN. Determine imperfection factor α A fy Ncr 8. Member buckling resistance example Background Cross-section Buckling Example Exercise Assess the suitability of a hot-rolled 244. From clause 3.0 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).5 mm = 10. Gardner 29 Member buckling resistance example Background Cross-section Buckling Example Exercise For a nominal material thickness (t = 10.y = 550000 mm3 I = 50730000 mm4 L.2.y = 415000 mm3 d Wpl. d t = 244.5×10 CHS in grade S 355 steel for this application.0 mm = 7370 mm2 t A Wel. Gardner 30 15 .6: E = 210000 N/mm2 L. 2.2): Background Cross-section Buckling ε= 235 / fy = 235 / 355 = 0.4): Nc .Rd = 7370 × 355 = 2616 × 10 3 N = 2616 kN 1.00 2616 > 2110 kN ∴ Cross − section resistance is OK L. sheet 3) Example Exercise d/t = 244. 2 or 3 cross .5/10.Member buckling resistance example Cross-section classification (clause 5.81 Tubular sections (Table 5. Gardner 32 16 .2.sections ∴ Nc .7 > 24.5.Rd = Afy γ M0 for Class 1.5 Limit for Class 1 section = 50 ε2 = 40. Gardner 31 Member buckling resistance example Background Cross-section Buckling Example Exercise Cross-section compression resistance (clause 6.5 ∴ Cross-section is Class 1 L.0 = 24. 0 where Φ = 0.0. Gardner for Class 1.Member buckling resistance example Background Cross-section Buckling Example Exercise Member buckling resistance in compression (clause 6. 2 & 3 cross .Rd = χ= χ A fy γ M1 1 Φ + Φ 2 − λ2 for Class 1.sections 33 Member buckling resistance example Background Cross-section Buckling Example Exercise Elastic critical force and non-dimensional slenderness for flexural buckling Ncr Ncr = π 2EI π 2 × 210000 × 50730000 = = 6571 kN 2 4000 2 L cr 7370 × 355 = 0.2 of EN 1993-1-1: For a hot-rolled CHS.sections but χ ≤ 1.63 6571 × 10 3 ∴λ = From Table 6.5 1 + α (λ .2 ) + λ2 Afy Ncr [ ] and λ = L.1): Nb . Gardner 34 17 . 2 & 3 cross . use buckling curve a L.3. 0 0 0.Buckling curve selection Background Cross-section Buckling Cross-section Limits Buckling about axis Buckling curve S235 S275 S355 S420 S460 hot finished Example Exercise Hollow sections any any a c a0 c cold formed Extract from Table 6. Gardner λ 36 18 .5 0.5 2 2.5 Non-dimensional slenderness L.88 Curve a0 Curve a0 Curve a Curve b Curve c Curve d 0.63 1 1.6 0.0 ≈0. Gardner 35 Graphical approach 1.4 0.8 0.2 0.2 of EN 1993-1-1: For a hot-rolled CHS.2 Background Cross-section Buckling Example Exercise Reduction factor χ 1. use buckling curve a L. 6 kN 4.88 × 7370 × 355 = 2297 × 10 3 N = 2297 kN 1 .Rd = 2297 > 2110 kN L. L.1 of EN 1993-1-1. The column has pinned boundary conditions at each end. 244.21(0.5 m The critical combination of actions results in a design axial force of 305.6 kN. Gardner 38 19 .74 1 = = 0.Member buckling resistance example Background Cross-section Buckling Example Exercise From Table 6.5 m.63 2 0.5[1 + 0. 37 Member buckling resistance exercise Background Cross-section Buckling Example Exercise A UC section member is to be used as an internal column in a multi-storey building.88 0.74 2 − 0.63 − 0.74 + 0. in grade S 355 steel is acceptable. α = 0. NEd = 305. and the inter-storey height is 4.5x10 CHS.21 Φ χ = 0.63 2 ] = 0. ∴ Nb .0 ∴Buckling resistance is OK.2) + 0. for buckling curve a. Gardner The chosen cross-section. Gardner Eurocode 3: Design of steel structures 40 20 . b z tw h d y r z y tf Exercise h b tw tf r A Iy Iz = 157.6 mm = 152.9 mm = 6. Gardner 39 Session 7 Background Cross-section Buckling Example Exercise Compression members Dr Leroy Gardner Senior Lecturer in Structural Engineering L.4 mm = 7.6 mm = 3830 mm2 = 17480000 mm4 = 5600000 mm4 Section properties for 152x152x30 UC L.Member buckling resistance exercise Background Cross-section Buckling Example Try a 152x152x30 UC in grade S 275 steel.5 mm = 9. Session 8 Background In-plane bending Shear Serviceability LTB Exercises Beams Dr Leroy Gardner Senior Lecturer in Structural Engineering L. Gardner 2 1 . Gardner Eurocode 3: Design of steel structures 1 Outline Background In-plane bending Shear Serviceability LTB Exercises Overview: • Background • In-plane bending • Shear • Deflections • Lateral torsional buckling L. M Wyfy Material yielding (in-plane bending) MEd MEd Elastic member buckling Mcr Lcr 1.Rd Cross-section check (In-plane bending) MEd ≤ Mb . that both cross-sectional and member bending resistance must be verified: M Ed ≤ M c .0 L. as with BS 5950. Gardner Non-dimensional slenderness λ LT 4 2 .Eurocode 3 Background In-plane bending Shear Serviceability LTB Exercises Eurocode 3 states.Rd Member buckling check L. Gardner 3 Non-dimensional slenderness Background In-plane bending Shear Serviceability LTB Exercises Beam behaviour analogous to yielding/buckling of columns. Gardner 6 3 .Rd = M el = L. Gardner Wel fy γ M0 5 Cross-sections in bending Background In-plane bending Shear Serviceability LTB Exercises • Class 4 cross-sections: Mc .Rd = Mpl = Wpl fy γ M0 • Class 3 cross-sections: Mc .Cross-sections in bending Background In-plane bending Shear Serviceability LTB Exercises • Class 1 & 2 cross-sections: Mc .Rd = Weff fy γ M0 L. L. elastic or effective section modulus Plastic modulus Elastic modulus z Wpl Wel (S in BS 5950) (Z in BS 5950) (Zeff in BS 5950) Effective modulus Weff The partial factor γM0 is applied to all crosssection bending resistances. The design shear resistance of a crosssection is denoted by Vc.0.Section moduli W Background In-plane bending Shear Serviceability LTB Exercises Subscripts are used to differentiate between the plastic.0 Vc . VEd ≤ 1.Rd Shear check L. Gardner 8 4 .Rd) or an elastic distribution of shear stress.Rd and may be calculated based on a plastic (Vpl. Gardner 7 Shear resistance Background In-plane bending Shear Serviceability LTB Exercises The design shear force is denoted by VEd (shear force design effect). and equal 1. Rd = A v (fy / 3 ) γ M0 9 L. this is essentially the area of the web (with some allowance for the root radii in rolled sections).Plastic shear resistance Vpl. Gardner 10 5 . L.Rd The plastic shear resistance is essentially defined as the yield strength in shear multiplied by a shear area Av: Vpl.Rd Background In-plane bending Shear Serviceability LTB Exercises The usual approach is to use the plastic shear resistance Vpl. Gardner Shear area Av Background In-plane bending Shear Serviceability LTB Exercises 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. Gardner 12 6 . • Rolled I and H sections. Gardner Av = 2A/π 11 Definition of terms Background In-plane bending Shear Serviceability LTB Exercises 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. load parallel to depth: Exercises Av = Ah/(b+h) • CHS and tubes of uniform thickness: L.2. load parallel to web: Av = A – 2btf + (tw + r)tf • Rolled RHS of uniform thickness.0 (from UK NA) L. load parallel to web: Av = A – 2btf + (tw + 2r)tf but ≥ ηhwtw • Rolled channel sections.Shear areas Av Background In-plane bending Shear Serviceability LTB Shear areas Av are given in clause 6.6(3). Gardner Shear resistance example Background In-plane bending Shear Determine the shear resistance of a rolled channel section 229x89 in grade S 275 steel loaded parallel to the web.3 mm r = 13. Shear buckling need not be considered provided: ε hw ≤ 72 η tw where ε = for unstiffened webs 235 .9 mm tw = 8. though this is unlikely to affect cross-sections of standard hot-rolled proportions.6 mm b = 88.Shear buckling Background In-plane bending Shear Serviceability LTB Exercises The resistance of the web to shear buckling should also be checked.0 (from UK NA) fy 13 L. Gardner Section properties for 229x89 rolled channel section 14 7 . b z tw h y r z tf y h = 228.7 mm A = 4160 mm2 Serviceability LTB Exercises L.6 mm tf = 13. η = 1. 6+13.3) + (8.3 mm and tw = 8.7)×13. 16 8 .00 L.6 Vpl.1 to be 275 N/mm2.Rd = L.Rd = 2092 × (275 / 3 ) = 332000 N = 332 kN 1.9×13.Shear resistance example Background In-plane bending Shear Serviceability LTB Exercises For a nominal material thickness (tf=13. Shear resistance is determined according to clause 6.3 = 2092 mm2 Vpl.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. Gardner A v (fy / 3 ) γ M0 15 Shear resistance example Background In-plane bending Shear Serviceability LTB Exercises Shear area Av For a rolled channel section. loaded parallel to the web. Gardner For the same cross-section BS 5950 (2000) gives a shear resistance of 324 kN.2. the shear area is given by: Av = A – 2btf + (tw + r)tf = 4160 – (2×88. Deflection checks should therefore be performed against suitable limiting values. except portal frames Columns in portal frame buildings. misalignments of crane rails. From the UK National Annex. 17 L.Serviceability Background In-plane bending Shear Serviceability LTB Exercises Excessive serviceability deflections may impair the function of a structure. for example. not supporting crane runways In each storey of a building with more than one storey Deflection limit Height/300 To suit cladding Height of storey/300 L. causing difficulty in opening doors. Gardner 18 9 . deflection checks should be made under unfactored variable actions Qk. etc. Gardner Serviceability Background In-plane bending Shear Serviceability LTB Exercises Vertical deflection limits Design situation Cantilevers Beams carrying plaster or other brittle finish Other beams (except purlins and sheeting rails) Purlins and sheeting rails Deflection limit Length/180 Span/360 Span/200 To suit cladding Horizontal deflection limits Design situation Tops of columns in single storey buildings. leading to cracking of plaster. Gardner 20 10 . then lateral torsional buckling will be prevented and failure will occur in another mode.4 in some cases) L. circular or square bar • Fully laterally restrained beams Exercises • λ LT < 0. 19 L. without continuous lateral restraint. If continuous lateral restraint is provided to the beam. generally in-plane bending (and/or shear).2 (or 0. SHS.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. Gardner Lateral torsional buckling Background In-plane bending Shear Serviceability LTB Can be discounted when: • Minor axis bending • CHS. • The second is a simplified assessment method for beams with restraints in buildings. and is set out in clause 6. Lateral restraint Lateral restraint Lateral restraint Lcr = 1.57.3 (for rolled sections and equivalent welded sections).2.2.4.3.3.4. Gardner 22 11 .Lateral torsional buckling resistance 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).3.0 L Beam on plan L.56 and 6.3. 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. given in clause 6.2. and is set out in clause 6. L.2 (general case) and clause 6. • The third is a general method for lateral and lateral torsional buckling of structural components. y or Wel.Rd = Eurocode 3 Mb = pb Sx (or Zx) Wy will be Wpl.y BS 5950 L.Rd of a laterally unrestrained beam (or segment of beam) should be taken as: Mb. Gardner Reduction factor for LTB Equivalence to BS 5950 Background In-plane bending Shear Serviceability LTB Exercises Lateral torsional buckling resistance: χLT Wy fy γM1 Mb.Rd = χLT Wy fy γ M1 23 L. Gardner 24 12 . The design buckling resistance Mb.Lateral torsional buckling Background In-plane bending Shear Serviceability LTB Exercises Eurocode 3 design approach for lateral torsional buckling is analogous to the column buckling treatment. Gardner 25 Buckling curve selection Background For the general case.5 [ 1 + αLT ( λ LT − 0.3 L.4: In-plane bending Shear Serviceability LTB Exercises Cross-section Rolled I-sections Welded Isections Other crosssections Limits h/b ≤ 2 h/b > 2 h/b ≤ 2 h/b > 2 - Buckling curve a b c d d L. refer to Table 6.2) + λ2 ] LT Plateau length Imperfection factor from Table 6. Gardner 26 13 .Buckling curves – general case Background In-plane bending Shear Serviceability LTB Exercises Lateral torsional buckling curves for the general case are given below: χLT = 1 ΦLT + 2 ΦLT − λ2 LT but χLT ≤ 1.0 Φ LT = 0. 0 0.5 2 2.2 0.76 Imperfection 0.21 0.5 Reduction factor χLT Curve a Curve b Curve c Curve d Shear Serviceability LTB Exercises 0.4 0.6 0.34 0. Gardner 27 LTB curves Background In-plane bending 4 buckling curves for LTB (a.49 factor αLT L.5 1 1. Gardner Non-dimensional slenderness λ LT 28 14 .2 L.Imperfection factor αLT Background In-plane bending Shear Serviceability LTB Exercises Imperfection factors αLT for 4 buckling curves: Buckling curve a b c d 0. c and d) 1.8 0.2 1.0 0 0. b. 4 29 L.5 [ 1 + αLT ( λ LT − λ LT .75 Plateau length Recommended value = 0.5 1 1.40 0.60 0.00 0.5 L.20 0.80 0.20 Reduction factor χLT Shear Serviceability LTB Exercises 1.5 General (h/b>2) Rolled (h/b>2) 2 2. Gardner Non-dimensional slenderness λ LT 30 15 .0 ⎪ 1 but ⎨ χLT ≤ ⎪ λ LT ⎩ Φ LT = 0.0 ) + β λ2 ] LT β factor Recommended value = 0.Buckling curves – rolled or equivalent welded sections case Background In-plane bending Shear Serviceability LTB Exercises LTB curves for the rolled or equivalent welded sections case are given below.00 0 0. Gardner LTB curves Background In-plane bending Comparison between general curves and curves for rolled and equivalent welded sections (I-sections – h/b>2) 1.5 is used to select buckling curve: χLT = 1 2 Φ LT + Φ LT − β λ2 LT ⎧ χLT ≤ 1. Table 6. Such simplifications do not appear in the primary Eurocode method. by inclusion of the geometric quantities ‘u’ and ‘v’ in section tables.Non-dimensional slenderness Background In-plane bending Shear Serviceability LTB Exercises • 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 L. for example. where determination of the elastic critical moment for lateral torsional buckling Mcr is aided. Gardner Elastic critical buckling moment Mcr Background In-plane bending Shear Serviceability LTB Exercises Designers familiar with BS 5950 will be accustomed to simplified calculations. L. Gardner 32 16 . loaded through the shear centre at the level of the centroidal axis.5 L. and with the standard conditions of restraint described.0 G IT Iw Iz Lcr π 2EIz = 2 L cr ⎡ Iw L cr 2GIT ⎤ ⎢ + 2 ⎥ π EIz ⎦ Iz ⎣ 0 . Gardner 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 LTB Exercises Numerical solutions have been calculated for a number of other loading conditions. Mcr may be calculated by: π 2EIz Mcr = C1 2 L cr ⎡ Iw L cr 2GIT ⎤ ⎢ + 2 ⎥ π EIz ⎦ ⎣ Iz 0 . Gardner 34 17 .Mcr under uniform moment Background In-plane bending Shear Serviceability LTB Exercises For typical end conditions. and under uniform moment the elastic critical lateral torsional buckling moment Mcr is: Mcr .5 L. For uniform doubly-symmetric cross-sections. C1 factor – end moments Background In-plane bending Shear Serviceability LTB Exercises For end moment loading C1 may be approximated by the equation below.046 = = L. though other approximations also exist. Gardner 36 18 .365 F 1.565 C L F 1.285 LTB Exercises F = = 1. Gardner 35 C1 factor – transverse loading Background In-plane bending Shear w C1 values for transverse loading Loading and support conditions w 1.52ψ2 but C1 ≤ 2.88 – 1. C1= 1.40ψ + 0.132 Bending moment diagram Value of C1 Serviceability F 1. L.70 where ψ is the ratio of the end moments (defined in the following table). the following simplified expressions result.0. Gardner C1 may be conservatively taken = 1. 37 Simplified assessment of λ LT Background In-plane bending Shear Serviceability LTB Exercises Substituting in numerical values for λ1 . λ 1 = π L.9 λ z = C1 E fy λ 1 0 . though the level of conservatism increases the more the actual bending moment diagram differs from uniform moment. λLTmay be conservatively simplified to: λ LT = 1 0 .0. Gardner As a further simplification. S235 λLT = 1 L / iz C1 104 S275 λLT = 1 L / iz C1 96 S355 λ LT = 1 L / iz C1 85 L.Simplified assessment of λ LT Background In-plane bending Shear Serviceability LTB Exercises For hot-rolled doubly symmetric I and H sections without destabilising loads. 38 19 . C1 may also be conservatively taken = 1.9 z λ1 C1 λ z = L / iz . Determine BMD and SFD from design loads 2.Rd portion L. 3 or 4) LTB Exercises 4. Calculate Mcr and Wyfy L. Check ≤ 1. Determine effective (buckling) length Lcr – depends on boundary conditions and load level 5. Gardner 39 Design procedure for LTB Background In-plane bending Shear Serviceability LTB Exercises 6.Design procedure for LTB Background In-plane bending Shear Serviceability Design procedure for LTB: 1. Classify cross-section (Class 1. Select section and determine geometry 3. 2. Calculate buckling reduction factor χLT 9. Design buckling resistance Mb. Gardner 40 20 .Rd = χLT M Wy fy γ M1 Ed 10.0 for each unrestrained Mb. Determine imperfection factor αLT Wy fy Mcr 8. Non-dimensional slenderness λ LT = 7. 3.4) Background In-plane bending Shear Serviceability LTB Exercises Simplified method for beams with restraints in buildings (Clause 6.2. • Requires determination of plastic and elastic (buckling) resistance of structure.4) Background In-plane bending Shear Serviceability LTB Exercises General method for lateral and lateral torsional buckling of structural components • May be applied to single members. Gardner 42 21 .Simplified method (Cl.4) This method treats the compression flange of the beam and part of the web as a strut: b b Compression h Tension Compression flange + 1/3 of the compressed area of web Strut 41 L. which subsequently defines global slenderness • Generally requires FE L.3. Gardner Beam General method (Cl.3. 6.2. plane frames etc. 6. Check moment resistance.2 mm = 229. shear and deflections.6 mm = 12.Restrained beam exercise Background In-plane bending Shear Serviceability LTB Exercises The simply supported 610×229×125 UB of S275 steel shown below has a span of 6.pl Iy = 612. Dead load = 60 kN/m Imposed load = 70 kN/m 6.9 mm = 19. Gardner 44 22 .0 m. Gardner 43 Restrained beam exercise Background b In-plane bending Shear Serviceability z tw h d y y LTB r Exercises tf h b tw tf r A Wy.1×106 mm4 z 610×229×125 UB L.7 mm = 15900 mm2 = 3676×103 mm3 = 986.0 mm = 11.0 m Beam is fully laterally restrained L. Gardner 46 23 . Gardner 45 LTB Example Background In-plane bending Shear Serviceability LTB Exercises General arrangement L. and full lateral restraint may be assumed at these points.LTB Example Background Description In-plane bending Shear Serviceability LTB Exercises A simply-supported primary beam is required to span 10. Select a suitable member for the primary beam assuming grade S 275 steel.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. L. LTB Example Background In-plane bending Shear Design loading is as follows: 425.1 kN A 319.6 kN D Serviceability LTB Exercises B C 2.5 m 3.2 m 5.1 m Loading L. Gardner 47 LTB Example Background 267.1 kN A B 52.5 kN SF 477.6 kN C D In-plane bending Shear Serviceability LTB Shear force diagram B C D A Exercises BM 1194 kNm 1362 kNm L. Gardner Bending moment diagram 48 24 LTB Example Background In-plane bending Shear Serviceability LTB Exercises 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. L. Gardner 49 LTB Example Background In-plane bending Shear b z tw Serviceability LTB Exercises h d y y r z 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 L. Gardner 25 LTB Example Background In-plane bending Shear Serviceability LTB Exercises 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. L. Gardner 51 LTB Example Background In-plane bending Shear Serviceability Cross-section classification (clause 5.5.2): ε = 235 / fy = 235 / 265 = 0.94 Outstand flanges (Table 5.2, sheet 2) cf = (b – tw – 2r) / 2 = 109.7 mm LTB cf / tf Exercises = 109.7 / 21.6 = 5.08 Limit for Class 1 flange = 9ε = 8.48 > 5.08 ∴ Flange is Class 1 L. Gardner 52 26 0 mm cw / tw= 686. 0 = 1642 kNm > 1362 kNm ∴ Cross-section resistance in bending is OK.3 = 48.5): Mc .0 Limit for Class 1 web = 72 ε = 67.0 / 14.LTB Example Background In-plane bending Shear Serviceability LTB Exercises Web – internal part in bending (Table 5. L. Gardner 53 LTB Example Background In-plane bending Shear Serviceability LTB Exercises Bending resistance of cross-section (clause 6. Gardner 54 27 .y fy γ M0 for Class 1 and 2 sec tions 6198 × 10 3 × 265 = 1642 × 10 6 Nmm 1.y.2.0 ∴ Web is Class 1 Overall cross-section classification is therefore Class 1.Rd = = Wpl. sheet 1) cw = h – 2tf – 2r = 686.2.8 > 48. L. 88 – 1.3.Rd = χ LT Wy fy γ M1 where Wy = Wpl.5 55 LTB Example Background In-plane bending Shear For end moment loading C1 may be approximated from: C1 = 1.52ψ2 but C1 ≤ 2.5 × 10 0 . Gardner π 2EIz Mcr = C1 2 L cr ⎡ Iw L cr 2GIT ⎤ ⎥ ⎢ + 2 π EIz ⎦ ⎣ Iz 0 .40ψ + 0.70 ψ is the ratio of the end moments = Serviceability LTB Exercises 1194 = 0.88 1362 ⇒ C1 = 1.5 = 5699x106 Nmm = 5699 kNm L.5 × 10 6 ⎦ ⎣ 68.5 × 10 6 3200 2 ⎡ 9390 × 10 9 3200 2 × 81000 × 2670 × 10 3 ⎤ + ⎢ ⎥ 6 π 2 × 210000 × 68. Gardner 56 28 .2.05 × π 2 × 210000 × 68.05 Mcr = 1.LTB Example Background In-plane bending Shear Serviceability LTB Exercises Lateral torsional buckling check (clause 6.2) – Segment BC: MEd = 1362 kNm Mb .y for Class 1 and 2 sections Determine Mcr for segment BC (Lcr = 3200 mm) L. 54 5699 × 106 Select buckling curve and imperfection factor αLT: From Table 6.3 of EN 1993-1-1: For buckling curve b.2) + λ2 ] LT L.0 where Φ LT = 0. αLT = 0.2/266. Gardner 58 29 .85 For a rolled I-section with h/b > 2.5 [ 1 + α LT ( λ LT − 0.7 = 2. use buckling curve b L. χLT – Segment BC: χ LT = 1 2 Φ LT + Φ LT − λ2 LT but χ LT ≤ 1.4: h/b = 762.LTB Example Background In-plane bending Shear Serviceability LTB Exercises Non-dimensional lateral torsional slenderness for segment BC: λLT = Wy fy Mcr = 6198 × 103 × 265 = 0.34 Calculate reduction factor for lateral torsional buckling. Gardner 57 LTB Example Background In-plane bending Shear Serviceability LTB Exercises From Table 6. Rd 1425 Lateral torsional buckling check (clause 6.2) – Segment CD: MEd = 1362 kNm Mb .0 ∴ Segment BC is OK Mb .3.LTB Example Background In-plane bending Shear Serviceability LTB Exercises ΦLT = 0.2.5[1+0.87 Lateral torsional buckling resistance Mb.Rd – Segment BC: Mb.0 = χ LT Wy = 0.87 × 6198 × 10 3 × = 1425 × 10 6 Nmm = 1425 kNm L.702 − 0.70 ∴ χLT = 1 0.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.96 ≤ 1.Rd fy γ M1 265 1 .54-0. Gardner 59 LTB Example Background In-plane bending Shear Serviceability LTB Exercises MEd 1362 = = 0.54 2 = 0.542] = 0.2) + 0.34(0.70 + 0. Gardner 60 30 . Gardner 62 31 .5 × 10 6 ⎡ 9390 × 10 9 5100 2 × 81000 × 2670 × 10 3 ⎤ + ⎢ ⎥ 6 5100 2 π 2 × 210000 × 68.62 4311× 106 λLT = = The buckling curve and imperfection factor αLT are as for segment BC.5 Determine ψ from Table: ψ is the ratio of the end moments = ⇒ C1 = 1.88 Mcr = 1.LTB Example Background In-plane bending Shear Serviceability LTB Exercises π 2EIz Mcr = C1 2 L cr ⎡ Iw L cr 2GIT ⎤ ⎥ ⎢ + 2 π EIz ⎦ ⎣ Iz 0 .5 × 10 0. 5 0 =0 1362 L.88 π 2 × 210000 × 68.5 × 10 6 ⎦ ⎣ 68. Gardner = 4311×106 Nmm = 4311 kNm 61 LTB Example Background In-plane bending Shear Serviceability LTB Exercises Non-dimensional lateral torsional slenderness for segment CD: Wy fy Mcr 6198 × 103 × 265 = 0. L. 62-0.5 [ 1 + α LT ( λ LT − 0.5[1+0.76 ∴ χ LT = 1 0. Gardner LTB Example Background In-plane bending Shear Serviceability LTB Exercises Lateral torsional buckling resistance Mb.76 + 0.2) + λ2 ] LT = 0.83 63 L.0 where Φ LT = 0.2) + 0.Rd – Segment CD: Mb.622] = 0.62 2 = 0.34(0.Rd = χ LT Wy fy γ M1 = 0. Gardner 64 32 .00 Mb .83 × 6198 × 10 3 × 265 1 .LTB Example Background In-plane bending Shear Serviceability LTB Exercises Calculate reduction factor for lateral torsional buckling.0 = 1360 × 10 6 Nmm = 1360 kNm MEd 1362 = = 1.76 2 − 0. χLT – Segment CD: χ LT = 1 2 Φ LT + Φ LT − λ2 LT but χ LT ≤ 1. L.Rd 1360 Segment CD is critical and marginally fails LTB check. Session 8 Background In-plane bending Shear Serviceability LTB Exercises Beams Dr Leroy Gardner Senior Lecturer in Structural Engineering L. Gardner Eurocode 3: Design of steel structures 65 33 . Gardner Eurocode 3: Design of steel structures 1 Introduction Introduction Cross-section Members Annex A & B Simple construction Beam-columns: • Cross-section check • Member buckling check L.Session 9 Introduction Cross-section Members Annex A & B Simple construction Beam-columns Dr Leroy Gardner Senior Lecturer in Structural Engineering L. Gardner 2 1 . Clause 6. Gardner 3 Beam-columns – member checks Introduction Cross-section Two philosophies: • Interaction method . as below: My.Cross-section checks Introduction Cross-section Members Annex A & B Simple construction Cross-section checks similar to BS 5950.Rd Mz.Clause 6. L. Gardner 4 2 .Ed NEd + + ≤ 1 NRd My.3.Rd More sophisticated expressions are also provided for Class 1 and 2 for greater efficiency.3 Members Annex A & B Simple construction • Interaction ‘k’ factors from Annex A or B. • General method .Ed Mz.4 • Not for hand calculations (requires FE or similar) L. including a simplified linear interaction.3. z .so far as it is possible • Correct by calibration for plasticity etc.z .62 must be examined and satisfied: My. both Eqs.Simple construction Introduction Cross-section Members Annex A & B Simple construction In general.y. 6.61 Eq. 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 .Ed ≤ 1 Nb .Rd My.with FE and test results Annex B (Method 2) – German-Austrian • Derived all coefficients from FE .Rd Mc . 6.Ed M NEd + k yy + k yz z .Rd Mb .Rd Eq.Rd Mc . Gardner 6 3 .Ed ≤ 1 Nb .Rd Mb .61 and 6. 6.Ed M NEd + k zy + k zz z .62 L.z . .Calibrated with test results L. 7 L.simple construction Introduction Cross-section Members Annex A & B Simple construction ‘Simple construction’ is commonly used for the design of multi-storey buildings (particularly in the UK).Good news . Gardner 8 4 . • Beams are designed as simply supported • Columns are designed for nominal moments arising from the eccentricity at the beam-tocolumn connection. Gardner Simple construction Introduction Cross-section Members Annex A & B Simple construction Multi-storey frame: Wk & NHF Wk & NHF Wk & NHF Wk & NHF Gkr & Qkr Gkf & Qkf Gkf & Qkf Gkf & Qkf L. Rd Mc .61 Eq.Rd Mb . the axial load) of both expressions (Eq. Therefore.Ed ≤ 1 Nb .Ed ≤ 1 Nb . Eq.Rd Mc .z.Ed M NEd + k zy + k zz z .e.Rd > Nb. Gardner Eq.62 will always govern.Rd Mb . 6.Rd My. both Eqs. Gardner 9 NCCI Simplification Introduction Cross-section Members Annex A & B Simple construction For columns in simple construction. Iy > Iz (usually around 3 times greater). 6.62) dominates. 6.y.z .Ed M NEd + k zy + k zz z .z . so Nb.y. 6.Ed M NEd + k yy + k yz z .z . 6.Rd Eq.Rd (greater difference for higher slenderness).Ed ≤ 1 Nb .Rd L.Rd Mb . For UC sections.z .Simple construction Introduction Cross-section Members Annex A & B Simple construction In general. My.61 and 6. the first term (i.62 L. 6.Rd Mc . for practical simple construction situations and UC sections.z .62 must be examined and satisfied: My.61 and 6.62 10 5 . 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. resulting in: kzy = 1.5 My.Rd Mb .5 z .NCCI Simplification Introduction Cross-section Members Annex A & B Simple construction Given that the moment components are small for simple construction.3.Ed ≤ 1 Nb . the interaction factors can be conservatively simplified without any significant overall loss of efficiency. 6. use: . Gardner Eq.z .62 11 Recommendations Introduction Cross-section Members Annex A & B Simple construction Recommendations: • For pencil and paper calculations.0 + 1.Rd Mc .Ed M NEd + 1 .Rd L.Clause 6.z . L. Gardner 12 6 .0 and kzz = 1. Gardner Eurocode 3: Design of steel structures 13 7 .Session 9 Introduction Cross-section Members Annex A & B Simple construction Beam-columns Dr Leroy Gardner Senior Lecturer in Structural Engineering L. Gardner 2 1 .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 Welded joints Overview: • Introduction • Bolted joints • Welded joints L. the coverage of Part 1. L. Gardner 3 EN 1993-1-8 Introduction Bolted joints Welded joints Essentially. the resistance of fillet welds etc.EN 1993-1-8 Introduction Bolted joints Welded joints Part 1.1. L. The role of connections in overall frame design (Section 5 of Part 1.8) covering the basic strength of bolts in shear.8). covering the various possible approaches to joint classification and global frame analysis. 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.8 focuses on 4 topics: 1. Gardner 4 2 .8 of Eurocode 3 is some 50% longer than the general Part 1. 2. Fasteners (Sections 3 and 4 of Part 1. 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 Bolted joints Welded joints Fv,Rd = 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 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 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 = L. Gardner k 1αb fudt γ M2 9 Bolted joints Definitions of terms: Introduction Bolted joints Welded joints 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 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. L. Gardner 10 5 3 of EN 1993-1-8.Rd L.0 11 Bolted joints Introduction Bolted joints Welded joints Spacing requirements Minimum bolt spacings and edge and end distances are as below.Rd ≤ 1.4 Ft . In general. EN 1993-1-8 provides the following interaction expression to deal with such cases: Fv.2d0 L. • Minimum spacing of bolts in the direction of load transfer p1 = 2.Ed Fv. where d0 is the fastener (bolt) hole diameter. bolt capacities would be expected to reduce when high values of shear and tension are coexistent. Gardner 12 6 .Bolted joints Combined tension and shear Introduction Bolted joints Welded joints In some situations.2d0 • Minimum end distance in the direction of load transfer e1 = 1. These values are defined in Table 3.Ed 1. Gardner + Ft . bolts may experience tension and shear in combination. Gardner 13 Bolted joint example Introduction Bolted joints Welded joints Description Calculate the strength of the bolts in the lap splice shown below assuming the use of M20 Grade 4.2d0 L.6 bolts in 22 mm clearance holes and Grade S275 plate. L. Gardner 14 7 .4d0 • Minimum edge distance perpendicular to the direction of load transfer e2 = 1.Bolted joints Introduction Bolted joints Welded joints • Minimum spacing of bolts perpendicular to the direction of load transfer p2 = 2. A = As = 245 mm2.Bolted joint example Shear resistance: Introduction Bolted joints Welded joints Bolts are in single shear.Rd: Fb . Gardner α v fub A 0.25 Shear resistance per bolt Fv. e2 = 40 mm.Rd = L.Rd = k 1α b fu dt γ M2 From geometry: p1 = 60 mm. e1 = 40 mm. and it is assumed that the shear plane passes through the threaded portion of the bolts: αv = 0. fu of plate (Grade S275. From EN 10025-2.0 kN γ M2 1.6. Gardner 16 8 .25 15 Bolted joint example Bearing resistance: Introduction Bolted joints Welded joints Bearing resistance per bolt Fb. d0 = 22 mm.Rd: Fv . t > 3 mm) = 410 N/mm2.6 × 400 × 245 = = 47040 N = 47. fub = 400 N/mm2. γM2= 1. L. 66 × 410 × 20 × 16 = = 173 . for inner bolt.98 3.7) or 2.25) = 0.61 × 410 × 20 × 16 = = 160 .Bolted joint example Introduction Bolted joints Welded joints For end bolts. the resistance of the tension splice as governed by the shear resistance of the bolts = 3 × 47.7) = fub/fu = 400/410 = 0.66 L. Fb .61.25 And. k1 is the smaller of (2. for end bolts.8 2 − 1.5 × 0. Gardner 18 9 . Therefore.0 = 141 kN.61 3d0 p1 = (60/66 – 0.5 d0 (2. and for inner bolts αb = 0. αd = For inner bolts.2 kN γ M2 1.66 3d0 e For edge bolts.0 For end bolts αb = 0.Rd = k 1α b fudt 2.5 αb is the smaller of: αd.Rd = k 1α b fudt 2. L. Gardner 17 Bolted joint example Therefore. αd = e1 = (40/66) = 0.25 Clearly the resistance of the joint is controlled by the strength in shear.4.5 × 0. fu/fub or 1.8×(40/22) – 1. Introduction Bolted joints Welded joints Fb . ∴ k1 = 2.1 kN γ M2 1. directional method . L.simplified method L. fy in tension or compression or fy/ 3 in shear) provided that suitable electrodes are used. reduced by 3 mm for most partialpenetration butt welds.Welded joints Introduction Bolted joints Welded joints Design of welded joints: • Butt welds • Fillet welds . Gardner 20 10 .e. Gardner 19 Welded joints Introduction Bolted joints Welded joints Butt welds Strength of butt weld taken as that of parent metal (i. Throat thickness taken as minimum depth of penetration. Gardner 22 11 .Welded joints Introduction Bolted joints Welded joints Two methods are permitted for the design of fillet welds: • the directional method.Ed ≤ Fw.Rd Bolted joints Welded joints Fw.Ed is the design value of the weld force per unit length Fw.d a L. in which the forces transmitted by a unit length of weld are resolved into parallel and perpendicular components. Gardner 21 Welded joints Simplified approach Introduction Check Fw. These approaches broadly mirror those used in BS5950: Part 1.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 . in which only longitudinal shear is considered.Rd = fvw . • the simplified method. L. Welded joints fvw,d is the design shear strength of the weld Introduction Bolted joints Welded joints a is the throat thickness of the weld fvw ,d = 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 23 Welded joints Introduction Bolted joints Welded joints Values for correlation factor βw Steel grade S235 Thickness range (mm) t≤3 3 < t ≤ 100 S275 t≤3 3 < t ≤ 100 S355 t≤3 3 < t ≤ 100 Ultimate strength fu (N/mm2) 360 360 430 410 510 470 Correlation factor βw 0.80 0.85 0.90 L. Gardner 24 12 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. L. Gardner 25 Welded joint example 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 fvw ,d = 410 = 223 N / mm2 0.85 × 1.25 × 3 26 L. Gardner 13 Welded joint example 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. L. Gardner 27 Session 10 Introduction Bolted joints Welded joints Joints Dr Leroy Gardner Senior Lecturer in Structural Engineering L. Gardner Eurocode 3: Design of steel structures 28 14 National Strategy for Implementation of the Structural Eurocodes (IStructE. they are of vital importance to both the design and construction sectors of the Civil and Building Industries’. European Union website L.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’. 2004) L. Gardner 30 15 . Gardner 29 Conclusions Introduction Bolted joints Welded joints • ‘The Eurocodes will become the Europe wide means of designing Civil and Structural engineering works and so. Gardner 32 16 . Gardner 31 Eurocode 3 Introduction Bolted joints Welded joints Thank you Dr Leroy Gardner Senior Lecturer in Structural Engineering Eurocode 3: Design of steel structures L.Conclusions Introduction Bolted joints Welded joints 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 L. Table 1: Values for yield strength fy and ultimate strength fu (from EN 10025-2) Yield strength 2 fy (N/mm ) t ≤ 16 mm S235 S275 S355 S450 235 275 355 450 Yield strength fy (N/mm2) 16 < t ≤ 40 mm 235 265 345 430 Ultimate strength fu (N/mm2) 3 < t ≤ 100 mm 360 410 470 550 Steel grade 1 . Table 2 (sheet 1): Maximum width-to-thickness ratios for compression parts (Table 5.2 of EN 1993-1-1) 2 . 2 of EN 1993-1-1) 3 .Table 2 (sheet 2): Maximum width-to-thickness ratios for compression parts (Table 5. 2 tf ≤ 100 mm tf > 100 mm tf ≤ 40 mm y tf tf > 40 mm hot finished any a a0 cold formed z any c c tf y tw Welded box sections generally (except as below) any b b h y b z thick welds: a > 0. T.2 b z tf ≤ 40 mm y–y z-z y–y z-z y–y z-z y–y z-z y–y z-z y–y z-z a0 a0 a a a a c c b c c d tw h y y 40 mm < tf ≤ 100 mm r z z Welded Isections y z Hollow sections y tf z y z tf h/b ≤ 1.2 of EN 1993-1-1) Buckling curve Cross-section Limits Buckling about axis S 235 S 275 S 355 S 420 a b b c b c d d b c c d S 460 Rolled I-sections h/b > 1.Table 3: Selection of buckling curve for a cross-section (Table 6.and solid sections any c c L-sections any b b 4 .5tf b/tf < 30 h/tw < 30 any c c U-. 21 b 0.5 1 1.4 of EN 1993-1-1) 5 .0 0 0.2 0.8 Curve b Curve c 0.13 a 0.1 buckling curves (Figure 6.5 Non-dimensional slenderness λ Figure 1: Eurocode 3 Part 1.2 1.1 of EN 1993-1-1) Buckling curve Imperfection factor α a0 0.49 d 0.76 1.6 Curve d 0.Table 4: Imperfection factors for buckling curves (Table 6.5 2 2.4 0.34 c 0.0 Curve a0 a0 Curve a Reduction factor χ 0.


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