Guide to the major amendments in BS 5950-1:2000 (SCI publication)

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P304: Guide to the major amendments in BS 5950-1:2000 Discuss me ...

SCI PUBLICATION P304

Created on 30 March 2011 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

Guide to the Major Amendments in BS 5950-1:2000

M Heywood

MEng PhD CEng MICE

Published by: The Steel Construction Institute Silwood Park Ascot Berkshire SL5 7QN Tel: Fax:

01344 623345 01344 622944

P304: Guide to the major amendments in BS 5950-1:2000


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 2001 The Steel Construction Institute Apart from any fair dealing for the purposes of research or private study or criticism or review, as permitted under the Copyright Designs and Patents Act, 1988, this publication may not be reproduced, stored or transmitted, in any form or by any means, without the prior permission in writing of the publishers, or in the case of reprographic reproduction only in accordance with the terms of the licences issued by the UK Copyright Licensing Agency, or in accordance with the terms of licences issued by the appropriate Reproduction Rights Organisation outside the UK. Enquiries concerning reproduction outside the terms stated here should be sent to the publishers, The Steel Construction Institute, at the address given on the title page. Although care has been taken to ensure, to the best of our knowledge, that all data and information contained herein are accurate to the extent that they relate to either matters of fact or accepted practice or matters of opinion at the time of publication, The Steel Construction Institute, the authors and the reviewers assume no responsibility for any errors in or misinterpretations of such data and/or information or any loss or damage arising from or related to their use. Publications supplied to the Members of the Institute at a discount are not for resale by them. Publication Number: SCI P304 ISBN 1 85942 131 8 British Library Cataloguing-in-Publication Data. A catalogue record for this book is available from the British Library.

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FOREWORD The design Standard for structural steelwork, BS 5950-1, is used in the design calculations for the majority of constructional steelwork in the United Kingdom. Structural engineers are very familiar with the 1990 issue of this Code and have used it routinely to design safe and efficient structures in a fast and cost-effective manner. BS 5950-1 has been amended recently and there has been genuine concern within the industry that unfamiliarity with the revised code could result in poor productivity, increased costs and design errors, seriously damaging the competitiveness of the construction industry. BS 5950-1:2000, as the revised code is known, came into effect on 15 August 2001.


The purpose of this publication is to ease the transition from BS 5950-1:1990 to BS 5950-1:2000, by highlighting Clauses that have undergone significant technical changes and explaining how these changes will affect the design of steel building structures. This should enable designers to adopt the new Code quickly and with the minimum of disruption, thereby minimising the potential cost to the industry resulting from reduced productivity. This guide is not intended as an in-depth commentary to BS 5950-1, as it deals only with the major changes, however it does cover all the changes with important safety implications, allowing structural engineers to continue to use BS 5950-1 with confidence. This publication was written by Dr Martin Heywood of The Steel Construction Institute, with contributions to the worked examples from the late Mr Paul Salter, Mr Abdul Malik, Mr David Brown and Mr Charles King. Funding for the preparation of this guide was gratefully received from the Department of the Environment, Transport and the Regions (DETR) and Corus.

The Steel Construction Institute has produced a comprehensive guide to the amendments in BS 5950-1:2000. Available on CD, the guide contains a Clause-byClause comparison of the 1990 and 2000 editions, a description of all changes, interactive design paths, worked examples, a keyword search facility, and the facility to print a paper copy of the Standard.

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Contents


Page No. FOREWORD

iii

SUMMARY

vii

1

INTRODUCTION 1.1 Background 1.2 Scope of this publication 1.3 Summary of the changes

1 1 1 1

2

LIMIT 2.1 2.2 2.3 2.4

4 4 4 7 8

3

PROPERTIES OF MATERIALS AND SECTION PROPERTIES 3.1 Grades of steel 3.2 Section classification 3.3 Effective plastic modulus 3.4 Slender cross sections

10 10 10 11 11

4

DESIGN OF STRUCTURAL MEMBERS 4.1 Members subject to bending 4.2 Lateral-torsional buckling 4.3 Plate girder webs 4.4 Design of stiffeners 4.5 Tension members 4.6 Compression members 4.7 Combined moment and axial force 4.8 Column bases

13 13 15 20 24 27 28 30 32

5

CONTINUOUS STRUCTURES 5.1 Column bases 5.2 Frame stability 5.3 Portal frames 5.4 Multi-storey frames

35 35 35 37 39

6

CONNECTIONS 6.1 Bolted connections 6.2 Pin connections 6.3 Welded connections

42 42 47 48

7

REFERENCES

51

STATES DESIGN Load factors Stability Brittle fracture Structural integrity

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WORKED EXAMPLES Sway stability Choosing a steel sub-grade Restrained beam Unrestrained beam Plate grider Web bearing and buckling Compression member Axial load and bending Baseplate

53 55 57 59 63 67 71 75 77 83

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SUMMARY BS 5950 Structural use of steelwork in building, Part 1: Code of practice for design Rolled and welded sections has undergone major amendment. Almost every Clause of this widely used Standard has changed in some way; some of the changes are technical in nature, others are editorial and do not alter the recommendations for building design. The revised Standard, referred to as BS 5950-1:2000, became effective on 15 August 2001.


The aim of this publication is to ease the transition to BS 5950-1:2000 by guiding designers through the major technical amendments to the Standard. A short description of each important change is provided, and simple worked examples illustrate the revised design procedures. The major amendments to BS 5950-1 include revised rules for checking the stability of frames, changes to the method for selecting an appropriate steel sub-grade and the introduction of the effective-area method for class 4 slender sections. On the subject of member design, the greatest change relates to lateral-torsional buckling, where the n-factor method has been removed. Changes have also been made to the Clauses on shear buckling, stiffener design, tension members, compression members, combined axial load and bending, and the design of column bases. Elsewhere, important changes have been made to the rules relating to the in-plane stability of portal frames, prying forces and the transverse strength of fillet welds. Recueil des principales modifications à la norme BS 5950-1 Résumé La norme BS 5950 Usage structural des constructions en acier dans les bâtiments, Partie 1: Code de pratique pour le dimensionnement - Sections laminées et soudées, a subi d’importantes modifications. Pratiquement tous les articles de cette norme, fort utilisée en pratique, ont subi des modifications; certaines ont un simple caractère éditorial mais d’autres, par contre, apportent des modifications techniques importantes. La norme révisée, référence BS 5950-1:2000, est devenue d’application le 15 août 2001. Le but de cette publication est de faciliter la transition vers la nouvelle norme en guidant les utilisateurs à travers les modifications majeures. Une courte description de chaque changement important est donnée; des exemples simples illustrent les procédures de dimensionnement révisées. Les modifications majeures incluent la vérification de la stabilité des portiques, la méthode de sélection des nuances d’acier appropriées et l’introduction d’une méthode d’aire effective pour les sections de classe 4. Concernant le dimensionnement des éléments, la modification la plus importante a trait au déversement, où la méthode du coefficient n a été supprimée. Des modifications sont aussi apportées aux articles traitant du voilement par cisaillement, du dimensionnement des raidisseurs, des éléments en traction, en compression et en combinaison charge axiale-flexion ainsi que du dimensionnement des pieds de poteaux. D’autre part, des modifications ont aussi été apportées aux règles relatives à la stabilité dans leur plan des portiques, aux forces de levier et à la résistance transversale des soudures d’angles.

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Leitfaden für die wichtigsten Änderungen in BS 5950-1 Zusammenfassung BS 5950 Stahlbauten, Teil 1: Vorschrift zur Berechnung - Gewalzte und geschweißte Querschnitte hat sich wichtigen Änderungen unterzogen. Fast jeder Satz dieser allgemein verwendeten Vorschrift hat sich in gewisser Weise geändert; manche Änderungen sind technischer Art, andere sind redaktioneller Art und ändern die Empfehlungen für die Berechnung nicht. Die überarbeitete Norm, jetzt mit BS 5950-1:2000 bezeichnet, ist seit 15 August 2001 gültig.


Das Ziel dieser Publikation ist es, den Übergang zu BS 5950-1:2000 zu erleichtern, indem der Ingenieur durch die wichtigsten Änderungen geführt wird. Eine kurze Beschreibung jeder wichtigen Änderung ist enthalten, und einfache Berechnungsbeispiele illustrieren die überarbeiteten Berechnungsverfahren. Die wichtigen Änderungen in BS 5950-1 beinhalten überarbeitete Regeln zur Prüfung der Stabilität von Tragwerken, Änderungen zur Auswahl einer geeigneten Stahlgüte und die Einführung der Methode der wirksamen Fläche bei schlanken Querschnitten der Klasse 4. Bezüglich der Bauteilberechnung ergibt sich die größte Änderung beim Biegedrillknicken, hier wurde die n-Faktor Methode gestrichen. Änderungen gibt es auch bei folgenden Themen: Schubbeulen, Berechnung von Steifen, Zug- und Druckglieder, Normalkraft und Biegung, Stützenfüße. An anderer Stelle wurden wichtige Änderungen vorgenommen beim Stabilitätsverhalten in Tragwerksebene von Portalrahmen, bei Stützkräften und bei der Festigkeit von Kehlnähten in Querrichtung. Guida alle principali modifiche della BS 5950-1 Sommario La norma BS 5950 Carpenteria strutturale negli edifici, Parte 1: regole progettuali Sezione laminate e saldate ha subito un importante aggiornamento. Quasi ogni punto di questa norma, ampiamente diffusa ed utilizzata, è stato in qualche modo variato. Alcune di queste modifiche sono di natura tecnica, altre editoriale e non alterano le raccomandazioni relative al progetto degli edifici. La norma revisionata, denominata BS 5950-1:2000, è entrata in vigore il 15 Agosto 2001. Scopo di questa pubblicazione è facilitare il passaggio alla nuova BS 5950-1:2000, guidando i progettisti attraverso le principali modifiche tecniche che sono state effettuate. Viene fornita una breve descrizione di ogni variazione rilevante, e semplici esempi applicativi illustrano le procedure di progettazione aggiornate. I principali emendamenti alla BS 5950-1 includono una revisione delle regole di verifica della stabilità dei telai, del metodo di selezione dell’idonea classe di acciaio e l’introduzione del metodo dell’area efficace per le sezioni snelle della classe 4. Per quanto concerne la progettazione degli elementi, le maggiori variazioni si riferiscono instabilità flesso-torsionale, ove è stato eliminato il metodo del fattore-n. Sono state apportate modifiche anche ai punti relativi all’imbozzamento da taglio, alla progettazione degli irrigidimenti, agli elementi tesi ed a quelli compressi, alla combinazione di azione assiale e flettente e al progetto delle basi delle colonne. In altre parti, sostanziali modifiche sono state effettuate alle regole riguardanti la stabilità nel piano dei portali, le forze di contatto e la resistenza trasversale delle saldature a cordone d’angolo.

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Guia de correcciones mas importantes a la BS 5950-1 Resumen La norma BS5950 Uso de acero estructural en edificios, Parte 1: Reglas de buena práctica para el proyecto Perfiles Laminares y soldados ha sufrido importantes correcciones. Casi cada Cláusula de esta popular Norma ha cambiado de algún modo en el aspecto técnico aunque también hay cambios en el aspecto editorial que no afecta las recomendaciones de proyecto. La Norma Revisada, titulada BS 5950-1:2000, se hizo efectiva el 15 de agosto de 2001.


El propósito de esta publicación es facilitar la transición guiando a los proyectistas a través de las correcciones más importantes de la Norma. Se da una breve descripción de cada cambio importante y los procedimientos revisados de proyecto se ilustran desarrollando ejemplos sencillos. Las correcciones a la BS 5950-1 incluyen reglas de comprobación de la estabilidad de pórticos, cambios al método de selección de una subclase de acero adecuada al problema y la introducción del método del área efectiva para las secciones esbeltas de clase 4. Sobre el tema de proyecto de piezas el cambio mayor se refiere al pandeo lateral y de torsión donde se ha eliminado el método del factor n. También se han hecho cambios en las cláusulas sobre la abolladura por cortante, proyecto de rigidizadores, piezas a tracción y a compresión, flexión combinada con axil y proyecto de basas de columnas. Además se han llevado a cabo importantes cambios en las reglas relativas al pandeo en su plano de pórticos fuerzas de retracción y resistencia transversal de cordones de soldadura.

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1

INTRODUCTION

1.1

Background


Since its introduction in 1985, BS 5950 Structural use of steelwork in building, has gradually (although not completely) replaced BS 449 in the design office and is now the established Standard for the design of steel-framed buildings in the UK and several other countries. Part 1 of BS 5950 (referred to here as BS 5950-1) is the Code of practice for design using rolled and welded sections. It gives recommendations for the safe design of general building structures, including the specification of the appropriate steel sub-grade, the classification of sections, design for stability, the design of members subject to bending, tension and compression, stiffener design and the design of column bases and connections. BS 5950-1 has undergone a major amendment, affecting the majority of the Clauses to some extent. The revised Standard, BS 5950-1:2000[1], became effective on 15 August 2001. Because this Standard is so widely used for the design of structural steelwork, it is hardly surprising that news of this amendment was greeted with some trepidation among designers, many of whom are very familiar with the recommendations of its predecessor BS 5950-1: 1990[2]. The Steel Construction Institute recognised that some guidance was required during the period of transition, as designers familiarise themselves with the content and layout of the amended Standard. This publication provides a concise guide to the changes, together with advice on the implementation of the revised Clauses.

1.2

Scope of this publication

The purpose of this publication is to guide designers through the major technical amendments to BS 5950-1, by means of a short description of each important change and simple worked examples. It is not a commentary to BS 5950-1: 2000 and does not, therefore, attempt to give the theoretical background to the Clauses or any justification for the amendments. This publication is limited to those Clauses from Sections 2 to 6 of BS 5950-1:2000 that have undergone a major or significant technical amendment. Section 7 (which deals with testing) and the Annexes are beyond the scope of this publication, as are the numerous minor changes. For convenience, guidance on the modified Clauses has been grouped together into five sections, with numbered headings matching those used in BS 5950-1:2000. The numbered sub-sections do not correspond to the Sub-sections in BS 5950-1:2000, but Clause numbers are stated in all cases.

1.3

Summary of the changes

The 2000 amendment to BS 5950-1 has affected almost every Clause in the Standard to some extent, even though many of the changes are only editorial in nature (i.e. the technical recommendations are unchanged). Users of BS 5950-1:2000 will notice immediately that the familiar two-column format has been replaced with full-width pages, giving the impression that this is a completely new document. This impression is reinforced by the renumbering of many of the Clauses and the extensive re-drafting of much of the text. 1


Fortunately, this impression is deceptive and many of the technical requirements are completely unchanged. Even where changes have been made to the values and equations in the Standard, the majority of design procedures are the same as in BS 5950-1:1990. However, there have also been a number of significant technical changes and designers will need to familiarise themselves with several new methods of design.


One of the most important changes is the extension of the scope of BS 5950-1 to include cold formed structural hollow sections. The Steel Construction Institute first recommended that BS 5950-1 could be used for design using cold formed structural hollow sections in Advisory Desk article AD185[3] and offered advice on how such sections could be designed using a Standard written principally for hot rolled steel. These recommendations have now been incorporated into BS 5950-1. However, designers must note that the inclusion of cold formed structural hollow sections in BS 5950-1 does not mean that they can be used in direct substitution for a similar-sized hot finished member, because there are important differences between the two types of section in terms of section properties and residual stresses. Designers wishing to substitute cold formed for hot finished structural hollow sections must redesign the members using the appropriate strut curves, d/t limits (for section classification) and section properties. Other types of cold formed section should still be designed according to BS 5950-5[4]. Within Section 2, important changes have been made to the rules for checking the stability of all types of framed structure, including braced frames. All of the stability rules, apart from those for portal frames, can now be found in Clause 2.4.2, reducing the risk of the common misconception that only continuous frames need be checked. In fact, most of the changes to this Clause have been made to clarify the intent of the Standard and the technical recommendations are largely unchanged. The rules for brittle fracture have also been amended, resulting in a revised method for calculating the maximum allowable thickness of steel. Compared with BS 5950–1:1990, the new Standard includes a greater variety of details and temperatures (down to –45°C). Other changes to Section 2 include a few new load factors and changes to the rules for structural integrity and disproportionate collapse. In Section 3, numerous minor changes have been made to the limiting width-tothickness ratios used in the classification of cross sections, although the general principle remains unchanged. By far the greatest change to Section 3 of BS 5950-1 relates to the treatment of class 4 slender sections. BS 5950-1:2000 recommends the use of the effective-area method, in which the reduction in capacity due to local buckling is allowed for by the use of effective section properties, as an alternative to the conservative approach of reducing the assumed design strength. Section 3 of BS 5950-1:2000 also introduces the effective plastic modulus, Seff, which may be used instead of Z for class 3 semicompact sections (the use of Z is over-conservative in many cases). Section 4, which deals with the design of structural members, has undergone many significant changes, particularly in relation to members subject to bending. The rules governing the design of both restrained and unrestrained beams have been modified, although most of the changes will only affect the design of class 3 semi-compact and class 4 slender beams. The only significant change relevant to the design of Universal Beams under pure bending (which are usually class 1 plastic or class 2 compact) is the removal of the n-factor 2


method for lateral-torsional buckling. Changes have been made to the methods for checking the shear buckling of plate girder webs in Sub-section 4.4 and the design of bearing and buckling stiffeners in Sub-section 4.5. The design of tension members has also been amended, but the rules for compression members are unchanged, unless the section is class 4 slender. Other amendments to Section 4 include new interaction equations for the buckling of members subject to combined compression and bending, and a new method for designing column bases.


The greatest change to the design of continuous structures in Section 5 relates to the in-plane stability of portal frames. The sway-check method has been extended to include a check for lateral load cases, but the use of this method is now restricted to frames within certain geometrical limitations. Frames that lie outside these limits must now be checked using either the amplified-moments method or second-order analysis. Tied portals must be analysed by secondorder analysis. Out of plane, the method for determining the distance from the restraint at a plastic hinge to the adjacent restraint has been modified to include a “more exact” option, although the old conservative method may still be used. Changes have also been made to Sub-sections 5.6 and 5.7, dealing with the elastic and plastic design of multi-storey frames. BS 5950-1 does not go into great detail on the design of connections. Nevertheless, there have been a number of important changes in Section 6, regarding the design of bolted and welded connections. The most significant change to the design of bolted connections relates to the design of bolts in tension. In BS 5950-1:1990, prying was allowed for implicitly by the use of reduced tension strengths in Table 32. This approach is still valid in BS 5950-1:2000, but designers now have the option of calculating the prying forces and allowing for their effects explicitly in the design. If this latter option is chosen, the full tension strength of the bolts may be used. BS 5950-1:2000 also introduces the concept of block shear and gives limitations on the use of packing. The major change to the design of welds is that BS 5950-1:2000 acknowledges that the transverse strength of fillet welds is greater than the longitudinal strength. Designers may take advantage of this higher strength by the use of the “directional method”, although the old approach, referred to as the “simple method”, may still be used. The Appendices have been renamed Annexes in BS 5950-1:2000 and each Annex has its status declared as either “normative” or “informative”. Within the Annexes, there have been a number of technical changes, particularly in relation to lateral-torsional buckling (Annex B) and web buckling (Annex H). There is also a new Annex on combined axial compression and bending (Annex I).

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2

LIMIT STATES DESIGN

2.1

Load factors

Clause 2.4.1 BS 5950-1:2000 contains load factors for several new load combinations. Values of ãf are now given for storage tanks (full and empty), earth and groundwater loads, exceptional snow loads and various combinations of dead, imposed, wind and crane loads. A new case of “dead load whenever it counteracts the effects of other loads” has also been added, with γf = 1.0, to take account of the fact that in some cases the dead loads are actually beneficial (similar to, but more general than, “dead load when restraining sliding, overturning or uplift” in BS 5950-1:1990).


In addition, the load combinations that the designer must consider are now given explicitly in the Code. Previously, these combinations were listed in Table 2 along with their ãf values, but there was no compulsion to consider any or all of them (although it was assumed that competent designers would know that they had to consider all load combinations to determine the worst case for their buildings). According to BS 5950-1:2000, the following principal load combinations need to be considered: For buildings without cranes: •

Load combination 1 – dead load and imposed load (gravity loads)

•

Load combination 2 – dead load and wind load

•

Load combination 3 – dead load, imposed load and wind load.

For buildings with overhead travelling cranes: •

Crane combination 1 – dead load, imposed load and vertical crane loads

•

Crane combination 2 – dead load, imposed load and horizontal crane loads

•

Crane combination 3 – dead load, imposed load, vertical crane loads and horizontal crane loads.

In BS 5950-1:2000, the wind loading on outdoor overhead travelling cranes that are not in operation is now obtained from BS 6399-2 instead of CP3. For all cranes under working conditions, reference should be made to BS 2573-1.

2.2

Stability

Clause 2.4.2 The requirements for stability, which were previously contained in several Clauses in different Sections of the Code, have been brought together into Clause 2.4.2 in BS 5950-1:2000. The basic requirements have not changed, but the entire Clause has been rewritten to clarify which checks are required and to distinguish between the various modes of failure that are covered by the term

4


“stability limit state”. There have also been several technical changes to the methods of analysis. BS 5950-1:2000 recommends that structures be checked for the following: •

Static equilibrium

•

Resistance to horizontal forces

•

Sway stiffness.


The requirement for static equilibrium, as set out in Sub-clause 2.4.2.2, is simply that the most unfavourable realistic combination of the factored loads should not cause the structure, or any part of it, to slide, overturn or lift off its seating. This is similar to the recommendation in Sub-clause 2.4.2.2 in BS 5950-1:1990, except that the emphasis used to be on overturning, with no mention of sliding. Sub-clause 2.4.2.3 outlines the requirements for providing resistance to horizontal forces. The purpose of this Sub-clause is to ensure that designers consider the possibility of incidental horizontal loads acting on the structure and provide a practical level of robustness against their effects. This is particularly important in cases where the structural actions are dominated by gravity loads and there is a risk that the need to resist horizontal loading will be overlooked, leaving the structure vulnerable to horizontal impact or other accidental loading. In load combination 1, the gravity loads should be accompanied by the notional horizontal forces (see Sub-clause 2.4.2.4) to allow for imperfections in the structure. In load combinations 2 and 3, the structure should be designed to withstand the horizontal wind loading, as in BS 5950-1:1990. However, in BS 5950-1:2000, there is now a minimum wind load of 1% of the factored dead load, to ensure that a minimum horizontal resistance is provided, even in cases where the wind load is very small or non-existent. The important difference between this minimum wind load and the notional horizontal force of 1% of the factored dead load in BS 5950-1:1990 is that the notional horizontal forces are not taken to contribute to the net reactions at the foundations, whereas the wind loads are. In BS 5950-1:1990, it was possible to design a structure with no allowance for horizontal foundation loads. In BS 5950-1:2000, this is no longer possible, because the 1% of the dead load considered in load combinations 2 and 3 is carried through to the foundations. The notional horizontal forces are not externally applied loads in the way that the dead, imposed and wind loads are, but are a convenient means of taking into account the effects of imperfections, such as columns being out of plumb, on the performance of a structure. In reality, such imperfections exist in all structures, causing lateral forces to be induced in the structure under the action of gravity loads. For this reason, the notional horizontal forces must always be applied simultaneously with the gravity loads. In BS 5950-1:1990, the notional horizontal forces are taken as the greater of either 1% of the factored dead load or 0.5% of the dead plus imposed vertical loads. In BS 5950-1:2000, the 1% of factored dead load has been removed (except as a minimum wind load, see above) and the notional horizontal force is always 0.5% of the dead plus imposed vertical loads. The final check in this Clause relates to the sway stiffness of the structure and in particular whether it is safe for the second-order (P-delta) effects to be 5


neglected. One significant change to this subject is that all of the recommendations have now been placed together in this one Clause (in Subclauses 2.4.2.5 to 2.4.2.8). In BS 5950-1:1990, sway stiffness was only referred to very briefly in Section 2, while the majority of the recommendations was in Section 5, along with the rules for continuous construction. This gave the impression that sway stiffness was only important in continuous structures while, in reality, it is something for which all structures should be checked. The degree of sway stiffness is obtained by applying the notional horizontal forces at the floor or roof level under consideration and calculating the deflection ä at this level relative to the storey below by elastic analysis. An approximation to the sway mode elastic critical load factor of the frame ëcr is then determined from ëcr =

h 200δ

where h is the storey height.


For clad structures in which the stiffening effect of the cladding has been neglected in the analysis, if ëcr 10, it is safe to assume that the second-order effects are small enough to be ignored and the frame may be classed as “nonsway”. If ëcr 0.6Pv) has also been revised. In BS 5950-1: 1990, the moment capacity was given as follows: For plastic and compact sections: Mc = where ρ1

py(S – Svρ1) =

but

1.2pyZ

2.5 Fv − 1.5 Pv

For sections with equal flanges, Sv is the plastic modulus of the shear area Av. For sections with unequal flanges, Sv is the plastic modulus of the gross section less the plastic modulus of that part of the section remaining after the deduction of the shear area. For semi-compact and slender cross sections, the moment capacity was the same as in the low shear case, i.e. Mc =

pyZ

using a reduced py for slender sections. This was clearly wrong, as it neglected the effects of high shear completely for class 3 and class 4 sections. In BS 5950-1:2000, this error has been corrected and the new equations are as follows: For class 1 plastic or class 2 compact sections: Mc =

py(S – ρSv)

For class 3 semi-compact sections: Mc =

py(Z – ρSv / 1.5)

or

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Mc =

py(Seff – ρSv)


For class 4 slender sections: Mc =

py(Zeff – ρSv / 1.5)

In all three equations: ρ

=

[2 (Fv / Pv) – 1]2

Note: Although the moment capacity equation for class 1 plastic and class 2 compact sections is unchanged, the expression for ρ is different in BS 5950-1: 2000. The design of restrained beams is illustrated by Worked Example 3 of this publication.

4.2

Lateral-torsional buckling

Clause 4.3.5


Clauses 4.3.5 and 4.3.6 of BS 5950-1:1990 have been merged to form Clause 4.3.5 of BS 5950-1:2000. This new Clause covers the effective lengths of both simple beams and cantilevers for lateral-torsional buckling. A number of important changes have been made and a new Sub-clause on double curvature bending has been added. Clause 4.3.5 of BS 5950-1:1990 dealt specifically with the effective lengths of simple beams. Most of these requirements are unchanged, although the text has been completely rewritten. However, there have been amendments to the effective length values given in Table 13 (formerly Table 9). Firstly, a new symbol LLT, standing for segment length, has been introduced to distinguish this dimension from the member length L. The segment length is the length between restraints, whether these are intermediate restraints or supports. Secondly, several new restraint conditions have been added to the Table, allowing designers to model their structures more accurately. These are “compression flange fully restrained against rotation on plan” and “compression flange partially restrained against rotation on plan”. Note that the effective lengths of a beam with lateral and torsional restraint at one end and both flanges partially restrained against rotation on plan at the other end have been reduced to 0.8LLT and 0.95LLT for the normal and destabilising loading conditions respectively. Clause 4.3.6 of BS 5950-1:1990 considers the effective lengths of cantilevers. The content of this Clause has undergone several significant changes, with important implications for the design of these members. The first change relates to cantilevers with intermediate lateral restraints. In BS 5950-1:1990, the lengths between restraints were treated as beams and their effective lengths obtained accordingly using the provisions of Clause 4.3.5. In BS 5950-1:2000, provided that the cantilever is restrained laterally and torsionally at both ends (i.e. cases c4 and d4 in Table 14), an effective length of 1.0L is given for the normal loading condition, where L is the length of the relevant segment between adjacent lateral restraints. However, for the 15


destabilising loading condition, the effective length should be obtained from Table 14. In this case, L is taken as the length of the cantilever, unless the there are intermediate lateral restraints to the top flange. Secondly, the rules for determining the effective length of cantilevers without intermediate restraints are unchanged (i.e. LE obtained from Table 14), apart from the case where the cantilever has a moment applied to its tip. BS 5950-1: 1990 deals with such moments by treating the cantilever as a beam, as for cantilevers with intermediate restraints. In BS 5950-1:2000, the effective length is obtained from Table 14 then increased by either 30% or 0.3L, whichever is greater. The effective length values in Table 14 of BS 5950-1:2000 are unchanged from those given in Table 10 of BS 5950-1:1990, except that a new category of restraint has been added. Finally, the new Clause 4.3.5 contains a new Sub-clause on the subject of beams with double curvature bending. This Sub-clause has been added to emphasise that special consideration needs to be given to beams that have both hogging and sagging regions. Design rules are given for beams with intermediate lateral restraints to each flange, beams with intermediate lateral restraints to the compression flange in the sagging region only and beams directly supporting a concrete or composite floor or roof slab. Created on 30 March 2011 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

Clause 4.3.6 Clause 4.3.6 contains the design rules, tables and equations needed to calculate the buckling resistance moment Mb of unrestrained beams susceptible to lateraltorsional buckling. Although the general method is unchanged from that in BS 5950-1:1990, there have been a number of significant changes to the individual steps in the procedure. Overall, this Clause has undergone a major technical amendment. The Clause begins by listing the situations in which there is no need to check for lateral-torsional buckling. These are: •

bending about the minor axis

•

circular hollow sections, square rectangular hollow sections or circular or square solid bars

•

rectangular hollow sections when LE/ry does not exceed the limiting value from Table 15

•

I, H, channel or box sections when λLT

λL0.

The first of these cases was not listed explicitly in BS 5950-1:1990, but this is not a technical change, because lateral-torsional buckling occurs about the minor axis as a result of major axis bending. The second and third cases were noted in BS 5950-1:1990, but the table of limiting slenderness values for rectangular hollow sections has been extended to include 12 values of D/B, compared with the four values given in Table 38 in Appendix B.2.6 of BS 5950-1:1990. In the final case, ëL0 is the limiting slenderness obtained from the last row of Table 16 or Table 17. The inclusion of the ëL0 values in these tables is new to BS 5950-1:2000, although a formula for ëL0 was included in Appendix B.2.5 of BS 5950-1:1990.

16


Every segment length of an unrestrained beam, subject to a major axis moment Mx, should satisfy the following conditions: Mx ≤

Mb / mLT

and Mx ≤

Mcx

where: Mcx

is the moment capacity of the section from Clause 4.2.5

Mb

is the buckling resistance moment

mLT

is the equivalent uniform moment factor for lateral-torsional buckling, which takes account of the fact that the theory from which Mb is obtained assumes a uniform moment throughout the segment.


These requirements are unchanged from BS 5950-1:1990, although they have been rewritten and there is no longer any reference to the equivalent uniform moment M . Except for hot rolled angles, which are considered in Clause 4.3.8, the buckling resistance moment Mb may be obtained either using the conservative method described in Clause 4.3.7 or from the expressions below. For class 1 plastic or class 2 compact cross sections: Mb =

pb Sx

For class 3 semi-compact cross sections: Mb =

pb Zx

or

Mb =

pb Sx,eff

For class 4 slender cross sections: Mb =

pb Zx,eff

In BS 5950-1:1990, Mb was taken as pbSx irrespective of the classification of the cross section. This was based on the false assumption that lateral-torsional buckling and local buckling do not interact. The term pb in these equations is the bending strength and is obtained from Table 16 for rolled sections and Table 17 for welded sections, for given values of design strength py and equivalent slenderness ëLT. These tables are equivalent to Table 11 and Table 12 in BS 5950-1:1990 and have been amended to include new columns corresponding to different values of py and a new row containing values of the limiting slenderness λL0. In all other respects, these tables are unchanged. Designers familiar with BS 5950-1:1990 will recall that there used to be two approaches to analysing lateral-torsional buckling, one using the slenderness correction factor n (for loading between lateral restraints), the other using the equivalent uniform moment factor m (for cases with end moments only). In BS 5950-1:2000, there is a major change in that the n factor method has been removed from this Clause, leaving the latter method to be used for all cases (n

17


is still used for tapered or haunched members in Annex B.2.5). Consequently, the equation for ëLT has changed. In BS 5950–1:1990, ëLT was given by λLT =

nuvλ

where: λ

is the minor axis slenderness LE / ry

u

is a buckling parameter

v

is a slenderness factor.

In BS 5950-1:2000, λLT is now given by λLT =

uvλ β W

where βW is the ratio of section moduli, as described below. For class 1 plastic or class 2 compact sections: βw =

1.0


For class 3 semi-compact cross sections: βw =

Zx / Sx

or βw =

Sx,eff / Sx

For class 4 slender sections: βw =

Zx,eff / Sx

The slenderness ratio, v, for sections with two plain flanges, may be obtained from Table 19 for various values of ë/x and ç where x is the torsional index of the section and ç is given by ç

I yc

=

I yc + I yt where Iyc and Iyt are the minor axis second moments of area of the compression flange and tension flange respectively. The only difference between Table 19 in BS 5950-1:2000 and its equivalent in BS 5950-1:1990, is that the two columns at either end of the ç range, i.e. those corresponding to ç = 1.0 and ç = 0.0 (T sections), have been deleted. In BS 5950-1:2000, designers wishing to use T sections must instead refer to Annex B.2.8. As an alternative to Table 19, the slenderness ratio v may be obtained from the equations in Sub-clause 4.3.6.7. In BS 5950-1:1990, similar expressions could be found in Appendix B.2.5. The buckling parameter u and torsional index x may be obtained from the formulae in Annex B.2.3 (B.2.5 in BS 5950-1:1990) or from published tables of section properties. Alternatively, for rolled I and H sections with equal flanges, the following conservative approximations may be used: u

=

0.9

and

x

= 18

D/T


Sub-clause 4.3.6.8 has not been affected by the recent amendments, other than the change in Sub-clause number. As noted above, the two approaches to analysing lateral-torsional buckling in BS 5950-1:1990 have been replaced by a single method using the equivalent uniform moment factor mLT. Previously, for the normal loading condition, if a member was loaded between adjacent lateral restraints, m was taken as 1.0 and the appropriate value of n was obtained from Table 15 or Table 16. If the member was not loaded between its restraints, n was taken as 1.0 and m was obtained from Table 18. Thus, the allowance for the shape of the bending moment was made using either the n factor or the m factor, depending on the location of the loading, but never both. The change in BS 5950-1:2000 to the use of mLT for all cases has necessitated a substantial extension to Table 18. In BS 5950-1:1990, m values were provided for segments with end moments only, i.e. only linear variations in bending moment were considered. In BS 5950-1:2000, values are also given for four specific cases with transverse loads applied between restraints and a general formula is provided from which mLT may be obtained for more complex bending moment diagrams.


Note: 1. Slight changes have been made to the values of mLT for segments with end moments only, compared with those in BS 5950-1:1990. 2. BS 5950-1:2000 distinguishes between mLT, the equivalent uniform moment factor for lateral-torsional buckling, and mx, my, myx, the equivalent uniform moment factors for flexural buckling. Values of mLT are obtained from Table 18, while values of mx, my, myx are given in Table 26. The design of unrestrained beams is illustrated by Worked Example 4 in this publication. Clause 4.3.7 Clause 4.3.7 gives a simple but conservative alternative approach to the method presented in Clause 4.3.6 for determining the buckling resistance moment of a plain rolled I, H or channel section with equal flanges. There have been two changes to this method compared with BS 5950-1:1990. Firstly, in BS 5950-1:1990, Mb was taken as pbSx irrespective of the classification of the cross section. This was based on the false assumption that lateral-torsional buckling and local buckling do not interact. This has been corrected in BS 5950-1:2000, in which Mb is only taken as pbSx if the section is class 1 plastic or class 2 compact and is taken as pbZx for class 3 semi-compact sections. The second change relates to the table of pb values (Table 20 in BS 5950-1: 2000 and Table 19 in BS 5950-1:1990). In BS 5950-1:1990, pb was given as a function of x (= D/T) and the slenderness λ (= LE/ry), whereas in BS 5950-1: 2000, pb is given in terms of D/T and β w L E / ry . However, the numbers in Table 20 are unchanged, so for all class 1 and class 2 sections, for which βw = 1, there will be no change at all to pb, or indeed Mb.

19


Clause 4.3.8 Clause 4.3.8, which deals with the buckling resistance moment for single angles, has been revised resulting in a new basic method and a significant change to the existing simplified method. In BS 5950-1:1990, the calculation of the buckling resistance moment for single angles was very straightforward, as it simply involved one of the following equations: Mb =

0.8pyZ for L / rvv ≤ 100

Mb =


Mb =


where: rvv

is the radius of gyration about the v-v axis

L

is the unrestrained length.


This simple approach has been shown to be non-conservative for some cases (for low L / rvv) and has been replaced in BS 5950-1:2000 by two alternative methods. The first method, known as the “basic method” in BS 5950-1:2000, is applicable to equal and unequal angles and involves resolving the applied moments into their components about the principal axes u-u and v-v. The buckling resistance moment should then be obtained using an equivalent slenderness λLT obtained from Annex B.2.9. The second method, referred to as the “simplified method”, provides a simple alternative to the basic method, but it is only applicable to equal angles with b / t ≤ 15ε. In this case, the buckling resistance moment is given by Mb =

0.8pyZx

when the heel of the angle is in compression or Mb =

 1350ε − L E rv  pyZx   but 1625ε  

0.8pyZx

when the heel of the angle is in tension.

4.3

Plate girder webs

Clause 4.4.4 If the web of a plate girder is susceptible to shear buckling (i.e. d / t > 62ε), the moment capacity of the cross section should be obtained using one of the methods given in Sub-clause 4.4.4.2. BS 5950-1:1990 presented three alternative methods of analysis. The first option was to assume that the moment and axial load are resisted by the flanges alone, leaving the web to resist only the shear force. In the second method, the moment and axial load are assumed to be resisted by the whole section and the web is designed for shear and longitudinal stresses, using the method in H.3. The third method was a combination of the first two. 20


The methods presented in BS 5950-1:2000 are similar, except that there is now an additional option for low shear. Where the applied shear is no greater than 60% of the simple shear buckling resistance obtained from Sub-clause 4.4.5.2, it is acceptable to obtain the moment capacity using the rules given in Clause 4.2.5. Where the applied shear exceeds 60% of the simple shear buckling resistance, the moment capacity should be obtained either by assuming that all of the moment is resisted by the flanges or by using the rules in H.3 to design the web for the applied shear plus any moment beyond the “flanges only” moment capacity. These two high shear methods are essentially the same as methods a and c in BS 5950-1:1990. BS 5950-1:2000 also contains a new Sub-clause relating to axial loads. This states that where a member is subject to an axial load combined with a moment, reference should be made to the design rules in Clauses 4.8.1 to 4.8.3. In this case, when using the “flanges only” method, it should be assumed that the moment and the axial force are both resisted by the flanges alone, with each flange subject to a uniform stress not exceeding pyf. Clause 4.4.5


Clause 4.4.5, which considers the shear buckling of plate girder webs, has been rewritten for clarity and replaces the over-conservative simple method with the one used in BS 449. This is a major technical change. Clause 4.2.3 of BS 5950-1:1990 states that if the d / t ratio of a web exceeds 63å, the shear buckling resistance of the web should be checked in accordance with Clause 4.4.5. In BS 5950-1:2000, this limit has been replaced by two new limits: 62ε for a welded section and 70ε for a rolled section. BS 5950-1:1990 and BS 5950-1:2000 both present two alternative methods for calculating the shear buckling resistance of a web. The first method, known in BS 5950-1:2000 as the “simplified method”, does not allow for the beneficial effects of tension field action and may be used for webs with or without intermediate stiffeners. The second method, referred to as the “more exact method” in BS 5950-1:2000, does make use of tension field action and its use is restricted to webs with intermediate transverse stiffeners. For webs with longitudinal stiffeners, or as an alternative for webs with intermediate transverse stiffeners, reference should be made to BS 5400-3[6]. The “simplified method” in BS 5950-1:2000 is very similar to the “Design without using tension field action” in BS 5950-1:1990, although the two methods give slightly different results. In BS 5950-1:1990, the shear buckling resistance, Vcr, of a stiffened or unstiffened panel is given by: Vcr =

qcr dt

where: d

is the depth of the web

t

is the web thickness

qcr is the critical shear strength from Table 21.

21


In BS 5950-1:2000, the shear buckling resistance Vb of a web is taken as the simple shear buckling resistance Vw given by: Vw =

dtqw

where: d


t


qw is the shear buckling strength of the web from Table 21. While these two methods may appear to be identical, apart from the choice of symbols, they give different results because the values of qw in Table 21 of BS 5950-1:2000 are not the same as the values of qcr in Table 21 of BS 5950-1: 1990. In the more exact method, the difference between the two versions of BS 5950-1 is more apparent. In BS 5950-1:2000, if the flanges are fully stressed, the shear buckling resistance, Vb, equals the simple shear buckling resistance, Vw, given by:


Vw =

dtqw

If the flanges are not fully stressed, Vb is taken as the sum of the simple shear buckling resistance, Vw, and the flange-dependent shear buckling resistance, Vf, i.e. Vb

=

Vw + Vf

where Vf is given by Vf

=

(

Pv d / a

)

(

1 − f / p f yf 

)

2

 

1 + 0.15 ( M pw / M pf )

where: ff

is the mean longitudinal stress in the smaller flange due to moment and/or axial force

Mpf is the plastic moment capacity of the smaller flange about its own equal area axis perpendicular to the plane of the web Mpw is the plastic moment capacity of the web about its own equal area axis perpendicular to the plane of the web Pv

is the shear capacity from Clause 4.2.3

pyf is the design strength of the flange pyw is the design strength of the web. By comparison, in BS 5950-1:1990, the shear buckling resistance of a stiffened panel is given by: Generally: Vb

=

qb dt

22


If the flanges are not fully stressed: Vb

=

( qb + qf

)

K f dt

but ≤ 0.6pydt

where: qb

is the basic shear strength obtained from Table 22

qf

is the flange-dependent shear strength factor from Table 23

Kf

is a factor given by Kf

=

M pf  1 − f 4 M pw  p yf

   

(The f term here has the same meaning as ff defined above.)


Note that the basic shear strength qb used in the design of stiffened panels, using tension field action, is higher than the critical shear strength qcr used when designing panels without tension field action, giving a higher value of Vb, even when the flanges are fully stressed. By contrast, in BS 5950-1:2000, both methods use qw and therefore yield exactly the same result in the case of fully stressed flanges. The tension field action induced in the plate girder web produces a horizontal anchor force, Hq, at the end of the girder, as shown in Figure 4.1.

Hq

Figure 4.1

Tension field action in a plate girder

It is generally necessary to provide an end anchorage to resist this horizontal force, however BS 5950-1:2000 gives the following two cases when an end anchorage is not necessary: 1. If the shear capacity, rather than the shear buckling resistance, is the governing design criterion, i.e. Vw = Pv. 2. If sufficient shear buckling resistance is available without forming a tension field. The existence of this condition is indicated by: Fv

≤ Vcr

where Vcr is the critical shear buckling resistance and Fv is the maximum shear force. Vcr may be obtained from Annex H.2 or using the formulae in this Clause. These exemptions are new to BS 5950-1:2000.

23


Note that the provisions dealing with the design of the end anchorage, which used to be in Clause 4.4.5, have been moved to Annex H.4 in BS 5950-1:2000. The design of plate girders is illustrated by Worked Example 5 in this publication.

4.4

Design of stiffeners

Clause 4.5.1 Clause 4.5.1 considers the design of web bearing and buckling stiffeners. Subclause 4.5.1.3, which deals with the stiff bearing length, has been amended and Sub-clause 4.5.1.5, on the subject of hollow sections, has moved to this Clause from Clause 4.5.12. In other respects, the technical content of this Clause is unchanged.


In BS 5950-1:1990, the stiff bearing length b1 (i.e. the length that cannot deform appreciably in bending) was calculated by taking the angle of load dispersion through the steel to be 45° (see Figure 8 of BS 5950-1:1990). BS 5950-1:2000 also assumes a load dispersion of 45°, but there is no longer a need to analyse the geometry of the section, because b1 may be obtained directly from the formulae given in Figure 13. Clause 4.5.12 of BS 5950-1:1990 is now Sub-clause 4.5.1.5 in BS 5950-1: 2000. There is an additional sentence referring designers to a design procedure given in Steelwork Design Guide to BS 5950-1:2000, Volume 1: Section Properties and Member Capacities[7] but the general requirements for the design of hollow sections subject to concentrated loads are unchanged. Clause 4.5.2 The provisions in BS 5950-1 relating to the bearing capacity of an unstiffened web have been revised. In BS 5950-1:1990, the bearing capacity of an unstiffened web was given by (b1 + n2)tpyw where: b1

is the stiff bearing length (see Clause 4.5.1)

n2

is the length obtained by dispersion through the flange to the flange/ web connection at a slope of 1 in 2.5

t


pyw is the design strength of the web. In BS 5950-1:2000, the bearing capacity of an unstiffened web is given by Pbw =

(b1 + nk)tpyw

where n is taken as follows: –

except at the end of the member: n

–

=

5

at the end of the member: n

=

2 + 0.6 be / k, but n 24

5


where be is the distance to the end of the member from the end of the stiff bearing. k is given by: –

for a rolled I or H section: k

–

=

T+r

for a welded I or H section: k

=

T.

Therefore, in BS 5950-1:2000, the bearing capacity of an unstiffened web depends on the proximity of the end of the stiff bearing to the end of the member, whereas in BS 5950-1:1990, the bearing capacity was always determined assuming a dispersion of 1:2.5 across the flange thickness.


Where the applied load or reaction exceeds the bearing capacity of the unstiffened web, bearing stiffeners should be provided. These should be designed to carry the applied force minus the bearing capacity of the unstiffened web. This requirement is unchanged from BS 5950-1:1990. The bearing capacity of a web is considered in Worked Example 6 of this publication. Clause 4.5.3 The provisions in BS 5950-1 relating to the buckling resistance of an unstiffened web and the design of load-carrying stiffeners have been revised. In BS 5950-1:1990, the buckling resistance of an unstiffened web was given by Pw =

(b1 + n1)tpc

where: b1

is the stiff bearing length

n1

is the length obtained by 45° dispersion through half the depth of the section

t


pc

is the compressive strength obtained from Table 27c.

In BS 5950-1:2000, the buckling resistance of the web, Px, is obtained directly from the bearing capacity, Pbw, and the geometry of the section. There is no longer a need to refer to the “strut” curve (i.e. Table 24c) in BS 5950-1:2000. Three equations are presented for Px, depending on the restraint of the flange and the location of the applied load relative to the end of the member. If the loaded flange is effectively restrained against rotation relative to the web and against lateral movement relative to the other flange, Px, is given by: When ae ≥ 0.7d: Px

=

25εt

(b1 + nk ) d

Pbw

25


When ae < 0.7d: Px

=

a e + 0.7 d 1.4d

25εt

(b1 + nk ) d

Pbw

where: ae

is the distance from the applied load or reaction to the end of the member

d


Pbw is the bearing capacity of the unstiffened web obtained from Clause 4.5.2 n

is a dispersion factor obtained from Clause 4.5.2

k

is taken as follows: –

for a rolled I or H section: k

–

T+r

for a welded I or H section: k


= =

T

If one or both of the restraint conditions given above is not met, a reduced buckling resistance must be used given by Pxy =

0.7 d Px LE

In BS 5950-1:1990 and BS 5950-1:2000, the buckling resistance of a loadcarrying stiffener is given by Px

=

As pc

where: As

is the effective area of a cruciform section, consisting of the stiffeners and an effective width of web on each side of the centreline of the stiffeners

pc

is the compressive strength from Table 27c in BS 5950-1:1990 or Table 24c in BS 5950-1:2000.

The important change here is that in BS 5950-1:1990 the effective width on each side of the stiffener was taken as 20t, whereas in BS 5950-1:2000 it is limited to 15t. In other respects, the buckling check for a load-carrying stiffener is unchanged. Perhaps more importantly, BS 5950-1:1990 also included a bearing check for load-carrying stiffeners (Sub-clause 4.5.4.2). This stated that load-carrying stiffeners should be designed to resist 80% of the total applied force, irrespective of the capacity of the unstiffened web, i.e. A>

0.8 Fx p ys

26


where: A

is the area of the stiffener in contact with the flange

Fx

is the applied load

pys is the design strength of the stiffener. As a result of the reduction in the effective width of the web from 20t to 15t, this bearing check has been removed from BS 5950-1:2000. This is a key change because, in most practical cases, the size of the stiffeners was governed by this rule. There is still a requirement to check the bearing capacity, as BS 5950-1:2000 states that load-carrying stiffeners should also be checked as bearing stiffeners. However, this requirement is not as onerous as the previous 80% rule, because bearing stiffeners are only designed to carry the external load minus the bearing capacity of the unstiffened web, not the full external load. The buckling resistance of a web is considered in Worked Example 6 in this publication.


Clause 4.5.4 Clause 4.5.4 (formerly Clause 4.5.7) has been expanded to provide greater detail on the design of tension stiffeners. Two cases are presented for which tension stiffeners are provided and a separate method of design is given for each case. If the applied load or reaction exceeds the tension capacity of the unstiffened web at its connection to the flange, the tension stiffener should be designed to carry that portion of the load which exceeds the tension capacity of the unstiffened web. If, on the other hand, tension stiffeners are needed because the applied load or reaction exceeds the tension capacity of the unstiffened flange, the proportion of the load assumed to be carried by the stiffener should be consistent with the design of the flange. This latter case was not considered in BS 5950-1:1990.

4.5

Tension members

Clause 4.6.3 New formulae have been introduced for calculating the tension capacity of a simple tie consisting of an angle connected through one leg only, a channel connected only through the web, or a T section connected only through the flange. In BS 5950-1:1990, the tension capacity of a single angle, a single channel or a T section was calculated using an affective area equal to the net area of the connected leg plus the area of the outstanding leg multiplied by 3a1 3a1 + a 2

where a1 is the net cross-sectional area of the connected leg and a2 is the crosssectional area of the unconnected leg. In BS 5950-1:2000, the tension capacity, Pt, for single angles, channels or T sections with bolted connections is given by 27


Pt

=

py(Ae – 0.5a2)

and for welded connections by Pt

=

py(Ag – 0.3a2)

in which Ae is the sum of the effective net areas ae (see Clause 3.4.3), Ag is the gross area of the cross section and a2

=

Ag – a1

where a1 is the gross area of the connected element.


The reason for this change is that the formulation in BS 5950-1:1990 is nonconservative for grade S460 steel. The 3a1 / (3a1 + a2) factor was derived from tests on mild steel and allows for the effects of eccentricity and strain hardening, assuming Us / Ys to be the same for all grades of steel. In BS 5950-1: 2000, eccentricity and strain hardening are considered separately, by first calculating the effective area in Clause 3.4.3 (multiplying the net area by Ke) then using the formulae in this Clause to allow for eccentricity. Consequently, compared with BS 5950-1:1990, there has been an increase in capacity for S275, a small decrease for S355 and a significant decrease for S460. For welded angles, there is a very small decrease in capacity for all grades of steel. Similarly, there are two new formulae for double angle, channel and T section members, where the components are connected to both sides of a gusset plate and are interconnected by bolts or welds. In this case, the tension capacity for bolted connections is given by Pt

=

py(Ae – 0.25a2)

and for welded connections by Pt

=

py(Ag – 0.15a2)

Note: If the components of the tie are both connected to the same side of the gusset plate or are not interconnected as described above, the member should be treated as if it were a single angle, channel or T section.

4.6

Compression members

Clause 4.7.1 An additional paragraph has been added to Clause 4.7.1 giving the design loading for bracing systems. As with the intermediate restraints for lateraltorsional buckling (see Clause 4.3.2), bracing systems that supply positional restraint to more than one member must be designed to resist the sum of all of the individual restraint forces from each member reduced by the factor kr, given by kr

=

(0.2 + 1 / Nr)0.5

where Nr is the number of parallel members restrained. Clause 4.7.2 Sub-clause 4.7.3.2 of BS 5950-1:1990 gave maximum slenderness values for different types of member. These have now been removed as it was felt that the limits presented in the Code were not universally applicable and in many 28


cases were inappropriate, leading to the rejection of perfectly acceptable members. However, it is important for designers to recognise that there may still be practical limits on the slenderness of a member, for example to limit sag due to self-weight, and these must be considered carefully. One additional requirement is that it is now necessary to increase by 20% the slenderness of single-angle struts that have lateral restraints to their two legs alternately. Clause 4.7.4 The means by which the compression resistance of members with slender cross sections is calculated have changed, resulting in the removal of the anomalies leading to the over-conservative design of slender cross sections, such as Universal Beams used as columns. In BS 5950-1:1990, the compression resistance Pc of a member with a slender cross section was given by Pc

=

Ag pcs

where:


Ag

is the gross cross-sectional area of the section

pcs is the reduced compressive strength based on the slenderness ë and the reduced design strength pyr. In BS 5950-1:2000, the recommended method for slender cross sections uses effective areas equal to the semi-compact limits, instead of a reduced design strength (see Sub-section 3.6 for details). Consequently, the compression resistance Pc of a member with a slender cross section is now given by Pc

=

Aeff pcs

where Aeff is the effective area obtained from Sub-section 3.6 and pcs is the compressive strength based on the design strength py and a reduced slenderness of A λ  eff  Ag 

   

0.5

The design of compression members is illustrated by Worked Example 7 in this publication. Clause 4.7.5 Changes have been made to Table 23 (formerly Table 25) regarding the choice of strut curve for different types of section and to Table 24 (formerly Table 27) from which the compressive strength pc is obtained. The most significant change to this Clause is the inclusion of cold formed structural hollow sections in BS 5950-1:2000, for which Table 24c should be used for buckling about both axes. In addition, changes have been made to the choice of strut curve for rolled I sections. In BS 5950-1:1990, the compressive strength pc of all rolled I sections about the major and minor buckling axes was obtained from Table 24a and Table 24b respectively. In BS 5950-1:2000, the

29


use of Table 24a and Table 24b is limited to sections with a maximum thickness not exceeding 40 mm. Table 24b and Table 24c should be used for rolled I sections with a maximum thickness greater than 40 mm. This change also applies to rolled I sections with welded flange cover plates within the range 0.25 < U / B < 0.8 (see Figure 14 of BS 5950-1:2000). The strut curves themselves are unchanged, although the range of design strengths in Table 24 has been amended. Compressive strengths pc for py values of 225 N/mm2, 305 N/mm2, 320 N/mm2, 340 N/mm2, 395 N/mm2, 415 N/mm2 and 450 N/mm2 have been deleted and replaced by values corresponding to design strengths of 235 N/mm2, 315 N/mm2, 345 N/mm2, 400 N/mm2, 440 N/mm2 and 460 N/mm2. In addition, a few of the other pc values have been revised slightly, although the majority is unchanged. No technical changes have been made to the text of this Clause, but much of it has been rewritten, with Figure 14 in BS 5950-1:2000 replacing Table 26 in BS 5950-1:1990.

4.7

Combined moment and axial force


Clause 4.8.3 Clause 4.8.3 deals with members that are subject to combined compression and bending. It has undergone a major technical change. The capacity of a member subject to combined compression and bending is dependent on the local cross-section capacity and the overall buckling resistance of the member. This Clause contains checks for both of these modes of failure. Although both checks have been amended, only the overall buckling check has undergone a major change. BS 5950-1:1990 presented two alternative methods for checking buckling resistance. In the simplified method, the following relationship had to be satisfied: mM y mM x F + + ≤1 Ag p c Mb py Z y where: F

is the applied axial force

Ag

is the gross cross-sectional area

pc

is the compressive strength

m

is the equivalent uniform moment factor

Mb is the buckling resistance moment Mx is the applied moment about the major axis My is the applied moment about the minor axis py

is the design strength

Zy

is the elastic section modulus about the minor axis.

30


Alternatively, designers could opt for the more exact method by satisfying the following relationship: mM x mM y + ≤1 M ax M ay in which Max and May are the maximum buckling moments about the major and minor axes respectively in the presence of axial load. BS 5950-1:2000 also presents designers with the option of using either a simplified or a more exact approach. In the case of the simplified approach, the single equation in BS 5950-1:1990 has been replaced by the following two expressions: Fc m x M x m y M y + + ≤1 Pc py Z x py Z y my My Fc m M + LT LT + ≤1 Pcy Mb py Z y


In applying these equations, the following points need to be noted. 1. In the first term of the first equation, Pc is the smaller of the compression resistance for buckling about the major axis Pcx and the compression resistance for buckling about the minor axis Pcy. However, in the second equation, Pcy is always used, whether it is smaller than Pcx or not. 2. In the second term of the second equation, MLT is used in place of Mx. MLT is the maximum major axis moment in the segment length between restraints against lateral-torsional buckling. 3. The equivalent uniform moment factor m used in BS 5950-1:1990 has been replaced by mx, my and mLT. mLT is the equivalent uniform moment factor for lateral-torsional buckling for the pattern of major axis moments over the segment length LLT. It is obtained from Table 18. mx and my are the equivalent uniform moment factors for flexural buckling about the major and minor axis respectively. Both are new and are obtained from Table 26 using the appropriate moment pattern between the relevant flexural buckling restraints. 4. For cantilever columns and members in sway sensitive frames, BS 5950-1:2000 includes the following special requirements: a. If sway mode in-plane effective lengths are used, mx and my should not be less than 0.85. b. If amplified-sway moments are used, only the non-sway moment should be multiplied by mx or my, compared with the total (sway + non-sway) moment in BS 5950-1:1990. The more exact approach has undergone an even more radical change, with the original single equation being replaced by sixteen new ones, seven for I and H sections, nine for circular hollow sections, rectangular hollow sections and box sections. Separate equations are given for members with moments about the major axis only, members with moments about the minor axis only and 31


members with moments about both axes. There are typically two or three equations for each type of loading. For example, an I section member with moments about both axes should be checked for major axis buckling, lateraltorsional buckling and interactive buckling. Note: In BS 5950-1:1990, the more exact method could, in theory, be used for any type of section, whereas in BS 5950-1:2000, this method is clearly limited to circular hollow sections, rectangular hollow sections, I sections, H sections and box sections with equal flanges. In addition to the major amendments to the buckling resistance checks described above, a minor change has also been made to the cross-section capacity check, to take account of the new method for dealing with class 4 slender sections. In BS 5950-1:2000, the following two equations are given for checking the crosssection capacity: (a) For class 1 plastic, class 2 compact and class 3 semi-compact cross sections My Fc M + x + ≤1 Ag p y M cx M cy


(b) For class 4 slender cross sections My Fc M + x + ≤1 Aeff p y M cx M cy where Aeff is the effective area obtained from 3.6. In BS 5950-1:1990, only the first of these two equations was given and the reduced capacity of slender sections was taken into account by the use of a reduced py. As before, BS 5950-1:2000 presents an alternative method for class 1 plastic and class 2 compact sections, based on the reduced moment capacities Mrx and Mry. This method is unchanged from BS 5950-1:1990. Note: Designers of members with combined compression and bending should also refer to Annex I. This new Annex contains an alternative method for stocky members (I.1), formulae for Mrx and Mry (I.2), alternative interaction expressions for asymmetric members (I.3), design methods for single angles (I.4) and a method for evaluating internal second-order moments (I.5). The design of members subject to combined compression and bending is considered in Worked Example 8 in this publication.

4.8

Column bases

Clause 4.13.2 The empirical method for determining the size of baseplates in BS 5950-1:1990 has been replaced by the effective-area method in BS 5950-1:2000.

32


In BS 5950-1:1990, the minimum thickness of a symmetrical rectangular baseplate, for a concentrically loaded column, was taken as the greater of the flange thickness of the column and the result of the following empirical formula: t

(

 2.5 w a 2 − 0.3b 2  p  yp

=

)

  

0.5

where: a

is the greater projection of the plate beyond the column

b

is the lesser projection of the plate beyond the column

w

is the pressure on the underside of the plate based on a uniform distribution (≤ 0.4 fcu)

pyp is the design strength of the plate (but ≤ 270 N/mm2).


Since the publication of BS 5950-1:1990, doubt has been raised as to the validity of this method, especially for columns with a large depth / width ratio. In fact, The Steel Construction Institute has advised designers that this empirical method should not be used for bases to Universal Beams. The effective-area method is an alternative general design method for use in designing baseplates subject to concentric loads. This method was first introduced in Joints in Simple Construction – Volume 1: Design Methods and has since become the standard method of baseplate design in the UK. The effective-area method has now been adopted by BS 5950-1:2000, but other rational approaches are also permitted. In the effective-area method, it is assumed that the applied load is uniformly distributed over part of the baseplate, with the remainder of the area being considered ineffective. The extent of the projection of the plate, beyond the edge of the column, is based on the area required to keep the bearing pressure below the limiting bearing strength of 0.6fcu, where fcu is the weaker of the concrete cube strength and that of the bedding material. Effective areas for typical columns are shown in Figure 4.2.

2c + T 2c + T

Figure 4.2

2c + T Effective portion

Effective area of a baseplate

33

Stiffener


The shaded areas in the figure represent the minimum size of the plate required to keep the bearing pressure within the necessary limit. In practice, the overall size of the plate can be made larger, for instance to utilise rounded dimensions or to accommodate holding down bolts. Limiting the moment in the baseplate to the elastic moment capacity, the thickness of the plate is obtained using tp

=

c(3w / pyp)0.5

where c is the largest perpendicular distance from the edge of the effective portion of the baseplate to the face of the column. This method is based on the assumption that the load is applied to the baseplate from a concentric column. If this is not the case, the moment resulting from the eccentricity of the axial load should be calculated and must not exceed pypZp, where Zp is the elastic section modulus of the baseplate. This is a change from BS 5950-1:1990, in which the maximum allowable moment in the baseplate was 1.2 pypZp.


There are three further technical changes relating to the design of baseplates. Firstly, the bearing strength of the concrete is taken as 0.6fcu in BS 5950-1: 2000, instead of 0.4fcu in BS 5950-1:1990. This change has been made in the light of new research and brings BS 5950-1 into line with the recommendations in Joints in Simple Construction – Volume 1: Design Methods[8]. Secondly, the rule limiting the design strength of the plate pyp to 270 N/mm2 has been removed, allowing designers to take advantage of the stronger grades of steel, such as S355, if they wish. Finally, in BS 5950-1:1990, the thickness of the baseplate could not be less than the thickness of the column flange. This limit has been removed from BS 5950-1:2000. In addition to the changes described above, three new Sub-clauses have been added to this Clause. Sub-clause 4.13.2.3 deals with applied moments and states that if moments are applied to the baseplate by the column, the moments in the baseplate should be calculated assuming a uniform pressure not exceeding 0.6fcu under the effective portion of the compression zone. This moment should not exceed pypSp, where Sp is the plastic modulus of the baseplate. Sub-clause 4.13.2.4 reminds designers that, where moments are applied to the base of the column, holding down bolts need to be checked for tension. Finally, Sub-clause 4.13.2.5 gives guidance on the design of stiffened baseplates. The design of column baseplates is illustrated by Worked Example 9 in this publication.

34


5

CONTINUOUS STRUCTURES

5.1

Column bases

Clause 5.1.3 Clause 5.1.3, which considers the stiffness of column bases, has been expanded with a few technical changes. In both versions of BS 5950-1, where the column has a nominally rigid base, the base stiffness is assumed to equal the column stiffness for the purposes of elastic analysis. However, BS 5950-1:2000 states that the base may be treated as rigid, when determining the deflections at the serviceability limit state.


BS 5950-1:2000 also considers plastic global analysis, stating that the base moment capacity may be taken as any value between zero and the moment capacity of the column, provided that the foundation is designed to resist a moment equal to this assumed moment capacity. This new requirement recognises the fact that in rigid-plastic analysis it is the capacity, not the stiffness, that governs the design. For elastic-plastic analysis, the assumed base stiffness should be consistent with the assumed moment capacity, but should not exceed the stiffness of the column. According to BS 5950-1:1990, the base stiffness of a column with a nominally pinned base is assumed to be equal to 10% of the column stiffness. In BS 5950-1:2000, this value of 10% is still correct when checking frame stability and determining in-plane effective lengths, but an increased value of 20% of the column stiffness may now be used when calculating deflections. While this is new to BS 5950-1:2000, it has been common practice in industry for a number of years, following the publication of SCI Advisory Desk article AD 090[9]. BS 5950-1:2000 also considers semi-rigid bases, stating that a base stiffness of up to 20% of the column stiffness may be assumed provided that the capacity of the foundation is adequate. As before, the stiffness of a base with an actual pin or a rocker should be taken as zero.

5.2

Frame stability

Clause 5.2.3 Clause 5.2.3 has been extended to include several new requirements relating to plastic analysis. In addition, the entire Clause has been rewritten with a few minor technical changes. The most important additions and changes are summarised below. BS 5950-1:2000 states that the in-plane stability of the members of a frame designed using plastic analysis should be established by checking the stability of the frame as a whole. For single-bay portal frames, there is no need to perform a separate check on the in-plane stability of the individual members. This has always been the case, but was not made clear in BS 5950-1:1990. However, a

35


separate member stability check is required for the internal columns of multibay portals and the rafters of tied portals. Sub-clause 5.2.3.5 (formerly Clause 5.3.4) has been extended significantly to include guidance on members with non-uniform cross sections. It is now stated that the compression flange adjacent to a plastic hinge location should be class 1 plastic for a distance along the member of at least twice the depth of the web or the distance to the point at which the bending moment has fallen to 80% of the plastic moment capacity, whichever is greater. It is also stated that the web thickness adjacent to a plastic hinge location is not reduced for a distance of at least twice the depth of the web at the plastic hinge location. Two new Sub-clauses have been added: Sub-clause 5.2.3.6 limits the area of bolt holes at or adjacent to a plastic hinge location; Sub-clause 5.2.3.8 simply states that haunches should be proportioned to avoid plastic hinges forming within their length. Clause 5.3.1


Clause 5.3.5 of BS 5950-1:1990, which deals with the out-of-plane stability of continuous frames, has been divided into Clauses 5.3.1, 5.3.2 and 5.3.3 in BS 5950-1:2000. Clause 5.3.1 provides a general introduction to out-of-plane stability. The most significant change to this Clause is that designers are now permitted to check the resistance of a member or segment to out-of-plane buckling using the forces and moments corresponding to the required load factor ër, instead of the plastic load factor ëp, whenever ëp is greater than ër. This may be achieved either by multiplying the moments and forces from plastic analysis by the ratio ër / ëp or by determining the forces and moments at ër directly using elasticplastic analysis. Previously, some designers had used the reduced loads, while others had argued that the collapse loads had to be used. BS 5950-1:2000 has now clarified this matter. Clause 5.3.3 Clause 5.3.3 considers the length of a segment adjacent to a plastic hinge and has been extended to include a modification to the existing method to allow for the moment gradient. In BS 5950-1:1990, the distance from the restraint at a hinge to an adjacent restraint was limited to the value given by: Lm =

38ry  f  c 130 

 py   x  2      +    36   275    2

0.5

where fc is the average compressive stress due to the axial load, py is the design strength, ry is the radius of gyration about the minor axis and x is the torsional index. In BS 5950-1:2000, this approach is still valid. However, the designer now has the option of making an approximate allowance for the moment gradient by multiplying the limiting length by the factor φ. 36


Note: This new method should only be used for uniform I section members with equal flanges and D/B ≥ 1.2, in S275 or S355 steel where fc does not exceed 80 N/mm2.

5.3

Portal frames

Clause 5.5.2 Clause 5.5.2 has been expanded to give greater guidance on the elastic design of portal frames, something that was lacking in BS 5950-1:1990. Unless the frame is independently braced, the in-plane stability should be verified by checking the cross-section capacity and out-of-plane buckling resistance of the members using amplified moments and forces, taken as the values given by linear elastic analysis multiplied by the required load factor ër.


Note the use of ër in BS 5950-1:2000. In plastic design, ër is the required value of the plastic collapse factor ëp, while in elastic design it is a factor by which the first-order forces and moments are multiplied to allow for second-order effects. Where second-order effects are negligible, ër will usually be equal to 1. However, where second-order effects are significant, ër will be greater than 1. For portal frames, ër is obtained from Clause 5.5.4. Clause 5.5.4 Clause 5.5.4, which considers the in-plane stability of portal frames, has undergone several major technical changes. The first major change to this Clause relates to the sway-check method in Sub-clause 5.5.4.2. Research conducted during the drafting of BS 5950-1:2000 demonstrated that this method is only valid when the frame geometry lies within certain limits. Consequently, in BS 5950-1:2000, the use of the sway-check method is restricted to frames that satisfy the following conditions: 1.

The span L does not exceed 5 times the mean height of the columns.

2. The height hr of the apex above the tops of the columns does not exceed 0.25L. 3. If the rafter is asymmetric, hr satisfies (hr / sa)2 + (hr / sb)2 Figure 18 in BS 5950-1:2000).

0.5 (see

There were no such limitations on the use of this method in BS 5950-1:1990. The sway-check method involves calculating the horizontal deflection at the top of each column, δi, due to the application of the notional horizontal forces. In the gravity load case, it can be assumed that it is safe to neglect the P-delta effects altogether and take ër equal to 1, so long as äi does not exceed hi / 1000, where hi is the height of that column. This is unchanged from BS 5950-1:1990. However, BS 5950-1:2000 states that the use of the stiffening effects of the cladding should be ignored when calculating the notional horizontal deflections δi for the gravity load case. As an alternative to calculating the sway deflections, the stiffness of the frame may be assessed using the formula given in Sub-clause 5.5.4.2.2. This formula was derived for regular frames with columns at every valley and with the roof 37


load as the only imposed load. Thus, the application of this method is restricted to frames that are not subject to loads from valley beams, crane gantries or other concentrated loads (other than the standard purlin loads). This method, as presented in BS 5950-1:2000, is almost identical to that in BS 5950-1:1990, except that the formula for the effective span of the bay Lb has changed. One major addition to this Clause is the extension of the sway-check method to consider load cases including horizontal loading, separate from the gravity load case. The P-delta effects can never be neglected under horizontal loading and must always be taken into account when designing portal frames for load combinations 2 and 3. Provided that the frame has satisfied the sway check for the gravity load case, λr for the horizontal loading is given by: λr

=

λ sc (λ sc − 1)

λsc is an approximation to the elastic critical buckling factor for the sway mode and is obtained from


λsc =

hi 200δ i

where hi is the storey height and δi is the horizontal deflection at the top of each column due to the application of the appropriate notional horizontal forces for the load case under consideration. As in the gravity case, provided that the frame is not subject to loads from valley beams, crane gantries or other concentrated loads larger than the purlin loads, there is a formula that may be used as an alternative to calculating δi. In this case, the formula allows the direct calculation of ësc without the need to perform an elastic analysis of the frame under the notional horizontal forces. Note: 1. When calculating the deflections for the horizontal load cases, the sway stiffness of the plan bracing and roof sheeting may be included. However, it should be remembered that certain sheeting types offer very little shear stiffness and the original sheeting may be replaced during the life of the structure. 2. The sway-check method is not applicable when ësc < 5.0. cases, second-order analysis must be used.

In such

The method for checking the snap-through stability of portal frame rafters is almost unchanged from BS 5950-1:1990, but designers should note that there has been a minor, yet important, change to the formula as shown below. 1990:

 22( 4 + L / h   Lb I c   275   ≤  1 + tan 2 èr   D I r   p yr   Ω ( Ω – 1)  

2000:

  22( 4 + L / h   Lb 1 + I c  ≤   D Ir   4 ( Ω – 1)  

38

 275    tan 2 èr  pyr   


Where the geometrical limitations for the sway-check method are not satisfied, BS 5950-1:2000 provides an alternative means of allowing for the P-delta effects, known as the “amplified-moments method”. This method, which is applicable to all portal frames, requires the accurate calculation of the elastic critical load factor ëcr (the approximate formula given in Clause 2.4.2 must not be used for this purpose). In this case, ëcr should be taken as the lowest elastic critical load factor for the bare frame, without any allowance for the stiffening effects of the cladding, because the 0.9 factor (see below) already allows for cladding stiffness and strain hardening. Once the lowest value of ëcr has been obtained for the load case under consideration, ër is determined as follows: if ëcr ≥ 10:

ër = 1.0

if 10 > ëcr ≥ 4.6:

ër =

0.9 λ cr λ cr – 1


This method is new to portal frame design in BS 5950-1:2000, but the same method did exist in BS 5950-1:1990 for plastic design of multi-storey frames. If ëcr < 4.6, the amplified-moments method must not be used and the in-plane stability of the frame should be checked by second-order analysis. It is recommended that such analysis be performed by computer software, but where such software is not available, hand calculation is possible and advice is given in In-plane stability of portal frames to BS 5950-1:2000[10]. Where second-order analysis is used, ër should be taken as 1.0. It is expected that second-order analysis will usually be more economical than the amplified-moments method and will often allow design of more slender frames than the sway-check method. Tied portals cannot be treated in the same way as normal portal frames, because they tend to have very high axial forces in their rafters. This is recognised in BS 5950-1:2000 by the introduction of a new Sub-clause specifically for this type of frame. This Sub-clause simply states that the in-plane stability of tied portals should be checked by second-order analysis with ër taken as 1.0.

5.4

Multi-storey frames

Clause 5.6.4 Clause 5.6.4, which considers the elastic design of sway-sensitive multi-storey frames, has been amended in line with the changes to Clause 2.4.2 regarding stability. These are significant technical changes. Sway-sensitive frames should initially be designed to resist gravity loads, as for independently braced frames, without taking account of sway. For this purpose, pattern loading must be considered. A separate check should then be carried out in the sway mode, by applying the notional horizontal forces, together with the gravity loads. In this case, pattern loading is not required. Sway sensitive frames should also be checked in the sway mode for load combinations 2 and 3. As in BS 5950-1:1990, the effects of sway may be allowed for either by the “effective-length method” (Annex E) or the “amplified-sway method”, but with several differences as outlined below.

39


When the effective-length method is used, BS 5950-1:2000 states that the beams should be designed to remain elastic, which is a slightly different limit to “90% of its reduced moment capacity” in Appendix E of BS 5950-1:1990. If the elastic critical load factor ëcr is less than 10, the frame is classed as “sway-sensitive” (see Clause 2.4.2) and the second-order sway effects should either be determined directly by second-order analysis or taken into account by modifying the first-order effects. In BS 5950-1:1990, the amplified-sway method involved the multiplication of the sway effects by an amplification factor given by λ cr λ cr − 1

BS 5950-1:2000 contains a similar method, but there have been three significant changes. Firstly, there are now two alternative expressions for the amplification factor:


1. For clad structures where the stiffening effect of the cladding is not taken into account: kamp =

λ cr 1.15λ cr − 1.5

but

kamp ≥ 1.0

2. For unclad frames or clad frames where the stiffening effect of the cladding is taken into account: kamp =

λ cr λ cr − 1

Secondly, BS 5950-1:2000 places a lower limit on the use of the amplified-sway method of ëcr = 4. If ëcr < 4, the amplification of the first-order results is not an acceptable means of dealing with the P-delta effects and a second-order analysis should be carried out. No such limit was placed on the amplified-sway method in BS 5950-1:1990. Finally, in BS 5950-1:1990, when the amplified-sway method was used, the effective length of the columns in the plane of the frame was taken as 1.0L. BS 5950-1:2000 recommends the use of the non-sway mode in-plane effective lengths obtained from Annex E. This approach gives effective lengths that are less than 1.0L. Clause 5.7.3 Clause 5.7.3 presents a simple check for the stability of plastically designed multi-storey frames that may be used as an alternative to the elastic methods described in Sub-section 5.6 or second-order elastic-plastic analysis. This check is identical to the method given in BS 5950-1:1990, but the following three additional conditions have been placed on its use:

40


•

The bases of the columns should be fixed.

•

It should be ensured that no localised beam or storey-height plastic hinge mechanisms can form at a lower load factor than the overall frame mechanism.

•

The storey height of the frame should nowhere exceed the mean spacing of its columns in that storey.

If ëcr < 4.6 for clad structures in which the effect of the cladding is not taken into account, or ëcr < 5.75 for unclad frames or clad structures in which the effect of the cladding is taken into account, the simplified method cannot be used and the frame stability must be checked either by elastic analysis or second-order elastic-plastic analysis.


BS 5950-1:2000 does not propose the use of concrete casing of the steel frame to increase the value of ëcr.

41


6

CONNECTIONS

6.1

Bolted connections

Clause 6.2.4 Clause 6.2.4 introduces the concept of block shear failure and provides a simple method for ensuring that this failure mode is avoided. Several examples of block shear failure are shown in Figure 6.1.

Fr

Fr Lv

Lv

Lt Lv


Lt

Lt Lv

Fr

Lt Fr

Figure 6.1

Block shear

In each case, failure occurs in shear along the line of bolts parallel to the applied force and, simultaneously, in tension along a perpendicular face. This mode of failure results in a block of material, represented by the shaded area in each example above, being torn out by the applied shear force, hence the name block shear failure. To avoid block shear failure, BS 5950-1:2000 states that it should be checked that the reaction force Fr does not exceed the block shear capacity Pr given by Pr

=

0.6pyt [Lv + Ke(Lt – kDt)]

where Dt is the hole size for the tension face, t is the thickness, the dimensions Lt and Lv are as shown in Figure 6.1 and k has the following values: For a single line of bolts parallel to the applied shear, k = 0.5 For two lines of bolts parallel to the applied shear, k = 2.5. Clause 6.3.2 Clause 6.3.2 has undergone three significant changes: new values for the shear strength of bolts, a new Sub-clause on packing and a new Sub-clause on kidneyshaped slots. In BS 5950-1:1990, Table 32 gave the shear strength, bearing strength and tension strength of grade 4.6 and grade 8.8 bolts and formulae for the strength of other grades. In BS 5950-1:2000, this Table has been replaced by three new 42


Tables: Table 30 for shear strength, Table 31 for bearing strength and Table 34 for tension strength. Table 30 in BS 5950-1:2000 contains several new shear strength values for grade 10.9 and HSFG bolts that were not previously considered explicitly. In addition, the formula for “other grades” has changed from 0.48Uf, but no greater than 0.69Yf, in BS 5950-1:1990 (where Uf and Yf are the specified minimum ultimate tensile strength and the specified minimum yield strength respectively) to 0.4Ub in BS 5950-1:2000 (where Ub is the specified minimum tensile strength of the bolt). Note that Ub is limited to 1000 N/mm2 in BS 5950-1:2000. A new Sub-clause, 6.3.2.2, has been added on the subject of steel packing. This states that the total thickness of the packing tpa at a shear plane should not exceed 4d / 3, where d is the nominal diameter of the bolts. In addition, the number of plies should preferably not exceed four. For cases where the total packing thickness exceeds d / 3, the full shear capacity (Ps = ps As) cannot be used and a reduced shear capacity should be calculated using


Ps

=

 9d As   8d + 3t pa 

   

The second new Sub-clause, 6.3.2.4, deals with kidney-shaped slots. It states that where a connection has two bolts, one in a standard clearance hole and one in a kidney-shaped slot, the shear capacity of each bolt should be taken as 0.8Ps, i.e. 80% of the shear capacity for a standard clearance hole. Clause 6.3.3 Clause 6.3.3 has been expanded to give improved guidance on the bearing capacity of connections using non-preloaded bolts. In assessing the bearing capacity of a connection, it is necessary to consider the bearing capacity of the bolt and the bearing capacity of the connected parts. Both are considered in this Clause, but only the latter has undergone a significant change. In BS 5950-1:1990, the bearing capacity of a connected ply was taken as Pbs =

dtpbs

0.5etpbs

where: d

is the nominal diameter of the bolt

e

is the end distance

t

is the thickness of the connected ply

pbs is the bearing strength of the connected ply, from Table 33. In BS 5950-1:2000, this formula has been modified by the introduction of a new coefficient, kbs, which takes into account the shape of the bolt holes. The bearing capacity of the connected part is now given by Pbs =

kbsdtp pbs but

Pbs

0.5kbsetp pbs

where pbs is obtained from Table 32. Provided that the size of the holes does not exceed the dimensions given in Table 33, the following values of kbs should be used: 43


•

for standard clearance holes:

kbs = 1.0

•

for oversized holes:

kbs = 0.7

•

for short slotted holes:

kbs = 0.7

•

for long slotted holes:

kbs = 0.5

•

for kidney-shaped slots:

kbs = 0.5.

Where standard clearance holes are used, this change has no effect whatsoever on the bearing capacity of the connected parts. However, for all other types of holes, the bearing capacity has been reduced greatly compared with BS 5950-1: 1990 (a 30% reduction for oversized holes and short slotted holes and a 50% reduction for long slotted holes and kidney-shaped slots). This is a very significant change. Note: 1. Some of the values of pbs given in Table 32 have been amended slightly (although not for steel grades S275 and S355).


2. BS 5950-1:2000 gives standard hole dimensions for non-preloaded bolts (Table 33) and preloaded bolts (Table 36). Previously, values were only given for preloaded bolts (Table 35 in BS 5950-1:1990). The bearing capacity of the bolt, Pbb, is unchanged from BS 5950-1:1990 and is given by Pbb =

dtp pbb

where pbb is the bearing strength of the bolt obtained from Table 31. Although the formula is unchanged, the range of pbb values has been expanded and now includes grade 10.9 and HSFG bolts, in addition to the usual grades 4.6 and 8.8. There have also been a few minor changes to some of the pbb values. Note that Ub is limited to 1000 N/mm2 in BS 5950-1:2000. Clause 6.3.4 Clause 6.3.4 has been amended to allow the full tensile strength of the bolts to be used in cases where prying is either avoided or allowed for explicitly in the analysis. This is a major technical change. In BS 5950-1:1990, there was no need to consider explicitly the effects of prying, so long as the tension strengths given in Table 32 were used. This was possible because these strengths had already been reduced to take account of prying. While this approach was very straightforward, it could also be conservative, because the same strengths had to be used for every case, irrespective of whether the connection was subject to significant prying or not. In BS 5950-1:2000, designers have the choice either to continue with the previous method and use reduced bolt strengths, or to use an alternative method in which the prying forces are calculated and allowed for explicitly in the analysis. If the latter option is chosen, designers may use the full tension strengths of the bolts as given in Table 34 of BS 5950-1:2000. Both methods are outlined below.

44


More exact method Consider the two situations shown in Figure 6.2. 2Ft

Ft

2Ft

Ft + Q

Ft

M1

Q

M1

a) Single curvature bending


Figure 6.2

Ft + Q

M1

M1

M2

M2

Q

b) Double curvature bending

Prying in bolted connections

In the first example, the connection is subject to single curvature bending and there is no prying force. In this case, each bolt is designed for a tension force equal to Ft, using a bolt tension capacity based on the tension strength pt from Table 34. The method in BS 5950-1:1990 was clearly conservative in this case, as it made an allowance for prying even though there was none. The new approach is less conservative, because it allows the use of the full tension strengths from Table 34, instead of the reduced tension strengths in Table 32 of BS 5950-1:1990. In the second example, the connection is subject to double curvature bending and the applied tension 2Ft induces a prying force Q at either end of the plate or flange as it bends under the applied load. Consequently, the total tension in each bolt is Ft + Q, denoted in BS 5950-1:2000 by Ftot. Designers wishing to take advantage of the full tension capacity of the bolts must calculate Q and design each bolt for the total tension Ftot. In this case, although the tension capacity is greater than before, the tension in each bolt has also increased. The overall effect will depend on the magnitude of the prying force relative to the applied tension. Simple method In the simple method, the prying force is neglected and the bolt force is simply taken as equal to Ft. However, in this case, the full tension capacity of the bolts cannot be used and, instead, the connection must be designed so that Ft does not exceed the nominal tension capacity of the bolt given by Pnom = 0.8ptAt There are two conditions on the use of the simplified method. Firstly, this method should only be used if the cross-centre spacing of the bolt holes, s, does not exceed 55% of the width of the flange or end-plate, as shown in Figure 6.3.

45


s ≤ 0.55B

s B

Figure 6.3

Maximum cross-centres of bolt lines for the simple method


This is to ensure that the prying force is kept within the limits allowed for by the use of 0.8pt in the calculation of the nominal tension capacity. A cross-centre spacing greater than 0.55B may result in a prying force in excess of that allowed for in the simplified method, making this method unsafe. In such circumstances, the prying forces must be taken into account explicitly by using the more exact method. Secondly, where the connected part is designed assuming double curvature bending, the moment capacity of the connected part per unit width should be based on the elastic capacity rather than the plastic capacity (i.e. the moment capacity per unit width should be taken as pytp2/6, where tp is the thickness of the connected part and py is its design strength). Note that Ub is limited to 1000 N/mm2 in BS 5950-1:2000 (Table 34). Clause 6.4.3 BS 5950-1:1990 recommended that the slip factor ì should either be obtained from tests as specified in BS 4604 or, in the case of general grade fasteners in connections with untreated surfaces, ì could be taken as 0.45. This placed designers in the awkward situation of not being able to calculate the slip resistance in their connections without first conducting tests. In BS 5950-1:2000, this situation has been rectified by the inclusion of Table 35, which contains a range of ì values for different surface conditions. Alternatively, ì may still be obtained from tests as specified in BS 4604. Clause 6.4.5 In BS 5950-1:1990, it was stated that all preloaded bolts subject to combined shear and tension should satisfy the following relationship: Fs F + 0.8 t PsL Pt

≤ 1

where: Fs

is the applied shear

Ft

is the external applied tension

46


PsL is the slip resistance Pt

is the tension capacity.

In BS 5950-1:2000, this relationship has been replaced by two new expressions. For connections designed to be non-slip in service: Fs Ftot + PsL 1.1Po

≤ 1

and for connections designed to be non-slip under factored loads: Fs Ftot + PsL 0.9 Po

≤ 1

where: Ftot is the total applied tension in the bolt, including the prying force


Po

is the specified minimum preload.

Note that the tension force in these new expressions is Ftot, the total tension in the bolt, not Ft, the applied tension. The difference is that Ftot includes the effects of prying, which may be allowed for explicitly in BS 5950-1:2000 (see Clause 6.3.4). By contrast, BS 5950-1:1990 allowed for prying implicitly, by reducing the tension capacity of the bolts.

6.2

Pin connections

Clause 6.5.3 Changes have been made to the methods of calculating the shear capacity, bearing capacity and moment capacity of pins. In each case, the single expression in BS 5950-1:1990 has been replaced by two expressions in BS 5950-1:2000. The choice of expression depends on whether the pin is required to rotate and whether it needs to be removable. In BS 5950-1:1990, the shear capacity of a pin was given as 0.6pypA, where pyp is the design strength of the pin and A is its cross-sectional area. In BS 5950-1: 2000, the shear capacity remains as 0.6pypA if rotation is not required and the pin is not intended to be removable. However, if rotation is required or if the pin is intended to be removable, a new lower value of 0.5pypA should be used. The bearing capacity of a pin was given in BS 5950-1:1990 as 1.2pydt, where d is the diameter of the pin, t is the thickness of the connected part and py is the lower of the design strengths of the pin and the connected part. In BS 5950-1: 2000, the bearing capacity of a pin that is neither required to rotate nor to be removable has increased to 1.5pydt. However, if the pin is required to rotate or is intended to be removable, a lower value of 0.8pydt must be used. A similar change has been made to the moment capacity, which used to be taken as 1.2pypZ, where Z is the elastic modulus of the pin. In BS 5950-1: 2000, the moment capacity may now be taken as 1.5 pypZ if rotation is not required and if the pin is not intended to be removable, but should be taken as 1.0 pypZ if rotation or the removal of the pin is required. This effectively allows the plastic moment capacity of a pin to be developed if the pin does not 47


rotate, but restricts bending to first yield if the pin does rotate or is intended to be removable. In addition, there has been a change relating to the assumptions on which the bending calculations are based. According to BS 5950-1:2000, the moments on the pin should be calculated on the basis that the connected parts form simple supports. It should also be assumed that the reactions between the pin and the connected parts are distributed uniformly along the length in contact on each part. However, as an alternative, if the thickness of one or more of the connected parts exceeds that needed to provide sufficient bearing capacity, it may be assumed that the reactions are distributed over reduced contact lengths adjacent to the interfaces, based upon the minimum thickness needed to provide sufficient bearing capacity. This alternative did not exist in BS 5950-1:1990.

6.3

Welded connections

Clause 6.8.5


BS 5950-1:1990 stated that where a member is connected to a plate by a symmetrical fillet weld, the strength of the weld can be taken as equal to the strength of the parent metal, if the following conditions are satisfied: 1. The weld is made with a suitable electrode that will produce all weld tensile specimens as specified in BS 709, having both a minimum yield strength and a minimum tensile strength not less than those specified for the parent material. 2. The sum of the throat sizes is not less than the connected plate thickness. 3. The weld is principally subject to direct compression or tension. In BS 5950-1:2000, this rule has been deleted, leaving designers to use the design strengths given in Table 37. These values are significantly lower than the design strengths of the parent material (pw = 220 N/mm2 for S275 and pw = 250 N/mm2 for S355). This reduction in capacity has been partly offset by the allowance in BS 5950-1: 2000 for the higher transverse strength of fillet welds (see Clause 6.8.7). For the case of two plates connected at right angles by symmetrical fillet welds, è = 45° and K = 1.25, resulting in the following transverse strengths: S275: 1.25 × 220 = 275 N/mm2 S355: 1.25 × 250 = 312.5 N/mm2 Clearly, where S275 steel is used, the transverse strength is still equal to that of the parent metal, but for S355 the strength is significantly lower. Note that the design strengths given in Table 37 of BS 5950-1:2000 have been amended slightly, compared with Table 36 of BS 5950-1:1990. Clause 6.8.7 Clause 6.8.7 has undergone a major amendment, resulting in a new method for calculating the capacity of fillet welds subject to transverse forces.

48


It has long been understood that the transverse strength of fillet welds is greater than the longitudinal strength and this has been recognised in other codes, such as BS 5400. However, this higher strength was not taken into account in BS 5950-1:1990, in which the capacity of a fillet weld was simply taken to be the product of the design strength pw and the throat size a, irrespective of the direction of loading. BS 5950-1:2000 presents two alternative methods for calculating the capacity of fillet welds. In the “simple method”, the capacity should be taken as sufficient if the vector sum of the design stresses, due to the forces and moments transmitted by the weld, does not exceed the design strength of the weld pw at any point along the weld. This method is the same as in BS 5950-1:1990. Alternatively, designers may choose to use the “directional method”, in which the design forces transmitted by the weld are resolved into their longitudinal and transverse components, designated FL and FT respectively, as shown in Figure 6.7. 2F T


FL FT

FT

FL FT

a) Welds subject to longitudinal shear FT

FT

b) Welds subject to transverse forces

a θ

Throat of the weld

c) Resultant transverse force on weld

Figure 6.4

The directional method for fillet welds

In the longitudinal direction, the capacity PL per unit length of the weld is given by PL =

pwa

The transverse capacity of the weld per unit length is given by PT =

KPL

where the coefficient K depends on the angle è between the weld throat and the direction of the transverse force that is transmitted by the weld. For the special 49


case of two mutually perpendicular members, connected by two equal leg fillet welds, with the applied force parallel to one leg, è = 45° and K = 1.25, giving a 25% enhancement in strength. To take account of the interaction between longitudinal and transverse forces, the following relationship should be satisfied throughout the length of the weld:


(FL / PL)2 + (FT / PT)2 ≤ 1

50



7

REFERENCES

1.

BRITISH STANDARDS INSTITUTION BS 5950 Structural use of steelwork in building BS 5950-1:2000 Code of Practice for design - Rolled and welded sections BSI, 2000

2.

BRITISH STANDARDS INSTITUTION BS 5950 Structural use of steelwork in building BS 5950-1:1990 Code of Practice for design of simple and continuous structures – hot rolled sections BSI, 1990, amended 1992

3.

Advice Note 185 Cold-formed structural hollow sections: interim guidance New Steel Construction, June/July 1996, p. 30

4.

BRITISH STANDARDS INSTITUTION BS 5950 Structural use of steelwork in building BS 5950-5:1998 Code of Practice for design of cold formed thin gauge sections BSI, 1998

5.

Building Regulations Approved Document A Part A: Structure The Stationery Office, 1992

6.

BRITISH STANDARDS INSTITUTION BS 5400 Steel, concrete and composite bridges BS 5400-3:2000 Code of practice for design of steel bridges BSI, 2000

7.

Steelwork design guide to BS 5950-1:2000, Volume 1: Section properties and member capacities, 6th edition (SCI P202) The Steel Construction Institute, in association with The British Constructional Steelwork Association and Corus, 2001

8.

M. HORDYK and A.S. MALIK (Editors) Joints in simple construction – Volume 1: Design methods (2nd edition) (SCI P205) The Steel Construction Institute and The British Constructional Steelwork Association, 1993

9.

Advice Note 090 Deflection limits – portal frames Steel Construction Today, Vol. 5, No. 4, July 1991, p. 203

10. KING, C. In-plane stability of portal frames to BS 5950-1:2000 (SCI P292) The Steel Construction Institute, 2001

51



Discuss me ...

52



WORKED EXAMPLES 1. Sway stability

55

2. Choosing a steel sub-grade

57

3. Restrained beam

59

4. Unrestrained beam

63

5. Plate girder

67

6. Web bearing and buckling

71

7. Compression member

75

8. Axial load and bending

77

9. Baseplate

83

53



Discuss me ...

54


Silwood Park, Ascot, Berks SL5 7QN Telephone: (01344) 623345 Fax: (01344) 622944

Job No.

CDS 138

Job Title

Example no. 1

Subject

Sway stability

Client

CALCULATION SHEET

SCI

Sheet

1

Made by

DGB

Date

Aug. 2001

Checked by

MDH

Date

Oct. 2001

of

2

Rev

Introduction Consider the braced frame shown below. An elastic analysis has been carried out on the bare frame under a set of unit point loads and the resulting deflections are shown in the figure below. Note that the building will be clad, although the stiffness of the cladding was not taken into account in the elastic analysis of the frame. Determine whether the frame is “non-sway” or “sway-sensitive”. If it is “swaysensitive”, what value of the amplification factor, kamp, must be applied to the bracing forces to allow for second-order effects? 7.5 m

7.5 m

7.5 m

1 kN

1 kN

3.5 m 3.5 m

0.65 mm

3.5 m

1 kN

0.84 mm

0.48 mm 3.8 m


1 kN

1.07 mm

Loading Every beam is subjected to an unfactored uniformly distributed load consisting of: Dead load:

21 kN/m

Imposed load:

30 kN/m

Solution Calculate the notional horizontal forces For the gravity load case (load combination 1), the design load is given by Factored load

=

(1.4 × 21) + (1.6 × 30) =

77.4 kN/m

Load per floor

=

77.4 × 7.5 × 3

1741.5 kN

Notional horizontal force

=

=

0.005 × 1741.5 =

8.7 kN

Note that the notional horizontal force is based on the factored load acting on the whole floor and not just on one beam length

55

2.4.2.4



Job No.

CDS 138

Job Title

Example no. 1

Subject

Sway stability

Client

CALCULATION SHEET

SCI

Sheet

2

Made by

DGB

Date

Aug. 2001

Checked by

MDH

Date

Oct. 2001

of

2

Rev

Calculate the lateral drift δ Multiplying the set of unit load deflections by the notional horizontal force per floor gives the following deflections at each floor level: 8.7 kN 3.5 m

0.65 x 8.7 =5.65 mm 0.48 x 8.7 = 4.2 mm

3.8 m


8.7 kN

3.5 m

8.7 kN

0.84 x 8.7 =7.3 mm

3.5 m

8.7 kN

1.07 x 8.7 =9.3 mm

The lateral drift is the relative deflection between two adjacent storeys. By inspection, the ground to first floor column is critical, with a lateral drift, δ, of 4.2 mm Determine δcr The lowest value of λcr is that for the lowest storey and is given by λ cr =

h 200δ

=

3800 200 × 4.2

=

4.52

2.4.2.6

Because λcr < 10, the frame is “sway-sensitive”, but as λcr > 4, the second-order effects may be allowed for by amplification of the first-order sway effects (i.e. there is no need for a second-order analysis) Calculate kamp The frame is a clad frame in which the stiffness of the cladding was neglected. Therefore, kamp is given by k amp =

λ cr 4.52 = = 1.22 1.15λ cr − 1.5 (1.15 × 4.52) − 1.5

56

2.4.2.7



Job No.

CDS 138

Job Title

Example no. 2

Subject

Choosing a steel sub-grade

Client

CALCULATION SHEET

SCI

Sheet

1

of

1

Rev

Made by

MDH

Date

Aug. 2001

Checked by

TCC

Date

Sept. 2001

Introduction An exposed steel structure is proposed. The steel is BS EN 10025 S355 and the thickest element is 30 mm. The beams are welded to the column flanges and the maximum tensile stress is 200 N/mm2. Choose an appropriate steel grade to avoid brittle fracture Solution The basic requirement is that, throughout the component, the thickness should satisfy the relationship:

2.4.4

t ≤ Kt1


Because the maximum thickness = 30 mm, the steel grade should be chosen such that t1 ≥ 30 / K The nominal yield strength Ynom = 355 N/mm2 The maximum tensile stress in the component = 200 N/mm2 Therefore stress > 0.3Ynom For welded connections to unstiffened flanges with stress > 0.3Ynom Table 3

K = 0.5 The requirement for t1 is therefore t1 ≥ 60 mm For external conditions (–15°C), the only suitable grade of steel is BS EN 10025 S355 K2 (t1 = 66 mm)

Table 4

Finally, check that the maximum thickness in the component does not exceed the limits for which the full Charpy impact value applies, as given in Table 6 For all “sections” of grade S355 to BS EN 10025, t2 = 100 mm t2 > 30 mm Therefore BS EN 10025 S355 K2 is suitable

57

Table 6



Discuss me ...

58



Job No.

CDS 138

Job Title

Example no. 3

Subject

Restrained beam

Client

CALCULATION SHEET

SCI

Sheet

1

Made by

MDH

Date

Sept. 2001

Checked by

TCC

Date

Oct. 2001

Introduction A 533 × 210 × 101 UB in grade S355 steel is simply supported as shown below and fully restrained along its length by a concrete floor slab. Check that the shear capacity and the moment capacity are adequate for the factored loading shown, which includes self-weight 225 kN

W = 25 kN/m

533 x 210 x 101 UB Grade S355


5m

5m

A

B

C

Solution Calculate the maximum moment and shear The maximum shear occurs at A and C Fv

=

25 × 10 225 + 2 2

The moment at A and C

=

238 kN

=

0

The maximum moment occurs at B Mx =

25 × 10 2 225 × 10 = + 8 4

The shear at B

=

225 2

=

875 kNm

113 kN

Classify the cross section From section property tables: Flange thickness T

=

17.4 mm

Flange b/T

=

6.03

Web d/t

=

44.1

59

of

3

Rev



Job No.

CDS 138

Job Title

Example no. 3

Subject

Restrained beam

Client

CALCULATION SHEET

Grade of steel

=

SCI

Sheet

2

Made by

MDH

Date

Aug. 2001

Checked by

TCC

Date

Sept. 2001

S355

ε

=

275 py

Rev

345 N/mm2 275 345

=

=

Table 11

0.89

Limit for outstand element of compression flange for a class 1 section

3.5.2 Table 11

b/T =

Table 11

9ε =

The actual b/T Created on 30 March 2011 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

3

3.1.1 Table 9

T > 16 mm Therefore py =

of

8.01 =

6.03, < 8.01

Therefore Flange is class 1 Limit for web with the neutral axis at mid depth for a class 1 section: d/t =

80ε

=

71.2

The actual d/t =

3.5.2 Table 11

44.1, < 71.2

Therefore Web is class 1 Both the flange and web are class 1, therefore the cross section is class 1 Check the shear capacity Pv

=

0.6 pyAv

py

=

345 N/mm2

Av

=

tD

t

=

10.8 mm

D

=

536.7 mm

Pv

=

0.6 × 345 × 10.8 × 536.7 × 10-3 =

From above, Fv

4.2.3

=

238 kN

Fvmax < Pv Therefore the section is adequate in shear

60

1200 kN



Job No.

CDS 138

Job Title

Example no. 3

Subject

Restrained beam

Client

CALCULATION SHEET

SCI

Sheet

3

Made by

TCC

Date

Aug. 2001

Checked by

MDH

Date

Sept. 2001

of

3

Rev

Check the moment capacity

Check whether the shear is “high”, i.e. Fv > 0.6 Pv, or “low”,

4.2.5

i.e. Fv < 0.6 Pv, at the point of maximum moment The shear at B 0.6 Pv

=

=

113 kN

0.6 × 1200 =

720 kN

113 < 720 Therefore the shear is low


For low shear, the moment capacity for a class 1 section is given by Mcx =

py Sx

Sx

2612 cm3 (from section property tables)

=

Mcx =

4.2.5.2

345 × 2612 × 10-3

=

901 kNm

Check limit to avoid irreversible deformation under serviceability limit state loads For a simply supported beam Mcx ≤ 1.2 pyZx Zx

=

4.2.5.1


1.2pyZx =

1.2 × 345 × 2292 × 10-3 =

Therefore Mcx

=

901 kNm

From above, Mx =

875 kNm

949 kNm

Mx < Mcx Therefore the section is adequate in bending

61



Discuss me ...

62



Job No.

CDS 138

Job Title

Example no. 4

Subject

Unrestrained beam

Client

CALCULATION SHEET

SCI

Sheet

1

Made by

MDH

Date

Aug. 2001

Checked by

TCC

Date

Sept. 2001

Introduction Consider the beam from Example 1 during construction, before the concrete floor slab has been constructed. In this case, the beam is unrestrained except at the location of the point load (B). Check the adequacy of the unrestrained beam between A and B for the factored loading shown 225 kN

533 x 210 x 101 UB Grade S355


5m

5m

A

B

C

Solution Calculate the maximum moment and shear The maximum shear occurs at A and C Fv

=

1.4 × self weight 225 + 2 2

=

112.5 + 7.1 = 120 kN

The maximum moment occurs at B Mx =

225 × 10 + Moment due to self weight 4

The shear at B

=

225 2

=

113 kN

Classify the section From section property tables: Flange thickness T

=

17.4 mm

Flange b/T

=

6.03

Web d/t

=

44.1

63

= 580 kNm

of

4

Rev



Job No.

CDS 138

Job Title

Example no. 4

Subject

Unrestrained beam

Client

CALCULATION SHEET

SCI

Sheet

2

Made by

MDH

Date

Aug. 2001

Checked by

TCC

Date

Sept. 2001

Grade of steel = S355

275 py

275 345

=

=

Table 11

0.89

Limit for outstand element of compression flange for a class 1 section

3.5.2 Table 11

b/T =

Table 11

9ε =

8.01

The actual b/T = 6.03 < 8.01 Created on 30 March 2011 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

Rev

345 N/mm2

Therefore py = =

4

3.1.1 Table 9

T > 16 mm

ε

of

3.5.2 Table 11

Therefore Flange is class 1 Limit for web with the neutral axis at mid depth for a class 1 section d/t =

80ε =

The actual d/t

71.2 =

44.1, < 71.2

Therefore Web is class 1 Both the flange and web are class 1, therefore the cross section is class 1 Check the shear capacity From example 3, Pv Fv

=

=

1200 kN

4.2.3

113 kN

Fv < Pv Therefore the section is adequate in shear 1.1.4

Check the moment capacity

From Example 3, Mcx = Mx =

901 kNm

4.2.5

575 kNm

Mx < Mcx Therefore the section is adequate in bending 64



Job No.

CDS 138

Job Title

Example no. 4

Subject

Unrestrained beam

Client

SCI

CALCULATION SHEET

Sheet

3

Made by

MDH

Date

Aug. 2001

Checked by

TCC

Date

Sept. 2001

4

Rev

4.3.6

Calculate the lateral-torsional buckling resistance Mb (i) Conservatively assume an effective length LE of 1.0L

of

=

5m

(ii) The slenderness is given by From section property tables, ry = 45.7 mm λ

=

LE/ry

=

5000 45.7

=

109

(iii) From section property tables, x = 33.2 and u = 0.873


λ/x =

109 33.2

=

3.3

(iv) The slenderness factor v is obtained from Table 19 v

=

4.3.6.7 Table 19

0.90

4.3.6.9

(v) For a class 1 section, βW = 1.0 Therefore, λLT (vi) For py = pb

=

=

uvλ

345 N/mm2

=

0.873 × 0.90 × 109 =

and λLT =

86

86, Table 16 gives

Mb =

pb Sx

Sx


Mb =

4.3.6.5 Table 16

170 N/mm2

(vii) For a class 1 section, the buckling resistance moment Mb is given by

=

4.3.6.7

170 × 2612 × 103 /106 =

4.3.6.4

444 kNm

Calculate the equivalent uniform moment factor mLT Between A and B, the bending moment diagram is as follows: M A= 0 MB = 580

Therefore, β =

0,

mLT =

4.3.6.6 Table 18

0.6

65



Job No.

CDS 138

Job Title

Example no. 4

Subject

Unrestrained beam

Client

CALCULATION SHEET

SCI

Sheet

4

Made by

MDH

Date

Aug. 2001

Checked by

TCC

Date

Sept. 2001

of

4

Rev

Check that the resistance is adequate Mb / mLT = Mx =

444 / 0.6

=

740 kNm

580 kNm

4.3.6.2

Therefore Mx ≤ Mb / mLT


The beam does therefore have adequate resistance to lateral-torsional buckling when unrestrained

66



Job No.

CDS 138

Job Title

Example no. 5

Subject

Plate girder

Client

SCI

CALCULATION SHEET

Sheet

1

Made by

MDH

Date

Aug. 2001

Checked by

TCC

Date

Sept. 2001

of

4

Rev

Introduction A plate girder consists of two 500 × 30 mm flange plates and a 1200 × 12 mm web plate, both in S275 steel 1. Find the moment capacity, assuming that the moment is carried by the flanges alone, and compare with the moment capacity based on the full section 2. Find the shear buckling resistance by the “simplified method”, assuming a stiffener spacing of 1.5 m 3. Find the shear buckling resistance by the “more exact method”, assuming ff = 0.1pyf


For this example, assume that the end post and stiffeners all have adequate capacity Solution Classify the cross section 3.1.1 Table 9

For the flange, grade of steel = S275 T > 16 mm Therefore py = 265 N/mm2

3.1.1 Table 9

For the web, grade of steel = S275 t < 16 mm Therefore py = 275 N/mm2 ε

=

Flange b/T

275 py =

275 265

=

=

1.02

 500 − 12    / 30 = 2  

8.13

From Table 11, the limit for a class 2 flange is 9ε. Therefore the flange is class 2 Web d/t =

1200 12

=

3.5.2 Table 11

100

From Table 11, the limit for a class 3 web subject to bending only (i.e. neutral axis at mid depth) is 120ε. Therefore, the web is class 3 Therefore the cross section is class 3 67

3.5.2 Table 11



Job No.

CDS 138

Job Title

Example no. 5

Subject

Plate girder

Client

SCI

CALCULATION SHEET

Sheet

2

Made by

MDH

Date

Aug. 2001

Checked by

TCC

Date

Sept. 2001

of

Calculate the moment capacity

4

Rev

4.4.4.2

The web depth-to-thickness ratio d/t > 62ε, so the web must be assumed to be susceptible to shear buckling. Consequently, the interaction of shear and moment must be taken into account when calculating the moment capacity of the cross section Because the flanges are not slender, the “flanges only” method may be used to obtain a conservative value of the moment capacity Mc =

4.4.4.2b

pyf Af hs

where pyf is the design strength of the flanges, Af is the area of one flange and hs is the distance between the centre of the flanges


Mc =

265 × 500 × 30 × 1230 / 106

=

4890 kNm

Assuming that this was the maximum shear on the section at the end of the girder and that, near the centre, the shear is reduced to less than 0.6 of this value, the moment capacity can be recalculated based on the full section. As the web is class 3, the elastic modulus is used to calculate Mc =

pyf Zx

Ix

=

 500 × 30 3  12 × 1200 3 + 500 × 30 × 615 2  + 2   12 12  

=

1.31 × 1010 mm4

=

(Ix / 630) × 10-3 =

20794 cm3

Mc =

265 × 20794 / 103 =

5510 kNm

Zx

4.4.4.2

(c.f. 4890 kNm obtained by using the flanges alone) Calculate the shear buckling resistance using the simplified method Using the “simplified method”, the shear buckling resistance is given by Vb

=

Vw =

d t qw

a/d =

1500 1200

=

d/t =

100

py

275 N/mm2

=

4.4.5.2

1.25

68



Job No.

CDS 138

Job Title

Example no. 5

Subject

Plate girder

Client

SCI

CALCULATION SHEET

Sheet

3

Made by

MDH

Date

Aug. 2001

Checked by

TCC

Date

Sept. 2001

of

139 N/mm2

Vb

Vw =

=

1200 × 12 × 139 / 103 =

Rev

4.4.5.2 Table 21

For these values of a/d, d/t and py, Table 21 gives qw =

4

2000 kN

Calculate the shear buckling resistance using the more exact method The “more exact method” permits the shear buckling resistance of the web to be increased above that given by the simplified method, for cases where the moment is low, i.e. towards the end of the girder Assume that in the end panel, the moment is not greater than 10% of the moment capacity and that ff / pyf = 0.1 Created on 30 March 2011 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

Using the “more exact method”, the shear buckling resistance is given by Vb

=

Vw + Vf

but

Vb ≤ Pv

The first step is to calculate the plastic moment capacity of the (smaller) flange about its own equal area axis perpendicular to the plane of the web Spf = = Mpf =

4.4.5.3

BT 2 / 4  500 × 30 2      4   pyf Spf

=

=

112.5 × 103 mm3

265 × 112.5 × 103 / 106

=

29.8 kNm

The plastic moment capacity of the web is then calculated Spw =

td 2 / 4

=

12 × 1200 2 = 4

Mpw =

pyf Spw

=

275 × 4320 / 103 =

4320 × 103 mm3 1190 kNm

The shear capacity is given by Pv

=

0.6 pydt =

=

2380 kN

4.2.3

0.6 × 275 × 1200 × 12 / 103

69



Job No.

CDS 138

Job Title

Example no. 5

Subject

Plate girder

Client

CALCULATION SHEET

SCI

Sheet

4

Made by

MDH

Date

Aug. 2001

Checked by

TCC

Date

Sept. 2001

The flange-dependent shear buckling resistance is then calculated using

Vf =

=

(

)

(

P v d / a 1 − f f / p yf 

[ 1 + 0.15 ( M

pw

)

/ M pf

2

 

)]

2380 (1.2 / 1.5 ) 1 − ( 0.1 ) 2   

[ 1 + 0.15 (1190

/ 29.8 )

]

=


and the total shear buckling resistance is given by Vb

=

Vw + Vf

Vb

=

2000 + 270 =

2270 kN, ≤ Pv

70

270 kN

of

4

Rev

4.4.5.3



Job No.

CDS 138

Job Title

Example no. 6

Subject

Web bearing and buckling

Client

CALCULATION SHEET

SCI

Sheet

1

of

3

Rev

Made by

MDH

Date

Aug. 2001

Checked by

TCC

Date

Sept. 2001

Introduction The same 533 × 210 × 101 UB used in Examples 3 and 4 is subjected to an applied point load, as shown below. Check that the bearing capacity and buckling resistance of the web are adequate at the location of the applied load and at the support 225 kN 152 x 152 x 23 UC


152.2

17.4

12.7

533 x 210 x 101 UB Grade S355

be

b1

238 kN

ae

be

=

50 mm

b1

=

100 mm

Solution At the location of the point load (i) Bearing capacity of the unstiffened web Pbw = (b1 + nk) t pyw n

= 5.0 (not at the end of the member)

k

= T + r for a rolled section

(b1 + nk)

=

152.2 + 5.0 (17.4 + 12.7)

=

302.7 mm

4.5.2.1

71



Job No.

CDS 138

Job Title

Example no. 6

Subject


Client

SCI

CALCULATION SHEET

Pbw =

302.7 × 10.8 × 345 × 10-3

Pbw =

1130 kN

Fx

225 kN

=

Sheet

2

of

3

Rev

Made by

MDH

Date

Aug. 2001

Checked by

TCC

Date

Sept. 2001

Fx < Pbw Therefore the bearing resistance of the web is adequate without the need for a bearing stiffener


(ii) Buckling resistance of the unstiffened web Px

=

ε

=

Px

=

25εt

(b1 + nk ) d 275 py

4.5.3.1

Pbw 275 = 345

=

25 × 0.89 × 10.8

(302.7 ) × 476.5

0.89

× 1130

= 714 kN Fx < Px Therefore the buckling resistance of the web is adequate without the need for a buckling stiffener At the support (i) Bearing capacity of the unstiffened web Pbw =

(b1 + nk) t pyw

n

=

2 + 0.6 be / k

be

=

50 mm

k

=

T+r

n

=

2 +

(b1 + nk)

=

but ≤ 5

17.4 + 12.7 =

0.6 × 50 30.1 =

4.5.2.1

=

30.1 mm

3.0

100 + (3.0 × 30.1)

=

190.3 mm 72



Job No.

CDS 138

Job Title

Example no. 6

Subject


Client

SCI

CALCULATION SHEET

Pbw = Fx

=

190.3 × 10.8 × 345 × 10-3 =

Sheet

3

of

3

Rev

Made by

MDH

Date

Aug. 2001

Checked by

TCC

Date

Sept. 2001

709 kN

238 kN

Fx < Pbw Therefore the bearing resistance of the web at the support is adequate (ii) Buckling resistance of the unstiffened web

4.5.3.1


Check whether ae < 0.7 d ae =

50 + 50

0.7d

=

=

100 mm

0.7 × 476.5 =

333.6 mm

Since ae < 0.7d, the buckling resistance Px is given by Px

=

a e + 0.7 d 1.4 d

=

100 + 333.6 25 × 0.89 × 10.8 × P 1.4 × 476.5 (190.3) × 476.5 bw

=

0.650 × 0.798 × Pbw

=

0.519 × 709

25εt

(b1 + nk ) d

Pbw

= 368 kN

Fx < Px Therefore the buckling resistance of the web at the support is adequate

73



Discuss me ...

74



Job No.

CDS 138

Job Title

Example no. 7

Subject

Compression member

Client

CALCULATION SHEET

SCI

Sheet

1

Made by

MDH

Date

Aug. 2001

Checked by

TCC

Date

Sept. 2001

of

2

Rev

Introduction A 356 × 368 × 129 UC in grade S275 steel is pin-ended about the x-x and y-y axes as shown below. Check that the compression resistance is adequate for the factored loading shown, which includes self-weight. The cross section is classified as not slender

356 x 368 x 129 UC

2500 kN


6m

Solution From section tables: T

=

17.5 mm

Ag

=

164 cm2

rx

=

15.6 cm

ry

=

9.43 cm

Grade of steel

=

S275 3.1.1 Table 9

T > 16 mm Therefore py =

265 N/mm2

The cross section is NOT slender therefore Pc

=

4.7.4

Ag pc

75



Job No.

CDS 138

Job Title

Example no. 7

Subject

Compression member

Client

SCI


CALCULATION SHEET

Sheet

2

Made by

MDH

Date

Aug. 2001

Checked by

TCC

Date

Sept. 2001

of

2

Rev

For buckling about the x-x axis, use strut curve b

Table 23

For buckling about the y-y axis, use strut curve c

Table 23

Slenderness λx

=

L EX rx

=

6000 15.6 × 10

=

38.5

λy

=

L EY ry

=

6000 9.43 × 10

=

63.6

For λx = 38.5 and py = 265 N/mm2, pcx =

243 N/mm2

Table 24b

For λy = 63.6 and py = 265 N/mm2, pcy =

188 N/mm2

Table 24c

pc is the lesser value of pcx or pcy Pc

=

164 × 102 × 188 / 103 =

=

188 N/mm2

3080 kN

Fc < Pc Therefore the compression resistance is adequate

76



Job No.

CDS 138

Job Title

Example no. 8

Subject

Axial load and bending

Client

CALCULATION SHEET

SCI

Sheet

1

Made by

MDH

Date

Aug. 2001

Checked by

TCC

Date

Sept. 2001

of

Introduction The top chord of a lattice girder supports purlins at A, B and C, as shown below, resulting in a factored point load of 11.2 kN at each location. The member is a 150 × 150 × 5 rectangular hollow section in S275 steel and is continuous at A, B and C. Check the adequacy of the top chord by considering its section capacity and buckling resistance, according to the requirements in Clause 4.8.3 1.8 m

1.8 m

11.2 kN

11.2 kN


A

11.2 kN

B

C

Loading The factored axial load in ABC =

598 kN

The factored bending moment diagram is 5.04

B A

5.04 C

5.04

Solution Classify the section From section tables: b

=

d

t

=

5 mm

d/t =

=

135 mm

27.0

Grade of steel

=

S275

t < 16 mm

77

Factored loads

6

Rev



Job No.

CDS 138

Job Title

Example no. 8

Subject


Client

SCI

CALCULATION SHEET

Therefore py = ε

275 py

=

Sheet

2

Made by

MDH

Date

Aug. 2001

Checked by

TCC

Date

Sept. 2001

275 N/mm2

=

135 / 5.0

=

27.0

The limitation for a class 1 flange


'Web'

=

Fc ≤1 2d t p y

r1

=

But –1.0 < r1 ≤ 1.0, therefore r1 Limit

=

28ε

The flange is therefore class 1

The limitation for a class 1 web =

r1

=

64å 1 + 0.6r1

=

Rev

1.0

Flange b/t

'Flange'

6

3.1.1 Table 9

275 = 275

=

of

3.5.2 Table 12

64ε 1 + 0.6 r1

598 × 10 3 2 × 135 × 5 × 275 =

3.5.2 Table 12

=

1.6

1.0

64 × 1 1 + 0.6 × 1.0

=

40

Therefore the web is class 1 Both the flange and the web are class 1, therefore the cross section is class 1 Check axial compression The first step is to evaluate the in-plane slenderness of the member given by λx

=

L Ex rx

In the plane of the girder, there is no restraint at B and the effective length is given by LEx =

0.85L

LEx =

0.85 × 3.6

r

5.9 cm

=

=

3.06 m

78

4.7.3 Table 22



Job No.

CDS 138

Job Title

Example no. 8

Subject


Client

CALCULATION SHEET

λx =

SCI

Sheet

3

Made by

MDH

Date

Aug. 2001

Checked by

TCC

Date

Sept. 2001

of

6

Rev

3.06 × 10 3 = 51.9 5.9 × 10

Similarly, the effective length of the member out of plane is given by λy

=

L Ey ry


Out of plane, the purlin at B provides restraint to the top chord and the effective length is given by LEy =

1.0L

LEy =

1.0 × 1.8

=

1.8 m

λy =

1.8 × 10 3 5.9 × 10

=

30.5

4.7.3 Table 22

The next step is to calculate Pcx and Pcy For λx = 51.9 and py = 275 N/mm2, pcx = Pcx =

Ag pcx

=

28.7 × 102 × 249 × 10-3

For λy = 30.5 and py = 275 N/mm2, pcy = Pcy =

Ag pcy

=

249 N/mm2 =

266 N/mm2

28.7 × 102 × 266 × 10-3

Pc is the lesser of Pcx and Pcy

=

The applied axial load Fc

598 kN

=

715 kN

=

763 kN

4.7.5 Table 24a 4.7.5 Table 24a

715 kN

Fc < Pc Therefore the member is adequate in axial compression Check the shear capacity The maximum shear force Fv

=

11.2 kN

Shear capacity Pv = 0.6 pyAv A

=

4.2.3

28.7 cm2

4.2.3

79



Job No.

CDS 138

Job Title

Example no. 8

Subject


Client

SCI

CALCULATION SHEET

Av

=

 AD    D + B

=

28.7 × 10 2 × 150 150 + 150

Pv

=

0.6 pyAv

=

0.6 × 275 × 1435 / 103

=

237 kN

=

Sheet

4

Made by

MDH

Date

Aug. 2001

Checked by

TCC

Date

Sept. 2001

of

6

Rev

1435 mm2

Fv < Pv Therefore the cross section is adequate in shear Check the moment capacity


Check whether the shear is “high”, i.e. Fv > 0.6 Pv, or “low”, i.e. Fv < 0.6 Pv, at the point of maximum moment 0.6 Pv

=

0.6 × 237

=

4.2.5

142 kN

Fv < 0.6Pv, therefore the shear is “low” For low shear, the moment capacity for a class 1 section is given by Mc =

pySx

S


=

Mc =

275 × 156 / 103

=

42.9 kNm

Also check Mc ≤ 1.5 pZ Z

=

4.2.5


1.5 pyZ =

1.5 × 275 × 134 / 103 =

Therefore Mcx

=

4.2.5

55.3 kNm

42.9 kNm

Mx < Mcx Therefore the cross section is adequate in bending

80



Job No.

CDS 138

Job Title

Example no. 8

Subject


Client

SCI

CALCULATION SHEET

Sheet

5

Made by

MDH

Date

Aug. 2001

Checked by

TCC

Date

Sept. 2001

of

6

Rev

Check interaction between axial load and bending (i) Section capacity The following relationship must be satisfied: 4.8.3.2

My Fc Mx + + ≤1 Ag p y M cx M cy 598 × 10 3 28.7 × 10 2 × 275

+

5.04 0 + = 42.9 M cy

0.76 + 0.117 + 0

=

0.88, < 1

Therefore the section capacity is adequate Created on 30 March 2011 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

(ii) Buckling resistance In the simplified method, the following relationships must be satisfied: my My Fc m M + x x + ≤ 1 and Pc py z x py z y my My Fc m M + LT LT + ≤1 Pcy Mb py z y mx is determined between restraints on the x axis, i.e. A and C according to the shape of the bending moment diagram

5.04 M1

0 M2

0 M4

5.04 M5

5.04 M3

mx =

0.2 +

0.1 M 2 + 0.6 M 3 + 0.1 M 4 M max

=

0.8

For square rectangular hollow sections, such as the member in this example, there is no need to consider lateral-torsional buckling and Mb is given by Mb =

pySx = 42.9 kNm

MLT is the maximum moment in the segment = 81

5.04 kNm

4.8.3 Table 26 4.3.6.1



Job No.

CDS 138

Job Title

Example no. 8

Subject


Client

CALCULATION SHEET

SCI

Sheet

6

Made by

MDH

Date

Aug. 2001

Checked by

TCC

Date

Sept. 2001

of

mLT is determined between restraints on the y-y axis, i.e. A and B (or B and C) M

=

5.04,

βM =

–5.04,

therefore β

=

-1

and mLT =

6

Rev

4.3.6.6 Table 18

0.44

Finally, the two interaction relationships for buckling are evaluated my My Fc m M + x x + Pc py z x py z y

0.8 × 5.04 × 10 6 598 0 + + 3 715 py z 275 × 134 × 10

=

my My Fc m M + LT LT + Pcy Mb py z y

=

598 0.44 × 5.04 0 + + 763 42.9 py z y

=

0.95, < 1

=

0.84, < 1


In both cases, the values are less than 1, indicating that the top chord is adequate under these loading conditions, according to the simplified method

82

4.8.3.3.1



Job No.

CDS 138

Job Title

Example no. 9

Subject

Baseplate

Client

SCI

CALCULATION SHEET

Sheet

1

Made by

MDH

Date

Aug. 2001

Checked by

TCC

Date

Sept. 2001

of

2

Rev

Introduction A 254 × 254 × 73 Universal Column carries a factored axial load of 1500 kN. The proposed baseplate is 500 × 500 mm in S275 steel, on C30 concrete with stronger bedding material. Determine the thickness of the baseplate Solution 4.13.2.1

Determine the size of the effective area


The dimensions of the column are: D

=

254 mm

B

=

254 mm

T

=

14.2 mm

The cross-sectional area

=

9290 mm2

and the perimeter =

1480 mm

Calculate the area required to carry the load in the column Area required

=

applied fo rce 0.6 f cu

=

1500000 0.6 × 30

=

83300 mm2

The effective area of the baseplate can be approximated to the following quadratic equation: Area

=

4c2 + (column perimeter) c + column area

where c is the distance from the column to the edge of the effective area, as shown below

2c + T

Equating the effective area to the required area gives 4c2 + 1480c + 9290 =

83300

Solving this equation, c

=

44.6 mm

83



Job No.

CDS 138

Job Title

Example no. 9

Subject

Baseplate

Client

CALCULATION SHEET

SCI

Sheet

2

Made by

MDH

Date

Aug. 2001

Checked by

TCC

Date

Sept. 2001

of

2

Rev

Check that there is no overlap between the effective areas associated with the two flanges Gap

=

D – 2T – 2c =

254 – (2 × 14.2) – (2 × 44.6) =

136 mm

The gap is positive, indicating that there is no overlap Finally, the overall size of the effective area is checked to ensure that it fits on the base 254 + (2 × 44.6)

=

343 mm

This is less than the baseplate size of 500 mm and is therefore acceptable 4.13.2.2


Determine the minimum required thickness of the baseplate The minimum thickness of the baseplate is given by tp

=

 3 × 0.6 × f cu c  p yp 

   

0.5

=

 3 × 0.6 × 30  44.6   275  

0.5

=

However, because tp > 16 mm, pyp must be reduced to 265 N/mm2 Recalculating, tp

=

 3 × 0.6 × 30  44.6   265  

0.5

=

20.1 mm

84

19.8 mm



Discuss me ...

85



Discuss me ...

Typeset and page make-up by The Steel Construction Institute, Ascot SL5 7QN Printed in Europe by the Alden Group, Oxford 1500 12/01 CDS138

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Guide to Health Informatics (Arnold Publication)

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The Major Transitions in Evolution

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Flux (Sci Fi)

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